angle trisection different modes
TRANSCRIPT
Angle Trisection
Angle trisection is the division of an arbitrary angle into three equal angles. It was one
of the three geometric problems of antiquity for which solutions using only compass
and straightedge were sought. The problem was algebraically proved impossible by
Wantzel (1836).
Although trisection is not possible for a general angle using a Greek construction, there
are some specific angles, such as and radians ( and , respectively), which
can be trisected. Furthermore, some angles are geometrically trisectable, but cannot be
constructed in the first place, such as (Honsberger 1991). In addition, trisection of
an arbitrary angle can be accomplished using a marked ruler (a Neusis construction) as
illustrated above (Courant and Robbins 1996).
An angle can also be divided into three (or any whole number) of equal parts using the
quadratrix of Hippias or trisectrix.
An approximate trisection is described by Steinhaus (Wazewski 1945; Peterson 1983;
Steinhaus 1999, p. 7). To construct this approximation of an angle having measure ,
first bisect and then trisect chord (left figure above). The desired approximation is
then angle having measure (right figure above). To connect with , use the law
of sines on triangles and gives
(1)
so . Since we also have , this can be written
(2)
Solving for then gives
(3)
This approximation is with of even for angles as large as , as illustrated
above and summarized in the following table (Petersen 1983), where angles are
measured in degrees.
( ) ( ) ( ) ( )
10 3.333333 3.333804 3.332393
20 6.666666 6.670437 6.659126
30 10.000000 10.012765 9.974470
40 13.333333 13.363727 13.272545
50 16.666667 16.726374 16.547252
60 20.000000 20.103909 19.792181
70 23.333333 23.499737 23.000526
80 26.666667 26.917511 26.164978
90 30.000000 30.361193 29.277613
99 33.000000 33.486234 32.027533
has Maclaurin series
(4)
(Sloane's A158599 and A158600), which is readily seen to a very good approximation
to .
SEE ALSO: Angle Bisector, Archimedes' Spiral, Circle Squaring, Conchoid of
Nicomedes, Cube Duplication, Cycloid of Ceva, Maclaurin Trisectrix, Morley's
Theorem, Neusis Construction, Origami, Pierpont Prime, Quadratrix of Hippias,
Tomahawk, Trisectrix
REFERENCES:
Bogomolny, A. "Angle Trisection." http://www.cut-the-
knot.org/pythagoras/archi.shtml.
Bogomolny, A. "Angle Trisection by Hippocrates." http://www.cut-the-
knot.org/Curriculum/Geometry/Hippocrates.html.
Bold, B. "The Problem of Trisecting an Angle." Ch. 5 in Famous Problems of Geometry
and How to Solve Them. New York: Dover, pp. 33-37, 1982.
Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag,
pp. 190-191, 1996.
Courant, R. and Robbins, H. "Trisecting the Angle." §3.3.3 in What Is Mathematics?:
An Elementary Approach to Ideas and Methods, 2nd ed. Oxford, England: Oxford
University Press, pp. 137-138, 1996.
Coxeter, H. S. M. "Angle Trisection." §2.2 in Introduction to Geometry, 2nd ed. New
York: Wiley, p. 28, 1969.
Dixon, R. Mathographics. New York: Dover, pp. 50-51, 1991.
Dörrie, H. "Trisection of an Angle." §36 in 100 Great Problems of Elementary
Mathematics: Their History and Solutions. New York: Dover, pp. 172-177, 1965.
Dudley, U. The Trisectors. Washington, DC: Math. Assoc. Amer., 1994.
Geometry Center. "Angle Trisection."
http://www.geom.umn.edu:80/docs/forum/angtri/.
Honsberger, R. More Mathematical Morsels. Washington, DC: Math. Assoc. Amer.,
pp. 25-26, 1991.
Klein, F. "The Delian Problem and the Trisection of the Angle." Ch. 2 in "Famous
Problems of Elementary Geometry: The Duplication of the Cube, the Trisection of the
Angle, and the Quadrature of the Circle." In Famous Problems and Other Monographs.
New York: Chelsea, pp. 13-15, 1980.
Loy, J. "Trisection of an Angle." http://www.jimloy.com/geometry/trisect.htm.
Ogilvy, C. S. "Solution to Problem E 1153." Amer. Math. Monthly 62, 584, 1955.
Ogilvy, C. S. "Angle Trisection." Excursions in Geometry. New York: Dover, pp. 135-
141, 1990.
Peterson, G. "Approximation to an Angle Trisection." Two-Year Coll. Math. J. 14, 166-
167, 1983.
Scudder, H. T. "How to Trisect and Angle with a Carpenter's Square." Amer. Math.
Monthly 35, 250-251, 1928.
Sloane, N. J. A. Sequences A158599 and A158600 in "The On-Line Encyclopedia of
Integer Sequences."
Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover, 1999.
Wantzel, M. L. "Recherches sur les moyens de reconnaître si un Problème de Géométrie
peut se résoudre avec la règle et le compas." J. Math. pures appliq. 1, 366-372, 1836.
Wazewski, T. Ann. Soc. Polonaise Math. 18, 164, 1945.
Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London:
Penguin, p. 25, 1991.
Yates, R. C. The Trisection Problem. Reston, VA: National Council of Teachers of
Mathematics, 1971.
CITE THIS AS:
Weisstein, Eric W. "Angle Trisection." From MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/AngleTrisection.html
angle trisection explanation
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Trisection of an Angle
© Copyright 1997 and 2003, Jim Loy
Under construction (just kidding, sort of).
This page is divided into seven parts:
Part I - Possible vs. Impossible
Part II - The Rules
Part III - Close, But No Banana
Part IV - "Cheating" (violating the rules)
Part V - "Cheating" (using other tools)
Part VI - "Cheating" (using curves other than circles)
Part VII - "Cheating" (choosing an arbitrary point)
Part VIII - Comments
Part I - Possible vs. Impossible
In Plane Geometry, constructions are done with compasses (for drawing
circles and arcs, and duplicating lengths, sometime called "a compass") and
straightedge (without marks on it, for drawing straight line segments
through two points). See Geometric Constructions. With these tools (see
the diagram), an amazing number of things can be done. But, it is fairly
well known that it is impossible to trisect (divide into three equal parts) a
general angle, using these tools. Another way to say this is that a general
arc cannot be trisected. The public and the newspapers seem to think that
this means that mathematicians don't know how to trisect an angle; well they don't, not
with these tools. But they can estimate a trisection to any accuracy that you want.
Certain angles (90° for example) can be trisected. A general angle cannot. In fact, a 60°
angle cannot. It is usually much more difficult to prove that something is impossible,
than it is to prove that something is possible. In this case, mathematicians had to show
just what kinds of lengths could be constructed. And then they could show that other
lengths could not be constructed, because they were not the right kinds of lengths.
What can be done with these tools? Given a length a, we can multiply this length by any
integer, and divided it by any integer. Together, these allow us to multiply this length by
any rational number. Given two lengths a and b (and sometimes a unit length), we can
add them together, subtract them, multiply them, divide one by the other. Given a length
a and a unit length, we can find the length that is the square root of a. These are the only
things we can do, with these tools. This has been proven, but I won't do that here. We
can do these operations in many many combinations of ways [like sqr(a+sqr(b+sqr(c))),
for example, where sqr() is the square root function]. But, if a construction involves a
length that cannot be expressed by some combination of ratios and sums and square
roots, then it cannot be constructed.
Towards a proof: If we begin with two points, the distance between them can be called
1. With those two points, and compasses and straightedge, we can produce any integer
(as we can do addition and multiplication with these tools). Since we can perform
division with compasses and straightedge, we can produce any fraction (rational
number). We can easily construct the square root of any number that we have already
constructed, including earlier square roots, using the Pythagorean theorem. We can
combine rationals and square roots with these tools. What more can we do? The answer
is, "Nothing." We can draw lines through points and we can draw circles. These involve
addition, subtraction, multiplication, division, and taking the square root (Pythagorean
theorem). OK, I can create any number of a certain kind (rationals and square roots and
their combinations) with these tools. I am now considering some points and I am going
to create a new point. I can do this by using previous lengths of our certain kind and
drawing arcs which intersect, or by drawing lines which intersect with lines or arcs. We
find that if the positions of our points are represented by numbers of our certain kind
(coordinates of some kind), and we find our new point using arcs with radii of our
certain kind, then we always produce new lengths of the same certain kind. If we
create a new point by the intersection of two lines or an arc and a line, which were
based upon numbers of our certain kind, then we always produce new lengths of the
same certain kind. We add no new numbers of any kind (like cube roots), by
connecting points or drawing circles. I still have not proved that, but we are much
closer. Our certain kind of number has other names, including "ruler and compass
number."
One of the classic problems that cannot be done, is to construct a square the same area
as a given circle. This is called squaring the circle. This problem amounts to taking a
unit length and constructing a length equal to pi. There is no way, in a finite number of
steps, to represent pi in the form of ratios and sums and square roots, as was shown in
1882, by Ferdinand Lindemann. Another classic problem is to construct a cube that is
twice the volume of a given cube. This is called doubling the cube. This problem
amounts to taking the cube root of a length. This too cannot be done in a finite number
of steps, using the above operations, as was shown by Wantzel in 1836.
The trisection of an angle involves the solving of a cubic equation, something that
cannot be done, in general, using the above operations. This too was shown by Wantzel
in 1836. After Wantzel published his discovery, Gauss claimed that he had proved it in
about 1800, but had never published. A surprising result of this was that a 60 degree
angle cannot be trisected (in other words, a 20 degree angle cannot be constructed), and
so a regular nonagon (a nine-sided polygon) cannot be constructed, as they involve the
construction of a cube root. Karl Friedrich Gauss (Gauß) then showed just which
regular polygons could be constructed, and he explained how to construct a regular 17-
gon.
In the diagram on the right, the length of the trisecting chord x is
the root of the equation (in which x^3 means x cubed) AB=3x-
x^3 (discovered, or maybe rediscovered from Arab sources, by
Pitiscus about 1600). In general, a cubic equation cannot be
solved by construction, with compasses and straightedge. This
equation can apparently be deduced from the triple angle
equation: sin(3a)=3sin(a)-4sin^3(a), a cubic equation which can
be derived from the sine of the sum of two angles formula [sin(a+b)=sin(a)cos(b)-
cos(a)sin(b)].
I said above that the trisection (using compasses and unmarked straightedge in a finite
number of steps) of a general angle was shown to be impossible, back in the 19th
century. While it is difficult to reproduce that proof (I may attempt it some day), that
should be good enough to disprove any future attempted trisection. We shouldn't have
to disprove each attempted trisection as they come up. They are already disproved. Of
course these attempted trisections may be very close to real trisections (within small
fractions of degrees). Some are very close for small angles. Others are very close for
other angles. Maybe they are close enough for practical purposes. But none of them can
be exact.
I received this email. I guess I am stifling innovation when I point out that some things
are actually impossible:
I read your article on angle trisection. You seem intelligent enough, but possibly too
pompous to be a mathematician. I solved this problem when I was 13. You certainly
don't have to disprove every new idea that comes along, you probably don't even need
to see most of them. But to rest on an incomplete 200 year old theory with holes in it
and thereby attempt to stifle innovative thought is exemplary of the kind of arrogance
which allows fools to call themselves educated. Good day.
Can't pompous people be mathematicians? Darn. I may indeed be pompous in other
ways, but I was just reporting what geometry of today says. We have a proof that angle
trisection (using certain simple tools) cannot be done. You cannot disprove a proof just
by calling it outdated. In fact, you cannot expect people to believe you when you say
you have done something considered impossible, without any proof whatsoever. Thank
you for informing me that you can do the impossible; forgive me for not believing you.
I told him, "Of course you did not solve this problem when you were 13." He seemed to
take this as a cruel blow, and vowed to never write to me again. Too bad.
Stifling innovation? I may have a different temperament from the person who sent me
that email. Personally, I am intrigued as all get out by a problem that I know to be
impossible. How do we know it is impossible? That must be hard to prove. Why doesn't
this idea work or this other idea? I've learned some geometry by working on these
impossible problems.
Other people have sent their trisections to me. Fortunately, most of the methods have
been fairly easy to prove to be incorrect. Some have been difficult to understand at first.
But, once I could use the method to draw the proposed trisection of an arbitrary angle,
the three angles have never been the same. Usually, it was easy to see that the trisection
method was not working, just by trying to trisect a large angle (maybe well over 90
degrees).
The difficult ones are the ones that require cheating (like Archimedes method below).
One person said he could trisect an angle using an ellipse. Well, you may be able to
draw some points of an ellipse, using compasses and straightedge, but you cannot draw
the ellipse with these tools. Anyway, I have never heard that an ellipse could help trisect
an angle, so I am somewhat skeptical.
All these trisection attempts are experimental geometry. You try a method that might be
close to a trisection, and try to measure the angles to see if they are the same. They seem
to be the same angle. Ah, I've found a trisection. Wrong, that is not how geometry
works. Mathematicians prove everything they do, because you cannot ever be sure that
something is true, unless you can prove it. And trisection is a good example of this.
Four books which prove that trisection is impossible with compasses and straightedge
are The Trisection Problem by Robert C. Yates, Geometric Constructions by George E.
Martin, Euclidean Geometry and Transformations by Clayton W. Dodge, and Ruler and
the Round: Classic Problems in Geometric Constructions by Nicholas D. Kazarinoff.
The algebraic theory concerning this is called Galois theory (a part of group theory),
and there are a lot of books about that. Also see:
Trisecting Angles and Squaring Circles Using Geometric Tools
Why Trisecting the Angle is Impossible
Math Forum's "Impossible" Geometric Constructions
Of course the proof is difficult. See An Analogy, at the bottom of this page, to get a
simplified flavor of what the proof is actually about.
Part II - The Rules
The Rules: In using the classic tools (straightedge (and pencil) and compasses), it is
legal to use them only in the simple and obvious ways. We deduce the rules of this
game mostly from the constructions of Euclid. These rules are seldom spelled out,
although we do usually spell out that we can't use marks on the straightedge. One of my
emails suggested holding the compasses against the straightedge while moving the
straightedge. This amounts to making a not so simple tool of the two pieces of
equipment, which is obviously illegal. As I said at the top, we use compasses for
drawing circles and arcs, and duplicating lengths, and we use a straightedge (without
marks on it) for drawing straight line segments through two points. We can do nothing
else. We cannot even use the compasses to verify if one distance is equal to another
distance. And we cannot eyeball a point and a line to tell if the point is on the line
(sometimes we have to go to great trouble to prove that a point is on a line). And, the
mistake which is repeated many times on this page, you cannot use compasses to
measure or duplicate the length of an arc on circles of different radii (see Part IV,
trisection #6).
You cannot use these tools to design other tools like the tomahawk shown below; in
other words, you cannot pick up parts of your drawing and move them around as tools
in their own right. While you can draw some points of a complicated curve with these
classic tools, you cannot actually draw these curves. We cannot fold up the plane like
origami.
Below, you will see bogus trisections which come within a hundredth of a degree of
being right. Isn't that close enough? You cannot measure 1/100 degree on your
protractor. And you can't even be that accurate with real compasses and straightedge.
Well, our crude diagrams are a poor imitation of the perfect constructions that we are
describing. When I get out my compasses and straightedge, and bisect an angle, the
result is inaccurate. But when I describe the bisection of an angle, by drawing arcs, what
I describe is perfect, with perfect lines and perfect arcs, using fictional perfect
compasses and straightedge. And I end up with a perfect bisection, regardless of the
crudity of my drawings. This mathematical perfection is also part of the rules. We
pretend that we are using perfect instruments which draw perfect lines and circles, so
we can then prove things about what we have constructed. So it makes no sense to
measure an angle with a protractor and say, "Yup, that's 20 degrees." We have to prove
it.
Our straightedge is often called a "ruler" (even in many books on geometry), but (if we
want to follow the rules) we must not use the markings on the ruler.
Part III - Close, But No Banana
Here we see some attempts at trisection which are close, but not exact.
1. Trisect the chord: This diagram shows the most common
attempt at trisection. Points are marked on the sides of the angle,
an equal distance from the vertex. A line segment is drawn
between these two points. This segment is trisected (easy to do,
see next paragraph), resulting in the marking of two points. Lines
are drawn from these two points to the vertex. These two lines almost trisect the angle.
It gets fairly obvious that this doesn't work, if the angle is fairly large. For a proof that
this is not a real trisection, see below. Many other attempts at trisection are just this one
in a more complicated guise (also see below).
By the way, this diagram shows how to trisect a line segment
(AB). Through A, draw an arbitrary line. On this line mark off
an arbitrary segment, starting at A. Duplicate this segment twice,
producing a trisected segment AC (see the diagram). Draw the
line BC. Through the other two points on the trisected segment,
draw lines parallel to BC. These two lines trisect the segment
AB. You can use the same method to divide a segment into any
number of congruent (equal lengthed) smaller segments. Also see Geometric
Constructions. There are other ways to trisect a
line segment.
Here is a proof that the attempted trisection is
not a real trisection. We are trying to trisect
angle A by trisecting the line segment EF. For
this proof, extend AC to B so that BC=AC, then
connect BE. Triangle ACD is an isosceles
triangle with congruent sides AC and AD [Triangles ACE and ADF are congruent by
SAS (see Congruence Of Triangles, Part I), and that makes angle ACD=angle ADC].
Triangle BCE is congruent to triangle ACD [SAS]. Angle ACD is an acute angle. Then
angle ACE is an obtuse angle.
Now, let's assume that this actually is a valid trisection (for the purpose of finding a
contradiction, and proving that it is not a valid trisection). Then angle CAE is congruent
to angle CBE. Then triangle EAB is isosceles, and AE=BE [base angles are =]. Then
triangle ACE is isosceles [AC=AE]. Triangle ACE is congruent to triangle ACD [SAS
or other]. Then angle ACE=angle ACD, which is acute. But above we showed that ACE
is obtuse. So we have our contradiction. And our attempted trisection is not a valid
trisection.
From this proof, we can guess that the attempted trisection is really close when AE is
about the same length as AC. This happens when the angle A is small. But we knew that
already, because to trisect angle A, we need to trisect the arc EF, not the line segment
EF. Well, when angle A is small, arc EF is very close (in more than one way) to
segment EF. To see just how accurate this method is for small angles, see method #10,
below.
Also, it is apparently easy to prove (in the same diagram) that if the angle is actually
trisected (instead of the line segment EF), then CD is shorter than DF.
Here is another proof that trisecting the line segment does not trisect the angle. With
that method, try to trisect an angle of 60 degrees. You should find that the three angles
are measurably different. The middle angle is about 2 or 3 degrees larger. So we have a
counterexample. Which proves that that method is not a real trisection. Angles larger
than 60 degrees would make this proof even more obvious. Using this method to trisect
60 degrees, the smaller angle is 19.1066 degrees, which is 0.8934 degrees off. I
calculated the angle with eight digits of accuracy, and then rounded off to six digits.
Using this method on a 120 degree angle, we get 30 degrees (which may be exact?),
which is 1/4 of the angle, not 1/3.
Calling the angle that we are trying to trisect a, and the smaller angle that is our
estimated trisection b, here is an equation that I get, relating a and b: b = (sin a)/sqrt(9-
8sin^2(a/2)), where sqrt() is the square root function, and sin^2(a/2) is the sine squared
of a/2.
1a. The above method disguised: Here is an
attempted trisection which was sent to me by
email, a couple of years ago. I have redrawn it
(and added some lines to help with my
comments). It was originally done with
compasses only. Two equal segments are
marked off on each of the two rays of the
angle out to B and C. For the purposes of my
comments below, I have extended the rays out
to I and J. Then the midpoint between B and C is determined with compasses. Then we
mark off the two points between I and J, again using the compasses. These two points
were then claimed to trisect the original angle A. I drew red lines to them (from A). It is
easy to show (using the many small congruent triangles that I have drawn, some of
which are mirror images of others) that this construction trisects line segment IJ. And
that makes it equivalent to the attempted trisection #1 above, and is not a real trisection
of angle A.
1b. Disguised further: Here is another attempted trisection that I received by email
recently. Lines AB, AC and BC are bisected. And then the
various other lines are drawn. It was hypothesized that
points G and H trisect angle A. There are a couple of
parallelograms (BDFE and CDEF). And the diagonals of a
parallelogram bisect each other. So, G and H are at the
midpoints of DE and DF respectively. We can compare this
diagram with the previous diagram. Many of the same
congruent triangles appear in both. It becomes obvious that these two methods define
the same lines (the ones that I have drawn in red). It is not immediately obvious that this
method trisects line BC. But comparing this diagram with the previous one, it becomes
obvious that this is the same trisection. Again, this is not a real trisection of angle A.
1c. And further: Here, with angle AOB and
AO=BO, we bisect segment OB at C, then bisect
segment AC at D, which nearly trisects the angle.
This produces the same exact angle as method #1
above, and is close to a true trisection for small
angles. That should not be difficult to prove. So if
we try it on a 60 degree angle, we will get 19.11
degrees, which is 0.89 degrees off, better than some.
1d. Still further: Here, with angle AOB and AO=BO,
we bisect segment OB at C, then trisect segment AO at D
(AD = AO/3). The intersection of AC and BD (E) nearly
trisects the angle. This produces the same exact angle as
method #1 above, and is close for small angles. Again,
trying it on a 60 degree angle, we will get 19.11 degrees,
which is 0.89 degrees off.
1e. Even further: I got this by email, the
sender's initials are A. D. . We are trisecting
angle A. We make AB = AC, bisect AB at D,
and draw the perpendicular bisector EF of CD
at E. We draw the arc with center at C and
radius CD, intersecting EF at F. We bisect CF
at G, and draw DG. DG intersects EF at H.
The bisector of angle DHE intersects BC at K,
which supposedly trisects angle A. Instead, K
trisects segment BC (and CD and DE). You might want to show that, before reading the
next paragraph.
OK, if you draw segment DF, then triangle CDF is equilateral. That makes triangle
DHE a 30-60-90 right triangle, or half of a small equilateral triangle; I have drawn the
rest of the equilateral triangle faintly. DE and HK are then medians of the equilateral
triangle, and medians trisect each other (Euclid). So K trisects DE, and CD, and BC.
And this trisection method is a well-disguised form of method 1 above.
2. Geometric series: Here is another interesting attempt at trisection of an angle. Divide
the angle into fourths, easily done by bisecting twice. To this fourth, add 1/16 of the
original angle (1/4 of the fourth). Then add 1/64, and 1/256, etc. forever. We have an
infinite sum that adds up to 1/3 of the original angle.
The equation is: 1/3 = 1/4+1/16+1/64+1/256... This is just a Geometric Series. It is easy
to show that the sum is 1/3.
This construction is not something that you can actually do, because it takes an infinite
number of steps. Such an approach is not normally stated as being against the rules. But,
it is taken for granted that it is against the rules. Notice that my article said the various
impossible problems could not be done, "in a finite number of steps."
You can stop well before you have done infinitely many steps, and your approximate
trisection can be very accurate, as accurate as you want. Using the first three terms of
the series to trisect 60 degrees, I get 19.6875 degrees. The first four terms give 19.9219,
which is 0.0781 degrees off. We can, of course, use more terms to get more accuracy.
A similar trisection is to construct 1/4+1/12 = 1/3. Of course you would have to trisect a
1/4 to get 1/12. But it is a smaller angle, and the approximation is probably more
accurate, depending on the approximate method of trisection that you choose (see
method #10, below). 1/4+1/16+1/48 is also similar, where you have to trisect a 1/16 to
get 1/48. And you can do the same with any number of terms of the above infinite
series. It always involves trisecting some very small angle, and so is never exact. But it
can be as accurate as you want.
There are other infinite series which converge to 1/3, including other geometric series:
1/3 = 1/2-1/4+1/8-1/16+...
3. Mechanix Illustrated: Here is
a compasses and straightedge
method (which I have simplified)
that was apparently given in
Mechanix Illustrated, Feb. 1966.
It is really really close for angles
less than 90 degrees, within a few
hundredths of a degree. We are
trying to trisect angle AOB
(having drawn a circle about O with A and B on the circle). OC bisects the angle. Then
we draw two lines parallel to OC, through A and B. EC=CD, and D is the center of the
large circle. The attempted trisection is defined as shown. Probably the easiest way to
show that it is not a real trisection is to use it on some angle, and then deduce what the
three smaller angles really are, showing that the center one is not equal to the two outer
ones. Attempting to trisect a 60 degree angle, I find that the middle angle (FOG) is
20.0226 degrees, which is 0.0226 degrees off.
4. Arcs with the same chord: Some
attempts at trisection involve something
like this diagram (I've divided the
diagram into two pieces, upper and
lower, to make it clearer). We are trying
to trisect the larger red angle (LAM). We
have not yet determinded where the
points L and M are.
We estimate the trisection (somehow,
maybe by eyeballing or by some
complicated construction) and get the
small red angle (CBD) in the lower half
of the diagram, with its arc CD. We then
triple this angle, producing angle CBF
(which should be approximately the
same size as LAM). We then duplicate
the chord CF inside the angle LAM, producing the chord LM and fixing the points L
and M. The upper left part of the diagram shows how this chord can be duplicated. Then
we draw the arc LM, with center A. Then we duplicate the small circle centered at D,
and place this duplicate centered at M. This gives us the point O, on this circle and on
the arc LM, which very nearly
trisects the angle LAM.
Chords CF and LM are of equal
length. But arcs CF and LM have
different centers and different
radii. Nevertheless, people seem
to think that the two arcs are of
equal length, because their
chords are of equal length. On
the right we see the area around point O, blown up about 20 times. I have moved angle
FBC so that chords FC and LM exactly coincide (the green line here). As you can see,
the two arcs do not coincide. They are very very close (and D nearly coincides with O),
but they do not coincide. After all, it would be quite a coincidence if my estimated
trisection fit the original angle exactly. I think that it is fairly easy to show that if the
chords are the same length, but the radii are different, that the arc lengths are different,
in situations similar to the above situation.
But you can see why mere measurement of
the angles would imply that a trisection
had been accomplished.
A site on the WWW used this method to
trisect a 60 degree angle by trisecting a 45
degree angle outside of it. Here we have my drawing of this. The red lines are the 60
degree angle and its attempted trisection. The blue lines are the 45 degree angle and its
actual trisection. The attempted trisection gives an angle of 19.87 degrees, which is 0.13
degrees off. 45 degrees is not very close to 60 degrees, a better angle will produce
better results.
5. A straight line?: I have
been corresponding with a
person (initials H. W. S. )
who came up with this
attempt at trisecting a 60
degree angle. In this diagram
(a 30-60-90 degree triangle),
point H is positioned so that
dividing the right angle C
into pieces of some
proportion will divide the
angle B into the same
proportion. If we position
point H just right, we can
bisect both angles. And if we
position H at some other place, we can trisect both angles. Since it is easy to trisect the
90 degree angle, then we can use H to trisect the 60 degree angle. We could also use
other positions of H to divide the angles into any other equal number of smaller angles.
This person had come up with a theorem which implied that all possible positions of
point H lie on a straight line (the line defined by A and the incenter of the triangle; see
The Centers of a Triangle). Using this theorem, it is then easy to trisect the 60 degree
angle.
Unfortunately, all possible positions of H do not lie on that line (in other words, this
person's theorem is false). In the first diagram above, I have drawn a green line
representing the path of H as angle BCH changes. And it appears to be a curve, as one
might guess. I think it can be proved that this curve is convex in one direction, and if so
no three points on that curve can lie on the same straight line. But this curve is very
close to being straight for much of its length, and so a person might be fooled into
thinking that it is straight. Pretending that the curve is a straight line, in the second
diagram above right, we find that angle EBC is 38.7940 degrees, which is 1.2060
degrees off.
This curve is a rather complicated trigonometric curve. A trigonometric curve depends
upon angles and lengths and trig functions, not just lengths as do the more familiar
curves (Conic Sections).
This attempted trisection can be used to attempt the trisection of other angles, by using
different triangles. Such attempts produce other curves which are also nearly straight
lines.
6. My method: I suspect that this method has been used by someone else, in ancient
times. But, so far, I have seen no evidence of
that. So, for the time being, this one is my
invention, and it is extremely accurate.
We are trying to trisect angle BAC, and we
choose a point D (by some method, either
eyeballing it, or through any of the other
trisection attempts) which might trisect the
angle. We then duplicate this angle (BAD)
twice, producing points E and F in the
diagram. Well, D was just a guess, and the real
trisection point lies somewhere between D and F (F can be on either side of D). In fact it
is closer to D than it is to F. An excellent guess is 1/3 of the way from D to F. A new
point G, 1/3 of the way from D to F is really close to a trisection, but it is not perfect.
What is wrong with G as a trisection point? Doesn't it seem like it should work? The
entire angle (BAC) is 3x (with x being the angle of our first estimate) plus a small angle
m: A = 3x+m. A true trisection is 1/3 that, exactly: A/3 = x+m/3. So if we add our first
estimate to one third of our small angle m, we should get an exact trisection. But we
chose point G by trisecting the chord DF, not the angle DAF. So we were just using
method #1 above, which we already proved was inaccurate. The length of segment DG
is close to the length of the arc DG, but it is never exact. We are using method #1 on a
very small angle, and so the result should be very accurate. We can use G as an
estimate, and repeat the above approximation, to get an even better approximation.
Using this method (with method #1 above as a first estimate) on a 60 degree angle, we
get an estimate of 19.9999276, which is only 0.0000724 degrees off, making this the
second best trisection construction that I have tested (see #25 below). We can't possibly
draw our lines with that kind of accuracy, so isn't that close enough to be called exact?
No, it's still not a real trisection, because it is not exact. We are assuming (pretending)
that we are using perfect tools here. This method is an extremely good estimate, but it is
still an estimate.
7. "New theory": Here we see an attempted
trisection from New Theory of Trisection by F.
C. (I have decided to use initials instead of
names, for people who think that they have
found the elusive trisection), a geometry teacher,
and he claims to be a mathematician. On the
back cover, he admits that the construction is
impossible, and then says that he has done it. We
are trying to trisect angle XOY. We draw circle
O, with radius 1, which determines point S on
line XO (on the other side of O from X). Draw
SO2 perpendicular to line OY. Bisect SO2 at M1.
Draw circle M1 with radius equal to segment
O2N, defining point M on line OY as shown.
Draw O1 as shown, so O1M = OM. Then angle OO1S is the trisection of angle XOY.
This attempt is very accurate for angles less than 110 degrees. But there is a flaw in F.
C.'s proof. In particular, his theorem #11 is sloppily done, and is false.
His method is exact for 90 degrees (and 0
degrees). Using simple trigonometry, a calculator
with trig functions, and the Pythagorean theorem,
testing his method on a 60 degree angle produces
an angle of 20.05512386 degrees, which is
0.0551 degrees off, close but no banana. He has
patented a tool based upon this method, and I
suppose it is pretty darned accurate. Of course
there are simpler tools that are more accurate (see
Part V below).
You might be interested in seeing how I
calculated that 20.05512386 angle. It's not very
difficult. On the right we see my diagram for a 60
degree angle. These lengths can be deduced
(mostly using the Pythagorean theorem):
OX = OS = 1
OO2 = 1/2
O2M1 = sqrt(3)/4
O2S = sqrt(3)/2
M1M = O2N = 3/2
OM = MO1= (sqrt(33)-2)/4
O1O2 = (sqrt(33)-1)/2
tan(O1) = sqrt(3)(sqrt(33)+1)/32 = 0.3650601618
tan(20 degrees) = 0.3639702343
Every value is exact, except the two decimal values of the tangents, which are accurate
to ten decimal places. Taking the arctan of 0.3650601618, I get 20.05512386 degrees,
which is not 20 degrees. I read the above tan(20) off a calculator. I have gotten an even
more accurate value by estimating the solution of the cubic equation for the triple angle
tangent function, and the above value is accurate to the last displayed digit.
This trisection is based loosely upon Archimedes' trisection (Part IV, method #1,
below). F. C. doesn't believe my estimate of his error (0.0551 degrees) because,
"mathematical reason has proved that an algebraic method is not appropriate method for
discussing trisection." Hm? He doesn't allow trigonometry, either. So, his method
cannot be disproved, even by counterexample.
Mr. F. C. makes some extraordinary claims, among these:
Proof by Autocad: His method has checked out perfectly, using Autocad.
Proof by "Have I ever lied to you?": He wrote to me, "If my theory is not true, it
would not be published."
Trigonometry is inherently inaccurate: He bases this idea on the truism that
sin(x)<x<tan(x) (with x in radians), while not explaining what this has to do
with anything. Apparently he thinks that sines and tangents do not have exact
values.
He also feels that people who believe as he does have been condemned by the
mathematical community.
8. Trisect a circle:
I received a description of this by email. It is a variation of method #6, above. We
divide our angle into four equal arcs (bisect it twice), then construct a square with the
chord CD (in the diagram) as a side. We circumscribe the square with a circle. We
pretend that the circumference of this circle is equal to the arc length BC of our angle.
We then trisect the circle, by inscribing an equilateral triangle. Then we transfer the side
of the equilateral triangle to our arc BC (B'C' on the right part of the above diagram),
again pretending that the circumference of the circle is equal to the arc length BC.
Trisecting a 60 degree angle like this, I get an 18.40 degree angle, which is 1.60 degrees
off. This method was inspired by rolling an angle into a cone, and then trisecting the
circular base. See paper folding, below.
9. Not very close: Here is a trisection that I got in my email.
This may be what that person meant. We want to trisect angle
AOB. We trisect the side AO, and make AR = AO/3. We draw
the circle with center O and radius R. Then we draw the circle
through A (I assume) tangent to line OB at B (this circle is not
very difficult to construct). These two circles intersect at a point
X (within the angle). And angle AOX "appears to be the
trisection of the angle." This is actually very close for small angles. Using this method
on a 60 degree angle, I get a "trisection" of 24.73 degrees, which is 4.73 degrees off,
not very close. Of course, I may have misunderstood, as this person did not say that the
circle tangent to OB should go through A. But without that condition, the angle AOX
could be almost anything.
10. My improvement: Here is my improvement of the
above method. Instead of using arc RX above, I draw
perpendicular CD (AC = AO/3), to intersect the same arc. Testing this out on a 60
degree angle, I get 21.61 degrees, which is 1.61 degrees off, a significant improvement.
And it too gets very close for small angles.
11. LA
Times: Here
is a
construction
(which I
have
simplified
slightly)
from the LA
Times
(advertiseme
nt by
D.W.A.,
Mar. 6,
1966). We
are trying to
trisect angle
BAC. We
draw the circle with center A and radius AC (and BC). We draw line segment BC and
bisect it at O. We draw line AO. We draw the circle with center O and radius OC. This
crosses segment AO at L. We draw lines BL and CL. We trisect segment BC at W and
J. We draw the circle with center C and radius CO. This intersects circle O at Z as
shown. Draw line LZ, which intersects circle L at M as shown. Draw the perpendicular
bisectors of MZ and MJ, which meet at S, as shown. Draw a circle with center S and
radius SZ. This crosses circle A at point T (as shown) which may trisect angle BAC.
Using this method on a 60 degree angle, I get an angle of 19.97 degrees, which is 0.03
degrees off, not bad.
Above, I said that I simplified the construction slightly. The author of this method
trisected the right angle BLC to get point M. Later he drew MZ. Well, it turns out that
LM and MZ are the same line. I merely noticed that point Z trisects the right angle
BLC. And the facts that angle BLC is a right angle, and line LZ trisects it did not need
to be mentioned during the construction. This construction also appears in Geometry by
Harold Jacobs.
12. Bisect and trisect: This one was in Martin Gardner's
Scientific American column in 1966, and he got it from
Mathematical Snapshots by Hugo Steinhaus. We want to
trisect angle AOB. We bisect it getting C. We trisect
segment CA, getting D, which "trisects" the angle. Using
this method to trisect a 60 degree angle, I get 20.11
degrees, which is 0.11 degrees off. This is just method
#1 above, trisecting a chord. But the chord is shorter, and so the trisection is
significantly more accurate.
13. Bisect twice: A natural extension of the previous
method is to bisect the angle twice and then subtract off
an estimated trisection. Here we trisect angle AOB. First
we bisect it, giving point C, then we bisect again giving
point D. Then we trisect the segment DC, giving point E,
which is our trisection point. Using this method on a 60
degree angle, I got 19.9872 degrees, which is 0.0128
degrees off. You can see that trisecting the chord of a
smaller angle produces much improved results. Of course the chord (CD) is here much
closer in length to that of the arc (CD).
14. An interesting method: This comes from
Survey of Geometry, by Howard Eves. We are
trying to trisect the angle AOB, with AO = BO.
We draw a semicircle on segment AB, facing
the point O. We trisect the semicircle (easy to
do), giving us points D and E. We bisect
segment DE twice, giving us point G (EG =
ED/4), which is our trisecting point. This
method works fairly well for small angles, but blows up for big angles. Trying it out on
60 degrees, I get 13.9 degrees, which is 6.1 degrees off, making it the worst method (for
that particular angle) of all the methods I have tested.
15. d'Ocagne's method: Unlike most of these
methods, this one (by M. d'Ocagne in 1934) was
designed only as an approximation, as the inventor
knew that an exact trisection was impossible. We have
angle AOB in circle O. We bisect the arc AB at point
C. We extend BO to D, also on circle O. We bisect OD
at E. And angle CEB is the approximately 1/3 of our original angle. Using this method
on a 60 degree angle, I get 20.10 degrees, which is 0.10 degrees off. This method is
found in a couple of places.
16. Durer's method: Here is
Durer's method (1525), which is
very accurate. We draw arc AB
with center O, and chord AB. We
trisect the chord AB, at C. Draw
CD perpendular to AB,
intersecting the arc at D. With center A and radius AD, we draw the arc DE, intersecting
the segment AB at E. We draw point F on segment EC, so EF = EC/3. With center A,
and radius AF, we draw the arc FG, intersecting the arc AB at G. line OG
approximately trisects the angle. Trying this method on 60 degrees, I get an error of
less than 0.01 degrees. I will try to get a better estimate of the error. This method is
found in several places. Durer was a famous painter and mathematician.
17. Karajordanoff's method: This is an
intentional approximation. We draw two circles
about O, one with A and B on it, and one with
twice the radius. We bisect the arc AB at C. We
draw line AC. We draw BD perpendicular to
OB, which intersects line AC at D. Se draw DE
parallel to OA, which intersects the larger circle at E, which approximately trisects the
angle. Using this method on a 60 degree angle gives us 19.97 degrees, which is 0.03
degrees off. This method is found in The Trisection Problem, by Robert C. Yates.
18. Kopf/Perron method: This is
another intentional approximation. We
bisect OC at D, and draw the
perpendicular DE (to DA). DF is 1/3
DE. We extend OD out to H, so CH =
OC. We draw a circle with center F and radius AF. Draw line CB which intersects this
arc at G. Then angle AHG is approximately 1/3 of the oringinal angle. Testing this
method on 60 degrees, it is approximately 0.01 degrees off. This method is found in
The Trisection Problem, by Robert C. Yates.
19. JJG method: This was
given in a book called The
Mathematical Atom, by J. J. G.
(a mathematician) in 1932, as a
serious trisection. We bisect the
angle with OC, then bisect
AOC with OD (as shown).
Then draw DE perpendicular to
OD, intersecting OC at E. We draw the circle with center E and radius EO. The line OD
intersects this circle at F. Draw EG parallel to OF, intersecting the large circle at G, as
shown. Draw lines EF and DG, which intersect at H. OH nearly trisects the original
angle. Trying this method on 60 degrees, I get 19.99 degrees, which is 0.01 degrees off.
This method is found in The Trisection Problem, by Robert C. Yates, who apparently
simplified J. J. G's. original diagram.
20. Between one-half and one-fourth: I received email from L. D. K. showing this
trisection. With centers at the vertex of our
angle, we draw arcs with radii of 1, 2, 3, 4,
5, etc., each intersecting the sides of our
angle. We can bisect the angle which gives
us a point on the arc with a radius of 2. We
can bisect each of those smaller angles,
giving us four points on the arc with a
radius of 4. These give us the green points
shown. With that info, we can draw the lines in our latice, and fill in the trisecting
points on the arc with a radius of 3, and even divide the arc with a radius of 5 into 5
equal arcs. We can continue out farther, and divide our angle into any number of equal
arcs. The originator of this method says that it doesn't work for angles greater than 45
degrees, which should be a clue that something is wrong. In fact, none of the lines
(except the rays of the angle itself) is a straight line. If they go through the proper
points, they are all curved. The curvature is just not very noticeable for small angles.
Erasing the unnecessary lines and points, this
method makes a nice approximation of a trisection.
Here I trisect a 30 degree angle, and get 10.04
degrees, which is not bad. The inventor of this
method was right, that it is worse for larger angles.
Trisecting a 60 degree angle, I get 20.33 degrees,
which is 0.33 degrees off, still not bad.
21. "Early folly": Above left is a method which can be found at the site Early Folly.
This is a method from back in 1951, when three men discovered a trisection method,
had it notarized, and sent it around to universities. The instructions were somewhat
vague, and it took me a while to come up with the diagram. The second diagram is my
simplification, which is magnified. Here are the steps in my own words:
We want to trisect angle AOB, with segments AO = OB. Draw the arc AB with center
O (this is done later in the original, but it makes more sense here, to make AO = OB).
Trisect AB, producing point C (much of the original method was involved in trisecting
this line). Draw the semicircle with diameter AB, on the side of AB away from O.
Trisect semicircle, getting point D as shown. Draw CD, and bisect it at E. Draw EF
perpendicular to CD. EF intersects the extended AB at some point F. Draw the arc CD
with center F. This intersects the arc AB at some point G, which we are told is a
trisection point. Draw OG.
Trying this on a 60 degree angle, I get a trisection of 20.05 degrees, which is 0.05
degrees off. The method is better for smaller angles, and not quite as good for obtuse
angles.
I think this method is one of those where someone started with a near trisection
(trisecting the chord), and then chose a point way off to the side (the farther the better,
actually). Then an arc with this point as a center is very nearly a trisection for quite a
ways along its length, so we just choose the best such point that we can come up with.
No proof was ever imagined. Method #15 above gives the same impression.
22. Four circles: I received this
by email. We are trisecting angle
ABC. We draw the first circle
with center B, and label the
intersections with the rays of the
angle A and C. We draw the chord
AC. We bisect the chord at D, and
draw BD. We draw a second circle with center B and radius 2AB, and we draw a third
circle with center D and radius 2AB (the two larger circles in the diagram). Line BD
intersects this last circle at E. With center E and radius AB, we draw a fourth circle,
which intersects the second circle at two points, F and G, the two trisectors. The
trisection not very good. Using 60 degrees, I get a trisection of 18.01 degrees, which is
1.99 degrees off. This method never gets particularly close to a trisection. Using 3
degrees, I get a trisection of 0.89 degrees.
23. Same as #1?: I got this one in my email. We are
trisecting angle AOB, with AO = BO. We bisect AB
at C. Then we draw circle C, with radius AC, and
circle A with the same radius. These circles intersect
at point D, on the side of AB away from point O. And
this may be a trisection point. Trying it on a 60 degree
angle, I get 19.11 degrees, which is 0.89 degrees off,
the same as method #1, above. So is this equivalent to method #1? No, only when the
angle is 60 degrees. This method works well for angles close to 90 degrees (it is perfect
for 90 degrees and 180 degrees), but gets worse and worse for small angles. As the
angles get close to zero, this method approaches 1/4, not 1/3.
24. Same as above: I got this one in the mail from A. R.
S. from India. In the diagram, we draw the arcs with
centers C and D and radius CD. These meet at E, outside the arc CD. We bisect CE at
X, which is our trisection point. A little bit of simple geometry shows that this point X
is the same as point D in the previous method, and trying it on a 60 degree angle gives a
trisection of 19.11 degrees, which is 0.89 degrees off.
25. Mark Stark's amazing
estimation: This one was
never meant to be a true
trisection, but an estimate,
and it is very very accurate.
Find the interactive Java
applet at Mark Stark's Angle
"Trisection". We are trying to
trisect angle AOB, with AO =
BO. We draw the circle with center O and radius OA, and we draw the segment AB. We
then choose an estimate of a trisection point D on line AB, somewhere between 1/2 and
1/4 of the way from A to B. Then we draw the circle with center A and radius AD,
giving us point E where this circle intersects the arc AB. We extend line AO out to F as
shown, so OF = 3AO. We draw the circle with radius OF and center O. Line DE
intersects this circle at a point G, as shown. We then draw line GO, which intersects arc
AB at E', which is a much better estimate of our trisection point than E was. Mr. Stark
says that repeating this estimate one more time (using E' to get a new point E'') will
have an error of less than 0.0000000001 degrees.
Trying this method on a 60 degree angle, and choosing D as AD = AB/3 (method #1
above), gives us an angle AOE' of 19.9999937 degrees, which is only 0.0000063
degrees off, making this the best method that I have tested.
One might assume that the length OG can maybe be almost any length. Above, OG =
3AO, but 4AO or 2AO would actually produce good estimates of our trisection point.
However, it turns out that G is almost ideally positioned. If you take several different
estimates D (between 1/2 and 1/4 on AB as described above) and draw their lines (DE),
they will intersect at several points very near to the position of G in the diagram.
26. Missing theorem: There is a book called The Missing Theorem by L. O. R. (a later
edition is called Angular Unity), about triangles which have one angle twice another
angle (see Archimedes' method, below). The author claims to prove that any trisection is
possible, using the classical tools in the classical way. But he cannot show us how that
might be done. His proof is a little bizarre.
27. A natural attempt: I recently received this in
my email (from J. L., who is not me), but I am sure
that I have seen it before. We want to trisect angle
AOB. We draw an arc AB with center at O. We
trisect segment AB at C. We draw CD perpendicular to AB, and intersecting the arc as
shown. And D is our trisection point.
When a person finds that C (in this diagram) is a rather poor trisection point, it is natural
to seek a better one. D is the next most natural candidate, and is a distinct improvement.
Using this method on a 60 degree angle, we get a trisection (angle AOD) of 20.40
degrees, which is 0.40 degrees off, still not very accurate. As with most of these
methods, the accuracy improves for smaller angles.
Angle AOC is too small. AOD is too large, but closer. The next guess might be some
point on segment CD, closer to D than to C. Any such point would be a better guess
than C or D. Some of the above methods (methods 11, 16, and 21, for example) are very
probably based on this reasoning.
28. A parallel angle: I received this in
my email. We are trying to trisect angle
AOB. We draw an arc AB. We make a
guess of a trisecting chord CD, centered
about the angle bisector as shown. We
duplicate CD giving CE and DG as
shown. We draw EF parallel to AO,
intersecting the angle bisector at F. We
draw FG. Angle EFG = angle AOB, and
we would seem to have trisected it with
the chord CD (and CE and DG). But
these chords and their arc are not chords
and arc belonging to angle EFG; the
center (O) is the wrong center, which should be F. And so, the trisection is surely not
exact.
The accuracy of this method depends on how good a guess we made when choosing
chord CD. If the guess was exact, then the trisection is exact. Using this method on a 60
degree angle, and choosing a CD that is 90% of a perfect guess, I get a trisection of
19.95 degrees, or 0.05 degrees off. So this is a fairly accurate method.
Part IV - "Cheating" (violating the rules)
See Part II - The Rules.
1. Archimedes' method: This method is exact,
and is attributed to Archimedes. But has a small
flaw. We are trying to trisect angle BAC. Draw a
circle with its center at the vertex. It intersects the
sides of the angle at some points B and C (as in the
diagram). Draw line BED, as in the diagram, so E
is on the circle, and D is on line AC, and DE is the same length as the radius of the
circle (It may help you to mark the distance DE on the straight edge, in order to line
these up). Angle BDC is a perfect trisection of angle BAC. Proof: Call angle BDC x.
Triangle DEA is an isosceles triangle, so angle DEA=180-2x. That makes angle
AEB=2x. Triangle EAB is an isosceles triangle, so angle EAB=180-4x. Angle EAD=x,
so angle DAB=180-3x. That makes angle BAC=3x. So, angle BDC is 1/3 angle BAC.
The only problem is that it is impossible to draw line BED without cheating (by making
marks on the straight edge).
I received email saying that the above method doesn't work. Apparently that person
measured segments EA and ED, and they weren't an exact fit. He said that assuming AE
= ED was "a big mistake." He apparently did not read my description of the
construction. Well the method does work, regardless of the accuracy of my drawing.
It doesn't matter how far apart the marks are on our
I see on the WWW, and from my email, that some people consider this method valid,
since the rules for use of compasses and straightedge are seldom spelled out. See Part II
- The Rules.
2. An ancient method: Here is an ancient
method shown in Heath's A History of
Greek Mathematics. It is not shown as a
construction of a trisection, but as a help
in the analysis of the problem. ABC is the
trisected angle, and DBC is the trisection.
We draw a rectangle ACBE and then the point F is the intersection of BD and EA. We
find that DF = 2AB. So if we can construct line BF so that DF = 2AB, then we can
trisect the angle. And we can do this with a marked straightedge, but not with the classic
tools. This method is equivalent to Archimedes' method, above.
3. Paper folding:We are told
that an angle can be trisected
using paper folding. The book
Amazing Origami by Kunihiko
Kasahara gives the method for
trisecting a 90 degree angle,
producing angles of 30 degrees
and 60 degrees, but seems to
give no general method. On the left, we see this trisection.
We start with a square paper (we can fold any irregular paper into a square), which we
fold in half (vertically in the diagram). We then fold one of the corners over onto this
fold. It is easy to show that this produces the second side of an equilateral triangle, and a
60 degree angle, and thus a 30 degree angle.
Above right is a questionable (probably) method by which trisection can be done, using
paper folding. We roll the angle into a cone, and slowly flatten it while making sure that
we approach two folds and three equal portions of the circumference. The angle does
not have to be cut from a circle, but that simplifies the process. With this method, we
can divide an angle into any number of equal angles. A variation of this method is to
roll the angle into a cone, trisect the circular base (with an equilateral triangle) which
trisects the arc of our angle, and then unroll the angle.
Here we see a good paper folding method found
in the book Geometric Constructions by George
E. Martin and attributed to Hisashi Abe in 1980.
Here we are trying to trisect angle ABC. We
find D the midpoint of AB. We fold line a
perpendicular at B to BC, then make two lines
(lines EA and c in the diagram) perpendicular to
line a, through A and D, as in the diagram. Then
(the part which we cannot do with compasses
and straightedge) we find the fold b so that E
lies on line AB producing E' the image of E, and
so that B lies on line c, producing B' the image of B. Line BB' trisects the angle ABC.
See Origami Trisection of an Angle for the proof. Here we have found a reflecting line
for two pairs of points. We can do that with compasses and straightedge only if we
know where all four points are beforehand. Here each of the points could have been
anywhere on a given line.
The above method suggests that we can also use the edge of the paper as a marked ruler.
So, you can use any number of methods, probably including Archimedes' method
above, to fold a trisection. Most such methods would involve folding the paper more
than once simultaneously, to get your ruler to anywhere on the paper. The method
shown at the link above moves the ruler to a specific place on the paper, because it
involved only one fold to
move the ruler.
Here is another paper
folding method shown in
Geometric Constructions
by George E. Martin, and
the method and proof are
attributed to Dayoub and
Lott, for a geometry tool
called a Mira, a mirror
which allows you to draw
reflections. Paper folding produces the same reflections. Let me describe the trisection
using the diagram. We start with an arbitrary angle ABC. We find D, the midpoint of
AB. We fold line a, through D, and perpendicular to BC. We fold line b, through D, and
perpendicular to line a. Now (the part we cannot do with compasses and straightedge)
we find line c by folding so that A folds onto line a and B folds onto line b, producing
A' the image of A, and B' the image of B. The line BB' trisects angle ABC. Apparently
there are three lines which meet the criteria that I described (folding A onto A' and B
onto B'), but only one of them intersects segment AD. See the above book for the proof.
4. "Euclid Challenge": At Euclid
Challenge, we come to a diagram similar
to this (I have supplied the question mark
next to point T). The two arcs have the
same center, point B, which is out of my
picture, to the south. The lower arc has 3/4
the radius of the upper arc. The lines DD',
CC', and PP' meet at a point B. In the
previous pages, the author constructed
these points and lines so that angle DBP is
3.75 degrees (1/16 of 60 degrees) and
angle DBC is 15 degrees (1/4 of 60
degrees). We now are going to find point
T, so that angle DBT is 5 degrees, an
unconstructible angle according to
mathematicians. Let me quote the author
on this (his steps 14 through 19, rephrased slightly), the italics are mine:
Grasp with the left hand a compass leg above the legpoint (identify as leg R). Place
legpoint of leg R on point D'. Grasp with the right hand the other compass leg above the
legpoint (identify as leg S). Place legpoint of leg S on D. Move both legpoints at a
uniform rate; legpoint of leg R along arc D'C' toards point C', and legpoint of leg S
along arc DC towards point C. When the legpoint of R reaches point P', stop both
legpoints. At the location of the legpoint of leg S on arc DC mark point T -- trisection
point.
The author further explains his "uniform rate":
Confirmation of the "Uniform Rate": The maximum travel on arc D'C' = 5.625 degrees
(1/8 of 45 degrees) and on arc DC = 7.5 degrees (1/6 of 45 degrees or 1/4 of 30
degrees). These two angles, 5.625 degrees and 7.5 degrees can be constructed with a
compass, and used for test points by continuing the movement of the two legpoints
beyond the 3.75 degrees on ard K'L' [arc D'C'], and 5 degrees on arc KL [DC], to see
that legpoint R reaches test point 5.625 degrees at the same time that legpoint S reaches
test point 7.5 degrees.
This clever sliding of compass points at a uniform rate is, of course, a great violation of
the rules, even if it were possible. The idea is that arc D'C' (and any of the smaller arcs
on it) is 4/3 as long as arc DC, which is true. And this author seems to think that you
can measure fixed distances along an arc using compasses, which you cannot. If you
could, then you could make a line segment the same length as pi (which I think he does
on another page). In my diagram above, the length of the arc D'P' is supposedly the
same as the length of the arc DT. If we could do this with compasses, then angle DBT
would indeed be 5 degrees. But, we can't. See Part II - The Rules and the next
"trisection" below.
If, instead of measuring arc lengths with our compasses, we make the chords D'P' and
DT of equal length, then we get an angle DBT very close to 5 degrees, less than 1/100
of a degree off. But, as we have seen several times above, chord length and arc length
are not related in such a simple way.
To summarize my objections:
1. This method violates the normal rules.
2. If we accept this method, there is a much simpler way to use it. See "5.
Duplicating an arc length," below.
3. It can't be done. There is no precise way to move two points at the same speed,
on two different arcs, simultaneously. In fact, there is no precise, finite way to
measure arc lengths without measuring angles.
5. Duplicating an arc length: Using the same idea
(duplicating an arc length) as the previous "construction,"
we can devise a simpler trisection. In this diagram, we want
to trisect angle AOB. We make segment OC three times
segment OA. Then the length of arc CE is three times the
length of arc AB. We duplicate arc length AB and get arc
CD, which trisects angle AOB.
I have never seen this "construction" given as a valid trisection. But this duplicating of
an arc length is the main flaw in many of the trisection attempts shown above. Of course
this trisection is perfect; it just can't be done with the classic tools. If we use chords
instead of arcs, this "trisection" gives the same lengths as the one in Part III, "trisection"
#1.
Just how are chord length and arc length related? With an angle A and a radius r, the
chord length = r sin A/cos (A/2), and the arc length = 2 pi rA/360. You may remember
from trigonometry or calculus that sines and cosines are not particularly simple
functions of angles. They can usually only be estimated, and to evaluate them we use
infinite series.
6. Tripling an angle: One common mistake is to choose an angle (or construct it), then
triple it, getting a larger angle, and then claim that you have discovered a way to trisect
the larger angle. Of course, you can easily triple any angle. But to go the other way, and
trisect an arbitrary angle is what we are trying to do here. If we triple some angle, then
the larger angle is not arbitrary in any way (at least not as mathematicians use the
word). The Trisection Problem, by Robert C. Yates tells of a university president (J. J.
C) who went public to the newspapers with such a trisection.
Here is that trisection:
Draw lines BC and DF parallel to each other. Choosing
arbitrary points C and D on the two lines, we draw the arc
CF with center at D as shown. We draw lines DC and BF.
With center at F, we draw the arc DB. We draw angle DCE
equal to angle DCB. We draw AD parallel to CE and DE
parallel to BF; these two lines intersect at a point E. Then lines DF and DC trisect the
angle ADE.
How true. But angle ADE is not arbitrary. All the above accomplished was to construct
an arbitrary angle DCB and then triple it, giving us angle ADE. Angle DCB is arbitrary
(depending on where we put arbitrary points D and C), but it would be difficult to draw
a really small angle, as points D and C would have to be very far apart. There are easier
(somewhat) ways to triple an angle. You might want to verify that all of the small
angles in the diagram are equal.
7. With a ruler: The tangent of 20 degrees is roughly 0.363970234. Using a ruler with
markings that fine (or less accurate, if you're not so picky), draw a right triangle with
one leg exactly 1, and the other leg 0.363970234, and you've got a 20 degree angle. Of
course, the tangent of 20 degrees is not exactly 0.363970234, but that is more accurate
than any ruler you will ever find. So isn't that good enough? Nope, it's still an
approximation, no matter how accurate it is.
Part V - "Cheating" (using other tools)
With compasses and straightedge, only circles (and circular
arcs) and lines can be drawn. Abandoning compasses and
straightedge, it is simplicity itself to trisect any angle. Use a
protractor (see the drawing). I assume CAD programs can
trisect angles at will. We can measure the length of an arc
with a curved string or piece of metal (like spring steel). A
person could design any number of mechanical devices which can trisect any angle.
And curves have been drawn, which make trisection
easy (Part VI below). Or hard.
1. A trisection tool: Here is a little tool (apparently
invented by C. A. Laisant in 1875) which trisects
angles. It is exact, assuming the line segments are
exact. The dots are hinges. The two hinges at the far
right slide along their trisection lines. We have two
rhombuses and two of their diagonals.
We cannot use compasses and a straightedge to trisect a general angle. That has been
proven. But we can make other tools (such as the device drawn here) to trisect such an
angle. In fact, we can use a pair of compasses and a straightedge to make this device.
All that this device does is triple the smallest angle.
2. A tomahawk: Here we have a clever drafting tool called a tomahawk (or
shoemaker's knife (there is another shoemaker's
knife in geometry), invented before 1835),
which can be used to trisect angles, as shown in
the diagram. For small angles, the handle may
have to be extended. The top end is marked off
in three equal pieces, one of the marks being
the center of the semicircle. We can prove that
the trisection is exact by drawing three
congruent right triangles, as shown here.
Why can't we just draw a tomahawk using
compasses and straightedge, and then move
that part of our diagram over to our angle,
adjust its orientation, and then trisect the angle.
Well that amounts to using a tool; we are
cutting out a part of the plane and moving it,
which is a complicated process and a violation of the rules. But also, there is no way to
position our drawn tomahawk with perfect precision; in other words, we cannot draw
our tomahawk in place, on top of the angle, and trisect the angle. That may be hard for
you to believe.
3. A carpenter's square: A similar
trisection can be done with a carpenter's
square shown here. First we draw the
horizontal line shown, which is parallel to
the horizontal side of the angle, and the
width of the square from that side of the
angle. Then we position the square as shown,
with equal lengths marked off on the
righthand edge of square itself. And two of
those points trisect the angle. We can verify
that with the same three congruent triangles
that we used with the tomahawk.
Part VI - Cheating (using curves other than circles)
There are several curves that can be used to trisect angles. Such a curve is called a
trisectrix. These trisections are exact. But of course, you cannot draw these curves with
compasses and straightedge. I drew most of these curves by drawing an angle and its
triple, and then generating a curve (locus of points) by plotting the trisection point while
varying the angle. Most of these curves are of the fourth degree or higher.
1. Limacon: On the
left, we see a special
kind of limacon
(limaçon), the purple
curve that you see.
Its polar equation is r
= 1/2+cos(theta),
where theta is the
central angle. The
origin is where the
curve intersects itself. A general limacon has the
equation r = b+a cos(theta). A limacon is also called a
limacon of Pascal (named after the father of the
famous Blaise Pascal). I have discovered five ways to
use this limacon to trisect angles:
(i) On the right above is a simple method, just using the little loop of the limacon. We
want to trisect angle ABC, we make AB the right length to fit in the loop, and make BC
= AB. We then draw line AC, and where it crosses the limacon is a trisection point. If
trisecting AC was a true trisection of the angle (part III, method #1, above), then this
curve would be a circle. This seems to be the standard, well-known method. But
limacon #4, below has also been published.
(ii) Above left, I am trying to trisect angle ABC, where AB is the right length to fit in
the limacon as shown.. Make BC = AB, and draw line AC. Trisect line AC at D. Draw
DE perpendicular to AC. The point E where this intersects the limacon is a trisection
point..
(iii) Above left is yet another way which I discovered for using a limacon to trisect an
angle. Again we want to trisect angle ABC. Here we adjust the size of AB so that
segment BD (half the length BA, as shown in the diagram) fits inside the small loop.
We then draw AC and bisect it twice, giving AE = AC/4. Where line DE intersects the
large part of the limacon is a trisecting point.
(iv) Above middle is yet another way to use a limacon to trisect an angle ( found in The
Trisection Problem, by Robert C. Yates). It is closely related to method #3. We trisect
the angle AOB merely by drawing BC. Then angle OBC is 1/3 angle AOB (angle BCA
is 2/3 the same angle).
(v) Above right is yet another way to use a limacon to trisect an angle. Here OA is the
same length as from O to the bottom of the limacon. Our angle cuts the limacon (as
shown) at C. We draw a line OD a distance of OC from point A (again, as shown), and
OD is our trisection.
2. Maclaurin's
trisectrix: On the left
is Maclaurin's
trisectrix. We want to
trisect the angle ABC.
We choose length AB
so that AB = AO/3. I then label the point where our angle
crosses the loop of the curve C. Then angle AOC is the
trisection of angle ABC. The triangle BOC can be seen in
Archimedes' method (part IV, method #1, above). But point
A is not part of that method.
(ii) Above right is another way to use the Maclaurin
trisectrix. We are trisecting angle AOB. Here B is chosen so
angle ABO is a right angle. CD the perpendicular bisector of
AB (at C) intersects the loop of the curve at a trisection point. Notice that it doesn't
matter where A is (somewhere on the horizontal line), as constructing the perpendicular
bisector of AB will always produce the same line CD.
4. Conchoid of
Nicomedes?: One
WWW site calls this
the conchoid of
Nicomedes, but (as
far as I know) that
looks quite different.
We are trying to
trisect the angle
AOB. We make the length AO the same as the radius of the circle O. The side OB of
the angle intersects with the part of the curve that curls off to the left at a point B. We
duplicate length OB as AC perpendicular to OA, and C is a trisection point for our
angle. The polar equation of this curve is r = sec(theta/3) = 1/cos(theta/3).
5. Quadratrix of
Hippias: This
curve is the
quadratrix of
Hippias, and it
allows us to divide
an angle into any
number of equal
parts. Here we want
to trisect angle
AOB. The left ray
of the angle crosses
the curve at point
B, and we draw the
horizontal line BC,
with C on the y-
axis. Then we divide the y value by 3, and we get the y value of point D, also on the y-
axis. Then we draw the horizontal line DE, which crosses the curve at E, which is the
trisection point.
An equation for this curve is x = y cot(pi y/2) with x being a function of y.
Above we divided the y value of point B by 3, to get the trisection point. If we instead
divided by 5, we would have found the point which divides angle AOB into five equal
parts. Divide by 7 and get seven equal parts, etc.
This diagram, and the next one, were drawn with Geometer's Sketchpad and Paint Shop
Pro.
6. Tschirnhausen cubic: Here is
Tschirnhausen cubic, also called the
Catalan's trisectrix, and L'hospital's
cubic. Its polar equation is a = r
cos^3(theta/3). I don't know how this
can be used to trisect an angle. This
curve is related to a parabola with focus at the origin, opening toward the left in the
diagram, and the two curves intersect at (1,0) in this diagram. Apparently, both the
parabola and the cubic curve are used to trisect an angle.
I will try to figure this one out.
7. A cubic parabola: This is a cubic parabola with the
equation y = (x^3)/2. We are trying to trisect angle AOB.
Along with the curve, we have two circles of radius 1
and 2 (the two larger circles in the diagram). BO
intersects the smaller circle at C. We draw CD perpendicular to AO. OE = OD, as in the
diagram. We draw EF with a slope of 3/2. It intersects the curve at G. We draw GH
perpendicular to AO. GH intersects the larger circle at H, which is the trisection point.
There are other cubic equations that can be used in this way, with different slopes of
FG. But this one has the simplest equation. In my diagram, it may look like OC is
parallel to FG; it is not.
I found this trisection in The Trisection Problem, by Robert C. Yates. I found other
cubic equations by botching the curve drawing.
8. The "cycloid" of Ceva: Here
is a curve discovered by Ceva in
1699. We want to trisect the angle
ABC, which we fit into the blue
circle (the radius of the circle is
determined by the smallest loop of
the curve). Then we draw CD
parallel to BA until it meets the
curve to the right for the last time,
giving us point D, which trisects the angle. The intersecting point can be deduced to be
x = sin(theta) and y = cos(theta/3). A polar equation is r = 1+2cos(2 theta).
I assumed that I was not the original discoverer of this curve. But most books don't
seem to mention it. I finally found it in The Trisection Problem, by Robert C. Yates. In
that book, the position of our angle is somewhat different.
9. A hyperbola: This one is
apparently a hyperbola. It was
apparently discovered by Pappus
of Alexandria, who lived in about
the year 300. O is the center of the
hyperbola. We are trisecting angle
ABC. AB = AO = OX. We draw
AC perpendicular to AB, and CE
perpendicular to AC. We draw AD = 2BC and intersecting the curve at D. We draw DE
perpendicular to CE, and intersecting CE at E. Draw line BE, which trisects angle ABC.
We didn't need the last couple steps, as we already had an angle (OAD) which trisects
angle ABC. I guess that if a hyperbola can be used to trisect an angle, then maybe an
ellipse can too.
10. A rose curve: A three-leafed rose curve has the right
equation for the job: r = a sin(3 theta). For angles less
than 180 degrees, we just use the upper right petal (leaf). We want to trisect angle AOB.
We draw a circle half the radius of the rose curve, with center O (with B on the circle).
We draw BC perpendicular to AO, at C. We draw a larger circle with radius OA+BC.
This circle intersects the curve at two points, and we choose one of them (D) depending
on which quadrant B is in.
11. A parabola: This is the parabola y=x^2. We
have drawn a circle about the origin with radius = 2
(the purple circle). We want to trisect the angle
AOB, with A on the circle. We draw AC
perpendicular to the y-axis. We draw ED
perpendicular to the y-axis, with E two units above
the origin and ED = AC/2, as shown. We draw a
circle with center D and radius OD. This intersects
our parabola at F. We draw FB perpendicular to
OB. This perpendicular intersects our original
circle at G, which is the trisection point.
This method is found in The Trisection Problem,
by Robert C. Yates.
The following are curves which I discovered while experimenting with trisections. I will
begin the numbering again:
1. A more complicated curve: On the left are
four pieces of an intersting curve that I
discovered, which can be used to trisect an
angle. We want to trisect angle ABC. We fit
segment AB into circle B. CD is perpendicular
to BC. Where CD intersects with the curve is a
trisection point. Above right is the same curve,
as seen from a greater distance. If angle BCD is
some other fixed angle, besides 90 degrees,
then we get different positionings of the same
curve.
Each branch of our graph may be
a hyperbola. I will have to deduce
an equation, in order to tell.
2. Another complicated curve: Here is another curve that I discovered, that can be
used to trisect an angle. Just the outside loop is used. We
are trisecting angle ABC. We make segment AB the right
length to fit into the curve, then make BC = AB. We
make rhombus ABCD. We draw AC and trisect it, giving
us point E. Line DE crosses the curve at a trisection
point, F. Like the limacon above, this curve is very
nearly a circle for small angles.
3. Yet another
complicated curve: Here
is another curve that I
discovered. We want to trisect angle ABC. We choose
length AB so that AB = BD/3. Then we draw AC
perpendicular to BD (we may already have this drawn
while graphing the curve), giving us C. We draw CE
perpendicular to AC, and where it crosses the right part
of the curve is the trisction point E. So we draw BE.
This one may be a hyperbola. The two curves on the left
side of the diagram may be related hyperbolas? I'll have
to figure out the equation, in order to tell.
4.
An
oth
er: Her
e is
anot
her
cur
ve.
We want to trisect angle AOB (the
entire curve is shown on the right; I think I've seen him before). We fit OA into the
curve as shown, and draw circle O with radius OA. Then we find point C where the
small loop of the curve intersects OB, drawing a circle with center B and radius BC.
This circle intersects circle O at point D, within
the angle, which is a trisection point.
5. A useless curve: Here is a complicated and
useless curve for trisecting angles. For angles
less than 120 degrees, we use the branch of the
curve shown. We want to trisect angle AOB, we
fit our angle into the diagram, which should already contain segment AO, and the circle
O. The extended segment OB intersects the correct branch of the curve at C. Line AC
intersects the circle at the intersection point D. This curve is almost unusable for most
angles, as line AC is almost parallel to the circle for small angles, making the
intersection point hard to choose, and point C is way out there (off the chart) for most
other angles.
Sometimes I prefer the useless methods.
6. Another complicated
curve: This one is based
on the previous one, but
is more useful. We want
to trisect angle AOB. We
will use the left part of
the curve which looks
like a hyperbola, and the
circle O with radius
OA/2. Segment OB
intersects the curve at C.
We draw segment AC,
which intersects the circle at D, which is the trisection point.
This graph is the previous graph, but with a smaller circle, which makes AC cross the
circle at a better angle.
7. Another useless one: We trisect angle AOB within
circle O by extending OB to C on the correct branch of
the curve. We draw CD perpendicular to OA,
intersecting the circle at D, which is the trisection point.
This one is useless for exactly the same reasons as is
method #5.
8. A better one: Here we use
the branch of the curve that is
concave to the left. We draw
circle O, as shown. We are
trying to trisect angle AOB,
and we label B as the point on OB that intersects the correct
branch of the curve. We draw circle B with radius BO. This
circle intersects the original circle O at C within the angle,
which is a trisection point.
How did I come up with all of these interesting curves? Well, I just drew a bunch of loci
(paths of points) which involved trisections in various ways. Sometimes I came up with
a well-known curve (often a limacon), and at other times I discovered new curves.
Let me give an example. In method #2 above, which
used this diagram. I drew the angle ABF (I'll define point
F later), and then tripled it to get angle ABC. Then I
drew the rhombus, and trisected diagonal AC as shown.
Drawing segment ED gave me point F. I then varied
angle ABF and drew the path of point F, which gave me
the curve. This was fairly easy to do, using Cinderella.
Then I did a screen capture with Paint Shop Pro, and
cropped the picture, added the labels, and saved it as a
gif image with a transparent background.
Many of the images on this page, not just in this section,
were drawn in a similar way.
Part VII - "Cheating" (choosing an arbitrary point)
When proving things, or constructing things, it is sometimes useful to "choose an
arbitrary point," perhaps on a line or between two points. This is perfectly valid. If we
"draw an arbitrary line" through a point, we are essentially choosing an arbitrary point,
and then drawing the line through the two points. Some trisections involve arbitrary
points, and these arbitrary points are usually chosen so well that the trisection comes out
looking good. Other choices for these points will probably not look very good at all. In
that case, "choose an arbitrary point" is a scam (the scammer may be scamming
his/herself, so don't blame him/her). You should try moving the arbitrary point, and see
what happens to the trisection. Rarely, the choice of the arbitrary point may have
nothing whatsoever to do with the trisection; it may just be part of the smoke screen.
This method of cheating amounts to drawing the diagram just right (a Baby Bear
diagram) so that the trisection comes out almost perfectly. There is probably one perfect
point (or angle or length) out there, which you cannot locate with compasses and
straightedge (by the way), which will perfectly trisect your angle. Get close to that
point, and you will not be able to tell that you have not trisected the angle. Miss it by a
mile, and the trisection falls apart. These methods can get fairly amusing. Here are a few
of those: [under construction]
1. A 56.60 degree
angle: Here we trisect
side AO giving us C,
so AC = AO/3. And
we bisect angle A, the bisector intersects segment BC
at D, the trisector. This method is fine tuned for an
angle of about 56.60 degrees. For any other angle (including small angles), the point D
is way off. There is an angle here (about 56.60 degrees) which can be exactly trisected
with this method, but we don't know what it is exactly. Above right, I have used this
method to "trisect" a smaller angle, and the trisection ends up being larger than a
bisection.
2. LJRH method: This one is found in The
Trisection Problem, by Robert C. Yates. He says
that this appeared in a science periodical, and
was invented by L. J. R. H. We are trying to
trisect angle AOB. We draw the arc AB with
center O. Then we draw arc AO with center B,
and choose an arbitrary point N on that arc.
Then, with center N, we draw the trisected arc
BT of "any" length (actually we trisect it by just
drawing the circle, then making three arcs BC,
CD, and DT with equal chords of any length
BC). We bisect the arcs AB (at Q) and BT (at P).
Then we draw lines AT and PQ, which intersect
at some point K. Draw lines KD and KC, which
intersect arc AB at the
trisection points E and F.
This looks very good for
most positions of N and
relatively small lengths BC
(within a fraction of a
degree). When BC gets
longer, then the "trisection"
gets really bad; see the
diagram above right. For
some angles, it is even possible that line KD could miss circle O completely, and then
where would that trisection point be?
If this method specified where N should be, and how long BC should be, then it would
be a good trisection attempt for angles less than 90 degrees.
Part VIII - Comments
I didn't mean to become a trisection buster (or debunker). I was willing to point out
obvious errors. But some of the methods sent to me were complicated or poorly
described. How do you bust a construction that you cannot understand? Well, this is
now one of my really fun projects. Having busted some of the more complicated
methods above, I find that I enjoy this. And I am certainly improving my trigonometry
skills.
There is a joke that there are two types of people, those who divide people into two
types, and those who do not. Well, there are several different types of people who
choose to believe that trisections (using the ancient tools) are possible:
1. The first kind are mathematically illiterate, and seem unsure if 2+2 always
equals 4. Most of humankind, mostly nonscientists and nonengineers, are
similarly mathematically illiterate. I told one of these people that simple
trigonometry shows that his trisection of 60 degrees made one of the angles
19.07 degrees, and he informed me that the trisection was a geometry problem,
not a trigonometry problem, a response that left me speechless (for a moment).
2. A second kind of people are mathematically adept (algebraically and
geometrically) to some extent, but seem to think that an answer that is close to
being right is right. They measure two angles and think that they have the same
measure, and they don't bother to prove that they have the same measure. There
are many such people.
3. A third kind are also adept at mathematics, and think that they have trisected an
angle, but have made some sort of mistake; and then they find out that it is
supposed to be impossible. These people react with indignation when told they
did not really trisect an angle. Instead of trying to find their errors, which would
be very educational, they are angry at me, as if it is my fault.
4. Is there a fourth kind of person, who has actually trisected an arbitrary angle,
with the appropriate tools? Well, there is apparently a mathematical proof that
such a person cannot exist. A mathematical proof meets a very high standard of
proof.
In the excellent book Eureka! (Math fun from many angles) by
David B. Lewis, Mr. Lewis says that one of the reasons that you
cannot square the circle, is that you cannot construct the square
root of an irrational number. This is not true. Here is a
construction that produces the square root of any length x. The
two lines are perpendicular. If that length is irrational (like the
square root of 2) then the length shown is the square root of it
(fourth root of 2, in this case where x is the square root of 2). It
is impossible to construct a line segment equal to pi. If you
could, then you could take its square root.
Of course some angles (90 degrees, for example) can be trisected. I have a truly
remarkable trisection of a 0 degree angle, which can be done without any tools,
whatsoever. There is the stange case of the angle 3pi/7 (540/7 degrees). This angle
cannot be constructed. But (if you managed to miraculously have it before you) it can be
trisected (Honsberger 1991).
Conway's test: John Horton Conway suggests that a trisection method be tested by
seeing what happens if you double the angle being trisected. If you really have a true
trisection method, then doubling the angle will double the trisection. In other words, we
use our method on angle A, and get some angle B (hopefully A/3). Then if we use our
method on 2A, then our trisection should be 2B. If we get some other angle than 2B,
then our method doesn't work. Of course we are very suspicious of methods which only
seem to work for small angles (or for some other range of angles).
An analogy:
Here is an analogy which may put the above into perspective for you:
We have a line segment, and our construction tool is a rubber band with a point in the
middle. All we can do with this tool is bisect a segment. Both ends must be on the line,
and so, with this tool, we cannot construct any point outside the line.
Let's say we want to trisect the line segment (or any smaller sub-segment of the
segment). All we can do is repeatedly bisect segments. If we are allowed to do this
infinitely many times, then we can easily trisect the line segment (1/3 = 1/4 + 1/16 +
1/64 + 1/256 + ...). But no finite sum will ever do the trick. You can do it to any
accuracy you want (certainly more accurate than your tools), but you can never be
exact. There is no sum of these fractions (1/2, 1/4, 1/8, ...) which can possibly add up to
1/3, without using infinitely many of them.
The situation with straightedge and compasses is more complicated because those tools
can easily produce square roots. But they can't do cube roots and the roots to other cubic
equations. And they can't do many other lengths.
Most of the above diagrams were drawn using Cinderella and Paint Shop Pro.
Return to my Mathematics pages
Go to my home page
Go to Fun_Math Content Table Trisecting an Angle
Trisection using Special Curves
1. Quadratrix-Hippias
Hippias's(about 460BC-about 400BC) Trisection method is shown in the figure shown
below.
The procedure for
Trisection is:
Step 1:Draw the
quadratrix , then select a
point "A" on this curve to
define angle AOB
Step 2:Draw a line from
"A" parallel to AB and
the intersection with line
PO is point "C".
Step 3:Find a point "D"
on line CO such that OD
= CO/3
Step 4:Draw a line from
"D" parallet to OB, and
find a point "E"
,intersecting this curve.
Angle EOB trisects angle AOB.
*********** quadratrix_tri_desc.dwg ***********
You can see the process in animation. For detail, go to the section Quadratrix -
Hippias.
2. The Conchoid-invented by Nicomedes,used by Pappus
Trisection using Conchoid is shown in the figure shown below.
This curve was first called "Cochloid", then later named "Conchoid". It was invented by
Nicomedes(about 280 BC- about 210 BC) to solve the "Delian Problem", but later
Pappus(about 290-about 350) found out that it can be used for Angle Trisection.
The Trisection procedure
is:
Step 1: Draw a
rectangular coordinate
Axis O-XY, and locate
Point "A" on OY. Draw a
line OB to define angle
AOB. Angle AOB is to
be trisected.
Step 2: Draw a unit
radiuds(a) circle at "o",
and "L" is the intersection
of this circle and line OB.
Draw a line from "L"
parallel to OB and it
intersects Y axis at point
"D".
Step 3:Draw a
"Conchoid" with "O" as a
pole, line "DL" as "ruler"
and fixed length = 2a.The resulting curve is NCM.
Step 4:Draw a line form "L" parallel to Y axis and it intersects this Conchoid at point
"C". Line OC trisects angle AOB.
******** conchoid_trisection_desc.dwg ********
You can see the process in animation. For detail, go to the section Conchoid -
Nicomedes.
3. Archimedes' Spiral
Archimedes(287 BC -
212 BC) used Spiral
(called Archimedes'
Sprial) for Trisection. His
idea is shown in the
figure shown below.
Trisection Steps:
1. 1.Draw
Archimedian Spiral r =
a
2. 2.Select point A
to define angle AOB to
be trisected.
3. 3.Find intersection point "P" of line OA and this curve.
4. 4.Divide line segment OP into 3 equal parts, and let OQ = (1/3)*OP.
5. 5.Draw a circle with center at "O" and its radius OQ.
6. 6.Find point "E", which is an intersection of this circle and the curve .
7. Angle EOB = (1/3) angle AOB
************* spiral_tri_desc.dwg *************
You can see the process in animation. For detail, go to the section Spiral - Archimedes.
4. The Hyperbola-Pappus
Pappus(280 - 350) showed that "Huyperbola" can be used for Angle Trisection. His idea
is shown in the figure shown below.
The trisection of angle F1-O-F2 is equivalent to dividing arc F1-F2 into 3 equal arc
length. This is accomplished by using the hyperbola with eccentricity equal to 2. It
measn that for all the points on the hyperbola the distance from the F1(called Focus) is
two times the normal distance to Y-axis(called Directrix). Because of symmetry the
intersecting points B & C divides arc F1-O-F2 into 3 equal length arcs.
******** Pappus_hyperbola_tri_desc.dwg ********
You can see the process in animation. For detail, go to the section Hyperbola - Pappus.
5. Limaçon- by Pascal
Limaçon was invented by Blaise Pascal (1623 - 1662) about 1650, and used by Isaac
Newton (1643 - 1727) for Angle Trisection. The idea is shown in the figure shown
below.
How to draw Limaçon
Point "P" moves along a
circle of unit radius CO.
Let b be a fixed length.
Extend CP and produce
point "A" such that PA =
b. Then the loucs of "A"
is a Limaçon. The type of
Limaçon used for
Trisection is b = 1 case.
The yellow colored curve
is the Limason Pascal
used to Trisection.
Trisection
Draw a unit circle with its
center at "O". Draw a
Limaçon with fixed
length b = 1.
Select a point "A" on the
Limaçon to define an angle AOB to be trisected.
Draw a line CA. Angle CAO is the angle trisected.
******** limason_tri_desc.dwg ********
You can see the process in animation. For detail, go to the section Limaçon - Pascal.
6. Parabola by Rene Descartes
Rene Descartes(1596 - 1650) ,who is called the founder of the "Analytic
Geometry",published the famous treatise "La Geometrie" in 1637. In this book he
showed that Angle
Trisection can be done by
using Parabola. His idea
is shown in the figure
shown below.
Trisection Procedure
1. Draw a circle with
radius=2,then define a
point A . angle AOB is to
be trisected.
2. Drop the perpendicular
from A to OB. Let the
length be 2a (in Trisection equation)
3. Draw a parabola y = x2. Draw a circle with its center at point C (a, 2)
4. Get intersecting point "P", and drop the perpendicular to OB. This will cut the circle
drawn at step 3 at point "T"
5. The line OT is a trisector of angle AOB.
******** parabola_tri_desc.dwg ********
You can see the process in animation. For detail, go to the section Parabola - Rene
Descartes.
7. The Cubic Parabola
The idea of using Parabola can be easily extended to "Cubic parabola". The idea is
shown in the figure shown below.
Trisection Procedure
1. Draw the circles of
radii 1 and 2. Draw a
cubic parabola y = (1/2) x 3.Draw a line y = 3x/2 + 3
2. Pick a point "A" on the
outer circle to define
angle AOB for trisection.
OA intersects the inner
circle at A'. Drop a line
from A'perpendicular to
OB. Intersection is G.
Note here that length OG
is "a" in Trisection
Equation.
3. Get a point H on Y-
axis, such that OH = OG.
Draw a line through point
H parallel to line EF.
. 4. This line will cut the cubic parabola at point K. Drop a line perpendicular to OB.
This intersects the inner circle at point M. Line MO is the trisector of angle AOB.
******** cubic_parabola_tri_desc.dwg ********
You can see the process in animation. For detail, go to the section Cubic Parabola.
8. The Cycloid of Ceva
Tomasso Ceva(1648 - 1737) ,brother of Giovanni Ceva(1647 - 1734) ,who is known for
Ceva's theorem, applied "Insertion method" by Archimedes to Angle trisection using a
special curve called "Cycloidum anomalarun", or "The Cycloid of Ceva".
The idea is shown in the figure shown below.
Trisection Process
1. Draw "Cycloid of
Ceva".See how to draw
in animation
2. Define angle AOB
3. get unit length OE on
line OA, and draw a
parallel line to OB
through point E.
4. This line cuts "Cycloid
of Ceva" at 6 places. P,S
and T are for trisectiong
angles 3 , 360 deg + 3 ,
and 720 deg + 3 ,
respectively.
The remaining 3 points
P', S' and T' are for 180 -
3,
******** cycloid_of_Ceva_tri_desc.dwg ********
You can see the process in animation. For detail, go to the section Cycloidum
Anomalarum - Ceva.
9. Trisectrix by
Maclaurin
Colin Maclaurin (1698-
1746),who is known for
Maclaurin Series, used a
curve called "Trisectrix"
to accomplish Angle
Trisection.
His idea is shown in the
figure shown below.
How to draw Trisectrix
1. Draw a unit radius semi-circle BC with its center at O. EF is a perpendicular bisector
of line segment CO.
2. Pick a point Q on EF, and P is the intersection of CQ and the semi_circle.
3. Locate a point R such that CR = CP - CQ is satisfied.
4. Locus of R as point Q moves along the line EF is the curve called "Maclaurin's
Trisectrix".
Note: This process is also interpreted as a "imaginary" link CQ-QO-OR as Q moves
along EF. It is "imaginary" because the length of the link is variable, and in reality, it
does not exist.
******** Maclaurin_tri_desc.dwg ********
Trisection Process
This curve has the property such that angle ROB is 3 times of angle QCB. So once the
curve is drawn,the trisection process is straight forward.
You can see the process in animation . For detail, go to the section Trisectrix -
MacLaurin.
Go to Fun_Math Content Table Trisecting an Angle
All questions/suggestions should be sent to Takaya Iwamoto
Last Updated Nov 22, 2006
Copyright 2006 Takaya Iwamoto All rights reserved. .
Go to Fun_Math Content Table Trisecting an Angle
Source of the contents for this section
Most of the contents of this section come from "The Trisection Problem" (ref. 1) by
Yates. Actually this book is the reason why I have decided to open a public web site for
demonstrating geometry in dynamic fashion using CAD software.I have added topics on
Origami(taken from ref. 5 ), Maclaurin's method (from ref. 3), and discussion of
equivalence of three methods(Tomahawk, Carpenter's square and Origami).
So for those who want to learn seriously about this topic, purchase of ref.1 through used
bookstore is highly recommended.
History of the problem
Problem Definition:
To divide an
"ARBITRARY" angle
into three equal angles.
It is important to note that
angles like 45,72, 90 &
180 degrees
can be divided into 3
equal parts by using only
compass and ruler.
******** angle_trisection_problem.dwg ********
Origin of the Problem :
The origin of this problem is not clear even among the experts on this subject.
The bisection of any angle can be done easily with compass and straight-edge, but if the
number
of division becomes three, Greek mathematicians encountered a difficulty.
Rule of the Game ( Platonian Rule ) :
Tools allowed
the compasses and unmarked straight-edge
Permitted use of these 2 tools
1. The drawing of a straight line of indefinite length through two given points.
2. The construction of a circle with center at a given point and passing through a second
given point.
Summary of Early Attempts:
Greek mathematicians ,although they could not prove it, knew that Angle Trisection is
not possible because their trisection attempts were all violation of the Platonian Rule.
Quadratrix of Hippias, Conchoid of Nicomedes, and verging solution by Archimedes
are their typical answers.
Proof of the impossibility:
Carl Friedrich Gauss (1777 - 1855) had stated without proof that the problems of
trisecting an angle and doubling a cube cannot be solved
with ruler and compasses. But the first published proof was given by Frenchman, Pierre
Wantzel (1814 - 1848), in 1837.
Comments on the References:
A very good introduction is the following web site.
Trisecting an angle
If you are looking for a book, ref(1) is the best choice. It is easy to read with brief and
clear explanation.
The problem is that there are only a limited number of used books available on the
market.
1. Yates,Robert C.:"The Trisection problem",first published in 1942.
2. Dudley,Underwood :"The Trisectors" The Mathematical Association of
America,1994.
3. Ogilvy,Stanley C. :"Excursions in Geometry" Dover,first published in 1969
4. Dorrie, Heinrich :"100 Great Problems of ELementary Mathematics-Their History
and solution", Dover 1965
The original was published in 1932.
5. Abe, Hisashi : "Amazing Origami"(In Japanese)
Go to Fun_Math Content Table Trisecting an Angle
All questions/complaints/suggestions should be sent to [email protected]
Last Updated July 9-th, 2006
Copyright 2006 Takaya Iwamoto All rights reserved.
Up: Geometry Forum Articles
Angle Trisection
Most people are familiar from high school geometry with compass and straightedge
constructions. For instance I remember being taught how to bisect an angle, inscribe a
square into a circle among other constructions.
A few weeks ago I explained my job to a group of professors visiting the Geometry
Center . I mentioned that I wrote articles on a newsgroup about geometry and that
sometimes people write to me with geometry questions. For instance one person wrote
asking whether it was possible to divide a line segment into any ratio, and also whether
it was possible to trisect an angle. In response to the first question I explained how to
find two-thirds of a line segment. I answered the second question by saying it was
impossible to trisect an angle with a straightedge and a compass, and gave the person a
reference to some modern algebra books as well as an article Evelyn Sander wrote about
squaring the circle. One professor I told this story to replied by saying, "Bob it is
possible to trisect an angle." Before I was able to respond to this shocking statement he
added, "You just needed to use a MARKED straightedge and a compass." The professor
was referring to Archimedes' construction for trisecting an angle with a marked
straightedge and compass.
When someone mentions angle trisection I immediately think of trying to trisect an
angle via a compass and straightedge. Because this is impossible I rule out any serious
discussion of the manner. Maybe I'm the only one with this flaw in thinking, but I
believe many mathematicians make this same serious mistake.
Why tell people it is impossible to trisect an angle via straightedge and compass?
Instead we could say it is possible to trisect an angle, just not with a straightedge and a
compass. When told that it is impossible to trisect an angle with a straightedge and
compass people then often believe it is impossible to trisect an angle. I think this is a
mistake and to rectify my previous error I will now give two methods for trisecting an
angle. For both methods pictures are included that will hopefully illuminate the
construction.
The first method, Archimedes' trisection of an angle using a marked straightedge has
been described on the Geometry Forum before by John Conway. First take the angle to
be trisected, angle ABC, and construct a line parallel to BC at point A.
Next use the compass to create a circle of radius AB centered at A.
Now comes the part where the marked straightedge is used. Mark on the straightedge
the length between A and B. Take the straightedge and line it up so that one edge is
fixed at the point B. Let D be the point of intersection between the line from A parallel
to BC. Let E be the point on the newly named line BD that intersects with the circle.
Move the marked straightedge until the line BD satisfies the condition AB = ED, that is
adjust the marked straightedge until point E and point D coincide with the marks made
on the straightedge.
Now that BD is found, the angle is trisected, that is 1/3*ANGLE ABC = ANGLE DBC.
To see this is true let angle DBC = a. First of all since AD and BC are parallel, angle
ADB = angle DBC = a. Since AE = DE, angle EAD = a, and so angle AED = Pi-2a. So
angle AEB = 2a, and since AB = AE, angle ABE = 2a. Since angle ABE + angle DBC =
angle ABC, and angle ABE = 2a, angle DBC = a. Thus angle ABC is trisected.
The next method does not use a marked ruler, but instead uses a curve called the
Quadratrix of Hippias. This method not only allows one to trisect an angle, but enables
one to partition an angle into any fraction desired by use of a special curve called the
Quadratrix of Hippias. This curve can be made using a computer or graphing calculator
and the idea for its construction is clever. Let A be an angle varying from 0 to Pi/2 and
y=2*A/Pi. For instance when A = Pi/2, y=1, and when A=0, y=0. Plot the horizontal
line y = 2*A/Pi and the angle A on the same graph. Then we will get an intersection
point for each value of A from 0 to Pi/2.
This collection of intersection points is our curve, the Quadratrix of Hippias. We will
now trisect the angle AOB. First find the point where the line AO intersects with the
Quadratrix. The vertical coordinate of this point is our y value. Now compute y/3 (via a
compass and straightedge construction if desired). Next draw a horizontal line of height
y/3 on our graph, which gives us the point C. Drawing a line from C to O gives us the
angle COB, an angle one third the size of angle AOB.
As I mentioned before this curve can be computed and plotted via a computer. The
formula to find points on the curve is defined as x = y*cot(Pi*y/2). Yes here the vertical
variable, y, is the independent variable, and the horizontal variable, x, is the dependent
variable. So once the table of values is found, the coordinates will need to be flipped to
correctly plot the Quadratrix.
Now a justification for the formula. In the following figure, B =(x,y) is a point on the
Quadratrix of Hippias. Let BO be a line segment from the origin to B and and BOC be
our angle A. If we draw in a unit circle, and drop a vertical line from the intersection of
the angle we get similar triangles and see that sin(A)/cos(A) = y/x, or tan(A) = y/x. But
earlier we defined a point (x,y) on the Quadratrix to satisfy y=2*A/Pi. So we get
tan(Pi*y/2)=y/x or equivalently x = y*cot(Pi*y/2).
So we now have two different ways of trisecting an angle. I learned about the
construction of the second method in Underwood Dudley's book: "A Budget of
Trisections". In the book Dudley describes several other legitimate methods for
trisecting an angle as well as compass and straightedge constructions that people have
claimed trisect an angle. The book also contains entertaining excerpts of letters from
these "angle trisectors".
Besides stating it is impossible to trisect an angle, I think other problems occur in
discussing angle trisection. One difficulty is in explaining what it means for something
to be impossible in a mathematical sense. I definitely remember in high school being
told that it was impossible to trisect an angle. But I think at the time it meant the same
thing to me as that it was impossible for me to drive a car. I was only 14 years old and I
could not get a license to drive a car for another two years so it was just not possible AT
THAT TIME. I do not remember being told that when something is impossible in
mathematics, it was not possible five million years ago, it is not possible now, and it
will never be possible in the future. Granted I may not have been ready for an
explanation of mathematical logic and proof, but a statement like, "It is impossible to
trisect an angle with a straightedge and a compass. This means it is no more possible to
trisect an angle with those tools, then it is to add 1 and 2 and get 4" would have been
much more powerful. (I am assuming here that we all count the same way: 1, 2, 3, 4, ...)
I did not intend to attack my high school teacher; I learned an incredible amount of
mathematics from her as well as a deep love for the subject. Maybe my teacher did
explain what the words "mathematically impossible" meant, and I just do not remember
her comments. Regardless, I think a discussion of impossible in the mathematical sense
would be an interesting and valuable topic to discover in high school. Are there any
teachers out there who have spent time talking about mathematical impossibilities?
Up: Geometry Forum Articles
The Geometry Center Home Page
Comments to: [email protected]
Created: April 12 1995 --- Last modified: Jun 18 1996
Why Trisecting the Angle is Impossible
Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
First-time Visitors: Please visit Site Map and Disclaimer. Use "Back" to return here.
A Note to Visitors
I will respond to questions and comments as time permits, but if you want to take issue
with any position expressed here, you first have to answer this question:
What evidence would it take to prove your beliefs wrong?
I simply will not reply to challenges that do not address this question. Refutability is
one of the classic determinants of whether a theory can be called scientific. Moreover, I
have found it to be a great general-purpose cut-through-the-crap question to determine
whether somebody is interested in serious intellectual inquiry or just playing mind
games. It's easy to criticize science for being "closed-minded". Are you open-minded
enough to consider whether your ideas might be wrong?
The ancient Greeks founded Western mathematics, but as ingenious as they were, they
could not solve three problems:
Trisect an angle using only a straightedge and compass
Construct a cube with twice the volume of a given cube
Construct a square with the same area as a given circle
It was not until the 19th century that mathematicians showed that these problems could
not be solved using the methods specified by the Greeks. Any good draftsman can do all
these constructions accurate to any desired limits of accuracy - but not to absolute
accuracy. The Greeks themselves invented ways to solve the first two exactly, using
tools other than a straightedge and compass. But under the conditions the Greeks
specified, the problems are impossible.
Since we can do these tasks to any desired accuracy already, there is no practical use
whatever for an exact geometrical solution of these problems. So if you think you'll get
headlines, endorsement contracts and dates with supermodels for doing so, it is my sad
duty to tell you otherwise.
Also, there are a few trivial special cases, like a right angle, where it is possible to
"trisect" the angle. More specifically, it is possible to construct an angle one-third the
given angle. For example, if you draw a diameter of a circle and mark off 60 degree
intervals on the circle, you "trisect" the straight angle. This isn't trisection in any
meaningful sense because it doesn't generalize to other angles.
"Impossible" is a welcome challenge to a lot of people. The problems are so easy to
understand, but the impossibility proofs are so advanced, that many people flatly refuse
to accept the problems are impossible. I am not out to persuade these people.
Mathematicians have spent years corresponding with some of them, and many are
absolutely immune to persuasion (The trisectors, I mean. The mathematicians are too,
but they have reason to be). But if you want a (hopefully) intelligible explanation of
why mathematicians regard the problems as impossible - as proven to be impossible -
then this site might help you.
Warning: you need trigonometry and an understanding of polynomials - that is, the
equivalent of a good high school math education - to follow this discussion.
What is a Proof?
Enough people have asked me to look at their constructions that one thing is very
obvious. Most people do not have any idea what constitutes a proof in mathematics.
Step by step instructions for doing a construction are not a proof. A traditional
geometric proof for a construction would require:
A justification for every step in the construction. Most of the time this is simple:
you can always bisect an angle, draw a line through two points, and so on. But if
you say "draw a line through points A, B and C," you have to be able to prove
that points A, B, and C actually do lie on a straight line. If you say draw a line
through a particular point, you have to be able to show that the point actually
exists and can be constructed using the classical rules of geometry. There are a
lot of interesting geometrical fallacy puzzles that rely on assuming a step is
possible when it is not. Sometimes it's subtle. If you draw a construction that
passes through a circle, for example, you have to be prepared to show what
happens if the line misses the circle. Does it matter? It may or may not, but you
have to be prepared to deal with it.
Second, you have to prove that the construction does what it claims to do. If the
proof isn't inherent in the construction you have to come up with a way to prove
it.
Detailed instructions and diagrams are not proof.
Measurement is not proof.
If you have any questions, get a (preferably old-school) geometry textbook and study
their proofs.
Proving Things Impossible
One of the major problems people have with angle trisection is the very idea that
something can be proven impossible. Many people flatly deny that anything can be
proven to be impossible. But isn't that a contradiction? If nothing can be proven to be
impossible, and you can prove it, then you've proven something to be impossible, and
contradicted yourself. In fact, showing that something entails a contradiction is a
powerful means of showing that some things are impossible. So before tackling the
trisection problem, let's spend some time proving a few things impossible just to show
that it can be done.
The Domino Problem
Imagine you have a checkerboard and
start covering it with dominoes so that
each domino covers two squares.
Obviously there are a vast number of
ways to do it.
Now imagine you remove two opposite
corner squares. Can you still cover the
checkerboard with dominoes?
No. Rather than try every possible way
to place dominoes on the board,
consider this: each domino covers both
a red and a white square. If you remove
the two corners shown, there will be 32
white squares but only 30 red squares.
There's no way to cover the board with
dominoes without leaving two unpaired
white squares. So it can't be done. We
have proven that something is
impossible.
I expect some die-hard to ask what about coloring a white square red so there are 31 of
each color. It doesn't matter. (Actually, it's easy to see that coloring a white square red
will create a situation where you have to cover two red squares with a domino. Once
you cover them, you have an unequal number of red and white squares remaining, and
then we're back to square one, literally.) You can paint the board psychedelic if you like
and it still can't be done. The traditional checkerboard coloring makes it easy to prove it,
but if it can't be done on a traditional checkerboard pattern, it can't be done, period. In
fact, a lot of proofs depend on marking, labeling, or grouping items in a certain way to
show that some particular arrangement either is, or is not, possible.
An Oldie but Goodie: The Largest Prime Number
Prime numbers like 2, 3, 5, 7, 11 .... are divisible only by themselves and one. As
numbers get bigger, primes get more rare. Is there a largest one?
The ancient Greeks showed there is not. Imagine there is a largest prime p. Now
calculate the number q, which is 2 x 3 x 5 x 7 x 11 x...(all the primes less than p) x p. It
will be a huge number. Now consider q+1. It's not divisible by 2, because q is divisible
by 2. Likewise, it's not divisible by 3, 5, 7, or any other prime up to p, because q is
divisible by all those numbers. So there are only two possibilities. Either q+1 is prime,
or it's divisible by primes bigger than p (q+1 will be vastly bigger than p - if p is only 19
than q is 9,699,690 - there's lots of room for bigger primes). Either way the initial
assumption leads to a contradiction. Hence it must be wrong. There is no largest prime.
It is impossible to find a largest prime. It's not that people have tried, failed, and given
up. It's impossible because the idea itself leads to a contradiction. This method of proof
- making an assumption and then showing that it leads to a contradiction - is called
reductio ad absurdum.
It is impossible to find a largest prime, but it is possible to find arbitrarily long stretches
of numbers without them. Our number q is composite - it is the product of smaller
numbers. It's easy to see that q+2 must be composite, as well as q+3, q+4 (divisible by
2) q+5, q+6 (divisible by 2 and 3) and so on up to q+p. We can find so-called "prime
deserts" of any desired length, but there are always primes after them. Indeed, we keep
finding pairs of primes, like 5 and 7, or 101 and 103, however high we go, though
nobody has yet shown there is an infinite number of them.
Another Golden Oldie: Square Root of Two
Can you represent the square root of two as a fraction? The ancient Greeks also found
out that this is impossible. Imagine that there is a fraction p/q, where p and q are whole
numbers and p/q is in lowest terms (that qualifier is important), that equals the square
root of two. We can conclude:
p2/q
2 = 2 and therefore:
p2 = 2q
2, therefore,
p is even, since p2 is even and only even numbers have even squares. Also,
p2 is divisible by 4, since all even squares are divisible by 4. Thus 2q
2 is
divisible by 4 which means q2 must be divisible by 2. Therefore q must also be
even.
Thus p and q must both be even. Since we assumed they were in lowest terms,
and have arrived at a contradiction, the assumption must be false. We have
another reductio ad absurdum. It is impossible to represent the square root of
two as a fraction of whole numbers.
The sect called the Pythagoreans believed that everything was ultimately based on
whole numbers. They were horrified by this discovery. Tradition has it that they decreed
death to any member who divulged the secret. They called numbers that could not be
represented as ratios as irrational, and to this day irrational carries a negative
connotation. Actually this proof is related to the proofs that trisecting the angle,
doubling the cube and squaring the circle are impossible.
Uniform Polyhedra
Reductio ad absurdum is a powerful way of showing that some things are impossible,
but not the only way. If we can succeed in showing all the things that are possible, then
anything else must be impossible.
In simple cases we can list the possibilities by brute force. For example, what kinds of
three dimensional shapes have regular polygon faces, with all faces and vertices
identical? Clearly you have to have at least three faces meeting at a vertex, and the sum
of all the angles meeting at a vertex can't equal or exceed 360 degrees. For triangles, we
can have three, four or five meeting at a vertex (six would equal 360 degrees). For
squares we can have only three, same for pentagons, and for hexagons and above, it
can't be done (three hexagons add up to 360 degrees). So that's it. There are five shapes,
and no others, shown below. (If you allow faces and edges to cross through each other,
it gets a bit more interesting.)
What Kinds of Plane Patterns are Possible?
When there are a potentially infinite number of possibilities, we have to use the
general properties of the problem to devise a proof. Here's a fairly simple example.
What kinds of repeating patterns can we have on a plane, for example, wallpaper? We
say something has n-fold symmetry if, when you rotate it 360 degrees, there are n
positions where it looks the same. For example, a honeycomb has six-fold symmetry -
you can rotate it 360 degrees and there will be six positions where the pattern looks the
same. A sheet of square graph paper has four-fold symmetry. So does a checkerboard if
we ignore the colors, but if we include the colors, then it only has two-fold symmetry.
Something with no symmetry at all has one-fold symmetry.
Now a
single
object
like a
flower
or
propelle
r can
have
any
kind of
symmet
ry at all.
But
what
about a
pattern
that
repeats?
Look at
the
pattern
at right.
The pattern consists of points, only some of which are shown here. All the points are
identical (that's what a repeating pattern means). So if we can rotate the pattern around
one point and see symmetry, we can rotate it around every identical point. If we rotate
the pattern by angle 360/n so it still looks the same, then a point originally at on the
bottom row will be rotated to the same position as some other point on the top row (or
on the bottom row if it rotates 180 or 360 degrees). Note that we can rotate in either
direction. The pattern shown here is the familiar honeycomb pattern, but notice we have
not made any assumptions about the angles in the pattern at all. The pattern shown is
just for illustration.
Let's assume each point is distance one from its neighbor (ao = ob = ov = ou) The
distance between the two points u and v in the upper row has to be a whole number, but
it need not be 1. If a = 90 degrees (4-fold symmetry), it could be zero. But it has to be a
whole number, because it's the distance between two points on a row of the pattern.
From elementary trigonometry, we can also see that uv = 2 cos (360/n).
Now the cosine can only have values between -1 and 1, so 2 cos a, which must be a
whole number, can only take on values -2, -1, 0, 1 and 2. Thus cos a can only be -1, -
1/2, 0, 1/2 and 1, or a = 180, 120, 90, 60 and 0 (or 360), respectively. Since the rotations
are symmetry rotations, a = 360/n, so n = 2, 3, 4, 6, or 1.
Thus, repeating patterns in the plane can only have 1, 2, 3, 4 or 6-fold symmetry. In
particular, repeating patterns in the plane cannot have five-fold symmetry.
What about the stars in the flag? The stars have five-fold symmetry, but the
overall symmetry of the pattern does not. Rotate a flag by 72 degrees and see.
In recent years some fascinating materials have turned up with five-fold
symmetry. These materials, termed quasicrystals, do not have repeating patterns.
These are related to beautiful patterns called Penrose Tilings, which also have 5-
fold symmetry but do not repeat regularly.
Notice, this is not a matter of people trying to find five-fold repeating patterns and
failing, then concluding it must be impossible. We devised a method to find all the
patterns that are possible. We showed that only certain patterns are possible; therefore
all the others are impossible. Many proofs of impossibility are based on the fact that we
show what is possible, thus ruling out everything else.
More (much more) on symmetry and patterns
What's With the Ruler and Compass Anyway?
Plato's famous analogy of the Cave gives a lot of insight into the Greek mathematical
mind. In this analogy, a prisoner in a cave knows the outside world only from shadows
cast on the wall of the cave. Needless to say, his knowledge of the world is pretty
imperfect. To many Greeks, this world, or at least our sensory picture of it, were crude
images of a perfect, real world. Thus, to Greek mathematicians, the lines we drew were
crude approximations of real lines, which were infinitely long, infinitely sharp,
infinitesimally narrow, and perfectly straight. In principle, lines intersected in infinitely
tiny, precise points. In geometry, "the truth is out there", the answer already exists, and
they sought constructions that would find the answer infallibly and with infinite
precision. Thus the Greeks rejected any technique that smacked of approximation or
trial and error. The problem with trial and error is that, regardless how closely the
construction seems to fit, we can never be sure there isn't some microscopic error below
the limits of our visibility. The Greeks wanted to find the correct point directly, the first
try, with absolute precision.
It is possible to build
tools that will solve
the problem, using a
ruler and compass.
One of the simplest,
at right, was dubbed
the "hatchet." It
consists of a right-
angled T as shown
with a semicircle on
one side. Points A, B,
C and D are all
equally spaced. Place
A on one side of the
angle with the long
arm of the T passing
through O, and slide
the hatchet so that the
semicircle just
touches the other side
of the angle at E.
Lines OB and OC
trisect the angle. The
upper diagram shows
how it works, the
lower diagram shows
how one might be
constructed from
cardboard.
Proof: Since AB=BC and angles ABO and CBO are right angles, triangles AOB and
BOC are identical and angles AOB and BOC must be equal. Also BC=CE and angle
CEO is a right angle, so triangles BOC and COE are also identical and angles BOC and
COE are identical. Thus angle AOB = BOC = COE.
So what's wrong with this device? It can be laid out with ruler and compass, but not
physically made. A real hatchet, however perfect, would have imperfections. Even if it
were absolutely perfect, the process of lining the device up would be one of trial and
error. So this device was unsatisfactory to the Greeks. It's tempting to try to use a
straightedge and compass to lay out the hatchet right on the desired angle, but it also
can't be done without trial and error.
It is possible to trisect the angle using a marked straightedge, but that's not allowed by
the ancient Greek rules, since a mark can't be lined up against another point, line or arc
without trial and error, and without some inherent error of alignment. Likewise, you can
solve the problem with two carpenters' squares, but that's also not allowed since it also
involves trial and error and since no carpenters' square will have an absolutely perfect
right angle.
There are subterfuges people have developed for seeming to get around these
restrictions. You can hold the compass against the ruler to measure distance; technically
you haven't "marked" the ruler but the effect is the same. There are a variety of
constructions that involve sliding the pivot point of the compass along a line or arc;
these all involve some measure of trial and error as well (though the people who devise
these constructions sometimes hotly deny it). Using the compass for anything but
drawing an arc around a fixed center is forbidden.
I have modified this page several times as people come up with evasions of the classical
restrictions. So let's clarify the general principle: all loopholes violate the rules.
This is perhaps the simplest
trisection using a marked
straightedge. It was discovered by
Archimedes. Given the angle
AOX, draw a circle of arbitrary
radius centered at O. Extend one
side of the angle through the
opposite side of the circle at D
(top).
Mark off interval BC on the
straightedge. BC = OX = radius
of the circle.
Slide the straightedge so that B
lies on line DOX, C lies on the
circle and the straightedge passes
through A. Angle CBD is one-
third of AOX. Note the element
of trial and error inherent in
positioning the straightedge.
Proof:
Since BC=OC, angles CBD and COD are equal and angle BCO = 180-2CBD.
And since OC = AO, angles OCA and OAC are equal.
OCA + COD = 180 so OCA = 2CBD.
The angles in triangle ABO sum to 180, so we have CBD + DOA + OAC = 180
= CBD + DOA + 2CBD.
Rearranging, we get DOA = 180 - 3CBD, and since DOA + AOX = 180, AOX =
3CBD.
You can eliminate the marking by setting a compass with radius BC and sliding the
pivot along the line until B, C, and A are on a straight line. However, if you do, there
are two additional points to consider: where the radius of the compass hits the circle
(call it C') and where the radius hits the ruler (call it C"). You can never be certain that
C' = C" = C, regardless of how well the points seem to agree. Even if the same pencil
line covers all three points, if they are different in the most microscopic degree, the
construction is not exact.
It also doesn't do to indulge in arm waving and say "slide the compass and ruler until B,
C, and A are on a straight line." We know point C exists somewhere. You have to prove
that you have actually found it. You have to be able to prove that C' = C" = C.
There are other curves besides circles
that can be drawn, and can be used to
trisect the angle. The simplest, shown at
right, is the Archimedean Spiral. The
radius of the spiral is proportional to
azimuth. Obviously an angle can be
trisected by drawing an Archimedean
Spiral over the angle, finding the radius
where the other side of the angle
intersects the spiral, then trisecting the
radius. Draw arcs from the trisection
points of the radius to the spiral, then
draw radii through those points on the
spiral. In fact, an angle can be divided
into any number of equal parts using
this method.
The problem with the Archimedean Spiral, or any other curve used to trisect the angle,
is that although you can construct an infinite number of points on the curve using a
straightedge and compass, you can't construct every point. However closely spaced the
points are, there will always be tiny gaps between them, so that any curve we draw will
always be approximate, not exact.
If you think about it, this whole business is actually pretty artificial. The Greek
compass, unlike modern ones, had a spring so that it snapped shut once it was lifted off
the page. Anyone who has ever had a compass change radius while drawing a circle can
picture the potential for error in such a device. Of course, we can do all the
constructions the Greeks did, in many cases a lot more simply, by using a compass that
stays fixed. To use something as error-prone as a spring-loaded compass, then worry
about possible imperfections in constructing a tool like the hatchet, or positioning a
marked straightedge, is completely arbitrary.
Also, if you think about it, the business of lining a straightedge up through two points to
draw a line also has a lot of trial and error about it - as much as the hatchet tool. You
line the straightedge up against one point, then position it against the other, then go back
and correct any shifting at the first point, and so on. Then, actually to draw the line, you
need to take into account the fact that any marking tool has finite width, so that as often
as not the drawn line doesn't pass exactly through the two points.
Renaissance instrument makers soon discovered this problem. They found out that
markings plotted using only compasses were more accurate than those made using
straightedges. They began devising alternative constructions that eliminated use of the
straightedge. Although the constructions were often more complex, they were still more
accurate than those that required straightedges. Mathematicians finally showed that
every construction that can be done with a compass and straightedge can be done with a
compass alone. The only qualification is that we define constructing two points on a
given line as equivalent to constructing the line itself.
Proving It Can't Be Done
Follow the Rules
Many people who "solve" the angle trisection problem inadvertently violate the rules:
You can only use a compass and a plain straightedge.
You cannot use the straightedge for measuring, or put marks on it.
You can only use the compass for drawing arcs around a fixed center. You
cannot slide the pivot.
You cannot use a straightedge and compass to construct some other tool.
You cannot use a straightedge and compass to construct some other curve.
All loopholes and evasions of these restrictions violate the rules.
Why It's Impossible
The triple angle formula in trigonometry for the sine is: sin 3a = 3 sin a - 4 sin3a.
We can rewrite this to 4 sin3a - 3 sin a + sin 3a = 0
In other words, trisecting an angle amounts to solving a cubic equation. That's why
nothing has been said about doubling the cube. Doubling the cube amounts to finding
the cube root of two, that is, also solving a cubic equation. So algebraically, the two
constructions are equivalent. Squaring the circle is a bit more complicated.
Recall how the Pythagoreans, to their horror, found out that there are other kinds of
numbers than integers (whole numbers) and rational fractions. The process of
discovering all the types of numbers that exist turns out to be directly related to the
proof that the three classic problems are unsolvable. Since the time of Pythagoras,
mathematicians have discovered that there are many types of numbers:
Integers (whole numbers), like 1,2 and 3
Rational Numbers, numbers that are fractions involving integers, like 2/3, 1/2,
and 19/10. Integers are also rational numbers, since they can be written as
fractions like 1/1, 2/1, 3/1, and so on.
Irrational Numbers, those that cannot be written as fractions involving integers.
Almost all roots, trigonometric functions and logarithms are irrational. So is pi.
Positive and Negative Numbers. Also zero, which was unknown to the ancient
Greeks.
Imaginary Numbers, square roots of negative numbers. All other numbers are
called Real Numbers
Complex Numbers, combinations of Imaginary and Real Numbers.
The final solution to the
three famous problems of
antiquity came from
studying classes and
properties of numbers.
Geometrical constructions
can be considered the
equivalent of mathematical
operations. For example,
using only a straightedge,
you can add, subtract,
multiply and divide:
If you can draw circles, you
can also construct square
roots.
Proof: Triangle ACD is a
right triangle since all
angles inscribed in a
semicircle are right angles.
So if one angle is p, q = 90
- p and the remaining
angles have the values
shown. So triangles ABD
and DBC are similar. Thus
we can write BD/1 = X/BD,
or BD squared = X. Hence
BD equals the square root
of X.
One of the first fruits of these studies was the discovery by the young Karl Friedrich
Gauss that it was possible, using a ruler and straightedge, to construct a polygon of 17
sides. This was something completely unsuspected. He also found that polygons of 257
and 65537 sides could be constructed. (Strictly speaking, Gauss only discovered it was
possible; other mathematicians devised constructions and showed that no other
polygons were possible, but Gauss made the pivotal discovery.) A number of books
give constructions for the 17-sided polygon; constructions for the other two have been
devised but are hardly worth the effort.
Numbers that can be expressed as combinations of rational numbers and square roots,
however complicated the combination, are called Constructible Numbers. Only
Constructible Numbers can be constructed using a compass and straightedge. We can
now pose (and answer without proof) the following questions:
Are all numbers Constructible Numbers? (No - there are other kinds of numbers,
just as the Pythagoreans found there were other kinds of numbers than integers
and rational fractions.)
Are cube roots Constructible Numbers? (No - hence we cannot trisect the angle
or duplicate the cube using straightedge and compass. In fact higher roots like
fifth roots, and so on, are also not surds. Of course we ignore special cases like
the cube root of 27 or the fifth root of 32, and so on.)
Is pi a Constructible Number? (No - in fact pi belongs to yet another class of
numbers called transcendental numbers that cannot be obtained as the solution
of any finite-sized polynomial. Obviously, if pi is not a Constructible Number,
then we can't square the circle, either.)
Gauss did more than just find new polygons. Building on his results, other
mathematicians showed that polygons of 2, 3, 5, 17, 257 ... sides are the only ones with
prime numbers of sides that can be constructed. (Of course you can also construct
polygons by repeatedly bisecting the angles of these polygons to construct polygons
with 4, 6, 8, 10, 12, 16, etc. sides. You can also construct polygons of 15 sides by
combining the constructions for 3 and 5 sides, etc.) By enumerating what was possible,
he ruled out many other things as impossible. In particular, 7 and 9-sided polygons
cannot be constructed using straightedge and compass. Constructing a 9-sided polygon
requires trisecting a 120-degree angle. Since this can't be done, obviously trisecting any
desired angle is impossible.
Note, by the way, that 2=1+1, 3=21+1, 5=2
2+1, 17=2
4+1, 257=2
8+1, 65537=2
16+1. The
numbers are all primes, and equal to some power of 2 plus one, and the exponents are
all powers of 2. 65537 is the largest prime of this type known, although extensive
searches for larger ones have been made.
"But You Have to Solve the Problem Geometrically"
Most people who still send solutions to the three classic problems to mathematics
departments don't have a clue how the problems were finally solved. Many seriously
think mathematicians just gave up and decreed the problems unsolvable.
The problems actually can't be solved because they require properties that a straightedge
and compass simply do not have. You can't draw an ellipse with a straightedge and
compass (although you can construct as many points on the ellipse as you like), so why
is it a shock that you can't trisect the angle, duplicate the cube, or square the circle? You
can't tighten nuts with a saw or cut a board with a wrench, and expecting a straightedge
and compass to do something beyond their capabilities is equally futile.
Here's another analogy: if you spend your entire life driving a truck, you might
eventually be lulled into thinking you can see the entire country from the highway. It's
only if you get out and walk or fly over the landscape that you discover there are a lot of
other places as well. Geometry with straightedge and compass creates a similar illusion;
eventually we believe the points we can construct are all the points that exist. It was
only when mathematicians began studying the properties of numbers that they found out
it wasn't so. Just as you have to get off the highway to see that other places exist, to find
the limitations of geometry you have to get outside of geometry.
More mathematically literate angle trisectors are sometimes aware of the number-theory
approach, but reject it because they think a geometrical problem can only be properly
solved geometrically. But if the problem is that the solution requires capabilities beyond
those of a straightedge and compass, how in the world can that be discovered from
within geometry? Anyway, who says geometrical problems can only be properly solved
geometrically? The only thing that would justify that rule is some demonstration that the
geometrical solution to a problem and the algebraic solution yielded different results -
and then you'd have to prove the geometrical approach was the correct one. But there
are no cases where this has ever happened, so there is no justification for rejecting the
algebraic solution to the three classic problems. Indeed, it has been shown that if there
is an inconsistency anywhere in mathematics, it is possible to prove any proposition
whatsoever.
In fact, one of the most famous mathematical proofs of all times, Kurt Godel's
Incompleteness Theorem, showed that there is always an "outside" to mathematics.
Once a rule system gets complex enough (and Euclidean geometry is plenty complex
enough) it is always possible to make true statements that cannot be proven using only
the rules of that system. Thus is possible to make true geometrical statements (like
"angles cannot be trisected using ruler and compass") that cannot be proven using the
rules of geometry alone.
Reference Angles
I've seen several constructions that start off by generating an arbitrary angle, then
trisecting some angle derived from it. In all the cases I've seen, the construction
amounts to constructing an angle A, then trisecting 180-3A. Since one third of 180-3A
= 60-A, of course it's possible to construct 60-A. Note that I didn't say "trisect 180-3A,"
since constructing a desired angle isn't the same thing as trisecting three times that
angle.
If your method involves a "reference angle," you can bet that it's not a true
trisection.
If you ever construct one angle three times the size of another, or any derivative
of it, it's not going to be a trisection.
One Final Note
You can construct a 20-sided polygon with central angles of 18 degrees. You can
construct a 24-sided polygon with central angles of 15 degrees. Obviously you can
superimpose the constructions, and the difference between the two angles is three
degrees. That's the smallest integer value we can get using straightedge and compass
constructions. Since an arbitrary angle can't be trisected, and we can't find any integer
angles except multiples of three degrees,
Therefore you cannot construct an exact one-degree angle with ruler and compass!
I had one correspondent who got rather worked up over this. But what's so special about
360 degrees? It's divisible by a lot of numbers, but so is 240. If we defined a circle as
having 240 degrees, each degree would be 1-1/2 of our degrees. And we can construct
1-1/2 degrees exactly.
But so what?
We can't construct a one-degree angle exactly using a ruler and compass, but we can
subdivide line segments into any number of intervals. In principle, we could construct a
right triangle with one edge 100 million kilometers in length and another edge
1,745,506.492821758576 kilometers long at right angles to it. The hypotenuse and the
long edge define an angle of one degree to an accuracy of 10-8
cm, the diameter of an
atom. You'll need a very big sheet of paper, a lot of patience, and a very sharp pencil,
but in Greek terms it's "possible". So we can construct a one-degree angle to precision
so fine it surpasses any possible practical need.
Reference
Lots of books cover the mathematics behind the impossibility proofs. A really good
recent one is Robin Hartshorne, Geometry: Euclid and Beyond, Springer, 2000.
Warning: this is not fireside reading. The book is very clear but it will take serious study
to master it.
Return to Pseudoscience Index
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Created 10 December 1999, Last Update 26 January 2012
Not an official UW Green Bay site
ANGLE TRISECTION
The so-called THALES method used by the companions (historical
craftsmen).
Let AÔB be an angle
Create a circle centered at O radius OB diameter CB
Create an arc of circle centered at B radius BC
Create an arc of circle centered at C radius CB
These arcs cross at D
Draw AD that crosses OB at E
On any straight line trace 3 equal segments EG, GH, HI
Draw BI and parallels to BI passing through H and G and crossing EB at J and K
Draw DJ and DK crossing the circle centered at O at N and M
The arcs BM, MN and NA are equals
So AÔN = NÔM = MÔB
Following page: the nonagon drawing
Top of the page
PRESUMED IMPOSSIBILITIES
(18 November 2004. To the reader: This page has become the most visited one on this website, and I therefore feel I should call to the attention of the visitor that I treat other what may be viewed as thought-provoking subjects in the book mentioned on the home page. On that page I now added at the start a parenthesized paragraph like the present one, giving more information about the contents of the book. Perhaps the visitor of this page will find those contents likewise stimulating.)
(16 January 2006. After further consideration of the present issues, I thought I should bring out some matters I see as quite important. These pages are mainly confined to what I see as falsely presumed impossibilities. However, I also find many other false conclusions reached in highly respected fields, as I explore them in my book. What can be especially startling is that these falsehoods abound particularly in mathematics, a science considered most exact. What is more, the mathematical community fails to recognize a single thing wrong in the field. Physical sciences at least admit, if reluctantly, that its accepted views are revisable theories, but in mathematics the closest thing to this uncertainty appear to be what are known as conjectures. Writing more fully on this, as indicated, elsewhere, I feel compelled to at least mention here areas containing the falsities. Here noted ones are in non-Euclidean geometry, which consistently redefines terms to so commit the fallacy of equivocation; others concern set theory [a fitting name perhaps], regarding which infinite magnitudes are contradictorily considered and paradoxes wrongly interpreted; and still others are arbitrary postulates on which reasoning is built. The questioning of this and more is expectably resented, but it should instead be looked at with at least somewhat open minds.)
In the past—going back often to antiquity—thinkers have tackled many problems pronounced more recently as insoluble. Most have occurred in mathematics, and especially in geometry, owing to its renowned Greek legacy. It is my sense that the continued futility in solving the problems, and the eagerness to prove one's mettle, led many to prematurely decide that, rather than finding the answer sought, they found it to be impossible. Maybe the most famous of these problems is a proof of what is known as Euclid's 5th, or the parallel, postulate. My own consideration of it is included in the book referred to on the home page, so I will not consider it directly at present. But I might note, shocking as the idea is today, that I find the claim of that postulate's unprovability false and, yet more shocking, offer a proof.
At this time I would like to turn to some other ancient problems in geometry. A well known threesome of them is the trisection of an arbitrary angle, the duplication of the cube, and the squaring of the circle. Related is the construction of a regular heptagon. These are of interest to me at this moment because I just recently wrote about some to Professor Robin Hartshorne of the University of California at Berkeley, whose book, Geometry: Euclid and Beyond, I previously obtained.
At issue in the above mentioned constructions is performing them by use of merely compass and unmarked ruler, in the Euclidean tradition. They were only found instead possible by using added devices such as marks on the ruler, and also involving a "sliding" or "insertion" to make the tool fit certain points and lines,
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rather than depending on points determined. Whatever action was used, however, the constructions were never accomplished by compass and unmarked ruler alone.
This is here my main interest, without considering any proof presented that the constructions are in some manner impossible. With regard to only the past inability of using compass and unmarked ruler alone however applied, I indeed found ways of using these tools for such purposes. In my letter to Professor Hartshorne I turned specifically to the trisection and the heptagon, which he dealt with in detail in his book and are treated in the below drawings I sent him.
Both of them concern what was previously an insertion of a line by means of a marked ruler. The first drawing is of a simplified trisection that can correspondingly not be constructed with such a ruler, which relies on a length already given. The second drawing will serve to illustrate this reliance.
It is about a heptagon construction shown on page 265 of the book by Professor Hartshorne. In it the distance CF is marked on a ruler by which that distance is then transferred to segment GH by "sliding" the ruler so that the mark for G falls
on arc CG and the mark for H falls on line HF while line HGC goes through C. What was not recognized is that the same is accomplished if a compass with a center G on arc GC is positioned to describe an arc from F to a point H lying on lines HF and HGC which goes through C. In this drawing an unmarked ruler pivots on C till H is on the horizontal. (To briefly describe what in the construction is done before, with the right-side intersection of the circle and its extended horizontal diameter as center, and the radius of the circle, an arc is marked on the circle at C and the point vertically below it; with the distance between these two points as radius, and the same center as of the circle, the bottom arc on the vertical is marked; from that point the line to C is drawn, and with its intersection F on the horizontal as center the arc CG is described.)
Similarly to the preceding, in the first drawing a ruler for line ADC pivots on A till from B, on arc BD with center A, an arc BC with center D is drawn so that C is also on horizontal BC. Unlike before and in other cases, there is no preexisting radius like DC that could be marked on a ruler, making this a simpler construction than for instance on page 262 of the above book.
How this construction trisects an angle, here ABC, how for that matter the other drawing leads to a heptagon, will presently be omitted, there being opportunity for these elsewhere. (It may be helpful though to note that in the trisection the angle DBC is a third of angle ABC, and for the heptagon its initial sides are to be found by, with center H and radius of the circle, marking arcs on it above and below the horizontal—the distance then between the intersections on the circle of either of them and of the horizontal makes a side of the heptagon, with the other sides easily following.) What I wish to include is the following response I received by e-mail from Professor Hartshorne.
From: Robin Hartshorne [mailto:[email protected]] Sent: Friday, November 15, 2002 6:40 PM To: [email protected] Cc: [email protected] Subject: constructions
Dear Mr. Vjecsner Thank you for your letter about constructions. In spite of your protestations, I find nothing controversial in your letter. You have given a mathematically correct construction of the trisection and the heptagon neusis step. These constructions are not classical "ruler and compass constructions" as required in the classical problems, because they make use of the ruler and compass in ways not allowed by the traditional constructions. (See explanation on p. 21 of my book). I would say you have invented a new tool, that is simultaneous use of the ruler and compass in a sliding manner, and with this tool you have solved these problems. I wonder if your tool is equivalent to the use of the marked ruler--that is can it solve all and only those problems that can be solved with marked ruler? Sincerely yours,
Robin Hartshorne
The in this e-mail cited p.21, as I mentioned in my reply, allows that "at any time one may choose a point at random, or [a point] subject to conditions such as that it should lie on a given line or circle", which is what my use of "neusis" or "sliding" consists in. And as I indicated, previous attempts at making any of the constructions with only unmarked ruler and compass were unsuccessful regardless of whether those movements are allowed or not. I am reiterating this to bring to attention that the present solutions mean more than the various ones produced with the aid of additional devices.
If this sounds like boasting, I might lessen the effect by saying that an intention of mine is to bring my capacities to people's awareness, so they will give a listen to other things I offer, which I feel are of benefit. The preceding puzzles, again, are among many others I confronted, and are not included in the book of mine spoken of on the home page. This does not mean I do not like to carry the preceding explorations farther, and I have in fact done so and will try to describe them subsequently.
14 December 2002
Viewing the first above construction once more, I can add that the pivoting of the ruler, too, can be avoided here, as well as in the other cases. One can start by drawing arcs from B with a center D on arc BD, till a straight line ADC can be drawn with the ruler.
At this point it may be useful to say that the proof of this trisection follows from Euclid's Proposition I.32. By it angle ADB is equal to angles DBC+DCB, which are equal to each other, because triangle DBC is isosceles; hence angle ADB is twice angle DBC, and since triangle ABD is likewise isosceles, angle ABD, also, is twice angle DBC.
In light of such needed complication in even this trisection, probably the simplest compared to those known, which largely depend on the same proposition of Euclid, it may seem implausible if I offer a trisection simpler by far, whose rightness is so obvious that the impulse may be that the process is not legitimate.
As to be seen, however, the process is none other than the "insertion" used on the many other constructions. What is more, the only tool for it in this case is a compass.
To trisect angle AOB, with center O draw an arc AB; with a center close to C at estimated third of the angle draw the arc close to AD beginning at A; with reached center close to D at two thirds of the angle draw the arc close to CB beginning at last used center. Then "slide" the compass at C for a more accurate radius till the second arc ends at B.
The job is made still easier than the previous ones by knowing that because thirds are concerned, of the reached difference from the goal of B can be taken an again estimated but much finer third by which the radius is rightly changed, since each small angle must change by that third to attain the correct three angles.
To almost end this session let me return to the heptagon. As was seen, the described construction of it, notwithstanding that the drawing left out parts unrelated to the present issue, is quite elaborate and had to avail itself of insertion regardless, with an added device to boot. Other past attempts, equally requiring insertion and additional tools, were similarly intricate. As the preceding drawing may suggest, however, a much simpler process is possible here, too, as the following drawing will illustrate.
In likeness to the last case, the vertices of the heptagon can be determined by progressive arcs, to end at a specific point. Only 3 arcs are needed; upon estimating the length of a side close to the distance between A and B, draw with center near B an arc from A to a point near C on the circle; with center at that point, draw from A an arc to a point near D on the circle; and with that point as center draw from the point at B an arc toward A to complete the cycle. If that arc ends at A, distance AB becomes a side of the heptagon. As before, of the difference on the circle from that goal of A can be taken an estimated finer seventh, by which the initial radius for BA can be changed, till the third arc ends at A.
One could, to be sure, keep repeating the first arc, with each successive one centered at the endpoint of the former, till completion. But reducing instead the number of arcs not only shortens the job of "insertion", but the fewer the steps the smaller the chance of physical inaccuracy. The added dashed arcs in the drawing suggest a way to locate the remaining vertices.
It can be observed that the method in the last two examples can be employed to divide an angle into any number of equal parts, and to construct any regular polygon. The steps can, further, be drastically reduced by the likes of doubling each successive arc as in the preceding. Thus a 100-gon is constructible with 7 arcs.
Estimating the length of its side as radius of the smallest arc in the present drawing, draw the arc from a point A; draw succeeding arcs with centers at preceding endpoints and starting from points shown, until an endpoint near C is reached; and with that point as center, draw an arc from the point at B toward A to complete the cycle. Then, since distance AB amounts to eight times the concerned first radius, take of an eighth of the difference—found by halving the difference 3 times—on the circle from the goal of A an estimated finer hundredth by which to change that radius, till the last arc ends at A. The first radius equals then a side of this polygon. The dashed arc with center A suggests a beginning for correspondingly locating the other vertices.
The viewer should not have too much trouble calculating that the length of the first radius is contained 100 times in the measure starting and ending at A. Counting from A, the first endpoint reached after B can be seen to be 14 times that radius, the next endpoint 27 times, C 54 times, and since B is from C less by 8, namely 46, so is on the opposite side A; and 46+54=100.
Where and to what extent arcs can in this manner be increased (which they needn't be) for a polygon, or for a divided angle, may not be always easy to detect, although there are normally several relations possible for such arcs. My objective was rather, again, to demonstrate that not only can these methods be economical, but that the process of "sliding" or "insertion" can often be performed with simplified structures and compass alone, let alone with an unmarked ruler besides.
It may be added that the above method of constructing any regular polygon or dividing an angle into any number of equal parts is more significant in geometry than it may seem. Geometers may be at a complete loss if asked to construct for example a regular polygon of some arbitrary number of sides. An aid like a protractor may be of assistance, if deficiently so, especially the larger the polygon. The same is the case when approximations for certain polygons exist. In contrast, the above method allows for more precision the larger the polygon, because the
relative difference in accuracy can continually be reduced by any practicable amount.
26 December 2002
The next drawing utilizes once more the increasing arcs, this time without a need for "insertion", but the drawing is only an expected approximation in regard to "squaring the circle", one of the threesome of ancient problems mentioned. Its aim has been to construct a square of the same area as a circle, a task hindered by the irrational nature of connected pi, the measure of a circle's circumference in relation to the diameter.
In those constructions, pi, which is 3.14159... to infinity, has been approximated to half a dozen and more correct decimal places, and it is not the intention here to trump these, but rather to present a good approximation of exceptionally simple construction, by again only an unmarked ruler and compass. The approximation is to almost five decimal places, 3.14158..., which will be seen to go very far.
Past such approximations relied on rather involved constructions which in the end usually yielded the side of the square of area near that of the circle, and separate from it. One may be at a loss to instead find a construction of the square centered over the circle. From ancient Egypt it may be one that is of some fame, known as
of the Rhind Papyrus, and in which the side of the square is 8/9 of the diameter of the circle. Its accuracy as to pi is only somewhere near two decimal places, 3.16..., which nonetheless can in practice be quite adequate. (A substantial reference book on the subject is Pi: A Source Book, published by Springer.) (See a somewhat better approximation with an extremely simple construction on the next page.)
How the Egyptian square was realized appears uncertain, but evidently it was not by use of straightedge and compass. There is such a way, but now to the present drawing. The measurements resulting in its above approximation are 23 units for AB, and 44 for AC, with A and C respectively at half the side and corner of the square, whose intersection with the circle is at B, the point at issue.
Here, too, it may be asked how this can be realized when only the circle is given. The steps used are as follows. With center O of the circle draw prospective diagonals of the square; bisect as shown; on the bisecting line measure off chosen units of 21 (44 minus 23) and 44, at the end of which draw the perpendicular to that bisecting line; with their meeting point as center draw the arc from end of 21 units to that perpendicular; from the point meeting it draw the line to O. At the line's intersection B of the circle and perpendicular to the above bisector draw the side of the square, etc.
The preceding description may sound more difficult than the drawing obviously is, and I think the steps are self-explanatory and need not be justified. I could remark that the counting process, here shown with arcs on the left, is as a rule not depicted in the approximations, with only the resulting structure shown. As noted, further, the measurements are here made by means of increasing and few arcs, with the first chosen bottom radius the unit, and the numbering, which designates distance from center O, should aid in seeing the placement of those arcs.
It remains to give the calculation of the degree of accuracy in this construction as regards pi, known by the Greek letter π. The area of a circle is pi times radius squared, or A = πr 2. Therefore A/r2 = π; that is, the area of the above square divided by the square of the circle's radius will yield the approximation to pi. The area of the square is 882 = 7744, and the square of the radius is found by way of the Pythagorean theorem, through the sum of AB squared, 232 = 529, and AO squared, 442 = 1936, for the square of BO, 2465. And 7744/2465 = 3.14158...
If the square roots of pi and of this approximation, 1.772453... and 1.772450... respectively, are considered as sides of corresponding squares, and either of them is divided by the difference, about 0.000003, then 1 mile for either is found to differ from the other by less than 1/8 of an inch.
21 January 2003
9 June 2005 As done elsewhere, I am sandwiching in another item here, of a new approximation that occurred to me of a circle squared. The approximation is fairly good considering that this is the simplest way I found of performing it with unmarked ruler and compass, utilizing the above "sliding" or "insertion" and my circular arcs for division.
On a circle with a suitably vertical radius OA (=1) and A as center draw an arc BC such that on drawing with center B arc AD, with center C arc AE, and with center E arc BF, the arc drawn with center F from B meets D. Any part of the circle spanned by radius AB is then 1/13 of the circle, with angle AOB 27.692...°. The cosine of that angle is 0.88545..., and, with a side of the square drawn through B and C, this cosine (=the vertical from O to that side) is half that side. The full side then is 1.7709..., and the square 3.1360..., compared to pi, 3.1415..., the area of the circle (pi times OA² or 1). The difference between that side, 1.7709... again, and the square root of the circle, 1.7724... (the side of the corresponding square), is less than a millimeter for a meter.
It may be noted again that the circular arcs I employ to find the ultimate point to be reached are far more accurate than many methods not depending on found points but on predetermined ones, as in various other constructions leading to squares approximating the area of a circle. In those cases, though the steps arrive mathematically at a good approximation, there is no way of checking—by straightedge and compass—that the steps are physically accurate enough. In contrast, the very act of adjusting my arcs for proper "insertion" assures as close an accuracy as physically enabled (mainly the first arc needs adjustment, the others depending on it, which is why I sometimes show it alone drawn twice).
Someone may argue that what is of account in the other cases is their mentioned mathematical reliability, not the physical limitations. However, the methods of "sliding" or "insertion" are equally mathematically correct. As in the preceding, if distances or other requirements are met by the "inserted" elements, then the accuracy at issue pertains. (The paragraph following here and the rest belong to previous matter on this page.)
7 January 2008. I am again squeezing in another item, and the preceding last sentence, in parentheses, refers now to the following matter that starts on a yellow panel. At this time I am depicting an approximation of a "rectified" circle, for an additional method of squaring the circle with straightedge and compass to a fairly good accuracy. The number 3 17/120 yields 3.14166... as an approximation of pi accurate to almost four decimal places, only by one less than the
first of the two constructions above and with the present construction simpler by far, at least in comprehension if not in execution.
As should be easy to follow from the image, one way to measure off 3 17/120, after first drawing a line 4 times the length of the circle's diameter, is to continually halve the last of these lengths until 15 units are reached, divide the second 15 by three for 5 units, and then count 2 more fifths for 17/120. The square root of the total (the dashed line discounted), by which to construct the square, is known to be found by a simple method utilizing continued proportion similar to one shown below at right. (In practice, it is only necessary to extend 3 17/120 at left to 4 17/120, use this total length as a diameter for a semicircle, and construct a perpendicular from the start of 3 17/120 to that semicircle. This perpendicular is then the square root sought.) 1 February 2008. I am replacing below a former image, because this new one represents a more general approximation of pi as 3.1416, which is better than the preceding 3.14166..., since the 6 alone is much closer to the actual 59... in pi than is the repeating 66.... In fact it is also closer than the 58... in the above first of the four cases depicted. The present approximation is given by 3 177/1250. Accordingly, pictured is the pertinent part of the "rectification" of this pi, similar to the preceding.
As seen, the denominator of the fraction, 1250, standing for the diameter of the circle, or a unit, is the main dimension shown, and above it is measured the numerator, 177, the part of that unit belonging to the length of the pi. Of the 3 units, the whole numbers, that are to precede that part a portion only is shown at left as if a panhandle. Explaining the measuring, 1250 is divided into fifths of 250, the first of which is divided into fifths of 50, with the fourth divided into fifths of 10; from this is obtained 45, divided into fifths of 9, to obtain its multiple 27. This plus the 150 shown then equal the needed 177 out of 1250, the dashed line again signifying the discounted portion.
11 February 2008. I thought of a way to "rectify" approximations of pi, of a circle's circumference, similarly to the preceding two but much easier and as usable for any fractional form of such approximations. Since this page is getting quite cluttered, I am illustrating the method on the next page.
So as to treat of each of the triad referred to, a look is here taken at the problem of duplicating the cube. The issue is doubling the volume of a given cube, by again use of only straightedge and compass. Ironically, doing so, by way of “sliding” or “insertion”, was in antiquity virtually accomplished, failing only to put the proverbial “two and two together”.
As described in the book The Ancient Tradition of Geometric Problems by W.R. Knorr, the solution was in different periods attempted both by means of a pivoting ruler similar to the one in the first two drawings here but without recognition that the compass can be of aid (pp.188-9, Dover edition, 1993), and by means of a compass while overlooking the role of a ruler (p.305). Complying with the present drawing (it may be noted that rectangle ABCD is twice as wide as high, the rest to become evident), by the first of these cases a ruler “able to pivot about [B]” would “be manipulated until it assume the position such that [EF] = [EG]”; by the second of the cases, “instead of a sliding ruler [FBG], there is conceived a circle centered at [E] such that the vertex [B] and the two intercepts [F], [G] lie in a straight line”.
It is really the second case where a sliding ruler, other than a pivoting one, can literally be employed. As depicted, the radius of an arc with center E can be adjusted until a line connecting the arc’s endpoints F, G and point B can with a sliding ruler be drawn.
There should be no need now to demonstrate the resulting duplication, the information if of interest obtainable elsewhere. It may be of use though to note that if BC is the edge, viz. the cube root, of the first cube, then CG is the edge of the cube of double the volume.
28 January 2003
6 August 2004 Recently I became aware of another duplication of the cube offered (included on page 270 of the book by Robin Hartshorne referred to on top of this page) which uses a marked ruler. In accordance with the construction below, segment AB makes a right angle with BC, which makes a 30° angle with BD. With distance AB marked on a ruler, it is by "sliding" transferred as distance EF on line AF such that
the line lies on A, E is on BC, and F on BD. I again found the following way of achieving this with only an unmarked ruler and compass, as held impossible.
With center B and radius AB draw circle; by "sliding" the compass, find a center G on the circle for a circular arc of radius BG, intersecting BC at a point E, with which as center and radius GE an arc intersects BD at a point F, such that A, E and F lie on a straight line, as determined by the ruler.
The critical first arc is once again shown in two tries, indicating that the appropriate center G must be located. Since the radii are constant, AB and EF are of equal lengths. In this construction, with AB the cube root, edge, of the first cube, AE is to be the cube root of the cube twice the size.
16 September 2004 It should be no surprise if I continue with these constructions, thought to require some device other than compass and unmarked straightedge alone, whether or not performed by way of "sliding" or "insertion", by the indirect way of finding unspecified points through which exact requirements are reached.
For instance, I have been able to so construct with straightedge and compass all forms of cube duplication and angle trisection I have encountered and which often are believed to need involved mechanical devices or geometric shapes ranging from special curves to combinations of solids.
First let me return to a previous cube duplication for which "insertion" is considered used, but the nature of which is not specified. As I noted there, the process can be of greatest simplicity with straightedge and compass alone. One need merely adjust the compass for the radius of the arc until F, B and G lie in a straight line as determined by the straightedge.
Nonetheless, the mentioned book by W. R. Knorr, which contains these, presents
also (p.306) a more elaborate version, depicted below first and said to be simpler to execute by "neusis" ("insertion"), without giving an explanation. I am including it here not only because I find it doable with straightedge and compass, though much more involved than the other, but because the result has a certain symmetric beauty.
The idea in this version is that if rectangle ABCD is, with center E, circumscribed by a circle, then upon its intersection of correct line FG at points I and B the segments FI and BG are equal. My way of attaining this with straightedge and compass is as follows. Rotate the straightedge on B so that when with B as center an arc with a radius BG is drawn, and from the midpoint of line FG or its part IB is drawn the perpendicular to meet that arc at a point H, and with H as center and same radius, now HB, is drawn an arc BI, and with I as center and same radius, IH, is drawn an arc HF, the arc will meet line FG at line DA extended. (A much shorter way occurred to me, but I leave the depicted, as perhaps interesting. Rotate line FG on B until a perpendicular through the midpoint of FG or its part IB meets E beneath. The perpendicular is usually drawn in locating that midpoint.)
The second preceding drawing is likewise for doubling the cube and is accepted as requiring a device like a marked ruler. A somewhat differing form of it is in the above mentioned book by Robin Hartshorne (p.263), but I found the present one more frequent, and it is credited to Newton.
In it an equilateral triangle ABC is constructed; AB is then extended for BD of equal length and line DC is drawn and prolonged, as is done to BC. Thereafter a ruler is marked with distance AB and applied so that one of the marks falls on line DC at a point E, the other mark falls on line BC at a point F, and the ruler goes through A. With AB the edge of the smaller cube, AE is then the edge of the doubled one.
The way I used unmarked ruler and compass instead is by first drawing an arc with center B and radius BA (one could draw a different arc, but this one nicely
preserves a circle used to get the equilateral triangle by Euclid's Proposition 1). Then a center G on the arc is found such that with radius GB an arc BE is drawn and with center E an arc GF, with AEF a straight line.
Following further are two depictions of ancient angle trisections, equally included in the book by Robin Hartshorne (pp.262, 269), and that have likewise held to require devices like marked rulers, or curves like the conchoid of Nicomedes (used also for above cube duplications and dealt with in the books by Hartshorne (pp.263-264) and Knorr (pp.219-226). They are similar to an above trisection of mine, which, as noted, is simpler by disposing of the need for marking distances.
First here depicted is a trisection, of angle ABC, attributed to Archimedes. Drawn is semicircle CAE on CB extended; distance AB is then marked on a ruler, which is thereafter so positioned that one of the marks falls on line DC (point D), the other on the semicircle (point E), and the ruler goes through A.
Keeping the arrangement, I dispense with the marked ruler by drawing an arc with a center E so positioned on the semicircle that the arc, drawn from B, meets line DC at a point D that forms a straight line with E and A.
It may have been bothersome to some in antiquity that in that arrangement angle ABC does not contain its sought after third, angle EDB. In any event, the second preceding depiction is of another, similar, trisection that had been given. There seem to have been two basic versions, describable here. In both, to the to be trisected angle ABC is added line AF parallel to BC. In one case, then, a perpendicular is dropped from A to BC; distance AB is then marked twice successively on a ruler, which is then so positioned that it lies on B, with the first mark on AC (point D) and the last mark on AF (point F); from central mark, E, a line is then drawn to A for elucidation. The result is that angle FBC, contained in angle ABC, is one third of it. In the second case the perpendicular AC is omitted; an arc with center A and radius AB is drawn; distance AB is marked on a ruler, which is positioned then so that the farther mark is on AF (point F), the nearer mark is on arc BE (point E), and the ruler passes through B.
My own solution, without marks, is, on drawing arc BE, position a center E on it for another arc, AF, so that B, E and F are aligned.
Let me in the first following depiction add another trisection, similar to an above one. One difference is that instead of moving the center of the initial arc, its radius
17 January 2008. This late insertion is about the figure directly below. I felt it would be good to give a simple, non-algebraic, proof of this cube duplication, having to do with observed continued proportionality and also applying to the other cases.
Noting again that in the three triangles ADF, AFE and AEC, AD is to AF as AF is to AE and AE to AC, it follows that with AD length 1, the length (for n x 1 or n) by which AF exceeds AD is the length (for n
2) by which AE exceeds AF,
and the length (for n3) by
which AC exceeds AE. That is to say, AF (n) is the cube
alone is adjusted. To trisect angle ABC, with B as center draw arbitrary arc CA; draw with center A an arc DE with a radius of an estimated third of angle ABD, and adjust the radius until with center D an arc CE meets on extended arc CA both arc DE (at E) and line BC (at C).
root of AC (n3), which, as 2,
stands for twice the cube 1, their edges as cube roots respectively AF and AD.
The second preceding drawing returns to cube duplication. I thought it of much interest because it illustrates a basic concept involved, called mean proportionality. In the drawing, for instance, line AE is the mean proportional between AF and AC, because in the right triangles ACE, AEF and AFD, AC is to AE as AE is to AF. It was found that if in a semicircle like the present one the radius AB, namely half the diameter AC, is the edge of the smaller cube, then regarding the same triangles the continued mean proportional AF between AD (=AB) and AE is the edge of the double cube.
Making the construction proved to be a problem, however. The issue is to find point E on the semicircle such that all perpendiculars meet as required (angle AEC is always right by Euclid's Proposition III.31). A famed ancient solution, described in the above book by Knorr (pp.50-54) was given by Archytas, an associate of Plato. He used elaborate intersections of a cylinder, a cone, and a torus, and has been hailed for his ingenuity. Another recent book, Geometry: Our cultural Heritage by Audun Holme (Springer), even utilizes his construction as cover. A much simpler solution should nevertheless be more welcome. The same book also gives a solution (pp.54-55) where two T-squares are used, one of them marking yet distance AD on the crossbar at top from its right inside corner toward the left; that corner is then on arc BD here placed to coincide with a point D, with the other mark on A, the crossbar along a line AE, and the long bar along a line DF; the second T-square has the crossbar at bottom with the inside edge aligned with diameter AC, and the long bar upward. The two T-squares are then so adjusted that the right edge of the long bar of the first meets the left inside corner of the second on AC, while the bottom right edge of the crossbar of the first meets the left edge of the long bar of the second on semicircle AEC.
This would be better visualized with a picture, as in the book, but let me turn to my construction of the same with straightedge and compass. To make it easier to
This text comes after the first three paragraphs at left, and is about the construction immediately above. It is done with straightedge and compass by rotating a line AE on A so that on dropping a perpendicular from E to AC, and from meeting point F to AE, the meeting point D falls on arc BD.
I may remark that for these perpendiculars it suffices mathematically to with center E draw an arc tangent to AC, and with center F an arc tangent to AE. But unlike the standard way, after Euclid's Proposition 12 or 11, this is physically not very practical (as is not at all practical the intersecting of three solids as noted at left).
28 August 2008. Regarding the figure directly above, I wrote an explanation above it of how the cube root is determined in it, and not being sure that I made myself clear enough, on the
follow the text and picture, I am using the top of the column to the right for the description.
5 October 2004 Presently I felt it of interest to sandwich between earlier items this further consideration of the ancient answer by Archytas to the preceding cube duplication. As noted, the answer availed itself of the intersection of three solids, a cylinder, a torus, and a cone. These were really used only partially, as to be seen. Viewing the drawing below, in correspondence to the last drawn semicircle's diameter and radius as lengths for which mean proportionals are sought the bottom oval below, depicting a circle in three dimensions, has its diameter AB given again as one length, and the chord AC as half that length. The chord is extended to D, meeting line BD, tangent to circle ACB. I may remark that some postulations, like that tangency, are not required. Triangle ABD is to be rotated on axis AB (to describe part of the cone) toward position ABD', for which BD needn't be tangent as said. Also, with arc CEF described by the rotation, line CF is needlessly posited as parallel to BD or at right angles to AB. It will be so by the rotation. To continue, a right cylinder or half of it is constructed on circle ACB, and a vertical semicircle on diameter AB. This semicircle is rotated about A horizontally (to describe part of the torus) toward position AGB', intersecting, like triangle ABD', the cylinder in the process. The same semicircle and triangle will meet on the surface of the cylinder at a point G. Through a series of deductions triangle AGB' is, as above, found to yield the sought after result.
next page I did the like for drawing a square root, maybe helping in the above.
What I want to observe is that this elaborate structure relies like others on "sliding" or "insertion", though it may not be at once evident. The "cone" and "torus" are "loci", signifying the continuous motions of the respective triangle and semicircle by tracing their routes. Accordingly one is by moving them searching for the point, here G, at which they will appropriately meet. In other words, one has to by, as described, rotating both the triangle and semicircle find the point at which they meet on the cylinder. In the drawing preceding this one the point at which instead the semicircle and vertex of the inscribed triangle appropriately meet, is E. As seen, furthermore, the last pictured method is hardly achievable in practice. It is difficult to imagine how the concerned semicircle and triangle are physically rotated so as to meet on the cylinder as required, not to mention a desired solution using only straightedge and compass.
28 September 2004 Possibly more interesting than the construction of double mean proportionality in the last two drawings is the next one. Because of the adjacency of the triangles in their clockwise arrangement, the mean proportionals are easily seen. AB is to BD as BD is to BE and as BE is to BC. A difference is that in the last case the proportionality can continue in principle indefinitely (in the present case this too is possible if supposing overlapping spirals), not necessary for present cube duplication. With AB here twice as long as perpendicular BC, which is the edge of the small cube, BE is the edge of the double cube.
Given ABC, to draw this construction with only straight edge and compass may be viewed as my coup de grâce to the presumed impossibility of the like. The drawing is well known to have been approached since antiquity with a fairly elaborate mechanical device, used in a "sliding" manner of equal involvement. The method is described in the above book by Knorr (pp.57-60) and also in the often referenced books Science awakening by B.L.van der Waerden (pp.163-165) and, especially, History of Greek mathematics (vol.1, pp.255-258) by the most relied on English author in the field, T.L. Heath. He writes:
"If now the inner regular point between the strut KL and the leg FG does not lie on [CB] produced, the machine has to be turned again and the strut moved until the said point does lie on [CB] produced, as [D], care being taken that during the whole of the motion the inner edges of KL and FG pass through [A], [C] respectively and the inner angular point at G moves along [AB] produced.¶That it is possible for the machine to take up the desired position is clear from the figure of Menaechmus, in which [DB], [EB] are the means between [AB] and [CB] and the angles [ADE], [DEC] are right angles, although to get it into the required position is perhaps not quite easy."
Instead, my use of straightedge and compass only, in likewise a "sliding" manner, can be seen to be child's play. Rotate the straightedge on A so that on drawing at meeting point D of CB extended the perpendicular to AD toward AB extended, the perpendicular to DE at meeting point E meets C.
The second preceding drawing is once again a version of a previous cube duplication. It is an admired one by a contemporary of Archimedes, Nicomedes, known also for his invention of the conchoid, a special curve. He used the curve for this construction, and created it with also a mechanical device he built for it. Moreover, the curve had to be specially created for any construction at hand. The mechanical device, the construction, etc., are again found in the books by Heath, van der Waerden, Knorr, or Holme. In their figure, given rectangle ABCD draw line from B through midpoint E of DC to extension of AD at G; from midpoint F of AD draw perpendicular FH of such length that AH equals DE; join GH and draw AI parallel to it; from H draw HIJ meeting AI and DA extended at J so that IJ equals AH (=DE); draw line JBK to meet DC extended in K.
For my construction with straightedge and compass, with center D draw arc EHL for distance DH to get equal AH; on that arc find center L for an arc from D to a point I on AI so that with center I an arc from L meets DA extended at a point J to make straight line HIJ.
5 September 2007 Again, I am sandwiching in with this text and the next image a cube duplication that will likely be the last one by me, inasmuch as I seemed to have managed to draw with straightedge and compass, using the method of "sliding", all the known variations of the figures (excepting the squaring of the circle) thought impossible to construct with these tools.
The present duplication is a corresponding solution of one that was likewise done with a mechanical device, by Eratosthenes, a contemporary of Archimedes and mentioned previously. He, as also described by T.L. Heath in History of Greek mathematics (vol.1, pp.258-260), built a frame represented here by the top (dashed) and bottom horizontals and the far-left vertical (AE); the horizontal parts had interior grooves in which panes like the leftmost rectangle (which remained fixed) could slide and overlap; the two other rectangles here represent the sliding panes, which were the same as the first but partly hidden; each pane also had the same diagonal (like AF); there was a rod probably attached, to rotate, at A.
Now, the edge of the small cube is marked on the last pane as distance DH, which is half of AE; the doubled cube's edge, which is CG, is again found through mean proportionals; to have AE, BF, CG and DH in required continued proportion, the two right panes were made to slide until B and C, the meeting points of the corresponding verticals and diagonals, formed a straight line with A and D.
Here then is my construction of the preceding with straightedge and compass only. With only the solid lines considered, draw lines EH, AE and, at a random distance from the last, BF; draw diagonal AF and auxiliary horizontal at height D. Now rotate a line AD until on drawing a diagonal BG parallel to AF, a vertical GC, and a parallel CH, the vertical from H meets D.
22 October 2004 Let me now return to one of the first problems on this page, the heptagon, drawn for a communication with professor Hartshorne, by whom an e-mail is reproduced. That drawing is about a solution by the sixteenth-century mathematician Viète, and it involves the use of a marked ruler. As was seen, I was able to do the same with an unmarked ruler and also offered a much simpler method, applicable to any regular polygon.
It may be of interest to include more here on the pentagon, in particular a discussion of an answer known as given by Archimedes. It is considered in most of the books cited above, in Van der Waerden (pp.226-7), Knorr (pp.178-182), Hartshorne (p.270), Holme (pp.90-93).
Viewing the first below figure, Archimedes was to observe that in the heptagon if line AB and triangle CDE are drawn, then, with intersections F and G, AG x AF equals GB squared, and FB x FG equals AF squared. On dividing a line AB into these segments then, a heptagon can be constructed. To attain this division, Archimedes, in accord with the second figure below, is said to draw a square ABCD with diagonal AC, and, with AB extended, rotate a straightedge on D for a line DE so that, with intersections F and H, triangles CDF and BEH are equal in area; by dropping the perpendicular FG to AB, the segments AG, GB, and BE
correspond to AF, FG, and GB in the previous figure.
To add a comment before proceeding, there is interestingly no indication of how on so sliding the straightedge the equal areas of those triangles are determined. What is more, it appears to me that one could instead similarly adjust in the first figure the three concerned segments until the required areas of the rectangles and squares are attained, dispensing with the second figure. In fact, while I have no knowledge how areas of dissimilar triangles, as in that figure, are with Euclidean tools found equal (in this case, if reversely the segments of AE are known, the triangles can be found), there is a way of finding those segments (of AB in the first figure) by Euclid III.36. If starting with e.g. a segment AF squared, one can find rectangles like FG x FB of the same area, and use "sliding" until GB squared equals in area rectangle AF x AG. (I found later that in the second figure the rectangle formed by the height and base of either triangle in question can be made into a square by Euclid II.14, which then can be tried to be made by preceding III.36 into the like rectangle for the other triangle, having to of course rotate DE again to find the required equal areas in this elaborate process.)
There is, however, an easier way. As noted in Knorr (pp.181-2, 204), the roots of Archimedes' procedure may be compared with those of Euclid's pentagon (IV.11), dealing with properties of isosceles triangles. It has been correspondingly known that, as in triangle ABC in the first figure below, the triangles can be divided into adjacent isosceles triangles with accordingly equal sides, here AB, BD, DE, and EC, as also discussed in Hartshorne (pp.266-7) in connection with the above heptagon by Viète. Now, I want to demonstrate how presently, too, I use only straightedge and compass to construct the triangle and with it the heptagon.
At the midpoint of chosen line AB for a side of the heptagon erect perpendicular FC; with B as center draw circular arc AD; rotate on A a line AC through a point D on arc AD toward FC so that, on completing from the meeting point on FC a triangle ABC, if with D as center an arc BE is drawn, then with center E on the right side of that triangle an arc DC will meet its vertex. (Dashed transversals BD and DE are not required for the construction. They were merely used here for preceding information.)
The next figure, above, replicates the first one in the preceding pair, but for the addition of some dashed lines. It will be recalled that a special relationship of areas had
to be found regarding the segments of line AB for their length. Those lengths, were they still needed for the
purpose, are given by the figure at left, without search for the areas. In it AD, BE, and EC correspond above
respectively to FG, AG, and GB. To demonstrate, draw HB; then line AFGB is duplicated in DKGC; let triangle
EDB in the figure at left be replicated above by GID, for
Triangle ABC can then be circumscribed with a circle (Euclid IV.5), and the dashed arcs with center C and through E and D suggest a way to complete the heptagon. It can also be completed without the circle, with an arc from B centered at A and meeting the lower dashed line at left, and so forth.
parallelogram GIDB since the sides are of equal length (note the parallelogram above it for length CG and GB); draw IJ parallel to HB and ED. Then IE = JD = KG = FG,
and of course DG = AG and CG = GB.
26 October 2007. This is again squeezing in an item, after the text following on the yellow panel. At the start of that text I mentioned solutions to well-known problems with some constructions by straightedge and compass. Less known is similar use of those tools for other purposes. Archimedes used for such "neusis", an indirect connection of points, further aids on preliminary constructions in his work On Spirals. Here I am showing one of these as related to that treated in T.L. Heath's A History of Greek Mathematics, Vol. II, pp.386-8. He speaks of the use of special curves, taking it that straightedge and compass fail (Archimedes may have used a marked ruler, as above). One can, however, again do the job without the aids, as I describe below the diagram at left.
It is stated, "Given a circle, a chord [AB] in it less than the diameter, and a point [C] on the circle the perpendicular from which to [AB] cuts [AB] in a point [D] such that [AD is greater than DB] and meets the circle again in [E], it is possible to draw...a straight line [CHG] cutting [AB] in [H] and the circle in [G] in such a way that [HG is] equal to [DE]."
My use of straightedge and compass only is as follows. On drawing the circle and perpendiculars AB and CDE, with E as center draw arc FD, and with an F on it as center draw an arc EG such that with G as center for arc FH, GHC make a straight line.
31 October 2007. It seems I never stop adding to these pages. "Neusis", the indirect connections on this page, was also employed by Apollonius of Perga, regarded as second to Archimedes among the great Greek geometers, and he in fact wrote two books, now lost, on the subject. Their content was described in later work, likewise dealt with by Heath in the volume mentioned in the preceding. On pages 190-2 he speaks about a "construction by Apollonius [that] can be restored with certainty". It contains a rhombus (a parallelogram with four equal sides, like a square), here depicted as ABCD. The object of this "neusis" is that given a length EF, a straight line DHI be drawn such that segment HI from BC to AB extended equal EF.
(5 November. I am indicating in parentheses here the complex way this "neusis" was to be accomplished. The diagonal DB, not shown here, was extended upward to locate a line of the same ratio to EF as AB is to DB, the diagonal; for that line, its square had to equal a rectangle formed by that diagonal extended to a point and the distance from that point to B; "then with [that point] as centre and radius equal to [the line sought, drawn is] a circle cutting [AB] produced in [I] and [BC] in [H]. [HI] is then equal to [EF] and...verges towards [D]". I may note that this equality, etc., may have as well been reached by the known method utilizing marking of a ruler.)
I show now how the neusis can again be simply performed by straightedge and compass. With E as center draw arc FG, and with a G on that arc as center draw an arc EH such that with H as center for arc GI, DHI make a straight line.
My preceding solutions to trisection, cube duplication, and the heptagon, by using only straightedge and compass are difficult for the profession to acknowledge, because of the reluctance to recognize that any of its widely held convictions could be revealed wrong. I consider myself unique in being able to resolve many difficult issues, and therefore the views by many others that a number of general mathematical or scientific claims is false are, although they may instinctively be justified, insufficiently supported and correspondingly dismissed by professionals en masse as coming from the ignorant.
That the above problems could be solved by straightedge and compass alone, even if—in what is called "neusis", "sliding", "insertion", or "verging"—one has to, in order to arrive at the desired construction, find points not given in advance, is, as seen, not held possible, in believing that only the described marked rulers, special curves, mechanical instruments, or other tools can do the task. In the above reproduced e-mail by professor Hartshorne he twice speaks of my method as a tool, in apparent unwillingness to admit that what it consists in is a bypassing of the tools. (14 September 2007. A while ago someone wrote on the web page http://en.wikipedia.org/wiki/Talk:Heptagon, a discussion page for the heptagon entry in Wikipedia, that the heptagon can be constructed with straightedge and compass only, the person giving a link to this page. The discussion continues
there, with someone else answering, and me responding. If the reader is interested, I explain there that Euclid did not place a limit on the use of straightedge and compass, such as what points can be connected, unlike contended in trying to prove the concerned constructions impossible, a proof not accomplished otherwise.)
There are in the fields, as I indicated, other assertions—quite a few of them—that I dispute, beside claiming that I can answer many further questions. The preceding is intended to contribute to any confidence I may engender in what I do.
30 November 2004 Some recent findings of mine made me decide to add a little more geometry, but I do it on the next page for which click the succeeding link, my not wanting to overload the present page. To propose to prove Euclid's parallel postulate, the 5th, is ridiculed almost as much as to propose to prove the existence of God. Consequently, if I enter this arena I can only remind the reader that the depictions on this page are in the main of solutions (constructions with only Euclidean tools) not achieved in over 2,000 years by the best of minds. The reader may recognize by this that I am competent and conscientious in my work and therefore do not contend a result lightly. The impossibility alleged of proving the 5th postulate is based on fallacious reasoning I consider elsewhere, and so I encourage the reader to click the following for the next page.
MORE ON PRESUMED IMPOSSIBILITIES
PRESUMED IMPOSSIBILITIES
(continued1)
30 November 2004 As noted on the last page, I am adding to the top of the present one this material on Euclid's 5th postulate. I mentioned the alleged impossibility of proving it and that this is based on fallacious reasoning. How so I go into thoroughly in my book and prefer not to be sidetracked by at this time. I want my book understandably to be read anyhow, because of what I claim is its abundant and significant other content as well, beside including the present explorations.
What I am about to offer is, of course, considered preposterous in its claim, and it regardless consists of not only one demonstration of the 5th postulate, but also of two further ones that came to me later, each of them rather simple. In connection with these demonstrations I will refer to some other authors' work, to perhaps make the ideas clearer.
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In the above figure the solid lines represent the 5th postulate, which, simply told, asserts that if any three straight lines intersect to form angles a and b that make together less than two right angles then the two upward lines if indefinitely prolonged meet.
The 17th-century mathematician John Wallis proposed moving the right line leftward to meet the left line for a triangle, and supposing similar triangles in any size, he inferred one for the initial state. The similarity, however, is itself an unproved postulate equivalent to the 5th.
One does better moving that line, after making it (dashed
The next proof is based on Euclid's Proposition 17, which states the converse of the 5th postulate, though thinkers
could not utilize it accordingly. It states "In any triangle two angles taken together in any manner are less than
two right angles". To merely infer the converse commits the fallacy of "affirming the consequent". "A implies B"
does not imply "B implies A". But else is possible.
Triangles of any size encompass all angles a and b, up to when the sides merge into one line, for two right angles
or no angle. That is, triangles include all angles of the discussed postulate. Now, it may be asked whether the
distance between a and b can be so large that when they make next to two right angles, the two lines will not meet.
line) intersect the other at left, rightward. It will intersect the other only one point at a time, since the two cannot enclose a space (Euclid I.4 and proof in my book, p.131) or meet along a segment, making the lines of different angles. (It is in fact recognized that the distance between two intersecting straight lines progressively increases.) In consequence, since the lines are infinite, each point of intersection reached on moving the (dashed) line to the right leaves behind a next point, and thus the lines never part, namely they always meet, as was to be proved.
The motion can also be explained via Euclid I.15, by which two intersecting lines have their vertical angles equal. The angle between the two upward lines at left in the figure is the same above their intersection as below it. I.e. the lines will at no point of their intersection pass each other.
The question seems justified if it is assumed that similar triangles, ones that have all angles equal, do not exist. Let us then postulate a small enough triangle (its right side represented by the dashed line above) in which a and b make next to two right angles. The angle at the
apex, the smallest next to none, can be duplicated indefinitely if lengthening the left side, for ever larger
triangles, and the question is will the distance between a and b, which make likewise next to two right angles
before merging into one line, also increase indefinitely. This can be settled by Euclid I.28, by which the equal
angles at the apex of the two triangles make their right sides parallel. Namely, they never meet, which, with the
triangles growing throughout in least amounts, is to say the distance between a and b will increase indefinitely,
making the two sides always meet to form a triangle.
The following third demonstration is related to what is known as the "Saccheri quadrilateral", after the logician and priest of the 17th to18th centuries who with it began a lengthy endeavor to prove the 5th postulate and was a precursor of non-Euclidean geometry. He constructed a quadrilateral ABCD like the two below, where all sides are straight and AD and BC are at right angles to AB. (The dashed line, DC, although meant to be straight, is, as frequent, shown curved for illustrative purposes.) Then in conformance with the 5th postulate all four angles should be right angles.
Saccheri embarked on the journey of trying to prove the postulate by showing via reductio ad absurdum that if the postulate is denied, it will lead to a contradiction. He assumed the angles at D and C to be either obtuse or acute (as suggested respectively by the dashed lines), and proceeded to investigate the consequences.
Here I turn to an author who preceded and influenced him, Giordano Vitale. He likewise constructed such a quadrilateral, in order to prove straight line DC to be equidistant from AB, which would again be equivalent to the 5th postulate. He dropped an intermediate perpendicular FE from DC to AB, and attempted to prove this perpendicular equal in length to the others.
This approach, too, can be seen a backward one, as find below.
Instead of assuming DC to be straight and trying to determine the length of EF, one can assume EF to be the length of the other verticals and try to determine whether DC is straight. For the answer I will use an unconventional way, by turning from two dimensions to three (consider their non-Euclidean use and an ancient one). The present plane can in three dimensions accordingly be viewed from a side, when appearing as if a straight line. Of interest now is so viewing lines, specifically the above assumed equal verticals and AB. AB would be seen as a point, and the verticals as a single line. (The view is isometric, when size does not change with perspective, as it should not for comparison.) Now because the verticals are of equal height, their upper ends will also coincide in a single point, signifying a straight line. Therefore DC is a straight line equidistant from straight line AB, in equivalence to the 5th postulate. (The corresponding equidistance of parallel lines is a feature implied by the postulate.)
I may add that the three-dimensional way does not somehow violate Euclid, because the issue is to prove the 5th postulate by any method at disposal. Furthermore, as noted, e.g., by T.L. Heath in A Manual of Greek Mathematics, pages 175 or 216, "Plato had defined a straight line as 'that of which the middle covers the ends' (i.e. to an eye placed at one end and looking along the line), and Euclid's 'line which lies evenly with the points on itself' may well be an attempt to express Plato's idea in terms excluding any appeal to sight".
16 May 2005. As time passes, new things occur to me that I feel I want to add to this website, as I've done previously, and this time I am inserting something about the pentagon. Its construction has been seen as one of Euclid's special achievements, because it is based on a great deal of preliminary work by him. It is in fact interesting how beside supplying proofs for his construction, its mere carrying out by him is very elaborate. The reader should not be misled thinking the two pictured below as his; they both are much simpler but again utilize means not Euclidean. Let me indicate the many steps he takes. In IV.11 he shows how to inscribe in a given circle a regular pentagon; for this he first uses intricate IV.10, which in turn uses constructions in II.11, IV.1 and IV.5; of these, II.11 uses I.46, and IV.5 uses I.10; among the last, I.46 uses I.11 and I.31; of these, I.11 uses I.3 which uses I.2, and I.31 uses I.23; the last then uses impressive I.22 which again uses I.3; going back to initial IV.11, it also uses IV.2 and I.9, the first of which uses III.16. The whole then results in the pentagon.
At the start of that construction a triangle with each of two angles double the third angle is inscribed in the circle. The triangle at issue would be the incomplete one below with angles at A, B and E, in resemblance to a preceding heptagon drawing. A pentagon construction like the below has been supposed (e.g. in the book Euclid—The Creation of Mathematics by Benno Artmann) to have been made before Euclid, and it involves a
However, there is yet a much simpler construction also using "verging" or "insertion", as shown on the
preceding page regarding, for instance, a heptagon again. Below, on a given circle draw with center A an arc BC such that after drawing with center C arc AD, the arc
drawn with center D from A will meet B.
Beside further simplicity, this construction, like the
marked ruler, for the here previously discussed "neusis" or "verging". The chosen distance AB is marked on the ruler, which is then so applied that one mark falls on a point E of extended line CD bisecting AB, and the other mark falls on a point F of arc ADH, while the ruler passes through A. The pentagon is then easily completed.
As seen on the preceding page, I have been able to use "verging" with only compass and unmarked ruler in all cases, not held possible without a marked ruler or other additional tool. In the present case, this is possible as follows. On arc ADH find a center F for arc BE such that EFA make a straight line.
Euclidean one, allows choosing in advance the size of the circle and hence of the pentagon. The construction at
left must instead postulate the length of AB and change to the desired size afterward. Correspondingly one has to
also either circumscribe the would-be triangle ABE for the circle, or determine vertices G and H otherwise.
Using a simplified construction of the pentagon, rather than the Euclidean one, makes of course a whole lot of difference. The design is used in countless forms to decorate buildings and so forth, as is used the related pentagram (five-pointed star), for example on the U.S. flag and on military vehicles. It may accordingly be of more than mathematical interest that the pentagon, and other regular polygons, can as seen here be drawn by simply utilizing the property of circles to facilitate measurement by using their radii with differing centers. 25 May 2005. This time is of interest to me the regular 17-gon, famously found by Gauss to be constructible with only unmarked ruler and compass. Below I drew a version shown in the book by Robin Hartshorne (p.257) and which he called a particularly simple one. To make clear the amount of drawing involved, I in truth added most of the relevant steps that have to be taken. On the circle with center O and extended diameters OA and OB at right angles, get OC equal to 1/4 OA, for which I added the means of twice halving distances; with center C and distance CB draw arc meeting OA at D and E; with center D and distance DB draw arc for F; with center E and distance EB draw arc for G; with center H the midpoint of AF, found as shown again, and with radius HA draw arc for J on OB; with K the midpoint of OG, as shown, draw with center J and radius OK an arc meeting OA in L; to apply that radius here, by Euclid I.2, a point M equidistant from O and J is found, then the extended lines MO and MJ are drawn, whereupon with center O and radius OK arc KP and with center M and radius MP arc PN are drawn, giving wanted radius JN and arc NL. KL is
then a side of an inscribed 34-gon.
To transfer KL into the circle would require the same Euclidean way just shown (the diameters at right angles require more steps likewise), and of course it has to be transformed into a side of the 17-gon, whereafter, by mainly the repeated use of the compass as measuring device, the entire polygon must be constructed.
But as I showed elsewhere here, the polygons are constructible by such a use of the compass alone, in the indirect way of "insertion", by which the initial radius as side of the polygon is adjusted so that the final required point is met. A big irony is that an above seen method turns out to in practice be more indirect yet, because in the end one has to rely on the physical accuracy of the measuring devices like the compass, and in view of their limitations one has to continually adjust them to reach the required result. Thus with the many many steps in the preceding example there is a corresponding multitude of possible inaccuracies to be adjusted. This possibility is considerably reduced in the more direct use of sides of the polygons as I show on these pages. Moreover, in the above example, and in others—like the preceding heptagon, only a side of the polygon is determined, necessitating the very adjustment for the in the first sentence of this paragraph mentioned "insertion" to meet the final point required.
Below is accordingly a simple way I suggest for drawing the 17-gon by requiring only 5 arcs, which bring adjustments down to a minimum. The beginning double arcs on top merely indicate adjustment to the right radius, estimated for the side of the polygon, by with center A first drawing an arc BC; as subsequently easily seen, with center C is then drawn arc AD, then with center D arc AE, with center E arc AF, until with center F the arc through A goes through B.
1 November 2005. I decided to squeeze in here another geometric subject, the Pythagorean theorem. As many readers no doubt know, it states that in a right triangle (one with a right angle) the square on the hypotenuse (the longest side) equals (in area} the combined squares on its other sides.
There exist many proofs of the theorem, but there may be less than a handful that are frequently shown, usually ones easily followed. The exception may be Euclid's own proof, at the end of his first book and one of some length and not quite obvious. It is regardless often repeated. Here I am taking a look at a couple of proofs sometimes called proofs without words. They in any event don't require too much comment, and no algebra, which is used for some others.
This proof is customarily shown in two diagrams, and in my habit of seeking simplified answers to everything, I aim for one diagram.
The slanted square is seen as on the hypotenuse, with four of the triangles surrounding it; and the squares at bottom left and top right are seen as on the other sides of
That arrangement, of the two smaller squares at bottom, occurs on this, another known, diagram, traceable back close to a dozen centuries. The usual proof to me is awkward. Starting by viewing the two bottom squares, the large triangles in them are taken and shown to yield the large square
the triangle, with the surrounding triangles also four.
A new thought may be to move e.g. the two bottom triangles to the top, rearranging squares.
when placed at the top.
But the bottom triangles are much less obvious to fit there than they are at the top, where the proof could start instead, in considering the triangles placed at bottom afterward.
It also occurred to me that showing the whole previous diagram is not needed, it being often easier to visualize something than cluttering a diagram with it.
The present diagram is just an intermediary, to show that by removing one previous shaded triangle it is easy to visualize it replaced by the empty large triangle.
Here in fact I am utilizing this advantage by including two visual proofs in the same diagram. The solid lines obviously depict the previous large square with the two top triangles. It is easily seen that moving those triangles to the opposite sides of that square yields the two smaller squares at bottom.
And if the numbered sections in the large square are moved outside it in accordance with the dashed lines, the smaller squares are seen joined diagonally.
18 July 2007. At right I am offering another particular depiction of a
proof, spurred by one given—at the start of the book The Pythagorean
Theorem by Eli Maor—as the simplest known. That proof,
however, although graphically economical, is quite difficult to
follow, relying on complex geometric understanding. Instead the proof at right hardly requires
more special knowledge than that the angles inside a triangle equal
two right angles. I added this time letters for the sides, in keeping
with the expression of the theorem as a² + b² = c². As easily seen,
when the two top triangles inside c² are moved to the bottom as shown,
the result is a² + b².
8 August 2007. I felt I should say more about the proof in the Maor book just discussed. My saying the proof is graphically economical had to do with the basic shape of the solid-line triangles at left, given in the book. That shape doesn't really depict a proof, even with the added squares in dashed lines on a, b and c. The proof must still be verbally explained and relies, as noted, on complex understanding. In Maor's words (p.115), "the areas of similar [alike angles and sides] polygons are in the same ratio to each other as the squares of their corresponding sides". At left, the solid large triangle is similar to the two smaller ones that equal it, and therefore the two smaller squares on corresponding sides equal the corresponding large square. This proof is evidently far more difficult than the preceding one above.
9 December 2005. Still more on the Pythagorean theorem, given as a² + b² = c², is added below. As indicated on top of the sketch, I ran an "ad" in that issue of Scientific American, showing the basic figure depicted three times in the below left column, and informing that the figure, which I called "duckling" (from which flow several proofs, like water flows off a duckling), presents a multiple pictorial proof of the theorem.
This sketch was prepared in response to some inquiries about how the proof is performed.
As seen here, the shaded areas in the left figures are transposed in the right figures, giving c² in the first instance, and a² + b² in the next three.
It may be of interest to note how perplexing the problems were to mathematicians throughout history. The first two figures in the right column here (without the shading) were used in a proof described in the well known book, A History of Mathematics by Florian Cajori (Chelsea Publishing edition, 1991, p.87), and attributed to the 12th-century Hindu mathematician Bhāskara. In it the second of the two figures is (similarly to ones above this sketch) derived from the first, and it is overlooked that in the first figure, while keeping c², its two lower triangles (and their forming the small square) are superfluous for that proof, with the two upper triangles transposed sufficient for the second figure, a² + b².
10 January 2006. As informed by the text under the figure at left, the whole of this addition is also about an ad in Scientific American. The "ad" shows the related construction of a heptagon by use of straightedge and compass, the construction also found on the preceding page here.
The basic reasoning behind the central triangles has been known since antiquity, but I haven't seen it unified in a figure like this, and felt it a good idea to present to any interested reader. I didn't go into all the details (which are easy to look up), like why the angles of a polygon (see e.g. What is Mathematics? by Courant and Robbins, p. 10), and this too is interestingly based on the mentioned Euclid proposition I.32, seen as very fruitful.
As noted on the preceding page, I am adding here a much handier method of "rectifying" an approximation of pi, the circumference of a circle, for which task I somewhat rearranged these pages. As indicated, at issue are fractional forms of approximations, and the point now is to easily construct them with straightedge and compass, presently simply as straight lines approximating the length of the circle's circumference. With these lines it then only takes a known brief way to the construction of the required square.
Perhaps the favorite fraction close to pi is 355/113, because of its "prettiness" and
accuracy to six decimal places, in 3.1415929.... There have in the past been given constructions of it that bring about some dimension instrumental in attaining the desired square, and they can to an extent be viewed, from ca. pages 240 through 282, in the mentioned Pi: A Source Book. They require, however, variously complicated procedures, which was my intention to avoid.
On the preceding page I showed two rectifications of pi which were done by measuring off on the intended straight lines the fraction approximating it, so as to get the fractional part of the unit considered. The present fraction, 355/113, can be written as 3 16/113, with its fractional part accordingly meant to be placed on a line so as to get 16 parts of 113, the unit. My previous method was to divide that unit as necessary, usually several times, until the needed part of it was obtained.
The difficulty was that the divisions, except for bisections, require each a separate line on which to measure these off, to then transfer them to the primary line, beside the possible difficulty of finding convenient divisions for the particular unit. Now it occurred to me that the entire measurement can be easily done on such a separate line, for a very simple reason. Unlike the line to be fitted, which must be of the diameter of the circle considered, the separate line has no requirement regarding its length. As a result, instead of dividing any of it into certain parts, one merely has to repeat those parts. Below then is illustrated the present case.
The unit, equal to the circle's diameter and to have 113 parts, is given by AC. From C is here then drawn an oblique line of indeterminate length on which the parts are measured off. (In this depiction AC is drawn after the measuring, since, started at the left end, the right end is not known. This sequence is here spatially fitting, but the measuring line can instead be connected at A, with AC already in place.) It should then not be hard to follow the process, seeing that the arcs on top of the oblique line determine the 16 parts, and on the bottom the 113. Connecting the beginning then
with a line to A, and a parallel to that line at 16 for B, the proportions are transferred to AC. AB is the fractional part of 3 16/113, with the BC segment again discounted by dashed lines, the 3 to-be-added units indicated to the left of A.
As seen, I utilized increasing radii for the arcs as before, to make the process shorter, and more accurate because of fewer steps. The process can be applied to any fraction nevertheless, since one can repeat smaller arcs as often as required to reach the appropriate numbers.
7 March 2008.
Below I am adding yet another rectification of the circle, based on an approximation of pi by the great Ramanujan. The approximation results from (92+192/22)1/4, the last superscript meaning root to the 4th power, and it is 3.141592652..., accurate to eight decimal places, two more than the preceding. Ramanujan made a corresponding construction, seen in the mentioned Pi: A Source Book, pages 254 and 303. The construction is again rather intricate, ending with the observation that the mean proportional between certain resulting line segments is nearly equal to a sixth of the circumference of the given circle. There is also no proof furnished.
The above formula (92+192/22)1/4 can be written as (97 9/22)1/4, with the number inside the parentheses correspondingly rectified as below, after which the approximation of pi is accomplished by in the known manner drawing the square root twice in succession on reaching the concerned lengths.
The measuring off of 97 9/22 is shown here in two parts, because, I'll admit, the line is quite long and in practice may require paper bought at an architects or art supply store.
In the top left corner, AC designates the unit length, for the diameter of the circle to be rectified, the object there being to find its fractional part, 9/22, marked by BC. On the oblique line meeting C, the 9 parts of 22 are measured off as before, now from right to left, with increasing radii for efficiency, some of the distances marked for clarity; the endpoint reached is then again connected to A, with a parallel for B.
The large figure contains the entire measure of 97 9/22, including 1 unit with the fraction, for a whole length of 98; that unit, corresponding to AC in the first figure, is the starting measure at left for the rest, the corresponding dashed AB discounted as in the other cases; again, some distances are marked for clarity. For the resulting length, as noted above, is then drawn the square root twice in a row to obtain the rectified circle.
14 May 2008.
It happens that I thought of an approximation of squaring the circle which is somewhat better than the ancient Egyptian one mentioned on the previous page, but is of utmost simplicity of construction by straightedge and compass not comparable with any other construction I have seen. It could merely be described in words, but I felt I should show a figure, to display the stark simplicity. It is achieved by making the diagonal of the square 2½ times the length of the radius of the circle, for the approximation of 3.125 to pi, 3.141..., the radius being the unit, and pi correspondingly the area of the circle.
To explain, the square of 2.5, the diagonal, is 6.25, which (by the Pythagorean theorem) is twice the square of a side of the depicted square; half of 6.25 is 3.125, of which the square root, namely the side of our square, is 1.767..., in comparison to the square root of pi, 1.772...; the difference is .0046..., e.g. for a foot it is almost exactly 1/32 of an inch.
The construction is of course quite easy. For instance, one may draw the horizontal and vertical diameters of the circle and then a bisecting diagonal of the required length.
28 August 2008
On the preceding page I illustrated the concept of continued proportionality, when in a figure the ratio of one side to another is continued in an adjacent figure. That illustration had to do with the ancient problem of doubling the cube, which concerns finding the edge of a cube of twice the volume of a given one. The length of that edge is the cube root of the large cube, leading to it.
I tried there to explain without algebra why the drawing led to finding that cube root, but I may not have been clear enough. Since there is a similar way by which the square root of a line segment has been historically drawn, I felt I should here explain that case, non-algebraically again, for easier comprehension of both.
Drawing that square root is related to my preceding ways of approximately
rectifying, making a straight line of, a circle, in order to square it. Once it is rectified, one can avail oneself of the present method of drawing the square root so as to get the square.
Below is the graphic I am using to explain such a drawing, and I should note that I found a more space-saving way (solid lines) than the historical one (dashed lines). I'll describe the basic construction on this yellow panel, and then explain the mentioned proportionalities to the right of the graphic. The diameter of the solid semicircle is the line segment whose square root is to be found; the rectified circle, pi, would here be 3 units-plus long, but I used 4 units for clarity.
In the traditional method that line segment is extended by 1 unit (horizontal dashes), the full length is made the diameter of the large (dashed) semicircle, and a perpendicular (vertical dashes) erected at the start of the extension to meet the semicircle. That perpendicular is then the root sought (the oblique lines will be explained later).
In my reduced method, 1 unit, AB, is taken inward from endpoint A of diameter AC, a vertical BD erected to meet the semicircle, and AD drawn, which is the root looked for.
Explaining, triangles ACD and ABD are similar, of the same angles and proportions of their sides, namely AC is to AD as AD is to AB; AD, n, is n x 1; hence AC is n x n = n²; and the root of n² is of course n, which is of interest. In the example, as 2 is to 1, 2 x 1, so 4 is to 2, i.e. 2 x 2 = 2², and 2 is our root.
In the older form shown with dashed lines, the triangles on opposite sides of the vertical being similar, the vertical is, as said, the square root.
8 October 2008
In the figure below, I am applying the preceding way of constructing the square root of a line segment to a rectified approximation of pi, so as to correspondingly square a circle, with the circle and square also shown. The approximation is a known 3 1/7, or 3.142857 (the decimals infinitely repeating), comparing very nicely to pi, 3.141592..., the difference between their square roots being almost 1/5000, e.g. a little over a foot per mile. Under the figure is its description.
24 October 2008. The small figure below uses minimal relative space for drawing a square of any approximation of pi, in area, to the pi of the given circle, as explained here; the bottom horizontal is the rectified approximation (one of 3 1/7 is shown).
The long horizontal AD is 4 units long, of the last of which, CD, is found a 7th, CF, by means of CE of unspecified length and a parallel to ED for getting F; this makes AF the approximated pi, 3 1/7 (to save more space, without CD, CE may be drawn in an opposite direction, to apply to the 3rd unit of AC, which is then extended to a transferred F for the AF of the pi approximation); its square root is by the preceding found by drawing on it as diameter semicircle AGF, and erecting at the end of first unit AB the perpendicular BG; incline AG is then the square root; now, since the area of a circle with unit radius is pi, its diameter is 2 units, as depicted, and the square on its square root is also pi (or its approximation); and incline AG was seen to be the square root of pi, so that by drawing with A as center a circular arc GH, we have the horizontal AH for a convenient drawing of the square. The like can of course be done with other approximations, by which the length of CF minutely differs.
14 January 2011
It occurred to me that instead of seeking an approximation of pi for a circle, of which to then find the square root and construct a square, one can start with an actual square root of pi up to some decimal, and easily depict that square root so as to construct the square. The amount of decimals used is in principle indefinite, limited only by the physical possibility of construction. Below I give the description.
As mentioned, since the area of a circle is pi multiplied by the radius squared, if the radius is made the unit then the area of the circle is pi, and the square root of that area is the square root of pi. Pi, 3.1415926..., has, from its first two decimals on, the square root 1.772..., the root depicted above as measured on the circle's diameter (because of space, only about half of the circle, and of the resulting square, is shown). Since the radius is 1 unit, the left radius shows unit 1 of the root; then on a line diverging from the diameter are marked 10 equal parts of convenient length to determine the first decimal, which is 7 out of 10 of a 2nd unit; the endpoint of the marks is accordingly connected by a line to the end of the diameter, and parallels to that line are drawn for 7 and 8 parts; the 7 parts of course determine the first decimal, and the 8th part is for a subsequent division into 10 of one of 10, into 100 for the second decimal; there the process of finding 7 parts is repeated as shown, with an 8th part added for the next decimal, which is so tiny that it, with the rest of the root, fits inside the next pixel here; accordingly I drew a vertical at that pixel together with the vertical at the starting point to construct the upper half of the square.
It is noteworthy that if the above radius is a kilometer long (making the diameter well over a mile) then the divisions into ten need only be done 6 times to reach them for a millimeter. At that stage one can hardly find a pointed instrument that marks a finer division, and for all practical purposes the circle is accurately squared there, the width of the instrument spanning the end of the square root and the line beyond. The like applies whatever the length of the diameter. It should be kept in mind that the ancient idea is to make a construction with physical tools, preferably a bare straightedge and compass, but possibly with marks on a ruler, or a mechanical device with sliding parts, as used for some cube duplications.
Let me bring in some further thoughts regarding infinitesimals. In a famous paradox by Zeno (I discuss it with other paradoxes in my book) swift Achilles never catches up with a tortoise, because he has to before pass an infinity of ever shorter distances. We know though that he does catch up soon, with a finite number of steps. The point is that a finite distance (between the two runners) comes to an end, with ultimately a distance small enough not to be divisible. Pi (or its square root) is such a finite distance (for instance 3.1416 exceeds pi by a
minute distance). Accordingly, at some above stage the remaining distance is no longer divisible even in principle, and the end is reached.
PRESUMED IMPOSSIBILITIES CONTINUED
PRESUMED IMPOSSIBILITIES
(continued2)
Inasmuch as the title of the present pages is PRESUMED IMPOSSIBILITIES, it may be in order to turn attention to other questions that customarily belong in that category. The most significant one is doubtlessly to prove the existence of God. Although the attempt has been made by a number of thinkers through history, many suppose that at least since Immanuel Kant such a demonstration is impossible.
That I don't accept that judgment could be surmised from my indication on the home page of a book I published and so forth. Alongside fruitful explorations of other logical matters, there was a notion in me that reality just may contain certain evidence from which a simple demonstration of a supreme being can be inferred. It should be a matter of correctly assembling well-known phenomena as premises from which to derive a conclusion, the same as deduction in mathematics or other aspects of life.
And it is indeed our lives and by extension all life, as alluded to on my mentioned home page, that provides an opening in this regard. At this time, though, I want to concentrate on this subject of a supreme being and on others as they came up in a review of my book, so that I can point out flaws that beset respondents to me. (Since writing this about reviewing of my book, I decided to add a discussion regarding the issue of proof of the existence of God. It follows below.)
First, let me reproduce a letter from the editor of the journal that published the review. I should note that I am not a professor, as addressed there, and have no academic affiliation. But, as seen, the letter is quite positive, while the opposite applies to the review. Sadly, editor Castañeda died before that publication. Perhaps he would have been further helpful. Below this letter here, then, I discuss the review.
27 October 2003
HOME
PRESUMED IMPOSSIBILITIES, continued1, 2
PHOTOGRAPHY, continued1, 2, 3, 4
PORTRAITURE, continued1, 2, 3
COMMERCIAL ART, continued1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25
INVENTION
AUTOBIOGRAPHY, continued1, 2, 3, 4
It appeared in the December 1992 issue of NOUS. For convenience, I will add parenthesized numbers as hyperlinks to my numbered comments below the review.
Paul Vjecsner, On Proof for the Existence of God, and Other Reflective Inquiries (New York: Penden), 258 pp., $20.00.
PHILIP L. QUINN University of Notre Dame
This book consists of an introduction, which outlines the author’s philosophical methodology, and four chapters. The first chapter is about language and concepts; it is divided into sections on the signification of language, concepts and their
constituents, and definition. The second chapter is devoted to metaphysics; it contains a section on the self and mind and one on external reality that focuses on extension and causality. The topic of the third chapter is logic and mathematics. The fourth chapter offers an argument for the existence of God and concludes with some thoughts about goodness and the prospects for immortality.
It is a peculiar work. For one thing, it lacks the customary scholarly apparatus of footnotes and bibliography. For another, it contains numerous uninformed and cranky opinions about scientific matters. The author’s attempt to criticize relativistic physics is a typical example. The argument goes as follows: “As to energy, it is defined as the capacity for doing work, and it is hence a causal property. It can at once be noted that energy cannot therefore be an independent entity, let alone the constituent of material reality, as proposed in physics” (pp. 74-75). Physicists will not be impressed by this argument; nor should philosophers be. (1)
The chapter on logic and mathematics is particularly rich in astonishing claims. Here is one: “As infinities, however, cannot be unequal in size, possessing none, so can they not be equal” (p. 122). But what of Cantor’s proof that the set of real numbers has a larger cardinality than the set of natural numbers? The author’s objection to it goes as follows: “The reasoning flaw in set theory lay in supposing completion of these sets, so that by assuming completion of one of them the other can be continued since infinite. Neither can be complete, for their infinity allows continuance by definition” (p. 119). Modern geometry too comes in for criticism. After setting forth what he takes to be a proof of Euclid’s fifth postulate, the author dismisses non-Euclidean geometry with these remarks: “Meanings play instead a misleading part in non-Euclidean geometry, founded on the rejection of the just proven fifth postulate or its companion. These are replaced by postulates on which are built new geometries, which correspondingly, while thought to be consistent, engage in contradiction” (p. 133). Comments of this caliber about set theory and geometry seem to me a clear indication that they are subjects the author is not competent to discuss. (2)
If one reads this book from beginning to end, one has to get through more than two hundred pages studded with such claims (3) before arriving at its discussion of natural theology. Judging by the book’s title, it is of special importance to the author, and so I now turn to an examination of his treatment of arguments for the existence of God.
The final chapter begins with criticism of the work of others. The following argument is taken to be representative of the recent revival of interest in ontological arguments:
“2.i. It is either necessary or impossible that God exist. But, it is argued,
2.ii. It is not known to be impossible that God exist. Hence, it is inferred,
2.iii. It is not impossible that God exist. In consequence and by 2.i it is concluded that
2.iv. It is necessary that God exist” (p. 221). This is indeed a bad argument; the inference from 2.ii to 2.iii is fallacious, as the author notes. But is it an argument any philosopher takes seriously? Though he mentions no one by name, the author presumably has in mind such philosophers as Hartshorne, Malcolm and Plantinga when he speaks of the ontological argument being recently revived. He makes absolutely no effort, however, to show that any of these philosophers or, for that matter, any philosopher has ever endorsed or even recommended serious consideration of this particular bad argument. It is so obviously fallacious that it would be silly to try to foist it on any good philosopher. (4)
The author offers three teleological arguments for the existence of God, all of which are variations on [a] theme. The simplest of them goes as follows:
“III.1. God equals the cause of purposive events of concerned scope.
III.2. Purposive events of concerned scope exist in nature.
III.3. Events that exist in nature imply the existence of a cause of them. Hence, by III.2 and III.3,
III.4. A cause of purposive events of concerned scope exists. Therefore, by III.1 and III.4,
III.5. God exists” (p. 231). The first premise is said to be a “starting definition” (p. 231), and purposive events are said to be “events that are, proximate or eventual, objects of purpose” (p. 223). About the restriction to purposive events of concerned scope, the author says only this: “It should be prefixed that when referring in the proof to the scope of the powers and of events caused by them, meant is a correspondence between these magnitudes, in concord with earlier observations on force as measured by results” (pp. 223-224). Though I find this statement somewhat opaque, it suggests that purposive events of concerned scope are being defined as those purposive events from whose occurrence we may infer the presence of a particular kind of cause, namely, divine power, just as forces are inferred from their results. After all, it would seem that not all purposive events warrant such an inference since some at least are caused by humans or other organisms, and so some such restriction appears needed if the first premise is to be plausibly regarded as true by definition.
If this is right, then I think the argument is not a successful piece of natural theology. The trouble comes from the second premise. It would be hard to deny that there are purposive events in nature, but this is not what it asserts. Once the first premise is granted definitional status, what has to be established in order to show that there are purposive events of concerned scope in nature is that there are in nature certain events whose cause is God. This is not obvious and would not be hard to deny. And I can think of no way to establish it antecedent to or independent of showing that God exists and exercises certain causal powers. So I conclude that the argument is question-begging because, given its first premise, one would have to assume the truth of its conclusion in order to justify its second premise. In this respect it resembles the argument to the conclusion that God exists from the premises that 2 + 2 = 4 and that God exists if 2 + 2 = 4. This argument is sound, provided God exists, but it is not a proof of God’s existence or a successful piece of natural theology.
I can find nothing to praise in this book. I think its efforts to criticize the views of others are uniformly unconvincing and its attempts to set forth constructive arguments are failures too. I do not believe any professional philosopher would profit from reading it. (5)
My comments follow.
1. I responded to several points after the above letter, and more to the reviewer after the published review, but other than for a weakness of his he may have feared would be exposed, he did nothing about it. That my work be peculiar to him should be no surprise. It was not written in the form he is accustomed to, particularly in the style in which academic philosophers communicate among themselves. Instead it is an entirely independent work which avoids interruption by footnotes and too specific references to other work through bibliographies.
That it contain numerous uninformed and cranky opinions about scientific matters can be itself understood as an opinion of one who adheres unquestioningly to science's views of the day, admittedly often tentative. I am not at all uninformed about the subjects I discuss, and challenging anything widely accepted is automatically viewed as coming from a crank. The reviewer says that my "argument [criticizing relativistic physics] goes as follows...". However, I present quite many arguments, going into the subject, as in other cases, very thoroughly. The argument of mine he quotes, furthermore, stands on its own, as anyone unbiased can tell by a careful reading. Professor Quinn says that physicists will not be impressed. This slavish approach to science commits the fallacy of "argumentum ad verecundiam", of appealing to a supposed authority, rather than using reasoned argument. (I do not hesitate pointing out particular fallacies, as in a recent response to a book, this likewise serving in reaching the truth, as do more noticeably constructive findings.) Back to the discussed
2. Of the three quotations in this paragraph two are "out of context", meaning they are isolated from their surrounding which explains them. The reviewer again mocks them without offering an opposing argument, merely voicing his opinion that it seems to him clear I am not competent to discuss the subjects. Why? Because I clash with received pronouncements, not because the reviewer knows much about it. The first quotation does not include my given explanation of why infinities have no size, and the third one, which by him "dismisses" alone non-Euclidean geometry, does not give my explanation of how the meanings are contradictory. And the second quotation, once again, is self-explanatory. If the reviewer had the decency to read it carefully, supposing an adequate intellect, he might recognize its validity. To the discussed
3. Thank you for the appreciative "studded". It happens that in these "more than two hundred pages" I explore very thoroughly and positively major issues in philosophy and beyond, offering new understandings in how we acquire knowledge, how language and formation of concepts get involved, how fundamental worldly realities can be ascertained, and how a complete system of logic can be constructed that should be an extensive improvement and simplification of what is extant in the field. In the process I include many carefully thought-out and designed diagrams, that either serve to demonstrate assertions in logic or to edify others. The reviewer doesn't bother to scrutinize any of this, contemptuously dismissing at the end all of my work offhand. Back
4. Just where, dear professor, do you think I picked up this argument? I certainly think myself capable of creative arguments, but I would hardly bother to invent this one in order to refute it. The subject is much too important to me. I did find it useful to cite some former arguments (not only this one) to point out the way they can be erroneous. And in the present case, in keeping with my quest for simplicity and clarity I had to distill it from arguments much worse. They are presented in The Encyclopedia of Philosophy by Macmillan, 1972 edition, volume 5, page 540, and they originate with the philosophers Hartshorne and Malcolm mentioned by the reviewer. Let me quote two passages.
Thus God’s existence is either logically necessary or logically impossible. However, it has not been shown to be impossible—that is, the concept of such a being has not been shown to be self-contradictory—and therefore we must conclude that God necessarily
exists.
and
(6) Nq v N~q Inference from (4,5) (7) ~N~q Intuitive postulate (or conclusion
from other theistic arguments): perfection is not impossible
(8) Nq Inference from (6,7)
The first passage sounds very much like my quoted form, does it not? So there goes the contention that no "philosopher has ever endorsed or even recommended serious consideration of this particular bad argument. It is so obviously fallacious that it would be silly to try to foist it on any good philosopher". Perhaps any "good" philosopher. That I make "absolutely no effort" to show that any philosopher has done those bad things is another unseemly accusation. It is enough that I give for the purposes a generalized example of these arguments advanced. I described above the amount of effort I put into the entire work, and there is no justification in expecting me to describe more specific attempts by particular philosophers. If there is an issue of insufficient effort, it applies to the reviewer in his perusing my writing.
Returning to the familiar sounding first passage above, it is yet worse in wording than is my citing of that bad argument. "Has not been shown" could be equated with my "is not known", but the qualification between the dashes confuses the passage still more. First I may note that the inference from either "has not been shown" or "is not known" commits the fallacy of "argument from ignorance". Not knowing something to be true does not imply its falsehood and so the truth of its alternative. But the injection between the dashes needlessly and falsely adds to the for whatever reason unknown. It is also false, because non-contradiction does likewise not make an inference valid. This may not be the place though to explain this.
The second above passage is worse even than the first. In the symbolism on the left side (I will simplify my definitions), "N" stands for "necessarily", "q" for "God exists", the tilde "~" for "not", and the "v" in line (6) for "or". That line then says "It is either necessary that God exist or necessary that he does not, namely impossible that he exist"; line (7) says "It is not necessary that God does not exist"; and line (8) "It is necessary that God exist". On the right side are the reasons for the inferences, and of interest is line (7), previously based on the like of "It is not known to be necessary that God does not exist". And this reason, at least clearly stated, is now substituted by a mysterious "intuitive postulate (or conclusion from other theistic arguments)". Back to the review
5. Here now came the most extended criticism of my work. The reviewer again groundlessly limits what I say to what he seems to see on first sight. He writes (after citing p.223 of my book) that about a certain subject I say "only this", and then quotes a sentence of mine, as see. And that sentence does not concern that subject as referred to by him in the deduction there from III.1 to III.5 (p.231 in my book). It concerns and precedes a deduction on a previous page (224), and there should be no wonder he finds it "opaque" applying it to the wrong place. The sentence has to do, as it says, with the correspondence of the scope, the magnitude, of a force with its result. The scope referred to by him
has to do instead only with events not as forces. And what the reviewer did mean to attack is my speaking of "concerned" scope.
Put simply, he charges me in that paragraph with defining "purposive events of concerned scope" as "those purposive events from whose occurrence we may infer...a particular...cause, namely, divine power", which would circularly assume conclusion III.5 in premise III.2. He says a statement of mine "suggests" this. Now what he sees as suggested does not therein constitute a definition. I do not so define those events anywhere. On the contrary, I describe them as ones observed in live organisms, without reference to any cause, let alone a special one.
The reviewer starts the next paragraph with "If this is right", referring to that suggested definition. This is typical hypothesizing by contemporary thinkers. I would expect a substantiated claim instead, namely pointing at the alleged definition. He continues, "then I think the argument is not a successful piece of natural theology". Again, "I think" is not successful reasoning itself. And if my "piece" is not successful, whose is? Natural theology has been notorious for getting nowhere.
He goes on saying, "The trouble comes from the second premise". Now how would you know? You only hypothesize and "think" something is true. How then do you also know there is trouble? I am glad you say that it would be hard to deny that there are purposive events in nature. This has been really the bone of contention in the most publicized theistic arguments, exemplified by the dispute between Darwinians and Creationists.
The critic proceeds saying that "what has to be established in order to show that there are purposive events of concerned scope in nature is that there are in nature certain events whose cause is God" and that he "can think of no way to establish it antecedent to or independent of showing that God exists...". Again, that he can think of no way does not imply someone else can't, but how again does he substantiate his first statement quoted in this paragraph? No, "what has to be established in order to show that there are purposive events of concerned scope in nature is that there are in nature" purposive events of, to be redundant, concerned scope—meaning a wide-ranging scope that goes beyond the purposive events "caused by humans or other organisms" he refers to, to extend to life in general. That's all.
The striven for refutation is fallacious simply because the presumed definition of the events of concerned scope in premise III.2 as ones caused by God is not part of the deduction. Rather, the cause of those events is established separately, first as requirement via III.3 and III.4, and only then as identified with God via his definition III.1, to lead to conclusion III.5. This demonstration, incidentally, is not the simplest I offer, unlike the reviewer states, and the simplest excludes reference to that scope.
That this reasoning resemble the reviewer's argument with 2 + 2 = 4 is quite a clumsy analogy. The professor would have to point out again where in my deduction this kind of reasoning takes place. As a matter of fact, he fails to show in any way how my claimed question-begging is performed.
The last paragraph is shamefully unkind, refusing a single good word if for no
more than the work put into the project. This although he knows nothing else about me, or maybe because of it, since he displayed a disgusting air of superiority when I once met him. It may be, as he puts it in his last sentence, that no "professional philosopher would profit" from my work, and I wonder if that would be my desire. Professionals, or more precisely academics, belong to a "club", which as in the preceding will look down on an outsider like me, but will lavish unbounded praise on in my eyes immeasurably less insightful writing of an insider. The person I would like to profit is anyone who is professionally or otherwise disinterested, but has an interest in genuine and beneficial learning.
31 October 2003 It seems to make sense for me to include here another review of my book, and my reaction to it. It appeared in the December 1989 issue of THOUGHT, published by Fordham University. As before, I am adding parenthesized numbers as hyperlinks to my numbered comments afterward.
ON PROOF FOR THE EXISTENCE OF GOD, AND OTHER REFLECTIVE
INQUIRIES. By Paul Vjecsner. New York: Penden, 1988.
ADVERTISED AS A “CARTESIAN REEXAMINATION OF BASIC
PRESUPPOSITIONS, old and new, in philosophy and sciences,” this is a painstaking effort. Its stated purpose is to disclose how various truths such as free will, and including some now considered undemonstrable, “can be reflectively demonstrated and thereby an actual rather than speculative edifice of existence revealed” (3). The author sedulously presents his “[process] of inquiry” (4-8) and then four chapters: on language and concepts; on self, mind, and external reality; on logic and mathematics[;] and finally on the existence of God (216-33), with further inquiries into “the goodness of God and man["] and “heaven and immortality” (233-51). The chapters are independent but are cross-referenced, and “premises on which conclusions there are based may be substantiated elsewhere” (8).
A prospective reader, then, should not be misled by the title into expecting the early chapters to simply underpin the “proof for the existence of God.” The caution that “sections can be read individually by individual readers” (8) could alert those interested expressly in that proof and not in the exercises of erudition, acumen, and reasoning skill that precede it. “The existence of God is uppermost in consideration, because of its momentousness” (4), yet it covers 18 of 251 pages of text, in contrast to 112 on logic and mathematics.
Three theorems with their two dozen parts, and proof and explanation of each part, run through pages 41-171. The first theorem defines definition; the second deals with self, the world, and causality; the third relates to existence, attributes, and laws of thought. Among parts of Theorem III e.g. are: “A thing does not both have and not have a given attribute at the same time” (144); “A quantity is either a given one, a smaller one, or a larger one” (146); “when some A is B then some B is A” (171). Stages of the proofs are illustrated by 85 diagrams, and postulates and propositions of Euclid are discussed (e.g. 123-34). (1a)
Thereafter Vjecsner examines and rejects “former arguments” for the existence of God (219-21). In the “recently revived["]:[?] ontological argument (“It is either necessary or impossible that God exist”) the “error lies in equivocating the two uses of ‘possible’, or of ‘not necessary’…”(221). Among cosmological arguments are three of Aquinas’s “five ways,” but against them “the general concept of God can be held to require some power additional to that of first causation.” And the “proposal that a primal cause would be of greatest perfection, as advanced by Aquinas in his fourth way…is derived neither through natural nor logical law” (219).
The fifth way, “the teleological[", "]may be of widest appeal….It reasons by analogy, comparing the purposiveness of organisms to the similar purposiveness of a watch or of an arrow in flight….Analogy is inadmissible, however, as a form of inference. It constitutes what is known as the fallacy of undistributed middle” (221). Incidentally, Vjecsner appears to treat the “five ways” as arguments to convince atheists, instead of ways for believers to see how reasonable is their faith. (2a)
The proof [by the author] for the existence of God, presented in various forms (224-31)[,] comes to this: “Purposive events of concerned scope exist in nature” and “imply the existence of a cause of them”; and since “God equals the cause of purposive events of concerned scope,” therefore “God exists” (231). The question is better and more fully treated in Hans Küng’s Does God Exist? (1980) and much better still in John Macquarrie’s In Search of Deity (l985). (3a)
This serious work should not have been published in its present form; it reads at times like a labored translation. The author deserves an able and informed copy editor, or a colleague versed in standard English. He says that “the ideas discussed…require no specialized knowledge” and he strives to “make acquaintance with much of what follows feasible for every contemplative reader” (8). After an “informally abbreviated” proof he adds: “All this can be held comprehended by a hearer of the demonstration, as were observed comprehended beside truisms complex meanings behind ordinary words” (226). But diction, syntax and usage sometimes are so foreign that the full meaning may elude a careful reader. Hence a thorough revision is needed. (4a)
Loyola College in Maryland William Davish, S.J.
1a. This is the first paragraph that runs into trouble. My theorems in the book are twenty-three, not three. The reviewer may have thought that my numbering of theorems as, for instance, II.1 and II.2 meant they are parts of a theorem II, whereas my prefixed numbers I, II, and III signify the first three chapters, containing these theorems.
My impression, however, is that he tried to belittle my work, as expressed by his ho-hum attitude. I should note that the reviewer, unlike the previous one, is not a philosophy professor but a theologian. Correspondingly I hope not to be too harsh in saying that he largely doesn't know what he is talking about in the review, and may have tried to impress the editor. Thus he says in that paragraph that the "third" theorem "relates to existence, attributes, and laws of thought", quoting some of its "parts". But these concern known laws of logic (which belongs to philosophy), referring only incidentally to existence and so on. Also, only a smaller portion of the diagrams deals with logic. But lacking most is a discussion of any of the content as to its merits, as expected of a review. To continuation
2a. There is again in these two paragraphs no discussion of the merits of the quoted. As to the last sentence, Aquinas presented his "five ways" as proofs, not as articles of faith, and they have been treated accordingly since. In my case, I direct my arguments as much toward believers, of whatever creed, as toward non-believers. I make a point of the dangers of conflicting dogmas, witnessed these days tragically, and hence of the value of approaching these issues in the same reasoned manner as in other matters of knowledge; at the same time I mean to show to those of only secular viewpoints that this manner is applicable in other areas as well. Back
3a. Interestingly, regarding the existence of God the reviewer gives more than twice the space to my discussions of former arguments than to those of my
own, dispensing with them in this shortest of paragraphs. He picked the same deduction of mine as did the previous reviewer, but in a disorderly fashion, so that the reasoning cannot really be followed. It is obvious to me that he does not understand the process of deduction, and he accordingly again does not discuss the cited one of mine, merely ranking it beneath some work by others, without explanation. That my "proof...comes to this..." and "The question is more fully treated" in that work fails to recognize that my deductions are extensively explained in what precedes and follows them, including the other chapters, and they should be judged on their own merits. Back to the review
4a. Although the reviewer compliments me somewhat in the first two and this last paragraph, he offends at the same time. His remark that I deserve "a colleague versed in standard English" is an insult which may be resented by anyone even if it applies. But because of my history as a past immigrant, I did never learn any but standard English. I studied its grammar as a youth before coming to the U.S., and am frequently reminded of mastering it better than many who are native-born. They often grow up speaking a colloquial version and don't pay much attention to, or even have not acquainted themselves with, formal grammar.
And I should note that these comments by the reviewer came after I furnished a brief biography of myself at his request. I mentioned in it that I was foreign-born, and he quickly seized on this to make those unpleasant remarks. He didn't know that I had then been in America for over 40 years, about twice the years of my previous life, giving me plenty time to add to my linguistic skills. It may be notable that the previous and longer review above, whose author knew mentioned little about me, had no complaint about my language.
To add more to this dissertation of mine, the now responded to reviewer, though quoting me abundantly (and I think he made me sound pretty good), points at no instance again where my language is to fail. What does happen in his review of barely one page are several compositional errors of his own, marked above between brackets, beside a superfluous "since...therefore" in the next to last paragraph—either of the two words will do without the other.
My justification for spending this much space on that review's last paragraph is that I feel I should fully, if rightfully, defend myself for the sake of my work, though in this case only language is involved. The reviewer might indeed concern himself more with the rightness of the content than of the language.
While discussing here in regard to the above two reviews, and elsewhere on this website, my reasoning concerning the existence of God, I do not try to explicate the idea more in this place. The reason is chiefly that the thought of demonstrating that existence is met with such powerful skepticism and sometimes ridicule that any effort in that direction is ignored no matter how sound. And past thinkers have in my view indeed not been able to penetrate the surface of the matter. They are hampered by through the times built up conceptions, beyond which they find it difficult to see. These conceptions, for example of what the semblance of God must be, may have to be set aside for a better understanding, and therefore I reserve further discussions for more conducive opportunities if arising.
28 April 2005. It appeared right to me to now further discuss the just mentioned subject. (The material on this page that begins with this paragraph was alone here earlier. For space reasons I moved the now preceding material on this page to here, and it seems an appropriate preliminary to the following.)
As elsewhere noted, I decided to add here an exposition of the demonstration of the existence of God I offer, my being fully aware of how preposterous this move is considered to be. I find that the demonstration can be so simple that I would be remiss in not presenting it to present readers, regardless of the kind of reception it does or fails to receive.
The title on my webpages is “Exploring possible human knowledge”, and I chose it for important reasons. I find that the possibility at issue has not even been close to being exhausted. Today it seems tacitly accepted that the only area of discovery, of any addition to human knowledge, lies in physics, despite the extensive advances in mathematics, a field with a conceptual base, relying mainly on deduction.
This last method is almost forgotten as itself a discovery in its conscious application. The ancient Greeks are particularly known for having made geometry a deductive enterprise, exemplified by Euclid’s enormously influential Elements. In the last two or so centuries thinkers in contrast came to regard many questions as unanswerable, prompting, beside more, the title of my present section, “PRESUMED IMPOSSIBILITIES”.
As remarked earlier, among attempted proofs held impossible—most of which are themselves mathematical—the most ridiculed are perhaps those concerning the existence of God, although like others they have been approached by many prominent thinkers. My assertions in this regard are accordingly of course likewise not paid attention to, especially since not coming out of academic circles, which assume to have exclusive competence in these matters.
But the subject itself, the existence of a higher power over all of us, suggests that it should not be comprehended only by a limited few, ones thought to be accorded insights by virtue of their official positions rather than personal ability. For that matter, it is my intention to show that despite thoroughly embedded convictions to the contrary the issue is not only as accessible to reason as are other problems of inference, but that it can easily be understood by anyone. It concerns some simple ideas which have been overlooked, and which correctly understood can be as commonplace as are many ordinary facts.
Above I referred to the examples of ancient geometry as known forms of deduction. But one can fall back on much simpler examples—if less familiar, because they are not concerned with the more widely taught mathematics but have to do with deduction in general, which, though it has its rules, is mostly done inadvertently.
In this light let me introduce a famed syllogism, used in connection with Aristotle’s logic:
1. All men are mortal;
2. Socrates is a man;
therefore,
3. Socrates is mortal.
From two premises is deduced a conclusion by what is known as transitivity. This is a process so common in our lives, even in that of animals, that one is unaware of using it. If one knows that A (Socrates) leads to B (man) and B to C (mortality), one instantly infers that A leads to C.
The Socrates example can be examined more closely. Premise 1 asserts an object of experience; we find that all men are mortal. Premise 2 asserts an object of definition; Socrates is a man by the meaning of “man”. The point is that one can draw a conclusion from premises known true by whatever criteria, even by what is meant by an expression.
Now it is time to turn to the existence of God. It may be noted that the present exposition relies on factors that should be easily accepted as correct without express substantiation. Many of them may want to be looked into in greater detail, as professional philosophers insist on to the extent of hopeless entanglement, often brought on by ingrained prejudices. Nevertheless I do in the book deal more with many particulars in this and other subjects, particulars that are of interest there, but would make the ideas only more obscure at this point.
In the preceding I mentioned definition as possible part in an inference, and in wanting a proof about God, his definition is especially critical. What sort of being do we want to know about? What are one’s hopes and expectations regarding a supreme being?
My discussion is accordingly guided by what the most universally held attributes of God may be. An indication is how he is most commonly referred to, and it appears to be as the “Almighty”, or more formally the “Omnipotent” and “Omniscient”, or the above “Higher Power”. The issue is that of general interest in a supreme being is not perhaps a certain physical appearance but a supreme power of a kind over our lives.
This recognition should fundamentally influence the approach to the question discussed. We are interested in the presence or absence of a certain power. We are not seeking a physical identity, as is often supposed. And a power is determined by outcome, which means that in order to determine the existence of the supreme being, one should not seek the presence of an image of a sort—whose powers would come into question regardless—but see if those powers are manifested in known reality, the place of interest.
What kind of power are we then speaking about? Evidently it is a power that can purposely affect our lives or destiny, preferably to good ends. Ends toward which a power is directed is indeed what I mean by it acting “purposely”. The question then is, are our lives directed toward ends?
The question of direction toward ends—of purpose—in nature is in fact with regard to the present subject hotly today debated. The dispute is known to take place between Darwinism and Creationism or Intelligent Design, and it
concentrates on whether organisms, their bodily structures, have come about by purpose or not. The argument for purpose has been represented by that of the 18th-century cleric William Paley, who compared organisms to objects like a watch, which is constructed for the purpose of timekeeping, as appears, for instance, an eye constructed for the purpose of seeing. Darwin on the other hand is well known for his argument for evolution, in the course of which organisms are said to gradually change in accordance with a random process of natural selection.
There occurs, however, a gross oversight on both sides of the dispute about purpose. What appears entirely overlooked is that, aside from the organism's bodily form, its live activities are themselves purposeful, directed toward ends. Paley only had lifeless objects as models, and Darwin accordingly likewise remained oblivious of life’s activities, in a neglect advantageous to his contention. But on considering the functioning of life, we know that each organism develops into its complex form individually during its lifetime—from fertilization to full formation—rather than acquiring that form through prolonged periods of time, as if the organism were a single object existing and changing through the ages. And each organism functions, to so develop and otherwise preserve itself and its species, purposely. Evolution’s concept of formation of the organism by accidental changes is consequently not a finding, but only a false denial of the purpose, seen in the organism's individually occurring formation and propagation.
And this occurrence is the evidence needed. All of life is characterized by purpose, by being what is sometimes called “goal-directed”, and the general goal is known to be preservation of self and the species, in what can be viewed as in the organism’s interest.
Returning now to the consideration of a power, it is easily seen that these occurrences in the living presuppose a corresponding power, specifically a power directed toward their ends. The logic is again that once the supreme being sought is understood as a pertinent power, the power can be verified by its manifestation in nature, and the preceding establishes that manifestation.
The simplicity of the reasoning can be put in a form similar to the above syllogism. Making the first premise the definition of God as the appropriate power, and the second premise the manifestation of that power as found in nature, the corresponding connection between the two is stated in the conclusion.
1. God is the power of purpose in the interest of living things.
2. Power of purpose in their interest is revealed in living things.
Therefore,
3. God is the power of purpose in their interest revealed in living things.
To put this plainly in the above sequence of A, B, and C, God (A) is identified by the purposeful power (B), revealed in life (C).
It may be observed that scientists as well as laymen are inclined to explain away any contention of action resulting from other than aimless laws of nature. These laws are, in fact, utilized as a means in any activity, including the carrying out by
humans of their own purposes, which purposes, incidentally, are themselves manifestations of disputed purpose in nature. Of account is that, with the aid of otherwise aimless forces, live organisms function in the larger picture with their preservation as aim, and to call that function aimless is in consequence a contradiction.
With an open mind to the preceding it should then be easy to grasp the simplicity of demonstrating in nature the higher power in question. This may make it evident why peoples so naturally gravitate toward faith in supreme guidance of their lives.
It seems worthwhile here to clarify more some ideas about deduction, which unless appearing in entertaining ways like a detective story strikes one usually as quite intimidating. As a matter of fact, to explain how exactly one thing is deduced from another has baffled the best of minds in history. In Euclid and other works to this day rules of deduction are somehow presented as unproven axioms, those by him, for instance, known as common notions. The first of them states: "Things which are equal to the same thing are also equal to one another", and it is the basis for the strings of equations in mathematics that are today accepted without much thought at all.
One of my endeavors in my book is to show how axioms like those in geometry can be established deductively, and my preceding treatment of the parallel postulate is part of this. This attitude angers many, because they insist that one cannot make the first deduction from other than something assumed true without proof. Correspondingly I dealt thoroughly with the problem of how basic logical truths are determined. We determine worldly truths by observation, why then can conceptual truths not be determined similarly? Logical truths, regarded as holding in "all possible worlds", have in fact been described as not conceivable otherwise, meaning they are somehow "conceived" as true, not having to be assumed.
A lengthy mathematical formula, of course, can hardly be merely conceived as true. But the formula follows from simpler ones, following from yet simpler ones, and so on, as knowledge in general grows from simple beginnings. And basic logical truths, by the means of which others are found, can indeed be found by conceptualization. It is part of the above referred to inadvertent comprehension of connections, and, since holding in all possible worlds, these can be depicted diagrammatically, in likeness to figures in geometry. I should note that using the last cited figures as proof has been vehemently attacked, with certain justification. Like the mentioned mathematical formulas, the figures can be intricate and subtle. But they, also, are built on simpler concepts, and so forth, and I want to emphasize that the simplest logical connections carry far more certainty conceptualized than assumed.
To use corresponding diagrams is in accordance with the preceding likewise resisted, which may be ascribed to pride in the contenders' purely mental prowess, notwithstanding that insistence on unproved axioms amounts to guesswork. Diagrams instead accurately represent understood connections that cannot be demonstrated with words alone (although thinkers have in failures of proof, and the tendency to dismiss any learning not physical, tried to convince themselves that logical truths are so by definition, are tautologies). Diagrams have been used in the past, notably by the 18th-century mathematician Leonhard Euler, but they have had limited application and were somehow used mainly for confirmation, not demonstration. I went into the subject extensively in my book again, and here want to focus only on the simple connections discussed above.
As said there, we, and even animals, know instantly that if A leads to B and B to C then A leads to C. The first diagram below depicts why this is so regarding the first given syllogism; since Socrates (A) is a man (B), all of whom are mortal (C), Socrates is seen as among the mortals. The second diagram depicts the other implication; since God (A) is identical with the purposeful power (B), shown in the living (C), God is shown in the living.
It may be appropriate to caution against responses relegating these observations to a particular camp in the disputes discussed. Darwinists would accordingly without examination sweep aside the preceding as coming from the Creationist camp. This response in general is classified as committing the "ad hominem" fallacy, of speaking about the opponent instead of the subject. In my case, I don't belong to any camp, and am ignored equally by those arguing for God's existence and those arguing against it. As indicated, academic thinkers, who, beside perhaps some theologians, are virtually the only participants in these debates, regard outsiders as not worthy of listening to, and they furthermore take offense at any criticism of ideas accepted in their own midst.
This applies in any field, including of course science, and my main point here is that my thinking is not based on allegiance to any group. As made evident, I am not a Christian, unlike the predominance of those arguing on the theist side, nor do I endorse another doctrine. At the same time I don't feel constrained by scientific statements made on the other side. Much of the scientific community cannot get over the conflict between religion and science in Galileo's time, and it is reluctant to allow that it too is fallible. My own inquiries led for these and other reasons to questioning convictions in any area, with corresponding results.
24 April 2009. The below image was influenced by a puzzle shown in a recent book (page 108) titled How To Be An Intellectually Fulfilled Atheist (Or Not) by William A. Dembski and Jonathan Wells, both authors proponents of "Intelligent Design", the currently well known theist approach.
The authors illustrate the puzzle, not a recent one, with the first two depictions here of nine dots forming a square; the question is what is the possible least number of continuous line segments connecting the nine dots; the authors point out that many people assume that the lines must be confined to the square formed by the dots, when the line segments can't be less than five; but the problem doesn't require such confinement, and the second depiction gives the solution with four segments.
The authors use this puzzle in analogy with theist versus atheist arguments, observing that the latter unjustifiably confine themselves to "materialist" approaches, ones intending to explain all of life by undirected causes, while it is advisable to explore other possibilities.
On wondering whether the puzzle could not somehow be solved with even less line segments, it occurred to me it also doesn't require lines to pass through a specific part of the dots, e.g. their center, a dot not lacking dimension; accordingly two lines passing each through a different row of three dots can meet by Euclid's famous 5th postulate, as illustrated in the third depiction.
I cannot vouch that this three-line solution has not been thought of before; I would welcome information about it. But I feel it may encourage a reader to be open to the possibility of demonstration of the kind put forward above on this page. The last geometric solution below may suggest that one can not only, as advocated by Intelligent Design, go beyond supposing undirected causes to supposing directed ones regarding the formation of living things, but one can go beyond that formation to their actual disclosure of goal-directedness in their activities.
Angle Trisection By Paper Folding
It is a classical result that, in general, Angle Trisection is impossible by the Euclidean
ruler and compass rules. Several methods that went beyond ruler and compass - the
neusis constructions - have been known yet to the ancient Greeks (see angle trisection
by Archimedes and Hippocrates and a more recent.) Axioms of Paper Folding also
supply the tools sufficiently powerful to successfully tackle the angle trisection
problem.
The following method of angle trisection is due to H. Abe (1980).
Let's consider an acute angle ABC:
By axiom O4, erect a crease through B perpendicular to AB.
(Note that, by O1, AB may be considered a crease.) Execute
two creases parallel to AB and at equal distances from each
other: BD = DE.
By O6, fold E into BC and B into L. Let B1, D1, E1 be
images of points B, D, and E, respectively. Let L1 be the
image of crease L. Refold it as to make a crease on the upper
part of the paper that remain at rest. Straighten the paper into
its original position and fold and unfold crease L1 again. We
next show that the crease passes through the point B and,
moreover, divides ∠ABC in the ratio 2:1 thus solving the
trisection problem.
The whole structure is symmetric with respect to the crease
OOB. In the isosceles trapezoid BDD1B1, the diagonals
intersect on the axis of symmetry. ΔDD1OD is isosceles by
construction. Therefore, the creases BD1 and D1OD coincide,
or, in other words, D1OD does pass through the vertex B.
Now, everything is pretty much straightforward. Just
recollect that the crease L is the line DB1 and BOB lies along
the given line AB and that, by construction L and AB are
parallel.
Thus we see that ∠B1BOB = ∠DB1B. Since ΔBODB1 is
isosceles, ∠ODBB1 = ∠ODB1B. BD1 is perpendicular to OB1
because OB is perpendicular to DB1. Also, E1D1 = D1B1.
This shows that ΔE1BB1 is isosceles and, therefore,
∠E1BD1 = ∠D1BB1.
References
1. G. E. Martin, Geometric Constructions, Springer, 1998
Paper Folding Geometry
An Interesting Example of Angle Trisection by Paperfolding
Angle Trisection by Paper Folding
Angles in Triangle Add to 180o
Broken Chord Theorem by Paper Folding
Dividing a Segment into Equal Parts by Paper Folding
Egyptian Triangle By Paper Folding
My Logo
Paper Folding And Cutting Sangaku
Parabola by Paper Folding
Radius of a Circle by Paper Folding
Regular Pentagon Inscribed in Circle by Paper Folding
Trigonometry by Paper Folding
Folding Square in a Line through the Center
Trisectrix: An Angle Trisection Curve
The Trisectrix is the curve, in green, with polar equation r = 1 + 2 cos(θ)
Angle PCD is trisected by the curve, because the measure of angle OPC is one third that
of PCD.
Proof
We will denote the measure of angle OPC by α, and show that PCD has a measure of
3α. Begin by drawing OP, intersecting the red circle at Q. Note that OCQ is isosceles,
because CO and CQ are radii of the red circle. Note, too, that CQP is isosceles, because
the rays OQ and OP differ in length by 1 (note their equations to see why), and so
QP=1.
Now in triangle CQP, we see if QCP=α, then QPC=α as well, so the external angle
OQC=2α.
In triangle OCQ, we see since OQC=2α, QOC=2α, and so its external angle QCD is 4α,
so the measure of PCD is 3α.
Internet references
Xah Lee's Visual Dictionary of Famous Plane Curves: Trisectrix
Wikipedia: Trisectrix
Related pages in this website:
Back to the Geometry and Trig home, or Triangles and Polygons
Law of Sines - Given triangle ABC with opposite sides a, b, and c, a/(sin A) = b/(sin B)
= c/(sin C) = the diameter of the circumscribed circle.
Circumscribed Circle - The radius of a circle circumscribed around a triangle is R =
abc/(4K), where K is the area of the triangle.
Inscribed Angle -- proof that an angle inscribed in a circle is half the central angle that
is subtended by the same arc
Triangle Trisection -- If a point, P, on the median of triangle ABC is the isogonal
conjugate of point Q, on the altitude of ABC, then ABC is a right triangle.
FREDERIC R. HONEY, Ph.B., Yale University.
This section is from "Scientific American Supplement". Also available from Amazon: Scientific
American Reference Book.
The Trisection Of Any Angle
The following analysis shows that with the aid of an hyperbola any arc, and therefore any
angle, may be trisected.
If the reader should not care to follow the analytical work, the construction is described in the
last paragraph - referring to Fig. II.
Let a b c d (Fig. I.) be the arc subtending a given angle. Draw the chord a d and bisect it at o.
Through o draw e f perpendicular to a d.
FIG 1.
We wish to find the locus of a point c whose distance from a given straight line e f is one-half
the distance from a given point d.
In order to write the equation of this curve, refer it to the co-ordinate axes a d (axis of X) and e
f (axis of Y), intersecting at the origin o.
Let g c = x
Therefore, from the definition c d = 2x
Let o d = D
[Hence] h d = D-x
Let c h = y
[Hence] (2x)² = y² + (D-x)²
or 4x² = y² + D²-2Dx + x²
[Hence] y²-3x² + D²-2Dx = o [I.]
This is the equation of an hyperbola whose center is on the axis of abscisses. In order to
determine the position of the center, eliminate the x term, and find the distance from the
origin o to a new origin o'.
Let E = distance from o to o'
[Hence] x = x' + E
Substituting this value of x in equation I.
y²-3(x' + E)² + D²-2D(x' + E) = o
or y²-3x²-6Ex'-3E² + D²-2Dx'-2DE = o [II.]
In this equation the x' terms should disappear.
[Hence] -6Ex' - 2Dx' = o
[Hence] -E = - D/3
That is, the distance from the origin o to the new origin or the center of the hyperbola o' is
equal to one-third of the distance from o to d; and the minus sign indicates that the
measurement should be laid off to the left of the origin o. Substituting this value of E in
equation II., and omitting accents -
We have
y² - 3x² + 2Dx - D²/3 + D² - 2Dx + 2D²/3 = o
[Hence] y² - 3x² = - 4D²/3
This is the equation of an hyperbola referred to its center o' as the origin of co-ordinates. To
write it in the ordinary form, that is in terms of the transverse and conjugate axes, multiply
each term by C, i.e.,
Let √C = semi-transverse axis.
Thus Cy² - 3Cx² = - 4CD² / 3. [III.]
When in this form the product of the coefficients of the x² and y² terms should be equal to the
remaining term.
That is
3C² = - 4CD² / 3.
[Hence] C = 4D² / 9.
And equation III. becomes:
(4D² / 9) y² - (4D² / 3) x² = 16D4 / 27
The semi-transverse axis = √4D² /9 = 2D / 3
The semi-conjugate axis = √4D² / 3 = 2D / √3
Since the distance from the center of the curve to either focus is equal to the square root of
the sum of the squares of the semi-axes, the distance from o' to either focus
= √4D²/9 + 4D²/ 3 = 4D / 3
The Trisection Of Any Angle 787 4 fig2
We can therefore make the following construction (Fig. II.) Draw a d the chord of the arc a c d.
Trisect a d at o' and k. Produce d a to l, making a l = a o' = o' k = k d. With a k as a transverse
axis, and l and d as foci, construct the branch of the hyperbola k c c' c", which will intersect all
arcs having the common chord a d at c, c', c", etc., making the arcs c d, c' d, c" d, etc.,
respectively, equal to one-third of the arcs a c d, a c' d, a c" d, etc.
Title Scientific American Supplement Volumes 643, 647, 664, 711, 717, 787, 794, 795, 799
and 803
Author Various Authors
Publisher Munn & Co.
Year 1888-1891
Copyright 1891, Munn & Co.
Amazon Scientific American Reference Book
~Source : Hyperbolic Trisection and the Spectrum of Regular Polygons
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A geometric proof of the impossibility of angle
trisection by straightedge and compass
10 August, 2011 in expository, math.AG, math.CV, math.MG, math.NT | Tags: angle
trisection, Galois theory, monodromy | by Terence Tao | 35 comments
One of the most well known problems from ancient Greek mathematics was that of
trisecting an angle by straightedge and compass, which was eventually proven
impossible in 1837 by Pierre Wantzel, using methods from Galois theory.
Formally, one can set up the problem as follows. Define a configuration to be a finite
collection of points, lines, and circles in the Euclidean plane. Define a construction
step to be one of the following operations to enlarge the collection :
(Straightedge) Given two distinct points in , form the line that
connects and , and add it to .
(Compass) Given two distinct points in , and given a third point in
(which may or may not equal or ), form the circle with centre and radius
equal to the length of the line segment joining and , and add it to .
(Intersection) Given two distinct curves in (thus is either a line or a circle
in , and similarly for ), select a point that is common to both and (there
are at most two such points), and add it to .
We say that a point, line, or circle is constructible by straightedge and compass from a
configuration if it can be obtained from after applying a finite number of
construction steps.
Problem 1 (Angle trisection) Let be distinct points in the plane. Is it always
possible to construct by straightedge and compass from a line through that
trisects the angle , in the sense that the angle between and is one third of
the angle of ?
Thanks to Wantzel’s result, the answer to this problem is known to be “no” in general; a
generic angle cannot be trisected by straightedge and compass. (On the other
hand, some special angles can certainly be trisected by straightedge and compass, such
as a right angle. Also, one can certainly trisect generic angles using other methods than
straightedge and compass; see the Wikipedia page on angle trisection for some
examples of this.)
The impossibility of angle trisection stands in sharp contrast to the easy construction of
angle bisection via straightedge and compass, which we briefly review as follows:
1. Start with three points .
2. Form the circle with centre and radius , and intersect it with the line
. Let be the point in this intersection that lies on the same side of as . (
may well be equal to ).
3. Form the circle with centre and radius , and the circle with centre
and radius . Let be the point of intersection of and that is not .
4. The line will then bisect the angle .
The key difference between angle trisection and angle bisection ultimately boils down
to the following trivial number-theoretic fact:
Lemma 2 There is no power of that is evenly divisible by .
Proof: Obvious by modular arithmetic, by induction, or by the fundamental theorem of
arithmetic.
In contrast, there are of course plenty of powers of that are evenly divisible by , and
this is ultimately why angle bisection is easy while angle trisection is hard.
The standard way in which Lemma 2 is used to demonstrate the impossibility of angle
trisection is via Galois theory. The implication is quite short if one knows this theory,
but quite opaque otherwise. We briefly sketch the proof of this implication here, though
we will not need it in the rest of the discussion. Firstly, Lemma 2 implies the following
fact about field extensions.
Corollary 3 Let be a field, and let be an extension of that can be constructed out
of by a finite sequence of quadratic extensions. Then does not contain any cubic
extensions of .
Proof: If contained a cubic extension of , then the dimension of over would
be a multiple of three. On the other hand, if is obtained from by a tower of
quadratic extensions, then the dimension of over is a power of two. The claim then
follows from Lemma 2.
To conclude the proof, one then notes that any point, line, or circle that can be
constructed from a configuration is definable in a field obtained from the coefficients
of all the objects in after taking a finite number of quadratic extensions, whereas a
trisection of an angle will generically only be definable in a cubic extension of
the field generated by the coordinates of .
The Galois theory method also allows one to obtain many other impossibility results of
this type, most famously the Abel-Ruffini theorem on the insolvability of the quintic
equation by radicals. For this reason (and also because of the many applications of
Galois theory to number theory and other branches of mathematics), the Galois theory
argument is the “right” way to prove the impossibility of angle trisection within the
broader framework of modern mathematics. However, this argument has the drawback
that it requires one to first understand Galois theory (or at least field theory), which is
usually not presented until an advanced undergraduate algebra or number theory course,
whilst the angle trisection problem requires only high-school level mathematics to
formulate. Even if one is allowed to “cheat” and sweep several technicalities under the
rug, one still needs to possess a fair amount of solid intuition about advanced algebra in
order to appreciate the proof. (This was undoubtedly be one reason why, even after
Wantzel’s impossibility result was published, a large amount of effort was still
expended by amateur mathematicians to try to trisect a general angle.)
In this post I would therefore like to present a different proof (or perhaps more
accurately, a disguised version of the standard proof) of the impossibility of angle
trisection by straightedge and compass, that avoids explicit mention of Galois theory
(though it is never far beneath the surface). With “cheats”, the proof is actually quite
simple and geometric (except for Lemma 2, which is still used at a crucial juncture),
based on the basic geometric concept of monodromy; unfortunately, some technical
work is needed however to remove these cheats.
To describe the intuitive idea of the proof, let us return to the angle bisection
construction, that takes a triple of points as input and returns a bisecting line as
output. We iterate the construction to create a quadrisecting line , via the following
sequence of steps that extend the original bisection construction:
1. Start with three points .
2. Form the circle with centre and radius , and intersect it with the line
. Let be the point in this intersection that lies on the same side of as . (
may well be equal to ).
3. Form the circle with centre and radius , and the circle with centre
and radius . Let be the point of intersection of and that is not .
4. Let be the point on the line which lies on , and is on the same side
of as .
5. Form the circle with centre and radius . Let be the point of
intersection of and that is not .
6. The line will then quadrisect the angle .
Let us fix the points and , but not , and view (as well as intermediate objects
such as , , , , , , ) as a function of .
Let us now do the following: we begin rotating counterclockwise around , which
drags around the other objects , , , , , , that were constructed by
accordingly. For instance, here is an early stage of this rotation process, when the angle
has become obtuse:
Now for the slightly tricky bit. We are going to keep rotating beyond a half-rotation
of , so that now becomes a reflex angle. At this point, a singularity occurs;
the point collides into , and so there is an instant in which the line is not
well-defined. However, this turns out to be a removable singularity (and the easiest way
to demonstrate this will be to tap the power of complex analysis, as complex numbers
can easily route around such a singularity), and we can blast through it to the other side,
giving a picture like this:
Note that we have now deviated from the original construction in that and are no
longer on the same side of ; we are thus now working in a continuation of that
construction rather than with the construction itself. Nevertheless, we can still work
with this continuation (much as, say, one works with analytic continuations of infinite
series such as beyond their original domain of definition).
We now keep rotating around . Here, is approaching a full rotation of :
When reaches a full rotation, a different singularity occurs: and coincide.
Nevertheless, this is also a removable singularity, and we blast through to beyond a full
rotation:
And now is back where it started, as are , , , and … but the point has moved,
from one intersection point of to the other. As a consequence, , , and have
also changed, with being at right angles to where it was before. (In the jargon of
modern mathematics, the quadrisection construction has a non-trivial monodromy.)
But nothing stops us from rotating some more. If we continue this procedure, we see
that after two full rotations of around , all points, lines, and circles constructed from
have returned to their original positions. Because of this, we shall say that the
quadrisection construction described above is periodic with period .
Similarly, if one performs an octisection of the angle by bisecting the
quadrisection, one can verify that this octisection is periodic with period ; it takes four
full rotations of around before the configuration returns to where it started. More
generally, one can show
Proposition 4 Any construction of straightedge and compass from the points is
periodic with period equal to a power of .
The reason for this, ultimately, is because any two circles or lines will intersect each
other in at most two points, and so at each step of a straightedge-and-compass
construction there is an ambiguity of at most . Each rotation of around can
potentially flip one of these points to the other, but then if one rotates again, the point
returns to its original position, and then one can analyse the next point in the
construction in the same fashion until one obtains the proposition.
But now consider a putative trisection operation, that starts with an arbitrary angle
and somehow uses some sequence of straightedge and compass constructions to
end up with a trisecting line :
What is the period of this construction? If we continuously rotate around , we
observe that a full rotations of only causes the trisecting line to rotate by a third of a
full rotation (i.e. by ):
Because of this, we see that the period of any construction that contains must be a
multiple of . But this contradicts Proposition 4 and Lemma 2.
Below the fold, I will make the above proof rigorous. Unfortunately, in doing so, I had
to again leave the world of high-school mathematics, as one needs a little bit of
algebraic geometry and complex analysis to resolve the issues with singularities that we
saw in the above sketch. Still, I feel that at an intuitive level at least, this argument is
more geometric and accessible than the Galois-theoretic argument (though anyone
familiar with Galois theory will note that there is really not that much difference
between the proofs, ultimately, as one has simply replaced the Galois group with a
closely related monodromy group instead).
Read the rest of this entry »
A geometric proof of the impossibility of
angle trisection by straightedge
and compass
10 August, 2011 in expository, math.AG, math.CV, math.MG, math.NT | Tags: angle
trisection, Galois theory, monodromy
One of the most well known problems from ancient Greek mathematics was that of
trisecting an angle by straightedge and compass, which was eventually proven
impossible in 1837 by Pierre Wantzel, using methods from Galois theory.
Formally, one can set up the problem as follows. Define a configuration to be a finite
collection of points, lines, and circles in the Euclidean plane. Define a construction
step to be one of the following operations to enlarge the collection :
(Straightedge) Given two distinct points in , form the line that
connects and , and add it to .
(Compass) Given two distinct points in , and given a third point in
(which may or may not equal or ), form the circle with centre and radius
equal to the length of the line segment joining and , and add it to .
(Intersection) Given two distinct curves in (thus is either a line or a circle
in , and similarly for ), select a point that is common to both and (there
are at most two such points), and add it to .
We say that a point, line, or circle is constructible by straightedge and compass from a
configuration if it can be obtained from after applying a finite number of
construction steps.
Problem 1 (Angle trisection) Let be distinct points in the plane. Is it always
possible to construct by straightedge and compass from a line through that
trisects the angle , in the sense that the angle between and is one third of
the angle of ?
Thanks to Wantzel’s result, the answer to this problem is known to be “no” in general; a
generic angle cannot be trisected by straightedge and compass. (On the other
hand, some special angles can certainly be trisected by straightedge and compass, such
as a right angle. Also, one can certainly trisect generic angles using other methods than
straightedge and compass; see the Wikipedia page on angle trisection for some
examples of this.)
The impossibility of angle trisection stands in sharp contrast to the easy construction of
angle bisection via straightedge and compass, which we briefly review as follows:
1. Start with three points .
2. Form the circle with centre and radius , and intersect it with the line
. Let be the point in this intersection that lies on the same side of as . (
may well be equal to ).
3. Form the circle with centre and radius , and the circle with centre
and radius . Let be the point of intersection of and that is not .
4. The line will then bisect the angle .
The key difference between angle trisection and angle bisection ultimately boils down
to the following trivial number-theoretic fact:
Lemma 2 There is no power of that is evenly divisible by .
Proof: Obvious by modular arithmetic, by induction, or by the fundamental theorem of
arithmetic.
In contrast, there are of course plenty of powers of that are evenly divisible by , and
this is ultimately why angle bisection is easy while angle trisection is hard.
The standard way in which Lemma 2 is used to demonstrate the impossibility of angle
trisection is via Galois theory. The implication is quite short if one knows this theory,
but quite opaque otherwise. We briefly sketch the proof of this implication here, though
we will not need it in the rest of the discussion. Firstly, Lemma 2 implies the following
fact about field extensions.
Corollary 3 Let be a field, and let be an extension of that can be constructed out
of by a finite sequence of quadratic extensions. Then does not contain any cubic
extensions of .
Proof: If contained a cubic extension of , then the dimension of over would
be a multiple of three. On the other hand, if is obtained from by a tower of
quadratic extensions, then the dimension of over is a power of two. The claim then
follows from Lemma 2.
To conclude the proof, one then notes that any point, line, or circle that can be
constructed from a configuration is definable in a field obtained from the coefficients
of all the objects in after taking a finite number of quadratic extensions, whereas a
trisection of an angle will generically only be definable in a cubic extension of
the field generated by the coordinates of .
The Galois theory method also allows one to obtain many other impossibility results of
this type, most famously the Abel-Ruffini theorem on the insolvability of the quintic
equation by radicals. For this reason (and also because of the many applications of
Galois theory to number theory and other branches of mathematics), the Galois theory
argument is the “right” way to prove the impossibility of angle trisection within the
broader framework of modern mathematics. However, this argument has the drawback
that it requires one to first understand Galois theory (or at least field theory), which is
usually not presented until an advanced undergraduate algebra or number theory course,
whilst the angle trisection problem requires only high-school level mathematics to
formulate. Even if one is allowed to “cheat” and sweep several technicalities under the
rug, one still needs to possess a fair amount of solid intuition about advanced algebra in
order to appreciate the proof. (This was undoubtedly be one reason why, even after
Wantzel’s impossibility result was published, a large amount of effort was still
expended by amateur mathematicians to try to trisect a general angle.)
In this post I would therefore like to present a different proof (or perhaps more
accurately, a disguised version of the standard proof) of the impossibility of angle
trisection by straightedge and compass, that avoids explicit mention of Galois theory
(though it is never far beneath the surface). With “cheats”, the proof is actually quite
simple and geometric (except for Lemma 2, which is still used at a crucial juncture),
based on the basic geometric concept of monodromy; unfortunately, some technical
work is needed however to remove these cheats.
To describe the intuitive idea of the proof, let us return to the angle bisection
construction, that takes a triple of points as input and returns a bisecting line as
output. We iterate the construction to create a quadrisecting line , via the following
sequence of steps that extend the original bisection construction:
1. Start with three points .
2. Form the circle with centre and radius , and intersect it with the line
. Let be the point in this intersection that lies on the same side of as . (
may well be equal to ).
3. Form the circle with centre and radius , and the circle with centre
and radius . Let be the point of intersection of and that is not .
4. Let be the point on the line which lies on , and is on the same side
of as .
5. Form the circle with centre and radius . Let be the point of
intersection of and that is not .
6. The line will then quadrisect the angle .
Let us fix the points and , but not , and view (as well as intermediate objects
such as , , , , , , ) as a function of .
Let us now do the following: we begin rotating counterclockwise around , which
drags around the other objects , , , , , , that were constructed by
accordingly. For instance, here is an early stage of this rotation process, when the angle
has become obtuse:
Now for the slightly tricky bit. We are going to keep rotating beyond a half-rotation
of , so that now becomes a reflex angle. At this point, a singularity occurs;
the point collides into , and so there is an instant in which the line is not
well-defined. However, this turns out to be a removable singularity (and the easiest way
to demonstrate this will be to tap the power of complex analysis, as complex numbers
can easily route around such a singularity), and we can blast through it to the other side,
giving a picture like this:
Note that we have now deviated from the original construction in that and are no
longer on the same side of ; we are thus now working in a continuation of that
construction rather than with the construction itself. Nevertheless, we can still work
with this continuation (much as, say, one works with analytic continuations of infinite
series such as beyond their original domain of definition).
We now keep rotating around . Here, is approaching a full rotation of :
When reaches a full rotation, a different singularity occurs: and coincide.
Nevertheless, this is also a removable singularity, and we blast through to beyond a full
rotation:
And now is back where it started, as are , , , and … but the point has moved,
from one intersection point of to the other. As a consequence, , , and have
also changed, with being at right angles to where it was before. (In the jargon of
modern mathematics, the quadrisection construction has a non-trivial monodromy.)
But nothing stops us from rotating some more. If we continue this procedure, we see
that after two full rotations of around , all points, lines, and circles constructed from
have returned to their original positions. Because of this, we shall say that the
quadrisection construction described above is periodic with period .
Similarly, if one performs an octisection of the angle by bisecting the
quadrisection, one can verify that this octisection is periodic with period ; it takes four
full rotations of around before the configuration returns to where it started. More
generally, one can show
Proposition 4 Any construction of straightedge and compass from the points is
periodic with period equal to a power of .
The reason for this, ultimately, is because any two circles or lines will intersect each
other in at most two points, and so at each step of a straightedge-and-compass
construction there is an ambiguity of at most . Each rotation of around can
potentially flip one of these points to the other, but then if one rotates again, the point
returns to its original position, and then one can analyse the next point in the
construction in the same fashion until one obtains the proposition.
But now consider a putative trisection operation, that starts with an arbitrary angle
and somehow uses some sequence of straightedge and compass constructions to
end up with a trisecting line :
What is the period of this construction? If we continuously rotate around , we
observe that a full rotations of only causes the trisecting line to rotate by a third of a
full rotation (i.e. by ):
Because of this, we see that the period of any construction that contains must be a
multiple of . But this contradicts Proposition 4 and Lemma 2.
Below the fold, I will make the above proof rigorous. Unfortunately, in doing so, I had
to again leave the world of high-school mathematics, as one needs a little bit of
algebraic geometry and complex analysis to resolve the issues with singularities that we
saw in the above sketch. Still, I feel that at an intuitive level at least, this argument is
more geometric and accessible than the Galois-theoretic argument (though anyone
familiar with Galois theory will note that there is really not that much difference
between the proofs, ultimately, as one has simply replaced the Galois group with a
closely related monodromy group instead).
— 1. Details —
We now make the argument more rigorous. We will assume for sake of contradiction
that for every triple of distinct points, we can find a construction by
straightedge and compass that trisects the angle , and eventually deduce a
contradiction out of this.
We remark that we do not initially assume any uniformity in this construction; for
instance, it could be possible that the trisection procedure for obtuse angles is
completely different from that of acute angles, using a totally different set of
constructions, while some exceptional angles (e.g. right angles or degenerate angles)
might use yet another construction. We will address these issues later.
The first step is to get rid of some possible degeneracies in one’s construction. At
present, nothing in our definition of a construction prevents us from adding a point, line,
or circle to the construction that was already present in the existing collection of
points, lines, and circles. However, it is clear that any such step in the construction is
redundant, and can be omitted. Thus, we may assume without loss of generality that for
each , the construction used to trisect the angle contains no such redundant
steps. (This may make the construction even less uniform than it was previously, but we
will address this issue later.)
Another form of degeneracy that we will need to eliminate for technical reasons is that
of tangency. At present, we allow in our construction the ability to take two tangent
circles, or a circle and a tangent line, and add the tangent point to the collection (if it
was not already present in the construction). This would ordinarily be a harmless thing
to do, but it complicates our strategy of perturbing the configuration, so we now act to
eliminate it. Suppose first that one had two circles already constructed in the
configuration and tangent to each other, and one wanted to add the tangent point to
the configuration. But note that in order to have added and to , one must
previously have added the centres and of these circles to also. One can then add
to by intersecting the line with and picking the point that lies on ; this way,
one does not need to intersect two tangent curves together.
Similarly, suppose that we already had a circle and a tangent line already constructed
in the configuration, but with the tangent point absent. The centre of , and at least
two points on , must previously have also been constructed in order to have and
present; note that are not equal to by hypothesis. One can then obtain by
dropping a perpendicular from to by the usual construction (i.e. drawing a circle
centred at with radius to hit again at , then drawing circles from and with
the same radius to meet at a point distinct from , then intersecting with to
obtain ), thus avoiding tangencies again. (This construction may happen to use lines or
circles that had already appeared in the construction, but in those cases one can simply
skip those steps.)
As a consequence of these reductions, we may now assume that our construction is
nondegenerate in the sense that
Any point, line, or circle added at a step in the construction, does not previously
appear in that construction.
Whenever one intersects two circles in a construction together to add another
point to the construction, the circles are non-tangent (and thus meet in exactly
two points).
Whenever one intersects a circle and a line in a construction together to add
another point to the construction, the circle and line are non-tangent (and thus
meet in exactly two points).
The reason why we restrict attention to nondegenerate constructions is that they are
stable with respect to perturbations. Note for instance that if one has two circles
that intersect in two different points, and one of them is labeled , then we may perturb
and by a small amount, and still have an intersection point close to (with the other
intersection point far away from ). Thus, is locally a continuous function of and
. Similarly if one forms the intersection of a circle and a secant (a line which
intersects non-tangentially). In a similar vein, given two points and that are distinct,
the line between them varies continuously with and as long as one does not
move and so far that they collide; and given two lines and that intersect at a
point (and in particular are non-parallel), then also depends continuously on and
. Thus, in a nondegenerate construction starting from the original three points
, every point, line, or circle created by the construction can be viewed as a
continuous function of , as long as one only works in a sufficiently small
neighbourhood of the original configuration . In particular, the final line
varies continuously in this fashion. Note however that the trisection property may be
lost by this perturbation; just because happens to trisect when are in
the original positions, this does not necessarily imply that after one perturbs ,
that the resulting perturbed line still trisects the angle. (For instance, there are a number
of ways to trisect a right angle (e.g. by bisecting an angle of an equilateral triangle), but
if one perturbs the angle to be slightly acute or slightly obtuse, the line created by this
procedure would not be expected to continue to trisect that angle.)
The next step is to allow analytic geometry (and thence algebraic geometry) to enter the
picture, by using Cartesian coordinates. We may identify the Euclidean plane with the
analytic plane ; we may also normalise to be the points
, by this identification. We will also restrict to lie on the unit
circle , so that there is now just one degree of
freedom in the configuration . One can describe a line in by an equation of
the form
(with not both zero), and describe a circle in by an equation of the form
with non-zero. There is some non-uniqueness in these representations: for the line, one
can multiply by the same constant without altering the line, and for the circle, one
can replace by . However, this will not be a serious concern for us. Note that any
two distinct points , determine a line
and given three points , , , one can form a circle
with centre and radius . Given two distinct non-parallel lines
and
their unique intersection point is given as
similarly, given two circles
and
their points of intersection (if they exist in ) are given as
where
and
and the points of intersection between and (if they exist in ) are given as
The precise expressions given above are not particularly important for our argument,
save to note that these expressions are always algebraic functions of the input
coordinates such as , defined over the reals ,
and that the only algebraic operations needed here besides the arithmetic operations of
addition, subtraction, multiplication, and division is the square root operation. Thus, we
see that any particular construction of, say, a line from a configuration will
locally be an algebraic function of (recall that we have already fixed ), and this
definition can be extended until one reaches a degeneracy (two points, lines, or circles
collide, two curves become tangent, or two lines become parallel); however, this
degeneracy only occurs in an proper real algebraic set of configurations, and in
particular for in a dimension zero subset of the circle .
These degeneracies are annoying because they disconnect the circle , and can
potentially block off large regions of that circle for which the construction is not even
defined (because two circles stop intersecting, or a circle and line stop intersecting, in
, due to the lack of a real square root for negative numbers). To fix this, we move
now from the real plane to the complex plane . Note that the algebraic definitions
of a line and a circle continue to make perfect sense in (with coefficients such as
now allowed to be complex numbers instead of real numbers), and the
algebraic intersection formulae given previously continue to make sense in the complex
setting. The point now is allowed to range in the complex circle
, which is a Riemann surface (conformal to the
Riemann sphere after stereogrpahic projection). Furthermore, because all non-
zero complex numbers have square roots, any given construction that was valid for at
least one configuration is now valid (though possibly multi-valued) as an algebraic
function on outside of a dimension zero set of singularities, i.e. outside of a finite
number of exceptional values of . But note now that these singularities do not
disconnect the complex circle , which has topological dimension two instead of one.
As mentioned earlier, a line given by such a construction may or may not trisect the
original angle . But this trisection property can be expressed algebraically (e.g.
using the triple angle formulae from trigonometry, or by building rotation matrices), and
in particular makes sense over . Thus, for any given construction of a line , the set of
in for which the construction is non-degenerate and trisects is a
constructible set (a boolean combination of algebraic sets). But is an irreducible one-
dimensional complex variety. As such, the aforementioned set of is either generic (the
complement of a dimension one algebraic set), or has dimension at most one. (Here we
are implicitly using the fundamental theorem of algebra, because the basic dimension
theory of algebraic geometry only works properly over algebraically closed fields.)
On the other hand, there are at most countably many constructions, and by hypothesis,
for each choice of in , at least one of these constructions has to trisect the angle.
Applying the Baire category theorem (or countable additivity of Lebesgue measure, or
using the algebraic geometry fact that an algebraic variety over an uncountable field
cannot be covered by the union of countably many algebraic sets of smaller dimension),
we conclude that there is a single construction which trisects the angle for a
generic choice of , i.e. for all in outside of a finite set of points, there is a
construction, which amongst its multiple possible values, is able to output at least one
line that trisects .
Now one performs monodromy. Suppose we move around a closed loop in that
avoids all points of degeneracy. Then all the other points, lines, and circles constructed
from can be continuously extended from an initial configuration as discussed
earlier, with each such object tracing out its own path in its own configuration space.
Because of the presence of square roots in constructions such as the intersection (1)
between two circles, or the intersection (2) between a circle and a line, these
constructions may map a closed loop to an open loop; but because the square root
function forms a double cover of , we see that any closed loop in , if doubled,
will continue to be a closed loop upon taking a square root. (Alternatively, one can
argue geometrically rather than algebraically, noting that in the intersection of (say) two
non-degenerate circles , there are only two possible choices for the intersection
point of these two circles, and so if one performs monodromy along a loop of possible
pairs of circles, either these two choices return to where they initially started, or
are swapped; so if one doubles the loop, one must necessarily leave the intersection
points unchanged.) Iterating this, we see that any object constructed by straightedge and
compass from must have period for some power of two , in the sense that if
one iterates a loop of in avoiding degenerate points times, the object must return
to where it started. (In more algebraic terminology: the monodromy group must be a -
group.)
Now, one traverses along a slight perturbation of a single rotation of the real unit
circle , taking a slight detour around the finite number of degeneracy points one
encounters along the way. Since has to trisect the angle at each of these points,
while varying continuously with , we see that when traverses a full rotation, has
only traversed one third of a rotation (or two thirds, depending on which trisection one
obtained), and so the period of must be a multiple of three; but this contradicts Lemma
2, and the claim follows.
Angle Trisection by the Cuadratrix of Hippias.
Trisecting an Angle
This is a simple straightedge and compass construction of a third part of an angle. In the
illustration for Step Four, the angles DOC and OCD are equal, and they add up to the
angle ODE. Angle ODE is equal to angle OED, and their sum is equal to the sum of
DOC and FOE. Using algebra, it is easy to see that angle DOC is a trisection of angle
FOE.
Step One: This is an
acute angle
AOB. If this
angle can be
trisected, the
construction
can generalize
to any
arbitrary
angle.
Step Two: Set the
compass to an
arbitrary
distance. The
compass
setting will
not be
changed
throughout
this
construction.
Using O as a
center, draw a
circle. Use the
straightedge
to extend the
line OA to the
left, and set
the compass
on the
intersection of
the circle and
line OA, with
the other point
on the
extended line.
Step Three: Hold the
straightedge
to the
compass, and
slide one end
of the
compass
along the line
OA, while the
other end of
the compass
moves up the
circle. Stop
when the
straightedge is
aligned with
the
intersection of
the circle and
OB.
Step Four: Mark point D
on the circle
with the
compass. The
angle DOC is
exactly one
third of the
angle AOB.
Geometry
Angle Trisection and Regular Polygons
Hubert Shutrick
The Classical Problems
The aim is to give a relatively elementary proof that angles cannot be trisected and that
cube roots cannot be extracted using a straight edge and compass.
Given two points in the Euclidean plane, the classical problem is to find out which other
points can be constructed using a straight edge and compass. A coordinate system can
be set up by considering the line through the points as the x-axis, letting one of them be
(0 0), the other (1 0) and letting the y-axis be the orthogonal line through (0 0). It is
obvious then how any point (m n) with integer coordinates can be constructed.
Real numbers are said to be constructible if they are one of the coordinates of a point
that we can construct. If s and t are constructible reals, then (s+t 0) and (s−t 0) are the
intersections of a circle of radius t with centre (s 0) with the x-axis, their product st is
the x-coordinate of where the line through (0 t) parallel to the line joining (0 1) to (s 0)
meets the x-axis and their quotient s t is the x-coordinate of where the line through (0
1) parallel to the line joining (0 t) to (s 0) meets the x-axis.
The real numbers form what is called a field which means that one can do the arithmetic
operations addition, subtraction, multiplication and division with pairs of reals and the
usual rules apply. A subset of the set R of real numbers is itself a field, a subfield of R,
if 0 and 1 are in it and the sum, difference, product and quotient with non-zero
denominator, of its members are again in the subset. From this, it is clear that any
intersection of subfields is itself a subfield. The subfield that is the intersection of all the
subfields that include a given subset, the least subfield that contains it, is said to be
generated by the subset.
The subfield generated by 0 1 is the field Q of rational numbers. It is the smallest
subfield of R and all rationals are constructible by the constructions described above.
Apart from getting new numbers by those constructions, there is one other possibility
and that is to construct the square root of a number in the subfield. It is, in fact, the only
possibility because any constructed points are:
1. the intersection of two lines, which just gives numbers in the subfield;
2. the intersection of a line with a circle, which gives the solution to a quadratic
equation where the usual formula introduces a square root;
3. the intersection of two circles but subtracting the equations of the circles gives
the equation of the line through their points of intersection returning to the
second case.
An easy way to get the square root of s is to draw the circle whose diameter is the
segment joining (−1 0) to (s 0) because it meets the y-axis at (0 ± s) .
Suppose that r is not in a subfield F. The larger subfield F(r) generated by F r is
called the extension of F by r. In the case when r= f with f in F, the extension will be
the set of all reals of the form s f+t where s and t are in F. It is obvious that the sum,
difference and product of two reals s f+t and s f+t can be denoted in this way
and, if s f+t is the denominator of a quotient, one uses the usual method of
multiplying both the numerator and denominator by −s f+t
This form is unique in the sense that, if s f+t is the same number then s =s and t =t
and, of course, if not (s −s) f+(t −t)=0 would imply that f was in F. We can then
say that the dimension of G over F is 2 as one does in linear algebra considering f
1 as a base. A usual notation is dimFG=2 and it can be called a quadratic extension.
If dimQF=n with a base a1 a2 … an−1 1 , then, in the expression s f+t ,
s=s1a1+s2a2+…+sn−1an−1+sn
t=t1a1+t2a2+…+tn−1an−1+tn
uniquely in both cases with the coefficients in Q. So a base for G over Q is
a1 f a2 f … an−1 f f a1 a2 … an−1 1
It has dimension 2n over Q. The general rule is
dimQF dimFG=dimQG
To conclude, any subfield of constructible reals starting from the given two points will
have dimension 2m over Q for some integer m.
Theorem 1. If a polynomial of degree 3 with rational coefficients has no rational root,
then its extension Q(r) by a real root r has dimension 3 over the rationals and therefore r
is not constructible.
Corollary 1. Squaring the cube. The cube root of 2 is not constructible since x3−2 has
no rational root.
Corollary 2. Trisecting the angle. The angle 60° cannot be trisected because
cos20° satisfies 4x3−3x−1=0, which has no rational root.
A rational polynomial is one whose coefficients are rational and two of their properties
are useful in this context.
1.The division algorithm works for rational polynomials u(x) and v(x) just as it does
for integers and gives
u(x)=q(x)v(x)+r(x)
where the remainder r(x) has lower degree than v(x) and this can be used to find the
greatest common factor of u(x) and v(x) by using Euclid's algorithm which gives it in
the form
h(x)=f(x)u(x)+g(x)v(x)
2. A primitive polynomial is one with coprime integer coefficients. It is clear that, if
u(x) is a rational polynomial and n is the highest common factor of the numerators of its
coefficients and m is the least common multiple of their denominators, then u(x)=m
np(x) where p(x) is primitive. The product of two primitive polynomials is also
primitive so u(x) can be factorized if and only if p(x) is the product of two primitive
polynomials.
From this second property, it can be shown that the primitive polynomials x3–2 and
4x3−3x–1 can't have a rational root m n since nx+m would be a factor with n dividing
the coefficient of x3 and m dividing the constant term. There are not many possibilities
to check.
Proof of the theorem. Suppose u(x) is the polynomial so u(r)=0. The claim is that 1 r r2
is a base for the extension by r so any real in the subfield can be expressed uniquely as
v(r)=ar2+br+c where a b and c are rational.
It is obvious how to add and subtract such expressions.
The product of two of them v(r) and v (r) gives a polynomial of degree (at most) 4 in r
which can be expressed
v(x)v (x)=w(x)u(x)+z(x)
by the division algorithm and the remainder z(x) of degree (at most) 2 evaluated at r is
the product in the field since u(r)=0.
To calculate the inverse of v(r), use Euclid's algorithm to obtain
f(x)u(x)+g(x)v(x)=q
where q must be a non-zero rational since u(x) has no factors. Evaluating at r gives that
g(r) q is the inverse of v(r)
Uniqueness is also obtained from Euclid's algorithm. If v (r) was the same number as
v(r) then the polynomial v (x)–v(x) of degree at most 2 would have a common root
with u(x) and the algorithm would give a common factor that u(x) can't have.
Regular Polygons
Gauss' Formula. A necessary and sufficient condition that a regular polygon is
constructible is that its number of vertices is of the form
p1p2 pm2n
where the pj are distinct Fermat primes.
Fermat primes are of the form 22k+1 but the only known examples are 3 5 17 257 and
65537.
Complex numbers are useful here. Consider points with coordinates (x y) as z=x+iy in
the usual way so that the vertices of the regular n-gon can be kn for k=0 1 … n−1 ,
where n=ei n and n=2 n . They are the roots of the polynomial xn–1 . In theorem 1,
we used subfields of the reals R, but the method works also with subfields of the field
of complex numbers C. The points with rational coordinates become complex numbers
whose real and imaginary parts are rational and they form a subfield of C, the extension
Q(i) of the rationals by i.
In Gauss' formula, we can concentrate on the prime factors because:
1. If the regular polygon with mn vertices is constructible, then the one with n is too
since we can take every mth vertex.
2. If regular polygons with m and n vertices are constructible and m and n are coprime,
then so is the one with mn vertices because Euclid's algorithm gives integers a and b
such that am+bn=1 which implies that a n+b m= mn.
3. It is obvious that, from n, one can get 2n by bisecting.
The important property of the third degree rational polynomials in theorem 1 was that
they were irreducible which means that they could not be expressed as a product of two
non-constant rational polynomials. It was this property that ensured that division was
possible in the extended subfield and gave the uniqueness of the linear combinations of
the base elements that described its elements. Therefore the theorem can be generalized
as follows.
Theorem 1 . If a is a root of an irreducible rational polynomial of degree n, then the
extension field Q(a) has dimension n with base 1 a a2 … an−1 .
For constructibility of Q(a), it is necessary that the dimension is a power of 2 as shown
above. The general proof that it is sufficient requires some results in Galois theory and
the Cauchy-Sylow theorem in group theory and will therefore be omitted.
Eisenstein's criterion. If all the coefficients of a primitive polynomial except that of
the highest power are divisible by the prime p and the constant term is not divisible by
p2, then it is irreducible.
Proof by induction. If it could be expressed as the product of two primitive
polynomials
(a0+a1x+ )(b0+b1x+…)
then a0, say, is divisible by p but not b0. Assume that all coefficients ak with k n are
divisible by p. The coefficient of xn is anb0+an−1b1+ and an is therefore divisible by
p. This contradicts the assumption that a0+a1x+ is primitive.
Lemma 1. The dimension over Q of Q( p), where p is prime, is p−1 so it is
constructible if and only if p−1 is a power of 2. Since this is satisfied by the Fermat
primes, it will establish that regular polygons with that number of vertices are
constructible
Proof. Consider the polynomial
((x+1)p–1) x=xp−1+pxp−2+C2pxp−3+…+p
which is satisfied by x= p–1 . All its coefficients except that of the highest power are
divisible by the prime p and the constant term is not divisible by p2.
Lemma 2. Regular polygons with p2 vertices, where p is an odd prime, are not
constructible, which is why the Fermat primes in the formula must be distinct.
Proof. The polynomial ((x+1)p2–1) is not irreducible since it is divisible by ((x+1)p–
1) but the quotient satisfies Eisenstein's criterion because calculating modulo p leaves
only the leading term xp(p−1) and the constant term is p2 p. It has degree p(p−1) which
can't be a power of 2.
Note that, for m odd and greater than 1,
2nm+1=(2n+1)(2n(m−1)−2n(m−2)+…−2n+1)
which is not prime so the primes of constructible polygons must be Fermat.
Although the general proof of sufficiency in the theorem is omitted, it is possible to see,
to some extent, how constructions can be done by examining the case of the regular 17-
gon. Denote 17 by so we know from the polynomial that
16+ 15+…+ +1=0
Let
x1= + 9+ 13+ 15+ 16+ 8+ 4+ 2
and let x2 be the sum of the other roots. Now, x1+x2=−1 and, if you multiply x1 and x2,
remembering that 17=1, the product is 4(x1+x2)=−4 so they are roots of x2+x–4 and
can be constructed. The work of multiplying them can be reduced by using Euler's
formulae since
x1=2(cos +cos2 +cos4 +cos8 )
with a similar formula for x2.
The next step in the construction would be to split x1 above into the sum of
x 1= + 13+ 16+ 4
and x 2, the sum of the other terms.
The reason why this works depends on some field and group theory. The
automorphisms of the extension field that leave the rationals pointwise fixed form a
group called the Galois group. These automorphisms leave the set of roots of a rational
polynomial unchanged so they must permute the roots. An automorphism is then
uniquely defined by the image of because, if it takes to k, it must take n to nk to
preserve multiplication. Consider, then, the automorphism that takes to 3. It will take
each root to it's third power so repeating it 16 times gives
( 3 9 10 13 5 15 11 16 14 8 7 4 12 2 6)
Therefore the group is cyclic. If the rationals are extended by a suitable square root, the
automorphisms that leave this extension pointwise fixed should be the subgroup of
order 8, whose elements form the cycle
( 9 13 15 16 8 4 2)
Since the subgroup should permute the roots in this cycle, it will leave a symmetric
function of them fixed which is why x1 was chosen.
The method works for any regular polygon with a Fermat prime number p of vertices
because the corresponding Galois group is always cyclic. To see why, suppose that m is
the least positive integer such that pnm=1 for all n. If m p−1 then the polynomial
xm−1 would have more roots than its degree so m=p−1 and there must be an element of
order p−1.
A more complete account can be found in, for instance, N. JACOBSON, Basic Algebra
1 or Conjecture and Proof.
References
1. N. Jacobson, Basic Algebra 1, Dover, 2009 (second edition)
2. M. Laczkovich, Conjecture and Proof, MAA, 2001
Angle Trisection using Limacon of Pascal Avni Pllana
Click here to download the PDF document of this paper
A dozen of angle trisection methods using limacon of Pascal[1] are known, and one
more such a method is shown in Fig.1.
Fig.1
Let be the given angle to be trisected. We draw line OC which bisects
, therefore . Line OC intersects the limacon ( red loop ) at
point D.
Limacon of Pascal[2] is defined in the polar coordinates by the equation
We use only the little loop of the limacon (1) , enlarged by factor 2, mirrored with
respect to the y-axis, and shifted 1 unit to the right along the x-axis. Such a transformed
limacon is defined in the rectangular coordinates by the following parametric equations
where , and is the bisector of the given angle to be
trisected.
At point D on both sides of line OC we draw lines DE and DF which make with
line OC . Points E and F trisect , and is equilateral.
In order to prove the above statements it suffices to show that . Let us
denote by , and by . The Law of Sines for yields
From equations (2) and relation , follows
From (3) and (4) follows
and finally we obtain .
The presented trisection method can be generalized for the angle n-section, see Fig.2.
Fig.2
In Fig.2 angle is the half of the vertex angle of a regular n-gon. The Law of
Sines for in Fig.2 yields
From (5) and Fig.2 we obtain following equations
Next we show that equations (6) represent a special case of an epitrochoid[3]. An
epitrochoid is defined by the following parametric equations
For the special case a = 1, and h = a+b, from (7) follows
Making in (8) the substitution
we obtain
Comparing (6) and (10), it follows that (6) can be obtained from (10) for
, or
and by scaling (10) with the factor
[1] Loy, J. "Trisection of an Angle" http://www.jimloy.com/geometry/trisect.htm
[2] Eric W. Weisstein. "Limacon" http://mathworld.wolfram.com/Limacon.html
[3] Eric W. Weisstein. "Epitrochoid" http://mathworld.wolfram.com/Epitrochoid.html
See also:
Avni Pllana "Approximate Angle Trisection and N-Section"
http://approxnsect.webs.com/
Avni Pllana "Approximate Angle Trisection" http://approxtrisect.webs.com/
Avni Pllana "Approximate Construction of Heptagon and Nonagon"
http://heptanona.webs.com/
Avni Pllana "Three Concurrent Lines" http://threeconclines.webs.com/
Avni Pllana "A Generalization of Ceva's Theorem for Tetrahedron"
http://tetraceva.webs.com/
Avni Pllana "A Generalization of the Nagel Point" http://mathnagel.webs.com/
Avni Pllana "An Interesting Triangle Center" http://interestingetc.webs.com/
Avni Pllana "A Derivation of Mollweide Equations" http://mollweide.webs.com/
Avni Pllana "Miscellaneous Results on Tetrahedron" http://misctetrahedron.webs.com/
Avni Pllana "A Proof of Pythagoras' Theorem"
http://mathforum.org/kb/servlet/JiveServlet/download/129-2020419-6926265-
581226/ProofPyth.pdf
Avni Pllana "Some results on tangential triangle"
http://mathforum.org/kb/servlet/JiveServlet/download/129-2210946-7316515-
658325/tangential_triangle.pdf
Avni Pllana "Some results on side-bisector reflected triangle"
http://mathforum.org/kb/servlet/JiveServlet/download/129-2249511-7424086-
673765/side_bisector_triangle1.pdf
Avni Pllana "Circle Connections"
http://mathforum.org/kb/servlet/JiveServlet/download/129-2378278-7815459-
759175/Circle_Connections.pdf
ANGLE TRISECTION
Angle trisection is one of the classic problems in mathematics. It is an
extremely essential factor concerning the precision of calculations.
Human being has not achieved finding a general solution for that by this
day. Since ancient times, it has been the aspiration of every
mathematician to be able to trisect an arbitrary angle. Nevertheless some
high profile intellectuals have recently closed the door on this crucial
problem by assuming it impossible based on weak inductions, boosted by
inability in finding a solution; bare theories without a slight clue
regarding an authority over straightedge and compass constructions. At
this moment, those theories are encountered with a critical challenge in
confrontation with the foundational geometrical theorems and
principles.(1)
We draw the circle with the center S and a desired radius. We draw the
two perpendicular diameters FR and GT and the bisector of the ninety
degree arc GSF and extend it so that it crosses the circumference of the
circle at the points R and M.
We draw the chord GF so that the bisector is crossed at the point H. The
central adjacent angles GSR and FSR are equal to one another and the
intercepted arcs. We
connect the point M to the points G and F. The inscribed adjacent angles
GMR and FMR are formed which are also equal to one another with the
measure of a half of the intercepted arcs.
We suppose the desired point D on the bisector (the axis of symmetry)
and connect that to the points G and F. The internal adjacent angles
GDR and FDR are also equal to each other since the two right triangles
GDH and FDH are equal one another. This is also the case for the
external adjacent angles GNR and FNR with the vertex outside of the
circle.
It is a very important and noticeable point that not only the central and
the inscribed adjacent angles with equal intercepted arcs are equal to
each other, also they are equal to one another as long as the vertex is
anywhere on the axis of symmetry (the bisector of the sum of the adjacent
angles).
The following three significant principles are concluded at this point:
A – The equal adjacent angles intercept equal arcs as long as the vertex
stands on the axis on symmetry.
B – The adjacent angles intercepting equal arcs are equal to one another
as long as the vertex stands on the axis on symmetry.
C – The proportions of the adjacent angles are constantly retained
through shifting the vertex along the axis of symmetry (locus).
We draw the circle with the center S and a desired radius. We draw the
two perpendicular diameters GT and FR and the bisector of the ninety
degree arc GSF and extend it so that it crosses the circumference of the
circle at the points R and M.
Using the still compass, we divide the ninety degree chord GF into three
equal thirty degree arcs GP, KP and KF.
The three central
adjacent angles GSP, PSK and KSP are equal to one another.
Now we suppose the desired point D on the locus and connect that to the
points G, P, K and F. The measure of the internal angle GDP is twice as
that of the internal angle PDR since the proportions of the adjacent
angles are constantly retained through shifting the vertex on the locus.
This time the ratio is two to one.
Therefore we conclude that the two internal adjacent angles GDP and
PDK are equal to one another. Thereby the three internal adjacent angles
GDP, PDK and KDF are equal to one another. This is also the case for
the external adjacent angles with their vertex at the point N.
We can also look at it this way: The three central adjacent angles GSP,
PSK and KSF are equal to one another. We connect the point M to the
points G, P, K and F. The inscribed adjacent angles GMP, PMK and
KMF are formed which are also equal to one another with the measure of
one half of the intercepted arcs.
The three adjacent
angles are equal at the center and on the circumference. Therefore the
countless internal angles enclosed in between the two trisected central
and inscribed angles with the common intercepted arcs, have to be
trisected as long as the vertex stays on the axis of symmetry.
Ultimately we can reason this way as well: The measure of the angle GDP
constantly equals that of the arc GP plus that of the arc YZ divided by
two, which is the same as the arc PK plus the arc XY divided by two,
which is also the same as the arc KF plus the arc WX divided by two.
It can not possibly be imagined that
someone acquainted with the basic principles in geometry, rejects these
sorts of reasoning. The underlying principles of geometry are more
reliable than any other deficient derivative method, now confronted with
a serious challenge since the classic problem of angle trisection is solved
generally, using straightedge and compass.(2)
The proportions of the adjacent angles are constantly retained through
shifting the vertex along the axis of symmetry.
(1) For further clarification, you may refer to the article “A Shot In The
Dark”.
(2) Considering the previous restricted trisections, inductive reasoning
also reconfirms the possibility of general trisection. An action occurred
once, can take place for n times.
A SHOT IN THE DARK
Geometrical theorems are the prototypes of many of the topics in
mathematics. Geometry is built on the foundation of superposition. It is
the ultimate reasoning in geometry. The instruments are straightedge and
compass exclusively.
The operation range of these two instruments is very restricted since one
sketches straight lines and the other one sketches curved lines. What
expands the range of operations, is deliberation in geometrical subjects
and creativity in using these two instruments towards constructing
calculable figures and discovering the associated principles.
The circumference of the circle is divided into three hundred and sixty
equal sections. Nobody knows about the procedure of the division since
no reasonable method has been reported from any reliable source.
Supposing that we can achieve sectioning a seventy two degree arc out of
the circumference by inscribing a regular pentagon and subtracting a
sixty degree arc from that, we come up to the twelve degree arc. Through
further bisecting the variation, we reach to the three degree arc.
Experiments have shown us that there is no way to get to the arc unit
which is the one degree, by the possessing angles such as the sixty degree,
forty five degree, fifteen degree… through the basic operations. More
clearly, we can not attain the two degrees, four degrees, five degrees…
and the chords either.
Consequently the process of calculating a major extent of the
trigonometric values is totally obscure. The issue inevitably concerns the
credibility and precision of the values which have been constantly relied
on. Through what reasoning have we managed to determine the sides of
the countless triangles that get inscribed in a half of the circle other than
generalization of our limited geometrical acquaintance? Considering that
geometry is the applied mathematics, how have the overlooked slight
errors affected the accuracy of operations in large scales?
Without the essential creativity and comprehension mentioned above, we
have no choice but to insert the presumption factor somewhere over the
course of our calculations and artificially expand conjectures. Struggling
to bypass the original complication, we are trapped by the sequence of
presumptions. The shadow of ignorance is constantly over us until we put
an end to the geometrical challenge by finding the ultimate geometrical
solution.(1)
Arithmetic and geometry are two topics which are absolutely different by
the nature. The former is an abstract mental concept and the latter is a
tangible dynamic phenomenon. A length segment does not naturally
allocate any number to itself. It is we who define geometrical units and
call them by different names. In arithmetic, we only have one number
three, but in geometry, any line segment can be called three. We always
simplify and omit the number one in algebra, but can we overlook a
length segment in geometry? The question is, to what extent can
operations performed on one entity, actually apply to another one?
Over the time, we have been constantly and inattentively crossing the line
by diverting and switching between the principles of these two discrete
entities in our procedures. Such an approach has resulted in making
serious mistakes and also giving rise to math phobia by confusing
students and enthusiasts. For instance, we have frequently taken a line
segment as a ratio and vice versa. Addition, subtraction, and division can
somehow be simulated in geometry, but how do we to multiply a line
segment by another line segment? Is it possible to suppose a square root
or a negative number for a length segment? How could a mere calculation
method such as algebra or trigonometry which is an offspring of
geometry itself, independently determine an original concept in
unexplored realms of geometry?
Let’s go over a couple of illustrations:
As we know, we can inscribe a
regular hexagon inside the circle by using the radius. Halving the radius,
we attain the regular dodecagon… and so on.
Therefore we conclude that as we proceed with dividing the diameter, the
number of the sides of the regular inscribed polygon increases. This way
we induce the following proportion:
Accordingly in order to attain the regular pentagon or the
regular heptagon, we use the fractions 5/6 and 7/6.
Applying superposition however, the circle does not correspond to the
algebraic induction and noticeably follows its own path. In other words,
the proportions of the cords do not vary by the conceived regular rate as
it is the case for the arcs. This example fairly demonstrates the essence of
geometrical challenges.
The most critical issue in many fields, has been developing self executing
methods without essential geometrical causes and assuming that they can
independently operate in any field regardless of the associated
parameters and principles; derivative methods taken for granted while
there is absolutely no evidence of a binding aspect regarding an
interference or convergence to geometrical phenomena such as lines and
arcs by any means.(2)
According to the next figure we have:
This way the following quadratic equation is generated:
Thereby we can determine the lengths of the line
segments GE and GC.
This method which has been initiated based on squaring the rectangle,
has been ironically applied in the case of trigonometric equations. In this
example, even though we input the data, the equation results in a pair of
numbers as x. Failing to anticipate the obscurity of dual answers, the
equation implies x2 to be equivalent to DF2 which equals 6 since both of
them are the roots multiplied according to the theorem. To make a long
story short, such false impressions reveal that algebra is blind and does
not have the capacity of leading us anywhere independently, unless we
take the compulsive steps mentioned earlier.(3) What emphasizes this
fact, is merely the possibility of algebraic sophistries, in spite any
justification. For instance, supposing a and b equal to 2 we can write:
Let’s not forget that failing to notice the
distinct line between geometry and arithmetic as mentioned, complicates
the issue even further. Note that in the previous example, the line segment
with the length of one unit, has a functional role throughout the entire
operation and can not get simplified and omitted.
Here is another example of the miscomprehensions. According to the
figure one can write:
Consequently we have:
That makes plane identical to length!
These sorts of irrationalities and misconceptions such as transcendence,
are evidences of being on the wrong track due to neglecting the
authenticity and superiority of the geometrical substances.(4) It has been
said that certain numbers can not be constructed. Let’s complete that
statement this way: It is absolutely impossible to construct numbers. A
number is an incorporeal nonfigurative notion which can only be
perceived conditionally or conventionally. Only under certain
circumstances, lines can appear as numbers.(5) Nobody is perfect!
With regards to the reliability and the solidity of the foundational
principles of geometry, how do we justify and eliminate the occurred
contradictions?(6) Through a little bit of concentration, one can sense the
independence of geometry from algebra. The pathless features indicated
in the next figure, highlight certain aspects of the nature of geometry;
substantial irregularities pointing out that an abstract algebraic
approach to a geometrical phenomenon is practically a shot in the
dark.(7)
To suppose that it is the time to close the deal on plane geometry having
entirely explored and illustrated every possible aspect, is an evident
deceit. There still remain many unsolved mysteries and crucial challenges
to be taken care of in this primary field; essential for taking the further
steps in mathematics.
Should there be further concerns, you may refer to our publications.
(1) Developing inscribed and circumscribed regular polygons in order to
calculate pi, unconditionally relies on trigonometry which is subject to
angle trisection. Yet we would have to compromise and deceive ourselves
into admitting that the cords and the curves can superpose onto each
other if they are supposed extremely small. As it can be perceived,
methods of these sorts are desperate measures in order to go around and
skip the principal problem.
(2) Relying on the series and the cubic equations in the case of calculating
pi and trisecting the angle, are forms of abduction. Retroductive
reasoning is not acceptable in mathematics which is considered as the
infrastructure of our knowledge.
(3) Resorting to alternatives such as series in order to find the value of pi,
is simply a modification of the original problem; the algebraic equation of
π = C/D in which only one out of the three components is determined.
(4) Lindemann has considerably come to the conclusion that no sequence
of algebraic equations can determine the value of pi.
(5) Algebraically, irrational values are ultimately undetermined, while the
beginning and the ending points of any line segment are perfectly
demonstrated in geometry. Therefore any supposed line segment can be
equivalent to the circumference of a circle, independent from any
conventional value. In other words, there exists no irrational measure in
geometry. It is merely the deficiency of algebraic representations when it
comes to geometrical substances that inevitably proposes ideas such as
irrationality, transcendence…
(6) Galois theory on abstract algebraic methods establishing the
possibility of certain straightedge and compass constructions
(7) For instance, the circumference of the circle is constantly closer to the
perimeter of the circumscribed polygon by an absolutely unknown
measure. Therefore there has been a significant probability of converging
to an absent circle through the preceding methods.
Angle Trisection
القدامى بمسألة تثليث الزواية وكانت احدى اهم المسائل الهندسية الى جانب مسألة تربيع اهتم اليونانيون
الدائرة]م[وتضعيف المكعب. والمقصود بتثليث الزاوية طريقة عملية لرسم مستقيم يقسم الزاوية الى جزئين
احدهما ثلث الزاوية. مسألة تنصيف الزاوية باستخدام الفرجار والمسطر )الغير مدرجة( مسألة بسيطة ومحسومة
ر كما هو في حالة التنصيف. فيما بعد اثبت . من الطبيعي بعد ذلك ان يتم التفكير في تثليث الزاوية , ولم يكن األم
وقدم االثبات عليه Gaussاستحالة تثليث الزاوية بشكل عام باستخدام حافة مستقيمة وفرجار, ذكر ذلك جاوس
م. 7381في العام Wantzelوينزل
حاالت خاصة من الزوايا]م[يمكن تثليثها بالمسطرة الغير مدرجة والفرجار. مثال في حالة الزاوية القائمة يمكن
تثليثها من خالل رسم دائرتين متساويتين في طول نصف القطر]م[
, مركز األولى على رأس الزاوية ومركز الثانية
على نقطة تقاطع]م[الدائرة األولى مع أحد ضلعي الزاوية. المثلث
]م[متطابق األضالع , ألن ABCالناشيئ
درجة بواسطة حافة مستقيمة 06عه أنصاف أقطار. بالمناسبة هذه الطريقة تضمنت إمكانية رسم الزاوية أضال
درجة ال يمكن رسمها بهذه الطريقة. 06وفرجار فقط وهذا ليس ممكن لي زاوية بشكل عام , فالزاوية
باستخدام مسطرة مدرجة وفرجار أمر ممكن إذا استبدلنا المسطرة بأخرى مدرجة فإن تثليث الزاوية )اي زاوية(
على الدوام. وهناك عدة طرق قدمها اليونانيون القدامي للوصول الى زاوية تعادل ثلث الزاوية المعطاه ونظرا ألن
هذه المسألة كانت من اهم المسائل في ذلك الوقت فسنعرض لها بشيء من التفصيل.
وتسير على النحو التالي: Hippocratesقراط الطريقة األولى : وتعود الى اليوناني ابو
ثم نكمل Dارسم العمودي على الضلع المقابل ويتقاطع معه في A. من 8x, وقياسها ABCليكن لدينا الزاوية
رسم المستطيل]م[
ADBE مد الضلع .EA بشكل كاف ليتقاطع معه المستقيمBH في النقطةH والذي رسمناه ,
. FHمنتصف القطعة G. لتكن FH=2BAبحيث
حيث طول المتوسط في المثلث القائم يساوي نصف طول الوتر]م[
نستنتج ان
AG = FG = HG
. كذلك AHيوازي BCألن فإن إذا فرضنا أن
ABGألن المثلث . إذا AGHألنها خارجية في المثلث وبالتالي
. BHبواسطة المستقيم ABCمتطابق الضلعينوبهذا استطعنا تثليث الزاوية
الذي عاش قبل الميالد بقرنين من الزمان والفكرة Archimedesالطريقة الثانية: وتنسب الى ارخميدس
مشابهة لما سبق لكن باالعتماد على الدائرة لنتمكن من رسم زاوية اخرى تعادل ثلث الزاوية المعطاه.
في BCعلى الدائرة ارسم مستقيم يلتقي A. من Bزاوية اختيارية. ارسم الدائرة التي مركزها ABCإذا كانت ف
مساو لطول نصف القطر وهذا أمر ممكن . اآلن قياس EDواختر هذه النقطة بحيث يكون طول القطعة Eالنقطة
.ABCيساوي ثلث قياس الزاوية AECالزاوية
مثلث متطابق الضلعين. وبالتالي DEBألن إذا ض أن للتأكد من ذلك افر
مثلث متطابق DABألن ألنها خارجية في هذا المثلث. وعليه فإن
. , إذا الضلعين. إذا
. . واضح أن EAمواز للمستقيم BPاآلن ارسم )باستخدام الفرجار والمسطرة(
الذي عاش في الفترة ما بين الثالث والرابع قبل الميالد Hippiasالطريقة التالية تنسب الى األغريقي هيباس
ق. م , ولتفريق بينه وبين شخص آخر يحمل نفس االسم يقرن اسمه باسم البلدة أو 056حيث يعتقد انه ولد عام
( . Hippias of Elisالمدينة التي ولد فيها فيقال )
ما لتقسميها إلى اي عدد من األجزاء المتساوية. الطريقة المنسوبة الى هيباس ليست فقط لتثليث الزاوية وان
وربما يكون هو أول منحنى عرف في Quadratrixاعتمد هيباس على رسمه لمنحنى يسمى كوادراتركس
الرياضيات بعد الدائرة ويمكن توصيف هذا المنحنى كما يلي:
يحيط بقوس ABCDارسم مربعا ]م[ ABكما في الشكل. إذا تحرك نصف القطر AEDمن دائرة )ربع دائرة(
تقع على F' فإن نقطة التقاطع B'Cبنفس النسبة ووصل الى الموضع BCوتحرك الضلع AEالى الموضع
الناتجة من التقاطع الناتج من Fمنحنى الكوادراتركس. إذا منحنى الكوادراتركس هو المحل الهندسي للنقطة
بنفس BCبنسبة معينة من القوس والوضع النهائي لحركة الضلع ABالوضع النهائي لحركة نصف القطر
. ABالنسبة من الضلع
m:nبنسبة Pكما في الشكل واقسمها بواسطة النقطة FHالقطعة قم برسم m:nبنسبة EADلتقسيم الزاوية
لنحصل على تقسيم QA. من هذه النقطة نرسم Qنرسم خط افقى يالقي الكوادراتركس في النقطة P. من
بالنسبة المعطاه. EADللزاوية
ومعادلة الكارتيزيةالجزء الذي رسمه هيباس هو جزء صغير من منحنى يحمل اليوم نفس االسم "كوادراتركس"
.ومعادلته القطبيه
طريقة ابولونيوس: ابولونيوس , واحد من اعظم رياضي االغريق, وللتفريق بينه وبين علماء اغريق آخرين
( نقل لنا آخر منجزاتهم في CONICS( وكتابة المخاريط ) Apollonius of Pergaبنفس اإلسم يطلق عليه )
parabola, ellipse and hyperbolaة , وأول ما من قدم المصطلحات علم المخروط والقطوع المخروطي
والتي تعني القطع المكافيء , القطع الناقص و القطع الزائد.
طريقة ابولونيوس في تثليث الزاوية نقلها آخر الرياضييين االغريق المعروفين ويدعى بابوس االسكندرية
Pappus of Alexandria قطع المكافئ ونعرضها فيما يلي. . تعتمد الطريقة على ال
نصف الزاوية PABوالذي يجعل الزاوية P. المحل الهندسي للنقطة ABفي القسم األيسر من الصورة قطعة
PBA هو عبارة عن قطع مكافئ بؤرتهB ودليله هو العمود المنصف للقطعةAB.
. حيث نبدا برسم دائرة مركزها AOBالقسم األيمن من الصورة يوضح طريقة ابولونيوس في تثليث الزاوية
والذي يقاطع 0واختالفه المركزي B. ارسم القطع المكافي الذي بؤرته A,Bراس الزاوية وتقاطع ضلعيها عند
طية في الدائرة واألولى نصف الثانية محي PBA ,PABكما هو واضح في الصورة. كال الزاويتين Pالدائرة في
وكل واحدة منهما تساوي نصف الزاوية المركزية التي تحصر القوس نفسه المحصور بين ضلعيها. إذا الزاوية
POB تعادل نصف الزاويةPOA اي أن .PO مستقيم تثليث للزاويةAOB.
ة والتي عليها نقطتين تحددان مسافة طريقة نيكوميدس : في طريقة ارخميدس مر معنا كيف استخدمنا المسطر
Nicomedesثابتة وكيف احتنا آنذاك ان تبقى احدى النقط ثابتة على الخط المستقيم . حاول اليوناني نيكوميدس
XYالذي عاش في القرن الثاني قبل الميالد, وضع مسألة تحريك المسطرة مع بقاء نقطة ثابتة منها على مستقيم
باستحداثه لمنحنى الكنشوئيد وهذه التسمية مشتقة من كلمة يونانية تعني الصدفة أو في وضع نظامي أو مقنن
المحارة .
ذات ACفي الصورة يبين منحنى الكنشوئيد الذي قصده نيكوميدس حيث المسطرة XYالمنحنى أعلى المستقيم
Bبحيث ال تغادر العالمة عند Aعليها والمسطرة تتحرك حول النقطة B,Cأو قل العالمتين BCالمسافة الثابتة
يمثل أحد أوضاع هذه الحركة AEالمستقيم والعالمة األخرى من المسطرة هي التي ترسم لنا المنحنى. المستقيم
عندها العالمة التي كانت Eوالنقطة Bتوجد العالمة على المسطرة التي كانت عند النقطة Dحيث عند النقطة
.DE=BC. إذا Cعند النقطة
واقعة على امتداد المسطرة )على Fهو المحل الهندسي الذي ترسمه نقطة XYالمنحنى الواقع أسفل المستقيم
أسفل المستقيم BCيساوي الطول XY( والتي بعدها عن العالمة من المسطرة التي على المستقيم AEالمستقيم
واآلخر BCأصغر من الطول XYعن Aيوجد منحنيين أحدهما ذو العقدة هو ما نحصل عليه عندما يكون بعد
. ربما لم يناقش الجز السفلي من المنحنى قديما وعلى العموم يطلق BCأقل من XYعن Aعندما يكون بعد
على هذين المنحنيين )أعلى الستقيم واسفل المستقيم( معا منحنى الكنشوئيد. وباختصار شديد لفهم الشكل العام
, اآلن حرك XYوضع عالمة في منتصفها , وخذ نقطة خارج المستقيم لمنحنى الكنشوئيد , احضر مسطرة
المسطرة بالدوران]م[
, طرفا المسطرة أثناء هذه الحركة يرسمان المنحنى العلوى والسفلي من الكنشوئيد. Aحول
استخدم نيكوميدس هذا المنحنى في حل مسالة تثليث الزاوية , مع العلم انه من الناية العملية مسألة تحريك
المطلوب أسهل عمليا من رسم كنشوئيد وتثليث الزاوية المسطرة في طريقة ارخميدس حتى نحصل على الوضع
بواسطته.
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Trisection of an Angle
© Copyright 1997 and 2003, Jim Loy
Under construction (just kidding, sort of).
This page is divided into seven parts:
Part I - Possible vs. Impossible
Part II - The Rules
Part III - Close, But No Banana
Part IV - "Cheating" (violating the rules)
Part V - "Cheating" (using other tools)
Part VI - "Cheating" (using curves other than circles)
Part VII - "Cheating" (choosing an arbitrary point)
Part VIII - Comments
Part I - Possible vs. Impossible
In Plane Geometry, constructions are done with compasses (for drawing
circles and arcs, and duplicating lengths, sometime called "a compass") and
straightedge (without marks on it, for drawing straight line segments
through two points). See Geometric Constructions. With these tools (see
the diagram), an amazing number of things can be done. But, it is fairly
well known that it is impossible to trisect (divide into three equal parts) a
general angle, using these tools. Another way to say this is that a general
arc cannot be trisected. The public and the newspapers seem to think that
this means that mathematicians don't know how to trisect an angle; well they don't, not
with these tools. But they can estimate a trisection to any accuracy that you want.
Certain angles (90° for example) can be trisected. A general angle cannot. In fact, a 60°
angle cannot. It is usually much more difficult to prove that something is impossible,
than it is to prove that something is possible. In this case, mathematicians had to show
just what kinds of lengths could be constructed. And then they could show that other
lengths could not be constructed, because they were not the right kinds of lengths.
What can be done with these tools? Given a length a, we can multiply this length by any
integer, and divided it by any integer. Together, these allow us to multiply this length by
any rational number. Given two lengths a and b (and sometimes a unit length), we can
add them together, subtract them, multiply them, divide one by the other. Given a length
a and a unit length, we can find the length that is the square root of a. These are the only
things we can do, with these tools. This has been proven, but I won't do that here. We
can do these operations in many many combinations of ways [like sqr(a+sqr(b+sqr(c))),
for example, where sqr() is the square root function]. But, if a construction involves a
length that cannot be expressed by some combination of ratios and sums and square
roots, then it cannot be constructed.
Towards a proof: If we begin with two points, the distance between them can be called
1. With those two points, and compasses and straightedge, we can produce any integer
(as we can do addition and multiplication with these tools). Since we can perform
division with compasses and straightedge, we can produce any fraction (rational
number). We can easily construct the square root of any number that we have already
constructed, including earlier square roots, using the Pythagorean theorem. We can
combine rationals and square roots with these tools. What more can we do? The answer
is, "Nothing." We can draw lines through points and we can draw circles. These involve
addition, subtraction, multiplication, division, and taking the square root (Pythagorean
theorem). OK, I can create any number of a certain kind (rationals and square roots and
their combinations) with these tools. I am now considering some points and I am going
to create a new point. I can do this by using previous lengths of our certain kind and
drawing arcs which intersect, or by drawing lines which intersect with lines or arcs. We
find that if the positions of our points are represented by numbers of our certain kind
(coordinates of some kind), and we find our new point using arcs with radii of our
certain kind, then we always produce new lengths of the same certain kind. If we
create a new point by the intersection of two lines or an arc and a line, which were
based upon numbers of our certain kind, then we always produce new lengths of the
same certain kind. We add no new numbers of any kind (like cube roots), by
connecting points or drawing circles. I still have not proved that, but we are much
closer. Our certain kind of number has other names, including "ruler and compass
number."
One of the classic problems that cannot be done, is to construct a square the same area
as a given circle. This is called squaring the circle. This problem amounts to taking a
unit length and constructing a length equal to pi. There is no way, in a finite number of
steps, to represent pi in the form of ratios and sums and square roots, as was shown in
1882, by Ferdinand Lindemann. Another classic problem is to construct a cube that is
twice the volume of a given cube. This is called doubling the cube. This problem
amounts to taking the cube root of a length. This too cannot be done in a finite number
of steps, using the above operations, as was shown by Wantzel in 1836.
The trisection of an angle involves the solving of a cubic equation, something that
cannot be done, in general, using the above operations. This too was shown by Wantzel
in 1836. After Wantzel published his discovery, Gauss claimed that he had proved it in
about 1800, but had never published. A surprising result of this was that a 60 degree
angle cannot be trisected (in other words, a 20 degree angle cannot be constructed), and
so a regular nonagon (a nine-sided polygon) cannot be constructed, as they involve the
construction of a cube root. Karl Friedrich Gauss (Gauß) then showed just which
regular polygons could be constructed, and he explained how to construct a regular 17-
gon.
In the diagram on the right, the length of the trisecting chord x is
the root of the equation (in which x^3 means x cubed) AB=3x-
x^3 (discovered, or maybe rediscovered from Arab sources, by
Pitiscus about 1600). In general, a cubic equation cannot be
solved by construction, with compasses and straightedge. This
equation can apparently be deduced from the triple angle
equation: sin(3a)=3sin(a)-4sin^3(a), a cubic equation which can
be derived from the sine of the sum of two angles formula [sin(a+b)=sin(a)cos(b)-
cos(a)sin(b)].
I said above that the trisection (using compasses and unmarked straightedge in a finite
number of steps) of a general angle was shown to be impossible, back in the 19th
century. While it is difficult to reproduce that proof (I may attempt it some day), that
should be good enough to disprove any future attempted trisection. We shouldn't have
to disprove each attempted trisection as they come up. They are already disproved. Of
course these attempted trisections may be very close to real trisections (within small
fractions of degrees). Some are very close for small angles. Others are very close for
other angles. Maybe they are close enough for practical purposes. But none of them can
be exact.
I received this email. I guess I am stifling innovation when I point out that some things
are actually impossible:
I read your article on angle trisection. You seem intelligent enough, but possibly too
pompous to be a mathematician. I solved this problem when I was 13. You certainly
don't have to disprove every new idea that comes along, you probably don't even need
to see most of them. But to rest on an incomplete 200 year old theory with holes in it
and thereby attempt to stifle innovative thought is exemplary of the kind of arrogance
which allows fools to call themselves educated. Good day.
Can't pompous people be mathematicians? Darn. I may indeed be pompous in other
ways, but I was just reporting what geometry of today says. We have a proof that angle
trisection (using certain simple tools) cannot be done. You cannot disprove a proof just
by calling it outdated. In fact, you cannot expect people to believe you when you say
you have done something considered impossible, without any proof whatsoever. Thank
you for informing me that you can do the impossible; forgive me for not believing you.
I told him, "Of course you did not solve this problem when you were 13." He seemed to
take this as a cruel blow, and vowed to never write to me again. Too bad.
Stifling innovation? I may have a different temperament from the person who sent me
that email. Personally, I am intrigued as all get out by a problem that I know to be
impossible. How do we know it is impossible? That must be hard to prove. Why doesn't
this idea work or this other idea? I've learned some geometry by working on these
impossible problems.
Other people have sent their trisections to me. Fortunately, most of the methods have
been fairly easy to prove to be incorrect. Some have been difficult to understand at first.
But, once I could use the method to draw the proposed trisection of an arbitrary angle,
the three angles have never been the same. Usually, it was easy to see that the trisection
method was not working, just by trying to trisect a large angle (maybe well over 90
degrees).
The difficult ones are the ones that require cheating (like Archimedes method below).
One person said he could trisect an angle using an ellipse. Well, you may be able to
draw some points of an ellipse, using compasses and straightedge, but you cannot draw
the ellipse with these tools. Anyway, I have never heard that an ellipse could help trisect
an angle, so I am somewhat skeptical.
All these trisection attempts are experimental geometry. You try a method that might be
close to a trisection, and try to measure the angles to see if they are the same. They seem
to be the same angle. Ah, I've found a trisection. Wrong, that is not how geometry
works. Mathematicians prove everything they do, because you cannot ever be sure that
something is true, unless you can prove it. And trisection is a good example of this.
Four books which prove that trisection is impossible with compasses and straightedge
are The Trisection Problem by Robert C. Yates, Geometric Constructions by George E.
Martin, Euclidean Geometry and Transformations by Clayton W. Dodge, and Ruler and
the Round: Classic Problems in Geometric Constructions by Nicholas D. Kazarinoff.
The algebraic theory concerning this is called Galois theory (a part of group theory),
and there are a lot of books about that. Also see:
Trisecting Angles and Squaring Circles Using Geometric Tools
Why Trisecting the Angle is Impossible
Math Forum's "Impossible" Geometric Constructions
Of course the proof is difficult. See An Analogy, at the bottom of this page, to get a
simplified flavor of what the proof is actually about.
Part II - The Rules
The Rules: In using the classic tools (straightedge (and pencil) and compasses), it is
legal to use them only in the simple and obvious ways. We deduce the rules of this
game mostly from the constructions of Euclid. These rules are seldom spelled out,
although we do usually spell out that we can't use marks on the straightedge. One of my
emails suggested holding the compasses against the straightedge while moving the
straightedge. This amounts to making a not so simple tool of the two pieces of
equipment, which is obviously illegal. As I said at the top, we use compasses for
drawing circles and arcs, and duplicating lengths, and we use a straightedge (without
marks on it) for drawing straight line segments through two points. We can do nothing
else. We cannot even use the compasses to verify if one distance is equal to another
distance. And we cannot eyeball a point and a line to tell if the point is on the line
(sometimes we have to go to great trouble to prove that a point is on a line). And, the
mistake which is repeated many times on this page, you cannot use compasses to
measure or duplicate the length of an arc on circles of different radii (see Part IV,
trisection #6).
You cannot use these tools to design other tools like the tomahawk shown below; in
other words, you cannot pick up parts of your drawing and move them around as tools
in their own right. While you can draw some points of a complicated curve with these
classic tools, you cannot actually draw these curves. We cannot fold up the plane like
origami.
Below, you will see bogus trisections which come within a hundredth of a degree of
being right. Isn't that close enough? You cannot measure 1/100 degree on your
protractor. And you can't even be that accurate with real compasses and straightedge.
Well, our crude diagrams are a poor imitation of the perfect constructions that we are
describing. When I get out my compasses and straightedge, and bisect an angle, the
result is inaccurate. But when I describe the bisection of an angle, by drawing arcs, what
I describe is perfect, with perfect lines and perfect arcs, using fictional perfect
compasses and straightedge. And I end up with a perfect bisection, regardless of the
crudity of my drawings. This mathematical perfection is also part of the rules. We
pretend that we are using perfect instruments which draw perfect lines and circles, so
we can then prove things about what we have constructed. So it makes no sense to
measure an angle with a protractor and say, "Yup, that's 20 degrees." We have to prove
it.
Our straightedge is often called a "ruler" (even in many books on geometry), but (if we
want to follow the rules) we must not use the markings on the ruler.
Part III - Close, But No Banana
Here we see some attempts at trisection which are close, but not
exact.
1. Trisect the chord: This diagram shows the most common
attempt at trisection. Points are marked on the sides of the angle,
an equal distance from the vertex. A line segment is drawn
between these two points. This segment is trisected (easy to do,
see next paragraph), resulting in the marking of two points. Lines are drawn from these
two points to the vertex. These two lines almost trisect the angle. It gets fairly obvious
that this doesn't work, if the angle is fairly large. For a proof that this is not a real
trisection, see below. Many other attempts at trisection are just this one in a more
complicated guise (also see below).
By the way, this diagram shows how to trisect a line segment
(AB). Through A, draw an arbitrary line. On this line mark off
an arbitrary segment, starting at A. Duplicate this segment twice,
producing a trisected segment AC (see the diagram). Draw the
line BC. Through the other two points on the trisected segment,
draw lines parallel to BC. These two lines trisect the segment
AB. You can use the same method to divide a segment into any
number of congruent (equal lengthed) smaller segments. Also see Geometric
Constructions. There are other ways to trisect a line segment.
Here is a proof that the attempted trisection is
not a real trisection. We are trying to trisect
angle A by trisecting the line segment EF. For
this proof, extend AC to B so that BC=AC, then
connect BE. Triangle ACD is an isosceles
triangle with congruent sides AC and AD
[Triangles ACE and ADF are congruent by SAS (see Congruence Of Triangles, Part I),
and that makes angle ACD=angle ADC]. Triangle BCE is congruent to triangle ACD
[SAS]. Angle ACD is an acute angle. Then angle ACE is an obtuse angle.
Now, let's assume that this actually is a valid trisection (for the purpose of finding a
contradiction, and proving that it is not a valid trisection). Then angle CAE is congruent
to angle CBE. Then triangle EAB is isosceles, and AE=BE [base angles are =]. Then
triangle ACE is isosceles [AC=AE]. Triangle ACE is congruent to triangle ACD [SAS
or other]. Then angle ACE=angle ACD, which is acute. But above we showed that ACE
is obtuse. So we have our contradiction. And our attempted trisection is not a valid
trisection.
From this proof, we can guess that the attempted trisection is really close when AE is
about the same length as AC. This happens when the angle A is small. But we knew that
already, because to trisect angle A, we need to trisect the arc EF, not the line segment
EF. Well, when angle A is small, arc EF is very close (in more than one way) to
segment EF. To see just how accurate this method is for small angles, see method #10,
below.
Also, it is apparently easy to prove (in the same diagram) that if the angle is actually
trisected (instead of the line segment EF), then CD is shorter than DF.
Here is another proof that trisecting the line segment does not trisect the angle. With
that method, try to trisect an angle of 60 degrees. You should find that the three angles
are measurably different. The middle angle is about 2 or 3 degrees larger. So we have a
counterexample. Which proves that that method is not a real trisection. Angles larger
than 60 degrees would make this proof even more obvious. Using this method to trisect
60 degrees, the smaller angle is 19.1066 degrees, which is 0.8934 degrees off. I
calculated the angle with eight digits of accuracy, and then rounded off to six digits.
Using this method on a 120 degree angle, we get 30 degrees (which may be exact?),
which is 1/4 of the angle, not 1/3.
Calling the angle that we are trying to trisect a, and the smaller angle that is our
estimated trisection b, here is an equation that I get, relating a and b: b = (sin a)/sqrt(9-
8sin^2(a/2)), where sqrt() is the square root function, and sin^2(a/2) is the sine squared
of a/2.
1a. The above method disguised: Here is an
attempted trisection which was sent to me by
email, a couple of years ago. I have redrawn it
(and added some lines to help with my
comments). It was originally done with
compasses only. Two equal segments are
marked off on each of the two rays of the
angle out to B and C. For the purposes of my
comments below, I have extended the rays out
to I and J. Then the midpoint between B and C is determined with compasses. Then we
mark off the two points between I and J, again using the compasses. These two points
were then claimed to trisect the original angle A. I drew red lines to them (from A). It is
easy to show (using the many small congruent triangles that I have drawn, some of
which are mirror images of others) that this construction trisects line segment IJ. And
that makes it equivalent to the attempted trisection #1 above, and is not a real trisection
of angle A.
1b. Disguised further: Here is another attempted trisection
that I received by email recently. Lines AB, AC and BC are
bisected. And then the various other lines are drawn. It was
hypothesized that points G and H trisect angle A. There are
a couple of parallelograms (BDFE and CDEF). And the
diagonals of a parallelogram bisect each other. So, G and H
are at the midpoints of DE and DF respectively. We can compare this diagram with the
previous diagram. Many of the same congruent triangles appear in both. It becomes
obvious that these two methods define the same lines (the ones that I have drawn in
red). It is not immediately obvious that this method trisects line BC. But comparing this
diagram with the previous one, it becomes obvious that this is the same trisection.
Again, this is not a real trisection of angle A.
1c. And further: Here, with angle AOB and
AO=BO, we bisect segment OB at C, then bisect
segment AC at D, which nearly trisects the angle.
This produces the same exact angle as method #1
above, and is close to a true trisection for small
angles. That should not be difficult to prove. So if
we try it on a 60 degree angle, we will get 19.11
degrees, which is 0.89 degrees off, better than some.
1d. Still further: Here, with angle AOB and AO=BO, we bisect segment OB at C, then
trisect segment AO at D (AD = AO/3). The intersection
of AC and BD (E) nearly trisects the angle. This
produces the same exact angle as method #1 above, and
is close for small angles. Again, trying it on a 60 degree
angle, we will get 19.11 degrees, which is 0.89 degrees
off.
1e. Even further: I got this by email, the
sender's initials are A. D. . We are trisecting
angle A. We make AB = AC, bisect AB at D,
and draw the perpendicular bisector EF of CD
at E. We draw the arc with center at C and
radius CD, intersecting EF at F. We bisect CF
at G, and draw DG. DG intersects EF at H.
The bisector of angle DHE intersects BC at K,
which supposedly trisects angle A. Instead, K
trisects segment BC (and CD and DE). You might want to show that, before reading the
next paragraph.
OK, if you draw segment DF, then triangle CDF is equilateral. That makes triangle
DHE a 30-60-90 right triangle, or half of a small equilateral triangle; I have drawn the
rest of the equilateral triangle faintly. DE and HK are then medians of the equilateral
triangle, and medians trisect each other (Euclid). So K trisects DE, and CD, and BC.
And this trisection method is a well-disguised form of method 1 above.
2. Geometric series: Here is another interesting attempt at trisection of an angle. Divide
the angle into fourths, easily done by bisecting twice. To this fourth, add 1/16 of the
original angle (1/4 of the fourth). Then add 1/64, and 1/256, etc. forever. We have an
infinite sum that adds up to 1/3 of the original angle.
The equation is: 1/3 = 1/4+1/16+1/64+1/256... This is just a Geometric Series. It is easy
to show that the sum is 1/3.
This construction is not something that you can actually do, because it takes an infinite
number of steps. Such an approach is not normally stated as being against the rules. But,
it is taken for granted that it is against the rules. Notice that my article said the various
impossible problems could not be done, "in a finite number of steps."
You can stop well before you have done infinitely many steps, and your approximate
trisection can be very accurate, as accurate as you want. Using the first three terms of
the series to trisect 60 degrees, I get 19.6875 degrees. The first four terms give 19.9219,
which is 0.0781 degrees off. We can, of course, use more terms to get more accuracy.
A similar trisection is to construct 1/4+1/12 = 1/3. Of course you would have to trisect a
1/4 to get 1/12. But it is a smaller angle, and the approximation is probably more
accurate, depending on the approximate method of trisection that you choose (see
method #10, below). 1/4+1/16+1/48 is also similar, where you have to trisect a 1/16 to
get 1/48. And you can do the same with any number of terms of the above infinite
series. It always involves trisecting some very small angle, and so is never exact. But it
can be as accurate as you want.
There are other infinite series which converge to 1/3, including other geometric series:
1/3 = 1/2-1/4+1/8-1/16+...
3. Mechanix Illustrated: Here is
a compasses and straightedge
method (which I have simplified)
that was apparently given in
Mechanix Illustrated, Feb. 1966.
It is really really close for angles
less than 90 degrees, within a few
hundredths of a degree. We are
trying to trisect angle AOB
(having drawn a circle about O with A and B on the circle). OC bisects the angle. Then
we draw two lines parallel to OC, through A and B. EC=CD, and D is the center of the
large circle. The attempted trisection is defined as shown. Probably the easiest way to
show that it is not a real trisection is to use it on some angle, and then deduce what the
three smaller angles really are, showing that the center one is not equal to the two outer
ones. Attempting to trisect a 60 degree angle, I find that the middle angle (FOG) is
20.0226 degrees, which is 0.0226 degrees off.
4. Arcs with the same chord: Some
attempts at trisection involve something
like this diagram (I've divided the
diagram into two pieces, upper and
lower, to make it clearer). We are trying
to trisect the larger red angle (LAM). We
have not yet determinded where the
points L and M are.
We estimate the trisection (somehow,
maybe by eyeballing or by some
complicated construction) and get the
small red angle (CBD) in the lower half
of the diagram, with its arc CD. We then
triple this angle, producing angle CBF
(which should be approximately the
same size as LAM). We then duplicate
the chord CF inside the angle LAM, producing the chord LM and fixing the points L
and M. The upper left part of the diagram shows how this chord can be duplicated. Then
we draw the arc LM, with center A. Then we duplicate the small circle centered at D,
and place this duplicate centered at M. This gives us the point O, on this circle and on
the arc LM, which very nearly trisects the angle LAM.
Chords CF and LM are of equal
length. But arcs CF and LM have
different centers and different
radii. Nevertheless, people seem
to think that the two arcs are of
equal length, because their
chords are of equal length. On
the right we see the area around
point O, blown up about 20 times. I have moved angle FBC so that chords FC and LM
exactly coincide (the green line here). As you can see, the two arcs do not coincide.
They are very very close (and D nearly coincides with O), but they do not coincide.
After all, it would be quite a coincidence if my estimated trisection fit the original angle
exactly. I think that it is fairly easy to show that if the chords are the same length, but
the radii are different, that the arc lengths are different, in situations similar to the above
situation. But you can see why mere measurement of the angles would imply that a
trisection had been accomplished.
A site on the WWW used this method to
trisect a 60 degree angle by trisecting a 45
degree angle outside of it. Here we have
my drawing of this. The red lines are the
60 degree angle and its attempted
trisection. The blue lines are the 45 degree
angle and its actual trisection. The
attempted trisection gives an angle of
19.87 degrees, which is 0.13 degrees off.
45 degrees is not very close to 60 degrees,
a better angle will produce better results.
5. A straight line?: I have
been corresponding with a
person (initials H. W. S. )
who came up with this
attempt at trisecting a 60
degree angle. In this diagram
(a 30-60-90 degree triangle),
point H is positioned so that
dividing the right angle C
into pieces of some
proportion will divide the
angle B into the same
proportion. If we position
point H just right, we can
bisect both angles. And if we
position H at some other place, we can trisect both angles. Since it is easy to trisect the
90 degree angle, then we can use H to trisect the 60 degree angle. We could also use
other positions of H to divide the angles into any other equal number of smaller angles.
This person had come up with a theorem which implied that all possible positions of
point H lie on a straight line (the line defined by A and the incenter of the triangle; see
The Centers of a Triangle). Using this theorem, it is then easy to trisect the 60 degree
angle.
Unfortunately, all possible positions of H do not lie on that line (in other words, this
person's theorem is false). In the first diagram above, I have drawn a green line
representing the path of H as angle BCH changes. And it appears to be a curve, as one
might guess. I think it can be proved that this curve is convex in one direction, and if so
no three points on that curve can lie on the same straight line. But this curve is very
close to being straight for much of its length, and so a person might be fooled into
thinking that it is straight. Pretending that the curve is a straight line, in the second
diagram above right, we find that angle EBC is 38.7940 degrees, which is 1.2060
degrees off.
This curve is a rather complicated trigonometric curve. A trigonometric curve depends
upon angles and lengths and trig functions, not just lengths as do the more familiar
curves (Conic Sections).
This attempted trisection can be used to attempt the trisection of other angles, by using
different triangles. Such attempts produce other curves which are also nearly straight
lines.
6. My method: I suspect that this method has
been used by someone else, in ancient times.
But, so far, I have seen no evidence of that. So,
for the time being, this one is my invention,
and it is extremely accurate.
We are trying to trisect angle BAC, and we
choose a point D (by some method, either
eyeballing it, or through any of the other
trisection attempts) which might trisect the
angle. We then duplicate this angle (BAD) twice, producing points E and F in the
diagram. Well, D was just a guess, and the real trisection point lies somewhere between
D and F (F can be on either side of D). In fact it is closer to D than it is to F. An
excellent guess is 1/3 of the way from D to F. A new point G, 1/3 of the way from D to
F is really close to a trisection, but it is not perfect.
What is wrong with G as a trisection point? Doesn't it seem like it should work? The
entire angle (BAC) is 3x (with x being the angle of our first estimate) plus a small angle
m: A = 3x+m. A true trisection is 1/3 that, exactly: A/3 = x+m/3. So if we add our first
estimate to one third of our small angle m, we should get an exact trisection. But we
chose point G by trisecting the chord DF, not the angle DAF. So we were just using
method #1 above, which we already proved was inaccurate. The length of segment DG
is close to the length of the arc DG, but it is never exact. We are using method #1 on a
very small angle, and so the result should be very accurate. We can use G as an
estimate, and repeat the above approximation, to get an even better approximation.
Using this method (with method #1 above as a first estimate) on a 60 degree angle, we
get an estimate of 19.9999276, which is only 0.0000724 degrees off, making this the
second best trisection construction that I have tested (see #25 below). We can't possibly
draw our lines with that kind of accuracy, so isn't that close enough to be called exact?
No, it's still not a real trisection, because it is not exact. We are assuming (pretending)
that we are using perfect tools here. This method is an extremely good estimate, but it is
still an estimate.
7. "New theory": Here we see an attempted
trisection from New Theory of Trisection by F.
C. (I have decided to use initials instead of
names, for people who think that they have
found the elusive trisection), a geometry teacher,
and he claims to be a mathematician. On the
back cover, he admits that the construction is
impossible, and then says that he has done it. We
are trying to trisect angle XOY. We draw circle
O, with radius 1, which determines point S on
line XO (on the other side of O from X). Draw
SO2 perpendicular to line OY. Bisect SO2 at M1.
Draw circle M1 with radius equal to segment
O2N, defining point M on line OY as shown.
Draw O1 as shown, so O1M = OM. Then angle
OO1S is the trisection of angle XOY. This
attempt is very accurate for angles less than 110 degrees. But there is a flaw in F. C.'s
proof. In particular, his theorem #11 is sloppily
done, and is false.
His method is exact for 90 degrees (and 0
degrees). Using simple trigonometry, a calculator
with trig functions, and the Pythagorean theorem,
testing his method on a 60 degree angle produces
an angle of 20.05512386 degrees, which is
0.0551 degrees off, close but no banana. He has
patented a tool based upon this method, and I
suppose it is pretty darned accurate. Of course
there are simpler tools that are more accurate (see
Part V below).
You might be interested in seeing how I
calculated that 20.05512386 angle. It's not very
difficult. On the right we see my diagram for a 60
degree angle. These lengths can be deduced
(mostly using the Pythagorean theorem):
OX = OS = 1
OO2 = 1/2
O2M1 = sqrt(3)/4
O2S = sqrt(3)/2
M1M = O2N = 3/2
OM = MO1= (sqrt(33)-2)/4
O1O2 = (sqrt(33)-1)/2
tan(O1) = sqrt(3)(sqrt(33)+1)/32 = 0.3650601618
tan(20 degrees) = 0.3639702343
Every value is exact, except the two decimal values of the tangents, which are accurate
to ten decimal places. Taking the arctan of 0.3650601618, I get 20.05512386 degrees,
which is not 20 degrees. I read the above tan(20) off a calculator. I have gotten an even
more accurate value by estimating the solution of the cubic equation for the triple angle
tangent function, and the above value is accurate to the last displayed digit.
This trisection is based loosely upon Archimedes' trisection (Part IV, method #1,
below). F. C. doesn't believe my estimate of his error (0.0551 degrees) because,
"mathematical reason has proved that an algebraic method is not appropriate method for
discussing trisection." Hm? He doesn't allow trigonometry, either. So, his method
cannot be disproved, even by counterexample.
Mr. F. C. makes some extraordinary claims, among these:
Proof by Autocad: His method has checked out perfectly, using Autocad.
Proof by "Have I ever lied to you?": He wrote to me, "If my theory is not true, it would
not be published."
Trigonometry is inherently inaccurate: He bases this idea on the truism that
sin(x)<x<tan(x) (with x in radians), while not explaining what this has to do with
anything. Apparently he thinks that sines and tangents do not have exact values.
He also feels that people who believe as he does have been condemned by the
mathematical community.
8. Trisect a circle:
I received a description of this by email. It is a variation of method #6, above. We
divide our angle into four equal arcs (bisect it twice), then construct a square with the
chord CD (in the diagram) as a side. We circumscribe the square with a circle. We
pretend that the circumference of this circle is equal to the arc length BC of our angle.
We then trisect the circle, by inscribing an equilateral triangle. Then we transfer the side
of the equilateral triangle to our arc BC (B'C' on the right part of the above diagram),
again pretending that the circumference of the circle is equal to the arc length BC.
Trisecting a 60 degree angle like this, I get an 18.40 degree angle, which is 1.60 degrees
off. This method was inspired by rolling an angle into a cone, and then trisecting the
circular base. See paper folding, below.
9. Not very close: Here is a trisection that I got in my email.
This may be what that person meant. We want to trisect angle
AOB. We trisect the side AO, and make AR = AO/3. We draw
the circle with center O and radius R. Then we draw the circle
through A (I assume) tangent to line OB at B (this circle is not
very difficult to construct). These two circles intersect at a point
X (within the angle). And angle AOX "appears to be the
trisection of the angle." This is actually very close for small angles. Using this method
on a 60 degree angle, I get a "trisection" of 24.73 degrees, which is 4.73 degrees off,
not very close. Of course, I may have misunderstood, as this person did not say that the
circle tangent to OB should go through A. But without that condition, the angle AOX
could be almost anything.
10. My improvement: Here is my improvement of the above method. Instead of using
arc RX above, I draw perpendicular CD (AC = AO/3), to intersect the same arc. Testing
this out on a 60 degree angle, I get 21.61 degrees, which is 1.61 degrees off, a
significant improvement. And it too gets very close for small angles.
11. LA Times: Here is a construction (which I have simplified slightly) from the LA
Times
(advertiseme
nt by
D.W.A.,
Mar. 6,
1966). We
are trying to
trisect angle
BAC. We
draw the
circle with
center A and
radius AC
(and BC).
We draw line
segment BC
and bisect it
at O. We
draw line
AO. We draw the circle with center O and radius OC. This crosses segment AO at L.
We draw lines BL and CL. We trisect segment BC at W and J. We draw the circle with
center C and radius CO. This intersects circle O at Z as shown. Draw line LZ, which
intersects circle L at M as shown. Draw the perpendicular bisectors of MZ and MJ,
which meet at S, as shown. Draw a circle with center S and radius SZ. This crosses
circle A at point T (as shown) which may trisect angle BAC. Using this method on a 60
degree angle, I get an angle of 19.97 degrees, which is 0.03 degrees off, not bad.
Above, I said that I simplified the construction slightly. The author of this method
trisected the right angle BLC to get point M. Later he drew MZ. Well, it turns out that
LM and MZ are the same line. I merely noticed that point Z trisects the right angle
BLC. And the facts that angle BLC is a right angle, and line LZ trisects it did not need
to be mentioned during the construction. This construction also appears in Geometry by
Harold Jacobs.
12. Bisect and trisect: This one was in Martin Gardner's
Scientific American column in 1966, and he got it from
Mathematical Snapshots by Hugo Steinhaus. We want to
trisect angle AOB. We bisect it getting C. We trisect
segment CA, getting D, which "trisects" the angle. Using
this method to trisect a 60 degree angle, I get 20.11
degrees, which is 0.11 degrees off. This is just method
#1 above, trisecting a chord. But the chord is shorter, and
so the trisection is significantly more accurate.
13. Bisect twice: A natural extension of the previous method is to bisect the angle twice
and then subtract off an estimated trisection. Here we
trisect angle AOB. First we bisect it, giving point C, then
we bisect again giving point D. Then we trisect the
segment DC, giving point E, which is our trisection
point. Using this method on a 60 degree angle, I got
19.9872 degrees, which is 0.0128 degrees off. You can
see that trisecting the chord of a smaller angle produces
much improved results. Of course the chord (CD) is here
much closer in length to that of the arc (CD).
14. An interesting method: This comes from
Survey of Geometry, by Howard Eves. We are
trying to trisect the angle AOB, with AO = BO.
We draw a semicircle on segment AB, facing
the point O. We trisect the semicircle (easy to
do), giving us points D and E. We bisect
segment DE twice, giving us point G (EG =
ED/4), which is our trisecting point. This
method works fairly well for small angles, but blows up for big angles. Trying it out on
60 degrees, I get 13.9 degrees, which is 6.1 degrees off, making it the worst method (for
that particular angle) of all the methods I have tested.
15. d'Ocagne's method: Unlike most of these
methods, this one (by M. d'Ocagne in 1934) was
designed only as an approximation, as the inventor
knew that an exact trisection was impossible. We have
angle AOB in circle O. We bisect the arc AB at point
C. We extend BO to D, also on circle O. We bisect OD
at E. And angle CEB is the approximately 1/3 of our original angle. Using this method
on a 60 degree angle, I get 20.10 degrees, which is 0.10 degrees off. This method is
found in a couple of places.
16. Durer's method: Here is
Durer's method (1525), which is
very accurate. We draw arc AB
with center O, and chord AB. We
trisect the chord AB, at C. Draw
CD perpendular to AB,
intersecting the arc at D. With
center A and radius AD, we draw
the arc DE, intersecting the
segment AB at E. We draw point F
on segment EC, so EF = EC/3. With center A, and radius AF, we draw the arc FG,
intersecting the arc AB at G. line OG approximately trisects the angle. Trying this
method on 60 degrees, I get an error of less than 0.01 degrees. I will try to get a better
estimate of the error. This method is found in several places. Durer was a famous
painter and mathematician.
17. Karajordanoff's method: This is an
intentional approximation. We draw two circles
about O, one with A and B on it, and one with
twice the radius. We bisect the arc AB at C. We
draw line AC. We draw BD perpendicular to
OB, which intersects line AC at D. Se draw DE
parallel to OA, which intersects the larger circle at E, which approximately trisects the
angle. Using this method on a 60 degree angle gives us 19.97 degrees, which is 0.03
degrees off. This method is found in The Trisection Problem, by Robert C. Yates.
18. Kopf/Perron method: This is
another intentional approximation. We
bisect OC at D, and draw the
perpendicular DE (to DA). DF is 1/3
DE. We extend OD out to H, so CH =
OC. We draw a circle with center F and radius AF. Draw line CB which intersects this
arc at G. Then angle AHG is approximately 1/3 of the oringinal angle. Testing this
method on 60 degrees, it is approximately 0.01 degrees off. This method is found in
The Trisection Problem, by Robert C. Yates.
19. JJG method: This was
given in a book called The
Mathematical Atom, by J. J. G.
(a mathematician) in 1932, as a
serious trisection. We bisect the
angle with OC, then bisect
AOC with OD (as shown).
Then draw DE perpendicular to
OD, intersecting OC at E. We draw the circle with center E and radius EO. The line OD
intersects this circle at F. Draw EG parallel to OF, intersecting the large circle at G, as
shown. Draw lines EF and DG, which intersect at H. OH nearly trisects the original
angle. Trying this method on 60 degrees, I get 19.99 degrees, which is 0.01 degrees off.
This method is found in The Trisection Problem, by Robert C. Yates, who apparently
simplified J. J. G's. original diagram.
20. Between one-half and one-fourth: I
received email from L. D. K. showing this
trisection. With centers at the vertex of our angle, we draw arcs with radii of 1, 2, 3, 4,
5, etc., each intersecting the sides of our angle. We can bisect the angle which gives us a
point on the arc with a radius of 2. We can bisect each of those smaller angles, giving us
four points on the arc with a radius of 4. These give us the green points shown. With
that info, we can draw the lines in our latice, and fill in the trisecting points on the arc
with a radius of 3, and even divide the arc with a radius of 5 into 5 equal arcs. We can
continue out farther, and divide our angle into any number of equal arcs. The originator
of this method says that it doesn't work for angles greater than 45 degrees, which should
be a clue that something is wrong. In fact, none of the lines (except the rays of the angle
itself) is a straight line. If they go through the proper points, they are all curved. The
curvature is just not very noticeable for small angles.
Erasing the unnecessary lines and points, this
method makes a nice approximation of a trisection.
Here I trisect a 30 degree angle, and get 10.04
degrees, which is not bad. The inventor of this
method was right, that it is worse for larger angles.
Trisecting a 60 degree angle, I get 20.33 degrees,
which is 0.33 degrees off, still not bad.
21. "Early folly": Above left is a method which can be found at the site Early Folly.
This is a method from back in 1951, when three men discovered a trisection method,
had it notarized, and sent it around to universities. The instructions were somewhat
vague, and it took me a while to come up with the diagram. The second diagram is my
simplification, which is magnified. Here are the steps in my own words:
We want to trisect angle AOB, with segments AO = OB. Draw the arc AB with center
O (this is done later in the original, but it makes more sense here, to make AO = OB).
Trisect AB, producing point C (much of the original method was involved in trisecting
this line). Draw the semicircle with diameter AB, on the side of AB away from O.
Trisect semicircle, getting point D as shown. Draw CD, and bisect it at E. Draw EF
perpendicular to CD. EF intersects the extended AB at some point F. Draw the arc CD
with center F. This intersects the arc AB at some point G, which we are told is a
trisection point. Draw OG.
Trying this on a 60 degree angle, I get a trisection of 20.05 degrees, which is 0.05
degrees off. The method is better for smaller angles, and not quite as good for obtuse
angles.
I think this method is one of those where someone started with a near trisection
(trisecting the chord), and then chose a point way off to the side (the farther the better,
actually). Then an arc with this point as a center is very nearly a trisection for quite a
ways along its length, so we just choose the best such point that we can come up with.
No proof was ever imagined. Method #15 above gives the same impression.
22. Four circles: I received this
by email. We are trisecting angle
ABC. We draw the first circle
with center B, and label the
intersections with the rays of the
angle A and C. We draw the chord
AC. We bisect the chord at D, and
draw BD. We draw a second circle with center B and radius 2AB, and we draw a third
circle with center D and radius 2AB (the two larger circles in the diagram). Line BD
intersects this last circle at E. With center E and radius AB, we draw a fourth circle,
which intersects the second circle at two points, F and G, the two trisectors. The
trisection not very good. Using 60 degrees, I get a trisection of 18.01 degrees, which is
1.99 degrees off. This method never gets particularly close to a trisection. Using 3
degrees, I get a trisection of 0.89 degrees.
23. Same as #1?: I got this one in my email. We are
trisecting angle AOB, with AO = BO. We bisect AB
at C. Then we draw circle C, with radius AC, and
circle A with the same radius. These circles intersect
at point D, on the side of AB away from point O. And
this may be a trisection point. Trying it on a 60 degree
angle, I get 19.11 degrees, which is 0.89 degrees off,
the same as method #1, above. So is this equivalent to method #1? No, only when the
angle is 60 degrees. This method works well for angles close to 90 degrees (it is perfect
for 90 degrees and 180 degrees), but gets worse and worse for small angles. As the
angles get close to zero, this method approaches 1/4, not 1/3.
24. Same as above: I got this one in the mail from A. R.
S. from India. In the diagram, we draw the arcs with
centers C and D and radius CD. These meet at E, outside the arc CD. We bisect CE at
X, which is our trisection point. A little bit of simple geometry shows that this point X
is the same as point D in the previous method, and trying it on a 60 degree angle gives a
trisection of 19.11 degrees, which is 0.89 degrees off.
25. Mark Stark's amazing
estimation: This one was
never meant to be a true
trisection, but an estimate,
and it is very very accurate.
Find the interactive Java
applet at Mark Stark's Angle
"Trisection". We are trying to
trisect angle AOB, with AO =
BO. We draw the circle with center O and radius OA, and we draw the segment AB. We
then choose an estimate of a trisection point D on line AB, somewhere between 1/2 and
1/4 of the way from A to B. Then we draw the circle with center A and radius AD,
giving us point E where this circle intersects the arc AB. We extend line AO out to F as
shown, so OF = 3AO. We draw the circle with radius OF and center O. Line DE
intersects this circle at a point G, as shown. We then draw line GO, which intersects arc
AB at E', which is a much better estimate of our trisection point than E was. Mr. Stark
says that repeating this estimate one more time (using E' to get a new point E'') will
have an error of less than 0.0000000001 degrees.
Trying this method on a 60 degree angle, and choosing D as AD = AB/3 (method #1
above), gives us an angle AOE' of 19.9999937 degrees, which is only 0.0000063
degrees off, making this the best method that I have tested.
One might assume that the length OG can maybe be almost any length. Above, OG =
3AO, but 4AO or 2AO would actually produce good estimates of our trisection point.
However, it turns out that G is almost ideally positioned. If you take several different
estimates D (between 1/2 and 1/4 on AB as described above) and draw their lines (DE),
they will intersect at several points very near to the position of G in the diagram.
26. Missing theorem: There is a book called The Missing Theorem by L. O. R. (a later
edition is called Angular Unity), about triangles which have one angle twice another
angle (see Archimedes' method, below). The author claims to prove that any trisection is
possible, using the classical tools in the classical way. But he cannot show us how that
might be done. His proof is a little bizarre.
27. A natural attempt: I recently received this in
my email (from J. L., who is not me), but I am sure
that I have seen it before. We want to trisect angle
AOB. We draw an arc AB with center at O. We
trisect segment AB at C. We draw CD perpendicular to AB, and intersecting the arc as
shown. And D is our trisection point.
When a person finds that C (in this diagram) is a rather poor trisection point, it is natural
to seek a better one. D is the next most natural candidate, and is a distinct improvement.
Using this method on a 60 degree angle, we get a trisection (angle AOD) of 20.40
degrees, which is 0.40 degrees off, still not very accurate. As with most of these
methods, the accuracy improves for smaller angles.
Angle AOC is too small. AOD is too large, but closer. The next guess might be some
point on segment CD, closer to D than to C. Any such point would be a better guess
than C or D. Some of the above methods (methods 11, 16, and 21, for example) are very
probably based on this reasoning.
28. A parallel angle: I received this in
my email. We are trying to trisect angle
AOB. We draw an arc AB. We make a
guess of a trisecting chord CD, centered
about the angle bisector as shown. We
duplicate CD giving CE and DG as
shown. We draw EF parallel to AO,
intersecting the angle bisector at F. We
draw FG. Angle EFG = angle AOB, and
we would seem to have trisected it with
the chord CD (and CE and DG). But
these chords and their arc are not chords
and arc belonging to angle EFG; the
center (O) is the wrong center, which should be F. And so, the trisection is surely not
exact.
The accuracy of this method depends on how good a guess we made when choosing
chord CD. If the guess was exact, then the trisection is exact. Using this method on a 60
degree angle, and choosing a CD that is 90% of a perfect guess, I get a trisection of
19.95 degrees, or 0.05 degrees off. So this is a fairly accurate method.
Part IV - "Cheating" (violating the rules)
See Part II - The Rules.
1. Archimedes' method: This method is exact,
and is attributed to Archimedes. But has a small
flaw. We are trying to trisect angle BAC. Draw a
circle with its center at the vertex. It intersects the
sides of the angle at some points B and C (as in the
diagram). Draw line BED, as in the diagram, so E
is on the circle, and D is on line AC, and DE is the same length as the radius of the
circle (It may help you to mark the distance DE on the straight edge, in order to line
these up). Angle BDC is a perfect trisection of angle BAC. Proof: Call angle BDC x.
Triangle DEA is an isosceles triangle, so angle DEA=180-2x. That makes angle
AEB=2x. Triangle EAB is an isosceles triangle, so angle EAB=180-4x. Angle EAD=x,
so angle DAB=180-3x. That makes angle BAC=3x. So, angle BDC is 1/3 angle BAC.
The only problem is that it is impossible to draw line BED without cheating (by making
marks on the straight edge).
I received email saying that the above method doesn't work. Apparently that person
measured segments EA and ED, and they weren't an exact fit. He said that assuming AE
= ED was "a big mistake." He apparently did not read my description of the
construction. Well the method does work, regardless of the accuracy of my drawing.
It doesn't matter how far apart the marks are on our
I see on the WWW, and from my email, that some people consider this method valid,
since the rules for use of compasses and straightedge are seldom spelled out. See Part II
- The Rules.
2. An ancient method: Here is an ancient
method shown in Heath's A History of
Greek Mathematics. It is not shown as a
construction of a trisection, but as a help
in the analysis of the problem. ABC is the
trisected angle, and DBC is the trisection.
We draw a rectangle ACBE and then the point F is the intersection of BD and EA. We
find that DF = 2AB. So if we can construct line BF so that DF = 2AB, then we can
trisect the angle. And we can do this with a marked straightedge, but not with the classic
tools. This method is equivalent to Archimedes' method, above.
3. Paper folding:We are told
that an angle can be trisected
using paper folding. The book
Amazing Origami by Kunihiko
Kasahara gives the method for
trisecting a 90 degree angle,
producing angles of 30 degrees
and 60 degrees, but seems to
give no general method. On the left, we see this trisection.
We start with a square paper (we can fold any irregular paper into a square), which we
fold in half (vertically in the diagram). We then fold one of the corners over onto this
fold. It is easy to show that this produces the second side of an equilateral triangle, and a
60 degree angle, and thus a 30 degree angle.
Above right is a questionable (probably) method by which trisection can be done, using
paper folding. We roll the angle into a cone, and slowly flatten it while making sure that
we approach two folds and three equal portions of the circumference. The angle does
not have to be cut from a circle, but that simplifies the process. With this method, we
can divide an angle into any number of equal angles. A variation of this method is to
roll the angle into a cone, trisect the circular base (with an equilateral triangle) which
trisects the arc of our angle, and then unroll the angle.
Here we see a good paper folding method found
in the book Geometric Constructions by George
E. Martin and attributed to Hisashi Abe in 1980.
Here we are trying to trisect angle ABC. We
find D the midpoint of AB. We fold line a
perpendicular at B to BC, then make two lines
(lines EA and c in the diagram) perpendicular to
line a, through A and D, as in the diagram. Then
(the part which we cannot do with compasses
and straightedge) we find the fold b so that E
lies on line AB producing E' the image of E, and
so that B lies on line c, producing B' the image of B. Line BB' trisects the angle ABC.
See Origami Trisection of an Angle for the proof. Here we have found a reflecting line
for two pairs of points. We can do that with compasses and straightedge only if we
know where all four points are beforehand. Here each of the points could have been
anywhere on a given line.
The above method suggests that we can also use the edge of the paper as a marked ruler.
So, you can use any number of methods, probably including Archimedes' method
above, to fold a trisection. Most such methods would involve folding the paper more
than once simultaneously, to get your ruler to anywhere on the paper. The method
shown at the link above moves the ruler to a specific place on the paper, because it
involved only one fold to
move the ruler.
Here is another paper
folding method shown in
Geometric Constructions
by George E. Martin, and
the method and proof are
attributed to Dayoub and
Lott, for a geometry tool
called a Mira, a mirror
which allows you to draw
reflections. Paper folding produces the same reflections. Let me describe the trisection
using the diagram. We start with an arbitrary angle ABC. We find D, the midpoint of
AB. We fold line a, through D, and perpendicular to BC. We fold line b, through D, and
perpendicular to line a. Now (the part we cannot do with compasses and straightedge)
we find line c by folding so that A folds onto line a and B folds onto line b, producing
A' the image of A, and B' the image of B. The line BB' trisects angle ABC. Apparently
there are three lines which meet the criteria that I described (folding A onto A' and B
onto B'), but only one of them intersects segment AD. See the above book for the proof.
4. "Euclid Challenge": At Euclid
Challenge, we come to a diagram similar
to this (I have supplied the question mark
next to point T). The two arcs have the
same center, point B, which is out of my
picture, to the south. The lower arc has 3/4
the radius of the upper arc. The lines DD',
CC', and PP' meet at a point B. In the
previous pages, the author constructed
these points and lines so that angle DBP is
3.75 degrees (1/16 of 60 degrees) and
angle DBC is 15 degrees (1/4 of 60
degrees). We now are going to find point
T, so that angle DBT is 5 degrees, an
unconstructible angle according to
mathematicians. Let me quote the author
on this (his steps 14 through 19, rephrased slightly), the italics are mine:
Grasp with the left hand a compass leg above the legpoint (identify as leg R). Place
legpoint of leg R on point D'. Grasp with the right hand the other compass leg above the
legpoint (identify as leg S). Place legpoint of leg S on D. Move both legpoints at a
uniform rate; legpoint of leg R along arc D'C' toards point C', and legpoint of leg S
along arc DC towards point C. When the legpoint of R reaches point P', stop both
legpoints. At the location of the legpoint of leg S on arc DC mark point T -- trisection
point.
The author further explains his "uniform rate":
Confirmation of the "Uniform Rate": The maximum travel on arc D'C' = 5.625 degrees
(1/8 of 45 degrees) and on arc DC = 7.5 degrees (1/6 of 45 degrees or 1/4 of 30
degrees). These two angles, 5.625 degrees and 7.5 degrees can be constructed with a
compass, and used for test points by continuing the movement of the two legpoints
beyond the 3.75 degrees on ard K'L' [arc D'C'], and 5 degrees on arc KL [DC], to see
that legpoint R reaches test point 5.625 degrees at the same time that legpoint S reaches
test point 7.5 degrees.
This clever sliding of compass points at a uniform rate is, of course, a great violation of
the rules, even if it were possible. The idea is that arc D'C' (and any of the smaller arcs
on it) is 4/3 as long as arc DC, which is true. And this author seems to think that you
can measure fixed distances along an arc using compasses, which you cannot. If you
could, then you could make a line segment the same length as pi (which I think he does
on another page). In my diagram above, the length of the arc D'P' is supposedly the
same as the length of the arc DT. If we could do this with compasses, then angle DBT
would indeed be 5 degrees. But, we can't. See Part II - The Rules and the next
"trisection" below.
If, instead of measuring arc lengths with our compasses, we make the chords D'P' and
DT of equal length, then we get an angle DBT very close to 5 degrees, less than 1/100
of a degree off. But, as we have seen several times above, chord length and arc length
are not related in such a simple way.
To summarize my objections:
This method violates the normal rules.
If we accept this method, there is a much simpler way to use it. See "5. Duplicating an
arc length," below.
It can't be done. There is no precise way to move two points at the same speed, on two
different arcs, simultaneously. In fact, there is no precise, finite way to measure arc
lengths without measuring angles.
5. Duplicating an arc length: Using the same idea
(duplicating an arc length) as the previous "construction,"
we can devise a simpler trisection. In this diagram, we want
to trisect angle AOB. We make segment OC three times
segment OA. Then the length of arc CE is three times the
length of arc AB. We duplicate arc length AB and get arc
CD, which trisects angle AOB.
I have never seen this "construction" given as a valid trisection. But this duplicating of
an arc length is the main flaw in many of the trisection attempts shown above. Of course
this trisection is perfect; it just can't be done with the classic tools. If we use chords
instead of arcs, this "trisection" gives the same lengths as the one in Part III, "trisection"
#1.
Just how are chord length and arc length related? With an angle A and a radius r, the
chord length = r sin A/cos (A/2), and the arc length = 2 pi rA/360. You may remember
from trigonometry or calculus that sines and cosines are not particularly simple
functions of angles. They can usually only be estimated, and to evaluate them we use
infinite series.
6. Tripling an angle: One common mistake is to choose an angle (or construct it), then
triple it, getting a larger angle, and then claim that you have discovered a way to trisect
the larger angle. Of course, you can easily triple any angle. But to go the other way, and
trisect an arbitrary angle is what we are trying to do here. If we triple some angle, then
the larger angle is not arbitrary in any way (at least not as mathematicians use the
word). The Trisection Problem, by Robert C. Yates tells of a university president (J. J.
C) who went public to the newspapers with such a trisection.
Here is that trisection:
Draw lines BC and DF parallel to each other. Choosing
arbitrary points C and D on the two lines, we draw the arc
CF with center at D as shown. We draw lines DC and BF.
With center at F, we draw the arc DB. We draw angle DCE equal to angle DCB. We
draw AD parallel to CE and DE parallel to BF; these two lines intersect at a point E.
Then lines DF and DC trisect the angle ADE.
How true. But angle ADE is not arbitrary. All the above accomplished was to construct
an arbitrary angle DCB and then triple it, giving us angle ADE. Angle DCB is arbitrary
(depending on where we put arbitrary points D and C), but it would be difficult to draw
a really small angle, as points D and C would have to be very far apart. There are easier
(somewhat) ways to triple an angle. You might want to verify that all of the small
angles in the diagram are equal.
7. With a ruler: The tangent of 20 degrees is roughly 0.363970234. Using a ruler with
markings that fine (or less accurate, if you're not so picky), draw a right triangle with
one leg exactly 1, and the other leg 0.363970234, and you've got a 20 degree angle. Of
course, the tangent of 20 degrees is not exactly 0.363970234, but that is more accurate
than any ruler you will ever find. So isn't that good enough? Nope, it's still an
approximation, no matter how accurate it is.
Part V - "Cheating" (using other tools)
With compasses and straightedge, only circles (and circular
arcs) and lines can be drawn. Abandoning compasses and
straightedge, it is simplicity itself to trisect any angle. Use a
protractor (see the drawing). I assume CAD programs can
trisect angles at will. We can measure the length of an arc
with a curved string or piece of metal (like spring steel). A
person could design any number of mechanical devices which can trisect any angle.
And curves have been drawn, which make trisection
easy (Part VI below). Or hard.
1. A trisection tool: Here is a little tool (apparently
invented by C. A. Laisant in 1875) which trisects
angles. It is exact, assuming the line segments are
exact. The dots are hinges. The two hinges at the far
right slide along their trisection lines. We have two
rhombuses and two of their diagonals.
We cannot use compasses and a straightedge to trisect a general angle. That has been
proven. But we can make other tools (such as the device drawn here) to trisect such an
angle. In fact, we can use a pair of compasses and a straightedge to make this device.
All that this device does is triple the smallest angle.
2. A tomahawk: Here we have a clever drafting tool called a tomahawk (or
shoemaker's knife (there is another shoemaker's
knife in geometry), invented before 1835),
which can be used to trisect angles, as shown in
the diagram. For small angles, the handle may
have to be extended. The top end is marked off
in three equal pieces, one of the marks being
the center of the semicircle. We can prove that
the trisection is exact by drawing three
congruent right triangles, as shown here.
Why can't we just draw a tomahawk using
compasses and straightedge, and then move
that part of our diagram over to our angle,
adjust its orientation, and then trisect the angle.
Well that amounts to using a tool; we are
cutting out a part of the plane and moving it,
which is a complicated process and a violation of the rules. But also, there is no way to
position our drawn tomahawk with perfect precision; in other words, we cannot draw
our tomahawk in place, on top of the angle, and trisect the angle. That may be hard for
you to believe.
3. A carpenter's square: A similar
trisection can be done with a carpenter's
square shown here. First we draw the
horizontal line shown, which is parallel to
the horizontal side of the angle, and the
width of the square from that side of the
angle. Then we position the square as shown,
with equal lengths marked off on the
righthand edge of square itself. And two of
those points trisect the angle. We can verify
that with the same three congruent triangles
that we used with the tomahawk.
Part VI - Cheating (using curves other than circles)
There are several curves that can be used to trisect angles. Such a curve is called a
trisectrix. These trisections are exact. But of course, you cannot draw these curves with
compasses and straightedge. I drew most of these curves by drawing an angle and its
triple, and then generating a curve (locus of points) by plotting the trisection point while
varying the angle. Most of these curves are of the fourth degree or higher.
1. Limacon: On the
left, we see a special
kind of limacon
(limaçon), the purple
curve that you see.
Its polar equation is r
= 1/2+cos(theta),
where theta is the
central angle. The
origin is where the
curve intersects itself. A general limacon has the
equation r = b+a cos(theta). A limacon is also called a
limacon of Pascal (named after the father of the
famous Blaise Pascal). I have discovered five ways to
use this limacon to trisect angles:
(i) On the right above is a simple method, just using the little loop of the limacon. We
want to trisect angle ABC, we make AB the right length to fit in the loop, and make BC
= AB. We then draw line AC, and where it crosses the limacon is a trisection point. If
trisecting AC was a true trisection of the angle (part III, method #1, above), then this
curve would be a circle. This seems to be the standard, well-known method. But
limacon #4, below has also been published.
(ii) Above left, I am trying to trisect angle ABC, where AB is the right length to fit in
the limacon as shown.. Make BC = AB, and draw line AC. Trisect line AC at D. Draw
DE perpendicular to AC. The point E where this intersects the limacon is a trisection
point..
(iii) Above left is yet another way which I discovered for using a limacon to trisect an
angle. Again we want to trisect angle ABC. Here we adjust the size of AB so that
segment BD (half the length BA, as shown in the diagram) fits inside the small loop.
We then draw AC and bisect it twice, giving AE = AC/4. Where line DE intersects the
large part of the limacon is a trisecting point.
(iv) Above middle is yet another way to use a limacon to trisect an angle ( found in The
Trisection Problem, by Robert C. Yates). It is closely related to method #3. We trisect
the angle AOB merely by drawing BC. Then angle OBC is 1/3 angle AOB (angle BCA
is 2/3 the same angle).
(v) Above right is yet another way to use a limacon to trisect an angle. Here OA is the
same length as from O to the bottom of the limacon. Our angle cuts the limacon (as
shown) at C. We draw a line OD a distance of OC from point A (again, as shown), and
OD is our trisection.
2. Maclaurin's
trisectrix: On the left
is Maclaurin's
trisectrix. We want to
trisect the angle ABC.
We choose length AB
so that AB = AO/3. I then label the point where our angle
crosses the loop of the curve C. Then angle AOC is the
trisection of angle ABC. The triangle BOC can be seen in
Archimedes' method (part IV, method #1, above). But point
A is not part of that method.
(ii) Above right is another way to use the Maclaurin
trisectrix. We are trisecting angle AOB. Here B is chosen so
angle ABO is a right angle. CD the perpendicular bisector of
AB (at C) intersects the loop of the curve at a trisection point. Notice that it doesn't
matter where A is (somewhere on the horizontal line), as constructing the perpendicular
bisector of AB will always produce the same line CD.
4. Conchoid of
Nicomedes?: One
WWW site calls this
the conchoid of
Nicomedes, but (as
far as I know) that
looks quite different.
We are trying to
trisect the angle
AOB. We make the length AO the same as the radius of the circle O. The side OB of
the angle intersects with the part of the curve that curls off to the left at a point B. We
duplicate length OB as AC perpendicular to OA, and C is a trisection point for our
angle. The polar equation of this curve is r = sec(theta/3) = 1/cos(theta/3).
5. Quadratrix of
Hippias: This
curve is the
quadratrix of
Hippias, and it
allows us to divide
an angle into any
number of equal
parts. Here we want
to trisect angle
AOB. The left ray
of the angle crosses
the curve at point
B, and we draw the
horizontal line BC,
with C on the y-
axis. Then we divide the y value by 3, and we get the y value of point D, also on the y-
axis. Then we draw the horizontal line DE, which crosses the curve at E, which is the
trisection point.
An equation for this curve is x = y cot(pi y/2) with x being a function of y.
Above we divided the y value of point B by 3, to get the trisection point. If we instead
divided by 5, we would have found the point which divides angle AOB into five equal
parts. Divide by 7 and get seven equal parts, etc.
This diagram, and the next one, were drawn with Geometer's Sketchpad and Paint Shop
Pro.
6. Tschirnhausen cubic: Here is
Tschirnhausen cubic, also called the
Catalan's trisectrix, and L'hospital's
cubic. Its polar equation is a = r
cos^3(theta/3). I don't know how this
can be used to trisect an angle. This
curve is related to a parabola with focus at the origin, opening toward the left in the
diagram, and the two curves intersect at (1,0) in this diagram. Apparently, both the
parabola and the cubic curve are used to trisect an angle.
I will try to figure this one out.
7. A cubic parabola: This is a cubic parabola with the
equation y = (x^3)/2. We are trying to trisect angle AOB.
Along with the curve, we have two circles of radius 1
and 2 (the two larger circles in the diagram). BO
intersects the smaller circle at C. We draw CD perpendicular to AO. OE = OD, as in the
diagram. We draw EF with a slope of 3/2. It intersects the curve at G. We draw GH
perpendicular to AO. GH intersects the larger circle at H, which is the trisection point.
There are other cubic equations that can be used in this way, with different slopes of
FG. But this one has the simplest equation. In my diagram, it may look like OC is
parallel to FG; it is not.
I found this trisection in The Trisection Problem, by Robert C. Yates. I found other
cubic equations by botching the curve drawing.
8. The "cycloid" of Ceva: Here
is a curve discovered by Ceva in
1699. We want to trisect the angle
ABC, which we fit into the blue
circle (the radius of the circle is
determined by the smallest loop of
the curve). Then we draw CD
parallel to BA until it meets the
curve to the right for the last time,
giving us point D, which trisects the angle. The intersecting point can be deduced to be
x = sin(theta) and y = cos(theta/3). A polar equation is r = 1+2cos(2 theta).
I assumed that I was not the original discoverer of this curve. But most books don't
seem to mention it. I finally found it in The Trisection Problem, by Robert C. Yates. In
that book, the position of our angle is somewhat different.
9. A hyperbola: This one is
apparently a hyperbola. It was
apparently discovered by Pappus
of Alexandria, who lived in about
the year 300. O is the center of the
hyperbola. We are trisecting angle
ABC. AB = AO = OX. We draw
AC perpendicular to AB, and CE
perpendicular to AC. We draw AD = 2BC and intersecting the curve at D. We draw DE
perpendicular to CE, and intersecting CE at E. Draw line BE, which trisects angle ABC.
We didn't need the last couple steps, as we already had an angle (OAD) which trisects
angle ABC. I guess that if a hyperbola can be used to trisect an angle, then maybe an
ellipse can too.
10. A rose curve: A three-leafed rose curve has the right
equation for the job: r = a sin(3 theta). For angles less
than 180 degrees, we just use the upper right petal (leaf). We want to trisect angle AOB.
We draw a circle half the radius of the rose curve, with center O (with B on the circle).
We draw BC perpendicular to AO, at C. We draw a larger circle with radius OA+BC.
This circle intersects the curve at two points, and we choose one of them (D) depending
on which quadrant B is in.
11. A parabola: This is the parabola y=x^2. We
have drawn a circle about the origin with radius = 2
(the purple circle). We want to trisect the angle
AOB, with A on the circle. We draw AC
perpendicular to the y-axis. We draw ED
perpendicular to the y-axis, with E two units above
the origin and ED = AC/2, as shown. We draw a
circle with center D and radius OD. This intersects
our parabola at F. We draw FB perpendicular to
OB. This perpendicular intersects our original
circle at G, which is the trisection point.
This method is found in The Trisection Problem,
by Robert C. Yates.
The following are curves which I discovered while experimenting with trisections. I will
begin the numbering again:
1. A more complicated curve: On the left are
four pieces of an intersting curve that I
discovered, which can be used to trisect an
angle. We want to trisect angle ABC. We fit
segment AB into circle B. CD is perpendicular
to BC. Where CD intersects with the curve is a
trisection point. Above right is the same curve,
as seen from a greater distance. If angle BCD is
some other fixed angle, besides 90 degrees,
then we get different positionings of the same
curve.
Each branch of our graph may be
a hyperbola. I will have to deduce
an equation, in order to tell.
2. Another complicated curve: Here is another curve that I discovered, that can be
used to trisect an angle. Just the outside loop is used. We
are trisecting angle ABC. We make segment AB the right
length to fit into the curve, then make BC = AB. We
make rhombus ABCD. We draw AC and trisect it, giving
us point E. Line DE crosses the curve at a trisection
point, F. Like the limacon above, this curve is very
nearly a circle for small angles.
3. Yet another
complicated curve: Here
is another curve that I
discovered. We want to trisect angle ABC. We choose
length AB so that AB = BD/3. Then we draw AC
perpendicular to BD (we may already have this drawn
while graphing the curve), giving us C. We draw CE
perpendicular to AC, and where it crosses the right part
of the curve is the trisction point E. So we draw BE.
This one may be a hyperbola. The two curves on the left
side of the diagram may be related hyperbolas? I'll have
to figure out the equation, in order to tell.
4.
An
oth
er: Her
e is
anot
her
cur
ve.
We want to trisect angle AOB (the
entire curve is shown on the right; I think I've seen him before). We fit OA into the
curve as shown, and draw circle O with radius OA. Then we find point C where the
small loop of the curve intersects OB, drawing a circle with center B and radius BC.
This circle intersects circle O at point D, within
the angle, which is a trisection point.
5. A useless curve: Here is a complicated and
useless curve for trisecting angles. For angles
less than 120 degrees, we use the branch of the
curve shown. We want to trisect angle AOB, we
fit our angle into the diagram, which should already contain segment AO, and the circle
O. The extended segment OB intersects the correct branch of the curve at C. Line AC
intersects the circle at the intersection point D. This curve is almost unusable for most
angles, as line AC is almost parallel to the circle for small angles, making the
intersection point hard to choose, and point C is way out there (off the chart) for most
other angles.
Sometimes I prefer the useless methods.
6. Another complicated
curve: This one is based
on the previous one, but
is more useful. We want
to trisect angle AOB. We
will use the left part of
the curve which looks
like a hyperbola, and the
circle O with radius
OA/2. Segment OB
intersects the curve at C.
We draw segment AC,
which intersects the circle at D, which is the trisection point.
This graph is the previous graph, but with a smaller circle, which makes AC cross the
circle at a better angle.
7. Another useless one: We trisect angle AOB within
circle O by extending OB to C on the correct branch of
the curve. We draw CD perpendicular to OA,
intersecting the circle at D, which is the trisection point.
This one is useless for exactly the same reasons as is
method #5.
8. A better one: Here we use
the branch of the curve that is
concave to the left. We draw
circle O, as shown. We are
trying to trisect angle AOB,
and we label B as the point on OB that intersects the correct
branch of the curve. We draw circle B with radius BO. This
circle intersects the original circle O at C within the angle,
which is a trisection point.
How did I come up with all of these interesting curves? Well, I just drew a bunch of loci
(paths of points) which involved trisections in various ways. Sometimes I came up with
a well-known curve (often a limacon), and at other times I discovered new curves.
Let me give an example. In method #2 above, which
used this diagram. I drew the angle ABF (I'll define point
F later), and then tripled it to get angle ABC. Then I
drew the rhombus, and trisected diagonal AC as shown.
Drawing segment ED gave me point F. I then varied
angle ABF and drew the path of point F, which gave me
the curve. This was fairly easy to do, using Cinderella.
Then I did a screen capture with Paint Shop Pro, and
cropped the picture, added the labels, and saved it as a
gif image with a transparent background.
Many of the images on this page, not just in this section,
were drawn in a similar way.
Part VII - "Cheating" (choosing an arbitrary point)
When proving things, or constructing things, it is sometimes useful to "choose an
arbitrary point," perhaps on a line or between two points. This is perfectly valid. If we
"draw an arbitrary line" through a point, we are essentially choosing an arbitrary point,
and then drawing the line through the two points. Some trisections involve arbitrary
points, and these arbitrary points are usually chosen so well that the trisection comes out
looking good. Other choices for these points will probably not look very good at all. In
that case, "choose an arbitrary point" is a scam (the scammer may be scamming
his/herself, so don't blame him/her). You should try moving the arbitrary point, and see
what happens to the trisection. Rarely, the choice of the arbitrary point may have
nothing whatsoever to do with the trisection; it may just be part of the smoke screen.
This method of cheating amounts to drawing the diagram just right (a Baby Bear
diagram) so that the trisection comes out almost perfectly. There is probably one perfect
point (or angle or length) out there, which you cannot locate with compasses and
straightedge (by the way), which will perfectly trisect your angle. Get close to that
point, and you will not be able to tell that you have not trisected the angle. Miss it by a
mile, and the trisection falls apart. These methods can get fairly amusing. Here are a few
of those: [under construction]
1. A 56.60 degree
angle: Here we trisect
side AO giving us C,
so AC = AO/3. And
we bisect angle A, the bisector intersects segment BC
at D, the trisector. This method is fine tuned for an
angle of about 56.60 degrees. For any other angle (including small angles), the point D
is way off. There is an angle here (about 56.60 degrees) which can be exactly trisected
with this method, but we don't know what it is exactly. Above right, I have used this
method to "trisect" a smaller angle, and the trisection ends up being larger than a
bisection.
2. LJRH method: This one is found in The
Trisection Problem, by Robert C. Yates. He says
that this appeared in a science periodical, and
was invented by L. J. R. H. We are trying to
trisect angle AOB. We draw the arc AB with
center O. Then we draw arc AO with center B,
and choose an arbitrary point N on that arc.
Then, with center N, we draw the trisected arc
BT of "any" length (actually we trisect it by just
drawing the circle, then making three arcs BC,
CD, and DT with equal chords of any length
BC). We bisect the arcs AB (at Q) and BT (at P).
Then we draw lines AT and PQ, which intersect
at some point K. Draw lines KD and KC, which
intersect arc AB at the
trisection points E and F.
This looks very good for
most positions of N and
relatively small lengths BC
(within a fraction of a
degree). When BC gets
longer, then the "trisection"
gets really bad; see the
diagram above right. For
some angles, it is even possible that line KD could miss circle O completely, and then
where would that trisection point be?
If this method specified where N should be, and how long BC should be, then it would
be a good trisection attempt for angles less than 90 degrees.
Part VIII - Comments
I didn't mean to become a trisection buster (or debunker). I was willing to point out
obvious errors. But some of the methods sent to me were complicated or poorly
described. How do you bust a construction that you cannot understand? Well, this is
now one of my really fun projects. Having busted some of the more complicated
methods above, I find that I enjoy this. And I am certainly improving my trigonometry
skills.
There is a joke that there are two types of people, those who divide people into two
types, and those who do not. Well, there are several different types of people who
choose to believe that trisections (using the ancient tools) are possible:
The first kind are mathematically illiterate, and seem unsure if 2+2 always equals 4.
Most of humankind, mostly nonscientists and nonengineers, are similarly
mathematically illiterate. I told one of these people that simple trigonometry shows that
his trisection of 60 degrees made one of the angles 19.07 degrees, and he informed me
that the trisection was a geometry problem, not a trigonometry problem, a response that
left me speechless (for a moment).
A second kind of people are mathematically adept (algebraically and geometrically) to
some extent, but seem to think that an answer that is close to being right is right. They
measure two angles and think that they have the same measure, and they don't bother to
prove that they have the same measure. There are many such people.
A third kind are also adept at mathematics, and think that they have trisected an angle,
but have made some sort of mistake; and then they find out that it is supposed to be
impossible. These people react with indignation when told they did not really trisect an
angle. Instead of trying to find their errors, which would be very educational, they are
angry at me, as if it is my fault.
Is there a fourth kind of person, who has actually trisected an arbitrary angle, with the
appropriate tools? Well, there is apparently a mathematical proof that such a person
cannot exist. A mathematical proof meets a very high standard of proof.
In the excellent book Eureka! (Math fun from many angles) by
David B. Lewis, Mr. Lewis says that one of the reasons that you
cannot square the circle, is that you cannot construct the square
root of an irrational number. This is not true. Here is a
construction that produces the square root of any length x. The
two lines are perpendicular. If that length is irrational (like the
square root of 2) then the length shown is the square root of it
(fourth root of 2, in this case where x is the square root of 2). It
is impossible to construct a line segment equal to pi. If you
could, then you could take its square root.
Of course some angles (90 degrees, for example) can be trisected. I have a truly
remarkable trisection of a 0 degree angle, which can be done without any tools,
whatsoever. There is the stange case of the angle 3pi/7 (540/7 degrees). This angle
cannot be constructed. But (if you managed to miraculously have it before you) it can be
trisected (Honsberger 1991).
Conway's test: John Horton Conway suggests that a trisection method be tested by
seeing what happens if you double the angle being trisected. If you really have a true
trisection method, then doubling the angle will double the trisection. In other words, we
use our method on angle A, and get some angle B (hopefully A/3). Then if we use our
method on 2A, then our trisection should be 2B. If we get some other angle than 2B,
then our method doesn't work. Of course we are very suspicious of methods which only
seem to work for small angles (or for some other range of angles).
An analogy:
Here is an analogy which may put the above into perspective for you:
We have a line segment, and our construction tool is a rubber band with a point in the
middle. All we can do with this tool is bisect a segment. Both ends must be on the line,
and so, with this tool, we cannot construct any point outside the line.
Let's say we want to trisect the line segment (or any smaller sub-segment of the
segment). All we can do is repeatedly bisect segments. If we are allowed to do this
infinitely many times, then we can easily trisect the line segment (1/3 = 1/4 + 1/16 +
1/64 + 1/256 + ...). But no finite sum will ever do the trick. You can do it to any
accuracy you want (certainly more accurate than your tools), but you can never be
exact. There is no sum of these fractions (1/2, 1/4, 1/8, ...) which can possibly add up to
1/3, without using infinitely many of them.
The situation with straightedge and compasses is more complicated because those tools
can easily produce square roots. But they can't do cube roots and the roots to other cubic
equations. And they can't do many other lengths.
Most of the above diagrams were drawn using Cinderella and Paint Shop Pro.
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The Angle Trisector
[Introduction] [Geometry] [Experiments] [Solutions] [Statistics] [References]
Introduction
Angle trisection is the division of an arbitrary angle into three equal angles. It was one
of the three geometric problems of antiquity for which solutions using only compass
and straightedge were sought. The problem was algebraically proved impossible by
Wantzel in 1836.
A variety of mechanisms have been devised for the solution of the trisection problem.
Some of these mechanisms draw the curves that aid in the solution of the Trisection
equation; others solve the equation directly, or are applicable to the immediate division
of the angle into three equal parts. An example of the latter sort is "Laisant's Compass",
shown in the figure above. It was proposed by M. Laisant in 1875 [Brocard 1875], and
is composed of four straight bars hinged together at one point and forced to make equal
angles with each other. See [Yates 1941] for an interesting review of the trisection
problem.
The benchmark is chosen to illustrate CUIK's performance on problems that are
unsolvable by ruler-and-compass solvers. While it is true this linkage can be simulated
with dynamic geometry software like Cinderella or Sketchpad, the resulting
construction can only solve the reverse problem, angle triplication, and cannot
sistematically trace all branches of the linkage motion.
Geometry
The lengths are chosen so that OB = OC, CS' = BS', OD = OA, and AS = DS, with S
and S' as joints permitted to slide in straight grooves along the two trisecting bars. The
triangles OBS', ODS' and OAS are congruent with equal angles at O. The bars OS' and
OS are extended beyond O so that the third part can be set off upon the same arc.
Experiments
Rigid Trisector: Equations
Mobile Trisector: Equations
Results on two test cases are provided. The first one - the rigid trisector - solves the
position analysis of the linkage when the angle of body 2 (containing OB above), θ2, is a
fixed, known input angle (30 degrees), making the linkage rigid. The second one - the
mobile trisector - solves the same problem but assuming that θ2 is a free variable.
Solutions
Rigid Trisector: Solutions
Mobile Trisector: Solutions
The rigid trisector experiment has 4 solutions, displayed in the following figure. Notice
that only the bottom-right solution corresponds to a configuration in which the
mechanism actually performs its "trisection" function.
The color coding of the above figures, referred to the drawing at the top of this page, is
depicted here:
The mobile trisector has a single, one-dimensional assembly mode. Interestingly, the
mode is a curve with one bifurcation point where the linkage folds to other
configurations where the trisection function is lost. Here's an animation of such folding.
Angle trisection using origami
It is well known that it is impossible to trisect an arbitrary angle using only a compass
and straightedge. However, as we will see in this post, it is possible to trisect an angle
using origami. The technique shown here dates back to the 1970s and is due to Hisashi
Abe.
Assume, as in the figure below, that we begin with an acute angle formed by the
bottom edge of the square of origami paper and a line (a fold, presumably), , meeting
at the lower left corner of the square. Create an arbitrary horizontal fold to form the line
, then fold the bottom edge up to to form the line . Let be the lower left corner of
the square and be the left endpoint of . Fold the square so that and meet the lines
and , respectively. (Note: this is the non-Euclidean move—this fold line cannot, in
general, be drawn using compass and straightedge.) With the paper still folded, refold
along to create a new fold . Open the paper and fold it to extend to a full fold (this
fold will extend to the corner of the square, ). Finally, fold the lower edge of the
square up to to create the line . Having accomplished this, the lines and trisect the
angle .
Let us see why this is true. Consider the diagram below. We have drawn in , which
is the location of the segment after it is folded, , the fourth side of the isosceles
trapezoid , and , the second diagonal of . We must show that
, where and .
Because and are parallel, , and because is the altitude of
the isosceles triangle , . Thus
. Now, is an isosceles trapezoid and is
an isosceles triangle, so and are congruent isosceles triangles. Thus
. It follows that
.
The geometric properties of origami constructions are quite interesting. Every point that
is constructible using a compass and straightedge is constructible using origami. But
more is constructible. As we’ve seen, it is possible to trisect any angle using origami
(I’ll leave the obtuse angles as an exercise). It is possible to double a cube. It is possible
to construct regular heptagons and nonagons. In fact, where the constructability of -
gons is related to Fermat primes, the origami-constructibility of -gons is related to
Pierpont primes. While the field of constructible numbers is the smallest subfield of
that is closed under square roots, the field of origami-constructible numbers is the
smallest subfield that is closed under square roots and cube roots. In fact, it is possible
to solve any linear, quadratic, cubic, or quartic equation using origami!
There are quite a few places to read about geometric constructions using origami, but a
good starting point is this online article (pdf) by Robert Lang.
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Archimedes angle trisection or tripling?
Pi-day today, so a nice occasion to write something about circles. π is the ratio between
the circles' perimeter (περίμετρος in Greek) and diameter.
In the previous post, I explored some angular tricks using circles. The circle is a key
geometrical object when dealing with angles. For example, Archimedes' construction to
trisect an angle uses a circle and a straight line with a marked unit. But for clarity's sake,
I prefer to draw a set of circles which show how a straight line can mark odd multiples
of a unit angle α on intersecting circles, see Figure 1. In that way, it is easier to see how
Archimedes' trisection circle relates to the multiplication and division of angles by odd
integers, see alternative Figure 2.
This makes me think that Archimedes' angle trisection is in fact the inverse of
"Archimedes' angle tripling".
Approaching the Impossible
Construction and trisection of any chance acute angle.
It is geometrically given as impossible to trisect the general angle using
only an unmarked straightedge and compass, as shown at
en.wikipedia.org/wiki/Angle_trisection and many other sites.
Should it be geometrically possible to locate a given acute angle within the
automatic trisecting model the result will be a form of solution to the
trisection problem
This will raise an interesting question as to a constructible general
trisection and the above-mentioned accepted general trisection
impossibility rule.
Diagram 1 below: “Construction and trisection of any chance acute angle”
explains the basic outline of the automatic trisection model.
Diagram 2 is a simplified model based on Diagram 1.
Diagram 3 is an alternative “Construction and trisection of any chance acute
angle”
©Joe Keating Oct 2011
Neusis Construction
A geometric construction, also called a verging construction, which allows the classical
geometric construction rules to be bent in order to permit sliding of a marked ruler.
Using a Neusis construction, cube duplication, angle trisection, and construction of the
regular heptagon are soluble. The conchoid of Nicomedes can also be used to perform
many Neusis constructions (Johnson 1975). Conway and Guy (1996) give Neusis
constructions for the 7-, 9-, and 13-gons which are based on angle trisection.
Conchoid of Nicomedes
A curve with polar coordinates,
(1)
studied by the Greek mathematician Nicomedes in about 200 BC, also known as the
cochloid. It is the locus of points a fixed distance away from a line as measured along a
line from the focus point (MacTutor Archive). Nicomedes recognized the three distinct
forms seen in this family for , , and . (For , it obviously
degenerates to a circle.)
The conchoid of Nicomedes was a favorite with 17th century mathematicians and could
be used to solve the problems of cube duplication, angle trisection, heptagon
construction, and other Neusis constructions (Johnson 1975).
In Cartesian coordinates, the conchoid of Nicomedes may be written
(2)
or
(3)
The conchoid has as an asymptote, and the area between either branch and the
asymptote is infinite.
A conchoid with has a loop for , where , giving
area
(4)
(5)
(6)
The curvature and tangential angle are given by
(7)
(8)
SEE ALSO: Conchoid, Conchoid of de Sluze
REFERENCES:
Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC
Press, p. 215, 1987.
Johnson, C. "A Construction for a Regular Heptagon." Math. Gaz. 59, 17-21, 1975.
Lawrence, J. D. A Catalog of Special Plane Curves. New York: Dover, pp. 135-139,
1972.
Loomis, E. S. "The Conchoid." §2.2 in The Pythagorean Proposition: Its
Demonstrations Analyzed and Classified and Bibliography of Sources for Data of the
Four Kinds of "Proofs," 2nd ed. Reston, VA: National Council of Teachers of
Mathematics, pp. 20-22, 1968.
Loy, J. "Trisection of an Angle." http://www.jimloy.com/geometry/trisect.htm#curves.
MacTutor History of Mathematics Archive. "Conchoid." http://www-groups.dcs.st-
and.ac.uk/~history/Curves/Conchoid.html.
Pappas, T. "Conchoid of Nicomedes." The Joy of Mathematics. San Carlos, CA: Wide
World Publ./Tetra, pp. 94-95, 1989.
Smith, D. E. History of Mathematics, Vol. 2: Special Topics of Elementary
Mathematics. New York: Dover, p. 327, 1958.
Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover, pp. 154-155, 1999.
Szmulowicz, F. "Conchoid of Nicomedes from Reflections and Refractions in a Cone."
Amer. J. Phys. 64, 467-471, Apr. 1996.
Wells, D. The Penguin Dictionary of Curious and Interesting Numbers. Middlesex,
England: Penguin Books, p. 34, 1986.
Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London:
Penguin, pp. 38-39, 1991.
Yates, R. C. "Conchoid." A Handbook on Curves and Their Properties. Ann Arbor, MI:
J. W. Edwards, pp. 31-33, 1952.
SEE ALSO: Angle Trisection, Conchoid of Nicomedes, Cube Duplication, Geometric
Construction, Heptagon, Mascheroni Construction, Matchstick Construction, Ruler,
Steiner Construction
REFERENCES:
Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag,
pp. 194-200, 1996.
Johnson, C. "A Construction for a Regular Heptagon." Math. Gaz. 59, 17-21, 1975.
Referenced on Wolfram|Alpha: Neusis Construction
Trisecting an Angle
In mathematics the problem of trisecting an angle is a very interesting one. Perhaps
more so, due to its seemingly impossible nature when compared to the similar problem
of bisection of an angle which is easily achieved using compass and straightedge.
Several literature have shown the problem to be impossible to solve using Euclidian
geometric construction tools. However the problem is not impossible in nature. I have
got a brilliant way of doing this using a very old technique originally attributed to
Archimedes, the famous Greek mathematician.
The construction does not follow the Euclidean rules of construction which were used to
prove its impossibility. However there is a bit out confusion about the rules themselves,
as they are nowhere clearly mentioned in the writings of Euclid. However, most of the
scholars agree on a set of rules apparent from the constructions described by Euclid.
The following construction circumvents the rules, but does manage to trisect an angle
accurately.
Archimedes’s method of Trisection of Angle
Let the angle to be trisected be AOB as shown in the figure. The trisected angle is PRA
where 3xPRA = AOB (will be proved).
The steps of construction:
Using compass draw a circle with O as center. The intersection of the circle with OB is
marked as P.
Draw like PR intersecting the circle at Q, such that QR = OP, where the point R is
collinear with O and A. This is achieved by marking the distance OP on the straightedge
and aligning the markings with the like OA extended backwards. Although it may seem
a difficult procedure, it is actually quite easy to do.
Angle PRO is the trisected angle of AOB
Proof of Correctness
The proof is quite easy. I have put it down in a step by step way:
OP = OQ, as P and Q are points on circumference and O is the center of the circle.
triangle_POQ is isosceles, so angle_PQO = angle_QPO
angle_QOP + angle_PQO + angle_QPO = 180
=> angle_QOP = 180 – 2*angle_QPO
angle_RQO = 180 – angle_PQO = 180 – angle_QPO
triangle_RQO is also isosceles as QR = OP = QO, so angle_QOR = angle_QRO
angle_RQO + angle_QRO + angle_QOR = 180
=> 180 – angle_QPO + angle_QRO + angle_QOR = 180
=> angle_QRO + angle_QOR = angle_QPO
=> angle_QPO = 2*angle_QRO
angle_AOB + angle_POR = 180
=> angle_AOB + angle_QOP + angle_QOR = 180
=> angle_AOB + (180 – 2*angle_QPO) + angle_QRO = 180
=> angle_AOB – 2*angle_QPO + angle_QRO = 0
=> angle_AOB – 2*(2*angle_QRO) + angle_QRO = 0
=> 3*angle_QRO = angle_AOB
=> angle_QRO = (1/3) * angle_AOB
Hence we conclude that angle_QRO is the trisection of angle_AOB
So what is the catch?
There is no catch, the proof is indeed correct. In the beginning I have mentioned that the
problem is impossible in nature, yet the proof that we showed is correct. The problem
lies with the point 2 of the construction. The step required us to do a marking on the
straightedge, which is not permissible in Euclidean constructions.
Bending the Rules !!
So you can actually intersect an angle after all, using the standard set of tools only. But
the construction is not Euclidian. The interesting part is that this construction is quite
old, but somehow this did not receive much publicity as the problem itself. Sometimes
to solve a problem you have to bend the rules a bit.
Post by Bigyan Bhar on Sunday, May 24, 2009
Labels: Interesting, Mathematics
Trisection by Successive Approximation
Note
I wrote this a couple years ago while I was taking geometry. I know it is not precise, but
it is nice to see.
At last! I have proved the trisection of an angle using successive approximations!
This is set up for a graphical browser. If you don't have a graphical browser, you MAY
be able to follow it, but it would be a good idea to download the picture.
Start with a given angle (ABC). Mark the rays into equal segments. Then, make a circle
with the vertex (B) as the center and a side as the radius. Next, bisect the minor arc from
the ends of the two sides. (A&C) You can do this by drawing a line from A to C and
construct the perpendicular bisector. Next, split the arc on each side of the midpoint into
3 equal segments that intersect on the arc. If you are picky, you can do this by moving
each segment of the line along a line perpendicular to it so that its endpoints are on the
circle. Then you draw a line from each endpoint of the middle segment to the near
endpoint of the nearest segment. Then extend the middle segment one third the length of
each of the two lines you just drew and the end segments two thirds the length of their
corresponding short segments.
If you wish to be more accurate, split the arc into 6 segments, then 9, etc. Then take the
endpoint of the segment one third the distance from the midpoint (with 3 segments it
would be the segment next to the midpoint, with 6 the second segment from the
midpoint, etc.) and connect it to the vertex.
Congratulations! You have just trisected an angle using successive approximations. The
exactness depends on how many segments you split the arc into.
Trisecting an Angle
This is a simple straightedge and compass construction of a third part of an angle. In the
illustration for Step Four, the angles DOC and OCD are equal, and they add up to the
angle ODE. Angle ODE is equal to angle OED, and their sum is equal to the sum of
DOC and FOE. Using algebra, it is easy to see that angle DOC is a trisection of angle
FOE.
Step One: This is an
acute angle
AOB. If this
angle can be
trisected, the
construction
can generalize
to any
arbitrary
angle.
Step Two: Set the
compass to an
arbitrary
distance. The
compass
setting will
not be
changed
throughout
this
construction.
Using O as a
center, draw a
circle. Use the
straightedge
to extend the
line OA to the
left, and set
the compass
on the
intersection of
the circle and
line OA, with
the other point
on the
extended line.
Step Three: Hold the
straightedge
to the
compass, and
slide one end
of the
compass
along the line
OA, while the
other end of
the compass
moves up the
circle. Stop
when the
straightedge is
aligned with
the
intersection of
the circle and
OB.
Step Four: Mark point D
on the circle
with the
compass. The
angle DOC is
exactly one
third of the
angle AOB.
Angle Trisection
A classical Greek problem was to find a way to trisect an angle using just a ruler and a
pair of compasses. However, this is impossible.
It is possible though, to trisect an angle using a carpenter's square, as demonstrated by
the interactivity below.
We haven't received any solutions to this one yet. Here's a copy of one of the diagrams
from the Hint, and a suggestion.
You need to show that the angle has been trisected, which would mean angles CBR,
RBQ and QBP are equal. Congruent triangles may help you to show this!
An Interesting Example of Angle
Trisection by Paperfolding
Sidney H. Kung
January 6, 2011
Hull has shown a method (by H. Abe) of trisecting an angle via paperfolding (origami,
in Japanese). Here we present a different one that was communicated to me years ago
by Dr. Swend T. Gormsen, a retired Prof. of V.P.I. at Blacksburg, VA. I was able to
prove the result only recently.
Let 2θ be the vertex angle of an isoceles triangle ABC (Figure 1) and let ll' be the
perpendicular bisector of BC. AE is the extension of side CA.
Figure 1
Fold the paper up (by Angle Trisection by Paper Folding, O6) so that points B and C
fall upon AE (at the point B') and ll' (at the point C'), respectively, and then unfold.
Denote by O the point of intersection of ll' and the line of crease zz' (Figure 2).
Construct line segment BO. Then we have ∠AOB = 2θ/3.
Figure 2
Proof
By symmetry we see that OB = OB' = OC = OC'. So, B, B', C, and C' lie on a circle
centered at O. Thus, arcBC = arcB'C'. Construct C'E' parallel to CE intersecting the
circle at N. Let M be the antipode of C'.
Then we have arcCN = arcB'C'. Further,
θ = ∠MC'N
= (arcMC + arcCN)/2
= (arcBC/2 + arcB'C')/2
= 3·arcBC/4.
Therefore, ∠AOB = arcBC/2 = 2θ/3.
References
T. Hull, Project Origami: Activities For Exploring Mathematics, A.K. Peters Ltd.
(2007) 49-50.
Paper Folding Geometry
An Interesting Example of Angle Trisection by Paperfolding
Angle Trisection by Paper Folding
Angles in Triangle Add to 180o
Broken Chord Theorem by Paper Folding
Dividing a Segment into Equal Parts by Paper Folding
Egyptian Triangle By Paper Folding
My Logo
Paper Folding And Cutting Sangaku
Parabola by Paper Folding
Radius of a Circle by Paper Folding
Regular Pentagon Inscribed in Circle by Paper Folding
Trigonometry by Paper Folding
Folding Square in a Line through the Center
Morley's Trisection Theorem
In Proposition 4 of Book IV of the Elements, Euclid inscribes a circle inside an arbitrary
triangle by showing that the bisectors of any two of the interior angles meet at a point
equidistant from the three edges. Since there is only one such point, it follows that the
bisectors of all three angles meet at the same point. Letting 2, 2, and 2 denote the three
interior angles of a triangle, the law of sines implies that the edge lengths are proportional
to the sines of these angles, so we can scale the triangle to make the edge lengths equal to
these sines as shown below.
Since + + = /2, the central angles are + /2, + /2, and + /2, and the law of
sines gives the ratios
Making use of the double-angle formula
we can substitute for the sines of 2, 2 and 2 in the previous ratios and simplify to give
Hence by the sine rule we see that the point of intersection is a distance of
from each of the three edges, which confirms that this point is the center of the inscribed
circle.
Now, since the bisectors of a triangle meet at a single point (a fact which is not entirely
self-evident), it seems natural to go on to consider how the trisectors of a triangle meet.
However, Euclid apparently didn't consider this question, nor did anyone else for over
2000 years. (Of course, it's impossible to trisect an arbitrary angle using Euclidean
methods, i.e., by straight-edge and compass, so Euclid obviously couldn't have used
trisectors in any constructions, but he could still have proven some interesting theorems
about trisectors if he had wished.) It wasn't until 1899 that Frank Morley (one time was
president of the American Mathematical Association) discovered that lines trisecting the
angles of an arbitrary plane triangle meet at the vertices of an equilateral triangle as
illustrated in the figure below, where the central triangle (in blue) is equilateral.
At first it may not be obvious why this proposition is true, but it can be proven very
simply by showing that, beginning with an equilateral triangle, we can construct an
encompassing triangle with arbitrary interior angles 3, 3, 3 (summing to ) such that
the lines from the vertices of this new triangle to the vertices of the original triangle trisect
the interior angles of the former. Since the construction is unique, this is sufficient to
prove that the trisectors of an arbitrary triangle meet at the vertices of an equilateral
triangle. The construction is fairly simple, and could easily have been found by the ancient
Greeks if they had considered such problems. We begin with the equilateral triangle
marked ABC as shown in the figure below.
Then we construct the line bAc parallel to BC, and aBf parallel to AC, and dCe parallel to
AB. Now, for any given angles ,,, draw a line through point A making an angle of
with the line Ab, and draw a line through B making an angle with the line Ba. Let P
denote the intersection of these two new lines. Likewise we can construct the point Q and
R using the angles , and , respectively, as shown in the above figure. By construction
the angles APB, ARC, and BQC are , , and respectively. Furthermore, the triangles
PbA and AcR are similar, and the segment Ab equals the segment Ac, which implies that
the corresponding edges are in the ratio of Pb to bA, and hence the bases are in this ratio,
so the triangle PAR is also similar to these two (noting that the angle at A is ).
As a result the angles APR and ARP are and respectively. By the same reasoning, the
angle BPQ and CRQ are and respectively, and the angles BQP and CQR are both , so
we have constructed the desired triangle whose angle trisectors meet at the vertices of the
equilateral triangle ABC. Naturally the result can be scaled arbitrarily, so the proof is
complete.
The above proof requires nothing beyond Euclidean constructions and propositions, but it
represents only an existence proof. It doesn't enables us, given an arbitrary triangle PQR,
to construct the equilateral triangle ABC. Instead, the proof works in reverse, which is
sufficient to prove that the equilateral triangle ABC exists for any PQR, even though we
can't construct it (by straight-edge and compass). The ancient Greeks certainly made free
use of indirect proofs (recall Euclid's non-constructive proof of the infinitude of primes),
and they could easily have proven Morley's Theorem, but they apparently never noticed it.
It's often said that the Greeks didn't consider propositions involving trisections because of
the impossibility of trisecting an arbitrary angle by Euclidean methods, but in fact the
Greeks developed numerous methods for trisecting angles, based on higher degree curves,
so this doesn't fully explain why they never noticed Morley's theorem.
From the modern perspective, we can use trigonometry to develop a direct constructive
proof as well. First, it's worthwhile to review why an angle trisection cannot be
accomplished by straight-edge and compass methods. Such methods are capable of any
quadratic constructions, i.e., constructions corresponding to the solutions of polynomials
of degree 2. Obviously these methods are adequate to bisect any arbitrary angle, because a
bisection corresponds to solving the trigonometric identity sin(2x) = 2sin(x)cos(x), which
can be done purely by quadratic steps by means of the equation
In contrast, given sin(3x), the determination of sin(x) involves the solution of the
irreducible cubic
so it cannot be done by quadratic methods. Incidentally, this expression implies the
existence of a triple-angle formula analagous to the double angle formula
discussed previously. The triple-angle equation can be analyzed as follows
The quantity in the square brackets is the difference between two squared sines if we
replace 3/4 with the square of sin(/3). Then we can use the identity sin(u)2 sin(v)
2 =
sin(u+v)sin(uv) discussed in the note on Napoleon's Theorem, and arrive at the result
In general, it's easy to show by means of the exponential forms that the value of sin(nx)
for any n can be expressed as the product of n sines
The product form for the triple angle is particularly useful in developing a direct
constructive proof of Morley's theorem. Let's begin with an arbitrary triangle with interior
angles 3, 3, and 3 as shown below.
We first apply the Rule of Sines to scale the overall triangle so that R1 = sin(3), R2 =
sin(3), and R3 = sin(3). Then, focusing on the triangle with edges R1, r1, and r2, we have
the relations
Solving for r1 and r2 gives
At this point we want to eliminate the triple angle, so we make use of the triple-angle
identity
where we have substituted for /3 . With this we can re-write the lengths r1 and r2
as
Likewise the lengths of the other trisecting segments are
Now, by the Law of Cosines we know that the squared edge length a2 of the triangle that
we claim is equilateral is given by
Notice that the angles of the triangle with edges r1, a, r2 are , + /3, and + /3, so the
edge lengths are proportional to the sines of these angles. Therefore, the quantity in the
square brackets is simply sin()2, and so we have
which is symmetrical in the three angles. Hence the squared lengths b2 and c
2, have this
same value, proving that the triangle is equilateral.
Incidentally, by drawing such figures it might appear visually that the center of the
equilateral triangle given by Morley's theorem coincides with the center of the inscribed
circle of the original triangle, but they are actually not exactly coincident.
Return to MathPages Main Menu
BI & TRISECTION DE L'ANGLE
Bissection
Trisection
Découper un angle quelconque en deux parts égales
Découper un angle quelconque en trois parts égales
Bissection
Découper un angle quelconque en deux parts égales est facile
Pourquoi est-ce si difficile pour trois?
ÉQUATION POUR trois
Idée de la démonstration avec un angle de 20°
Calculons en général cos(3a) = cos(a)cos(2a) - sin(a)sin(2a)
= cos(a)(cos2(a) - sin
2(a)) - 2sin
2(a)cos(a)
= cos(a)(2cos2(a) - 1) - 2(1 - cos
2(a))cos(a)
= 4cos3(a) - 3cos(a)
Prenons le cas particulier
de
a = 20o
cos(3a) = cos(60o) = 1/2
L'équation, dans ce cas,
devient
1/2 = 4cos3(a) - 3cos(a)
0 = 8cos3(a) - 6cos(a) - 1
En remplaçant cos(a) = x
0 = 8x3 - 6x - 1
Avec v = 2x 0 = v3 - 3v - 1
Voir Équation
Solutions rationnelles ?
Supposons que Oui, alors v = p/q fraction minimale (simplifiée)
En remplaçant dans l'équation 0 = (p/q)3 - 3(p/q) - 1
En multipliant par q3 0 = p
3 - 3pq
2 - q
3
En reformulant q3 = p
3 - 3pq
2
= p (p² - 3q²)
On déduit que p est divisible par q3
Conséquence p est divisible par q
Impossible p/q est une fraction irréductible par
hypothèse
Et en factorisant avec p3 p
3 = 3pq
2 + q
3
= q (3p + q²)
On déduit que q est divisible par p3
Conséquence q est divisible par p
Impossible p/q est une fraction irréductible par
hypothèse
La supposition est fausse v n'est par rationnel
En généralisant
Il n'est pas possible diviser un angle par construction
Démonstration en 1837 par Pierre Laurent Wantzel (1814-1848)
Règle et compas
CONSTRUCTION À L'ÉQUERRE
La construction à la règle et au compas n'est donc pas possible
Mais, voici une construction assez pratique
1) Posez des repères sur l'équerre
On marque Q en prolongement du bord intérieur et R tel que PQ = QR
2) Préparation
L'angle à partager en trois est l'angle BAC - on note BAC
On construit la droite D
3) Trisection
On oriente l'équerre pour avoir
A sur le bord de l'équerre
P sur la droite D
et R sur la droite AB
Alors CAP = PAQ = QAR = 1/3 CAB
Il existe aussi une
CONSTRUCTION avec Conchoïde de Nicodème
Elle est plus compliquée et assez théorique
Voir Géométrie - Index
Sites Trisection de l'angle selon Thomas Ceva Serge MEHL (avec animations) Panoplie du constructible Pierre Delezoïde - Lycée Buffon - Paris XV Trisecting an angle by J J O'Connor and E F Robertson Trisecting the Angle by Steven Dutch Angle Trisection The Geometry Center Impossible Geometric Constructions Ask Dr. Math
Livre Le Dictionnaire Penguin des curiosités géométriques – David Wells -
Eyrolles
The Problem
What will you find on this site?
Things you will find on no other trisection site: a journal-published construction fully endorsed by the international mathematical fraternity, and many other methods of trisecting angles exactly using only a compass and an unmarked ruler — a challenge that has exercised the minds of geometry lovers for two and a half thousand years. Although my simple methods work perfectly, have been accepted by prominent mathematicians, and were published in the March 2007 issue of the Mathematical Association of America’s College Mathematics Journal, Euclidean trisection with only a compass and ruler is — quite correctly — still considered impossible. Puzzled? All will be explained. Rest assured in the meantime that practical compass-and-ruler non-Euclidean trisection does not depend upon making marks on the side of the ruler in the manner made famous by the great Archimedes of Syracuse.
The impossibility of trisection
As early as the fifth century B.C. Greek geometricians searched for a universal method for trisecting any arbitrary angle exactly. They had already found various ways to trisect many specific angles such as 45 degrees and 90 degrees. However, despite prodigious efforts spread over many centuries, they failed utterly in their quest to find a method that would work with any arbitrary angle.
For over two thousand years countless others faced similar frustration. Finally, in 1837 P.L. Wantzel explained their inevitable frustration by publishing the first completely rigorous proof that compass-and-ruler trisection within the classical constraints is impossible. This did little to discourage the world’s would-be trisectors: claims were still frequently made that the ancient challenge had finally been met. And they are still made today. The claimants, unfortunately, are deluded — every last one of them. Some are deluded because their methods simply don’t work. Others are deluded because they have broken the unwritten rules of what is essentially a mathematical game. These rules are really quite simple. In practical lay terms they can perhaps be summed up in three short sentences...
Many readers will know that things are a little more complicated than that. It is necessary to define just what compasses and straightedges are, and to establish the simple rules governing arc-drawing and line-drawing and the finding of construction points. But it is enough for our purposes here to say that the classical constraints
demand extreme simplicity — and the absolute accuracy that comes with it.
Needless to say, I don't abide by all the constraints imposed by the demands of Euclidean geometry, for they do indeed render the challenge impossible. Nor did the great Archimedes abide by them. He cheated by marking his ruler. I cheat too — but without having to mark it. You may wonder if there is anything noteworthy in being able to trisect by violating the strict rules of this ancient challenge. The answer is simple: yes, there is. Even in non-Euclidean constructions, it has widely been considered impossible, unless one used marks, to trisect using only a compass and ruler, the standard tools that are available to would-be trisectors in the real world. As the classical constraints demand the use of a straightedge (a single edge of infinite length), and I have found this instrument rather difficult to come by, my methods conveniently employ the two real-world tools usually stipulated when it is said that trisection is impossible — and they employ them successfully. That is why they are noteworthy.
Striving to achieve the impossible
Although it has been known for nearly 200 years that compass-and-ruler trisection within the classical constraints is impossible, thousands have continued stubbornly to embark on the ancient quest of finding a way to do it. I am one of them — sort of. I have two excuses for such an apparent waste of time. Neither involves the traditional trisector's disdain for the pronouncements of mathematics, which is akin to the perpetual motion machine designer's disdain for the pronouncements of science. Although I am neither a mathematician nor a scientist, I have the utmost respect for both disciplines — so much so that I believe our threatened world would be a very much better place if the scientific approach to solving problems were universally applied by both institutions and individuals. Science works. It works because its entire focus is the pursuit of truth by observation and reason. It does not object to being challenged. In contrast to many other areas of human thought, it demands that it should be, for its respect for the truth is so steadfast that it welcomes the exposure of its own inadvertent falsehoods.
So my first excuse is that science and its related disciplines do make mistakes. It is quite sensible for fresh minds to revisit their pronouncements in case they have misconstrued or overlooked anything.
My second excuse is that the intellectual challenge of wrestling with the impossible can be a rewarding one — as long as it is not done out of ignorance and contempt for the realities of mathematics and science and logic. One invariably develops a deep understanding of the truths underlying the insoluble problem confronted. And occasionally — just very occasionally — one might glean an original and perhaps useful insight. Over the years I have gleaned a number of such insights in various fields, both technical and artistic. My first super-simple method of non-Euclidean trisection is one of them — as are the new ones that have followed.
Another method soon to be announced
Although you will find most of my methods on this site, the latest — which takes the challenge a surprising step closer in one important respect to meeting the impossible aims of the ancient quest — is still under wraps (August 2009). It will be added in due course, but probably only after it has also been published in an academic journal. This
process is presently under way, for it has already gained the stamp of approval of prominent mathematicians, and is being discussed with the editor of a respected British mathematical journal.
Like its namesake once swung by the Sioux, The old Tomahawk trisector’s through,
For it seems it’s been matched By the ruler unscratched,
Which can paddle its own brave canoe.
I do not, Archimedes, know why You put marks on your ruler. Sir, I
Know that trisection tricks Give a trisector kicks,
But why fudge when it's easy as pi?
Old Methods
The marks of Archimedes
Many methods of trisection-by-cheating have been developed, but not using only a compass and unmarked ruler as I do. Some of them are simple and elegant. Others are complicated and cumbersome. Some of them use only the permitted geometrical tools in an illicit manner. Others use additional tools in an ingenious manner. I shall give you just one example of each category. You will find many others on the internet.
I am particularly fond of Archimedes’ method, partly because it is one of the many I discovered independently before I chose to survey the work of others.
Angle ABC is the angle to trisect. Extend AB to the right as shown. Draw a semicircle with centre B and radius AB. With the compass still open to that radius, hold its legs against the ruler. These two points will become D and E. Adjust both the ruler and compass until the edge of the ruler passes through C while D lies on the semicircle and E lies on the extended baseline. Angle AEC is exactly one third of angle ABC.
Archimedes succeeds to the extent that he uses only a compass and ruler, but he either puts marks on the ruler or uses the compass as a measuring instrument — and consequently fails to honour this explicitt part of the challenge.
The Tomahawk trisector
The example I have chosen for demonstrating trisection by the use of specialised tools is also one that I developed independently only to discover that it was an old idea. The edge DF on the head of the
blue Tomahawk is divided into three equal segments. Only segment DE is shown (an extension of the lower edge of the Tomahawk’s handle would pass through the missing point). The arc has its centre at E and passes through D.
Angle ABC is the angle to trisect. The Tomahawk is positioned so that its corner F lies on BC while the lower edge of its handle lies on B and its arc is tangential to AB. Angle ABE is exactly one third of angle ABC.
The Tomahawk method is ingenious, but it breaks the most basic rule of the challenge in that it introduces a specialised tool. Although it is interesting, it does nothing to advance the cause of angle
trisection using only a compass and ruler. My own methods do advance this cause.
My Methods
First solution
Like the Archimedes and Tomahawk methods, mine also impose the need to cheat. What sets them apart is that they require only the compass and unmarked ruler that are usually demanded by the terms of the challenge. They are, needless to say, mathematically exact methods for trisecting any arbitrary angle. This is the method I described in the March 2007 issue of the Mathematical Association of America’s College Mathematics Journal.
Angle ABC is the angle to trisect. D bisects BC. An arc is drawn with centre C and radius CD. DE is perpendicular to AB. The ruler is positioned so that its short left edge passes through B while its left upper corner meets DE and its upper edge is tangential to the arc. Angle ABF is exactly one third of angle ABC.
Although my method’s greatest strength is that it uses only a compass and ruler and requires no measurements along the ruler’s edge, it also has a neatness and simplicity that perhaps surpasses the elegance of Archimedes' method in that it places the trisecting line directly where it should be: inside the original angle. Also, without using any tools other than the two traditionally specified, it achieves exactly what the Tomahawk and other such contrived methods do. On the face of it, it meets the exact terms of the challenge as it is commonly, although too simply, expressed by teachers of geometry:
Other two-edge solutions
Here are a few of my many other methods of unmarked-ruler non-Euclidean trisection using a second edge of the ruler rather than marking it. I may add further methods from time to time.
1. Perhaps the simplest method of all This construction, like most of those that follow, uses the width of the ruler.
Angle ABC is the angle to trisect. D bisects BC, whose length is twice the width of the ruler. DE is perpendicular to AB. The ruler is positioned so that its corner F lies on DE, its corner G on the arc, and its edge passes through C. Angle ABF is exactly one third of angle ABC.
2. Also using the width of the ruler But the construction relies on somewhat different yet related principles.
Angle ABC is the angle to trisect. The distance from B to D is twice the width of the ruler. E bisects BD. EF is drawn parallel to AB, and an arc is drawn, with centre E, through B and D. The ruler is positioned so that its corner G lies on the arc, its corner H on EF, and its edge passes through D. Angle ABG is exactly one third of angle ABC.
3. Using a circle the width of the ruler The construction is very similar to the previous two but less elegant.
Angle ABC is the angle to trisect. The distance from B to D is twice the width of the ruler. E bisects BD. EF is drawn perpendicular to AB and EG parallel to AB. A circle is drawn with centre D and a radius half the width of the ruler. The ruler is positioned so that its corner H lies on FE, its corner J on EG, and its edges are tangential to the circle. Angle ABH is exactly one third of angle ABC.
4. A related but rather different approach Aspects of my pending one-edge solution are used in this construction.
Angle ABC is the angle to trisect. XXX is at XX degrees to XX, its length twice the width of the ruler. A circle with centre X is drawn through X. XX is perpendicular to XX and XX to XX. An arc is drawn, its centre at X, through the midpoint of XX. The ruler is positioned so that its corner X lies on XX, its long edge passes through X, and its short edge is tangential to the arc. Angle XXX is exactly one third of angle ABC.
5. A very different simple construction It still relies on the width of the ruler but the principles are different.
Angle ABC is the angle to trisect. The radius of arc DE is equal to the width of the ruler. BF bisects angle ABC. The ruler is positioned so that its corner G lies on the arc DE, its corner H on BF, and its edge passes through E. Angle ABG is exactly one third of angle ABC.
6. A similar but more complex approach Here, as in my main construction, the ruler’s width is irrelevant.
Angle ABC is the angle to trisect. The arc DE can have any radius. FG would be perpendicular to a bisector of angle ABC. DH is perpendicular to FG. FH is the same length as DF. The arc drawn through B has its centre at E. The ruler is positioned so that its corner K lies on FG, its short edge passes through H, and its long edge is tangential to the arc that passes through B. Angle ABJ is exactly one third of angle ABC.
7. The first of three wholly different trisections Only one construction line and one circle are needed.
Angle ABC is the angle to trisect. DE is drawn parallel to a bisector of angle ABC that is not shown. The circle, whose centre is at B, has a diameter equal to the width of the ruler. The ruler is positioned so that its top corners lie on AB and DE and its edges are tangential to the circle. Angle ABE is exactly one third of angle ABC.
8. The second of three wholly different trisections Here two construction lines and one circle are needed.
Angle ABC is the angle to trisect. DE is drawn parallel to AB and is tangential to a circle whose centre is at B and whose diameter is equal to the width of the ruler. The ruler is positioned so that its top corners lie on DE and BF and its edges are tangential to the circle. The angle between AB and the ruler is exactly one third of angle ABC.
9. The third of three wholly different trisections Again only one construction line and one circle are needed.
Angle ABC is the angle to trisect. DE is drawn parallel to BC and is tangential to a circle whose centre is at B and whose diameter is equal to the width of the ruler. The ruler is positioned so that its top corners lie on AB and DE and its edges are tangential to the circle. The angle between AB and the ruler is exactly one third of angle ABC.
All these trisections have relied on using more than just one edge of the ruler. Now look at my New Method to read about a construction, my latest and most challenging, that will use only one edge —
yet will still work without having to rely on any marks made on the ruler.
New Method
A one-edge solution
As you have seen, all my methods of trisection have depended on using a second edge of an ordinary ruler, thus dispensing with the need to make marks on it. Recently I started wondering if it would be possible to use only one edge of an unmarked ruler to trisect. This being impossible, of course, without cheating, I realized that the only possibility would almost certainly hinge on violating the Euclidean demand that the straightedge be of infinite length. In other words, the length of the ruler would have to be usefully employed. I still wasn't at all sure that it could be done, but it seemed like a worthwhile intellectual challenge. If I could do it, I would be able to show that it is indeed possible — with some clever cheating again — to not only trisect with just a compass and an ordinary unmarked ruler, but to use only one edge of it, which on the face of it would bring us closer to meeting the terms of the ancient challenge in practical terms because it would use a simple straightedge, just not one of infinite length (which does not in any case exist — and cannot).
I have indeed now found a simple and elegant way to do exactly that. As I have already noted on the first page of this site, my method will be announced here as soon as it has been published in an academic journal, an event that is already being discussed with the editor of a respected British peer-reviewed journal. As the wheels of publication often grind slowly, it may be some time before I can update this page.
Math 218:
History of Mathematics
Instructor:
Andrew Leahy
Office: SMC E-211
Office phone: 341-7439
Home phone: 342-1961
E-Mail: [email protected]
Office hours:
Monday 3rd and 5th Hour
Tuesday 5th and 6th Hour
Wednesday 2nd and 5th Hour
Thursday 4rd Hour
Friday 5th
You can always set up an appointment (by phone or e-mail) or stop by and see if I'm
available. I teach 1st MWF and 6th MWThF, but I'm free many other times (apart from
meetings, ugh) and I'm in or near my office most days from around 9:00 - 5:00.
Course Mechanics:
Meetings: MWF Period Four, SMC A-210
Required Texts:
A History of Mathematics: An Introduction (3rd edition) by Victory Katz. Addison-
Wesley, 2009. ISBN: 978-0-321-38700-4.
Euclid: The Thirteen Books of the Elements v. 1 (2nd edition). Translated by Sir
Thomas Heath. Dover Publications, 1956. ISBN: 978-0486600888.
Euclid: The Thirteen Books of the Elements v. 2 (2nd edition). Translated by Sir
Thomas Heath. Dover Publications, 1956. ISBN: 978-0486600895.
The Works of Archimedes. Translated by T.L. Heath. Dover Publications,2002. ISBN:
978-0-486-42084-4.
A copy of each of these texts (though another edition of Katz) will be placed on closed
reserve in the science library when they are available. These will be placed on two-hour
closed reserve, so ask for them at the front desk.
We will cover selected portions of the Katz book in class, beginning at Chapter One and
ending (approximately) with Chapter Sixteen. We will also be reading selections from
the three source books (Euclid and Archimedes). Some of the chapters will be covered
only briefly (or perhaps not at all). The daily class routine will consist primarily of
lectures, but students will occasionally be expected to give presentations on the readings
(see below). Students will also cover selected sections of the books as independent
research projects (to be discussed later).
Assessment:
Midterm and Final Exams (approximately 40%) There will be two mid-term examinations and a final examination. The midterm will
occur during fourth week and the seventh or eighth week of the term (specific date
TBA), and probably will not be held during regular class times. The final exam will be
held at the time specified by the registrar. Both exams will consist of mathematics
problems, short answer, and essay questions about the material covered in class and in
the texts. Essay questions will be distributed beforehand.
Homework and Quizzes (approximately 30%) Homework consisting of problems (usually from the text), reading assignments, etc.
will be assigned on a daily basis. Due dates will be given at the time of assignment, but
are most likely to be the next class day. Reading assignments from the text will also be
given on a daily basis, and I reserve the right to make reading quizzes to check that you
have completed the reading.
Presentations and Papers (approximately 30%) In addition to homework problems, papers and presentations will also be required
during the course. Papers will be be used to explore in greater depth a topic mentioned
only tangentially in class. A brief paper will be assigned during the beginning of the
term; a longer paper looking at the history of a modern topic in mathematics will be
assigned later in the term. Presentations will also be focued on material not covered in
my lectures. In particular, portions of Euclid's elements will be given as student
presentations. Some of these projects may be done within small groups.
Resources
This web page. It will be updated throughout the term. This link will have some
resources of interest.
Me. There will often be opportunities to ask questions during class. You can also see
me outside of class if you have any questions about the material or the mechanics of the
class.
The web in general. There are numerous online resources devoted to the history of
mathematics. Click here for some examples and reviews which were generated by the
students in a previous history of mathematics class.
Mathematical Reviews. This is an online searchable index or articles related to
mathematics. I have had some luck finding information related to history of
mathematics in this database.
The library. Believe it or not, the library is still a good placed to find information on
the history of mathematics. Apart from the textbook, numerous other works on the
history of mathematics have been written and are available on campus. Our library also
contains several journals which feature articles on the history of mathematics (e.g.,
Historia Mathematica and Mathematical Intelligencer) as well as the collected works of
selected mathematicians and other primary sources. (More later.)
Course Goals:
In addition to learning new mathematical results and techniques and informing you
about the background of some of the mathematical results you are already familiar with,
I also hope that this class will give you a broad overview of just what mathematics is all
about. Since writing and speaking assignments are a part of the course, you will also
learn to present mathematics both orally and in writing.
First Paper
Overview:
Below are suggested paper topics for the first major paper. This paper is to be
approximately 1000 words long and should be focused on one of the subjects below (or
another subject that you have discussed with me in advance). Note that this is to be a
research paper with (1) an introduction (with thesis), (2) conclusion, and (3) proper
citations. In addition, you must explain (in fact the majority of the body may be devoted
to) a significant piece of mathematics (most likely a proof of a result) in your own
words. This paper will be due on February 2. (Two weeks from the date of
assignment.)
Topics:
A biographically-focused paper. You should select one of the mathematicians in the
Greek tradition not covered in class and discuss their importance and/or their major
work. (Again, the primary goal should be to present a major piece of mathematics in a
context appropriate for an essay of this length.) Here are some suggested subjects:
Nicomachus
Eratosthenes
Heron of Alexandria
Menelaus
Diophantus
Pappus
Hypatia
Further exploration of another book of Euclid. Our study of Euclid will be devoted to
Books 1-6. However, there are seven other books, each devoted to a specific subject.
Your goal would not be simply to summarize the book, but to highlight and explain its
importance while giving a presentation of one or more of its major results.
Further exploration of concepts or results from Euclid. As we've read through Euclid,
we've briefly mentioned deeper issues stemming from the book. The goal for this topic
would be to explore these issues more deeply. Some possible subjects:
Missing axioms and/or flaws in Euclid
Alternative approaches to the Pythagorean Theorem
Alternative proofs of Propositions
Problems and alternatives to Euclid's Postulate Five.
Resources:
You should properly cite your work, and make use of multiple online and textual
resources. Some suggested starting points:
The course textbook, and the references it cites.
The extensive notes in the edition of Euclid's Elements that is required for the course.
The MacTutor history of Mathematics web site here. This website has extensive
resources on individual mathematicians. More importantly, it also has an extensive list
of references for each subject that it covers.
MathSciNet here. This is a search engine for mathematics-specific content. While it is
generally intended for use among reseach-level mathematicians, it might also be useful
for finding further resources on your subject.
The Dictionary of Scientific Biography here. This is a great starting point for
biographical research on mathematicians and other scientists. It is located in the
reference room of Seymour library.
Other library books. There are multiple books on the history of mathematics in general
and specific topics in the history of mathematics that can be found in the library. Please
make use of them.
Advice:
This paper should be approximately 1000 words and should be a coherent essay with a
controlling thesis statement and supporting citations.
To succeed you will undoubtedly need to narrow your subject to something that can be
addressed in a paper of this length. Talk with me as early as possible if you have
questions about what you intend to study. It would really be a good idea to make your
thesis as early as possible.
Wikipedia and other online sources that a search engine like google.com finds can be a
great starting point for research, but you shouldn't stop there. Check for yourself the
references that these pages use to make sure that you've been given an accurate picture.
Some additional Problems (due February
16)
(An angle trisection problem.) The figure given below
can be used to solve the angle trisection problem. (Here, PQ = QR = RS, PTR is a
semicircle with PR as a diameter, and UR is perpendicular to PS.) If <AOB is the tangle
to be trisected, place the tomahawk on the angle so that S lies on OA, the line segment
UR passes through O, and the semicircle with diameter PR is tangent to OB at T. Prove
that the triangles OTQ, ORQ, and ORS are all congruent, whence <ROA is one-third of
<AOB.
(Another angle trisection problem) Suppose (in the picture given below) that angle ABC
is to be trisected. Draw AC perpendicular to BC and complete the rectangle ADBC.
Extend DA to the point E which has the property that if BE meets AC in F, then the
segment FE is equal to twice AB.
Let FE be bisected at G. Show that FG = GE = AG = AB.
Show that angle ABG = angle AGB = 2 angle AEG = 2 angle FBC.
Show that angle FBC is 1/3 of angle ABC.
Prove that in the geometric figure given below
the area of the lunes I+II+III+ the semicircle on the diameter OD (where O is the
midpoint of the line AD) is equal to the area of the isosceles trapezoid ABCD (where
the length of AD is twice the length of each of AB, BC, and CD).
Online Resources
Note: These sites were generated by the students enrolled in Math 218 during Fall Term
of 2004. Their abstracts of the pages are included.
The MacTutor History of Mathematics Archive. This is probably the most extensive site
devoted to the history of mathematics on the web. Hosted by the School of Mathematics
and Statistics at the University of St. Andrews in Scotland and maintained by two
members of the faculty there, the site features numerous types of resources on the
history of mathematics which are available directly from the home page. The
"Biographies Index" is one of the most prominently featured resources. It contains brief
biographies of hundreds of mathematicians. These biographies usually include
photographs and a list of further references as well as links to other pertinent pages on
the site (e.g., maps and chronologically related mathematicians).. The site has received
numerous awards and is constantly being updated. In fact it features a list of newest
additions prominently on the home page and authors of the web pages are usually given
at the bottom of the page. (Submitted by Andrew Leahy.)
Some Pages within this site:
Babylonian and Egyptian Mathematics.
(Review One.) The School of Mathematics and Statistics host this site at the University
of St. Andrews Scotland. The site is broken into two main areas: Babylonian
mathematics, and Egyptian Mathematics. Both areas present you with a variety of topics
and a brief overview. The Babylonian index includes Babylonian numerals,
Pythagorean theorem in Babylonian math, as well as their history of zero. On the other
hand the Egyptian index includes the numerals, equations written on papyrus, and their
history of zero.
(Review Two.) This site is hosted by the School of mathematics and Statistics at St.
Andrews, Scotland. It deals almost exclusively with Babylonian and Egyptian
mathematics with a brief passage on the history of Zero. The site gives a broad
overview of the mathematics found in each civilization followed by a lengthy
discussion of respective number systems. The site also features a discussion of special
topics as they occur in Babylonian or Egyptian mathematics.
Mayan Mathematics. This site highlights what is known about Mayan mathematics. It
explains the rediscovery of the ancient mathematics after the Spanish missionaries
destroyed many of the relics. It contains links to other web sites with other information.
This site shows the Mayan numerical system and explains how the Mayans probably
developed their base ten system.
Chinese Mathematics. This site is hosted by the School of mathematics and Statistics at
St. Andrews, Scotland. It is primarily concerned with ancient China. It features a very
broad overview of Chinese mathematics, a discussion of both the nine chapters on the
mathematical art and The Ten mathematical classics. The site also features an
introduction to Chinese numerals and provides an extensive index of biographies.
Greek Mathematics. This page, provided by the School of Mathematics and Statistics at
St. Andrew's University in Scotland, links you to many articles about Greek
Mathematics and Mathematicians. The articles about the Mathematicians include short
biographies and some excerpts of what remains of their work including quotations and
proofs. There are also articles independent of the biographies investigating three of the
most important mathematical problems of the ancients Greeks: squaring the circle,
doubling the cube, and trisecting the angle.
David Joyce's History of Mathematics site.
(Review One.) This site does not go too in depth, but gives the chronological order of
how mathematics changed, what time period this happened and also where it happened.
It does not give information on the different mathematicians this site is maintained by
David E. Joyce from Clark University.
(Review Two.) This site is maintained by David Joyce of Clark University. As a viewer
of the website, you can view extensive lists of mathematicians based on chronological
order or the region from which they came from. If you're looking for information based
on a particular branch of mathematics, you can find information and sources of
information based on subject as well. This site also provides nice maps that show where
the regions listed actually are in relation to our world now. For example, the maps on
this site show that Babylonia was very near the current Iraq. In addition to providing
these maps and lists, this website provides links to other sites and books where you can
find more in-depth information not covered in the website.
(Review Three.) This is an archive of mathematical history, organized by region,
individual subject, and chronology. Regions include Babylonia, Egypt, China, Greece,
India, Japan, and Europe. Topics are separated into numerals and counting, algebra,
geometry, arithmetic and number theory, mathematical analysis, and probability and
statistics. Also, it offers a mathematical timeline from 1700 BCE up to 1940. It offers
links to MacTutor in several places.
(Review Four.) Includes maps, timelines, chronologies, archives, links to other sites and
more.
(Review Five.) This site, maintained by David Joyce of the Department of Mathematics
and Computer Science at Clark University in Worcester, MA, contains links to
information on the history of mathematics organized by geographical region, subject,
and chronology. It also provides hundreds of links and listings for related websites,
books, and journals.
Some Pages within this site:
The History of Mathematics: China. This site was made and is maintained by David E.
Joyce of the department of Mathematics and Computer sciences at Clark University.
This site is a detailed list of the writings and mathematicians of ancient china. The site
starts with a list of important events, books, and people, with a brief description of what
was significant about them. Later there is a very detailed list of these events, books, and
people, in chronological order, which covers everything from ancient traditional texts to
the 19th century.
David Wilkins History of Mathematics site
(Review One.) This site, maintained by Dr. David Wilkins of the School of Mathematics
at Trinity College of Dublin, is a brief history of the mathematicians of the seventeenth
and eighteenth centuries, with emphasis on William Hamilton, Georg Riemann, Isaac
Newton, George Boole, and Georg Cantor. In all, over eighty mathematicians are
discussed, with short biographies and contributions in a chronology taken from W. W.
Rouse Ball's A Short Account of the History of Mathematics. Information is also
available in detail about the Analyst Controversy, which doubts the fidelity of analysis
methods used by Mathematicians.
(Review Two.) This site is maintained by Dr. David Wilkins from Trinity College,
Dublin. It gives extensive details about Sir William Rowan Hamilton, George Reimann,
and smaller sections on Sir Isaac Newton, George Boole and Georg Cantor.
<="" em=""> This website is relevant when looking for mathematicians from the 17th
and 18th centuries. It provides a long list of mathematicians from this time period. In
addition to this list, it also provides information based on these individuals. While this
website provides little information for mathematics outside of the 17th and 18th
centuries, it does provide links to other websites that can help you find information prior
to or beyond the 17th and 18th centuries.
(Review Four.) This site contains biographies of famous mathematicians and tells about
many of their important discoveries. It also has links to other potentially useful sites
about math history.
(Review Five.) Dr. David R. Wilkins of the School of Mathematics at Trinity College in
Dublin maintains this website devoted to the history of mathematics. Although the site
focuses largely on two mathematicians (Hamilton and Riemann), Wilkins also discusses
a number of other subjects and includes a list of links to a vast supply of other
information on the history of math. Each of his pages also contains links which are
specifically relevant to the subject of the page.
Some Pages within this site:
Leonhard Euler. This site gives information about Leonard Euler.
Mathematicians of the Seventeenth and Eighteenth Centuries This site offers a
chronological list (by birth date) of the great mathematicians of that era. The great list
includes Descartes, Pascal, Rolle, Newton, Leibniz, L'Hospital, Euler, Lagrange,
Laplace, Fourier, and many others.
Mesopotamian Mathematics.
(Review One.) This site was designed to help Duncan J. Melville's History of
Mathematics students. The political history of Mesopotamia is briefly outlined to put
the mathematical achievements in context. The site is well organized and covers a wide
arrangement of topics such as multiplication, the evolution of their number system,
square roots, and pre-algebraic problem solving.
(Review Two.) This site is hosted by Duncan J. Melville, professor of Ancient and
Classical mathematics at St. Lawrence University, New York. This is probably the most
extensive source of information on Mesopotamian mathematics and it is still being
updated. It starts of by giving a rather detailed summary of the political history of
Mesopotamia and the main periods relevant to the development of mathematics
followed by a discussion of mathematics as it occurred in the various periods. The site
also features links to tablets on the web.
(Review Three.) This is a fairly large site devoted exclusively to ancient Babylonian
mathematics. It includes background information on Babylonian political history, the
main periods of Mesopotamian mathematical history, as well as chronological
summaries of their numerical system, and specific developments during 2 different time
periods. Specific topics covered include Cuneiform numbers, multiplication tables,
reciprocals, and quadratic equations. There are many links to other related sites and to
bibliographical information.
Donald Allen's History of Mathematics site.
(Review One.) This site is a series of lectures on the History of Mathematics from G.
Donald Allen of Texas A & M University. The topics include Egyptian, American,
Chinese, Hindu, and Greek ancient mathematics, as well as focusing individually on
Thales, Anaxagoras, Pythagorus, Eudoxus, Euclid, Archimedes, Diophantus, Pappus,
Newton, Leibnitz, Barrow and Reimann.
(Review Two.) Includes many good topics
Some Pages within this site:
Lectures on the History of Mathematics.
(Review One.) Maintained by G. Donald Allen from Texas A&M University gives this
professors lectures on the history of Mathematics.
(Review Two.) Includes good overview of material, and in-depth readings.
Egyptian Mathematics.
Mathematicians of the African Diaspora present The Ancients.
(Review One.) This page is dedicated to the math of Africa other than that of Egypt. The
page includes the Truths and Myths about African mathematics, and gives histories of
mathematics in several different regions of the African continent. The page is hosted by
the Mathematics Department of the State University of New York at Buffalo and
contains links to other areas on the Buffalo site of mathematical interest.
(Review Two.) The Mathematics department of The State University of New York at
Buffalo presents "Mathematics of the African Diaspora." The aim of this site is quite
specific. It intends to educate the reader about the mathematical contributions of the
peoples of Africa that are very rarely covered in most histories of math. There are pages
on both the most common lies about this history and with truths, and the rest of the
pages are specific treatments of the geographical areas of Africa, organized by their
modern-day nations.
Some Pages within this site:
A Modern History of Blacks in Mathematics.
(Review One.) The main page offers a timeline of great African American
Mathematicians from 1849 up to 1997. Most names mentioned offer links to
biographical sites for those individuals.
(Review Two.) This site talks about some of the most influential African American
mathematicians.
Women in mathematics.
(Review One.) This is a fascinating site dedicated strictly to the biographies of female
mathematicians. It is provided by the department of mathematics at Agnes Scott
University. The essays it links to are primarily written by students of the college, but
there are a great many to choose from ranging from the fourth and fifth centuries CE to
the 21st. Unfortunately, the site does not always go into much more depth about the
women it features. Some entries are little more than a list of qualifications and a
mention of what type of math they dealt with.
(Review Two.) This is a site that contains biographies on women mathematicians. Over
140 women are profiled, some with very basic biographies, and others that have been
researched more extensively (all biographical essays were written by students at Agnes
Scott College in Atlanta, GA). There is a fairly broad spectrum of mathematicians and
the site is continuously being updated as new biographies are added or current
biographies expanded. This is not an extremely large site, but it is useful because so
much information about women in mathematics is centralized in one place.
(Review Three.) This site tells about some famous female mathematicians.
Math Forum. This website, called the Math Forum, contains hundreds of links to
websites related to the history of mathematics including the early uses of various
notations and symbols, biographies of mathematicians and ancient math contributors,
information on Euclid, the Mayans and their contributions to the field, university
studies, and much more. You can search the sites to find what you want, and the links
are to credible and useful sites.
Some Pages within this site:
Links to Biographies. This site contains links to various sites which contain biographies
of some important mathematicians.
Famous Problems. While this is part of the Math Forum site mentioned above, it is very
interesting and therefore worth mentioning on its own. This site is full of famous
problems in the history and development of math, such as finding the value of phi,
understanding and finding prime numbers, developing probability techniques and the
binomial theorem, proving that e is and irrational number, and several more. This is a
very interesting and informative site.
Mathematics archives.
(Review Two.) This site give links to other mathematical sites of interest, like article
about the abacus, female mathematicians, logarithmic tables, etc. The site is maintained
by Earl Fife and Larry Husch.
(Review Two.) This site provides an alphabetical index of other sites on topics ranging
from the use of an Abacus in various number systems to Women's contribution to
Mathematics. Other interesting topics include Archimedes and the computation of pi,
Fibonnaci - his rabbits and his numbers and Kepler, and most intriguing of all, great
Polish mathematicians.
(Review Three.) The Math Archives provide links to web pages related to forty different
topics about the history of mathematics. The site also includes information on the Math
Archives themselves, which are devoted to "provid[ing] organized Internet access to a
wide variety of mathematical resources," a list of awards the site has received, and also
links to many other similar sites.
Abacus: The Art of Calculating with Beads. This is a site dedicated to the Abacus, with
a history of the device in several worldwide cultures and its modern use and variations.
The site also contains tutors on addition, subtraction, multiplication and division, square
root and cube root calculations using the abacus, and online abacus simulations written
in Javascript. There are also resources comparing modern abacus use with mental
calculation, electric calculators, the Roman hand-abacus, the Mesoamerican abacus, and
the Incan string and knot Khipu. The site is the winner of the Scientific American 2003
Sci/Tech Web Award.
Ethnomathematics on the Web. This site is a collection of links to sites specializing in
the mathematics of different ethnic, geographic and social backgrounds ranging from
African fractals to the space concepts of street children in Brazil. The site is maintained
by the International Study Group on Ethnomathematics, a group dedicated to the
mathematical practices of identifiable cultural groups.
Leibniz. This site is completely dedicated to man after whom our Math Server is named,
Leibniz. It offers sections on his life and works, his texts, philosophy, link to other
Leibniz pages, research, conferences, articles, books, and the work of the author of this
site. Though some of it is in Finnish, the English writings are a good read.
Mathematical Timeline. This page is a very simple and direct timeline of important
mathematical achievements throughout history. It is quite comprehensive, reaching
from the depths of prehistory circa 30000 BC all the way to 2000. While not of great
value for researching a topic, it is quite handy for pinning down exactly when who did
what and in what order.
Surveying in Ancient Egypt. This site is a very short one, because it deals with one very
narrow topic: Surveying in Ancient Egypt. Surveyors were some of the earliest people
to have to apply mathematics. This site details the different tools surveyors used, the
different jobs they had to do, and the ways they go around a less advanced level of
technology to achieve mathematical accuracy in measurement.
Mathworld. This website is a great resource for many different areas of mathematics. It
also includes various mathematical historic topics.
Math-Net. This site contains information on many topics in mathematical history:
mathematical phenomena during the Renaissance, the birthplaces of famous
mathematicians, the work of Archimedes and the Greeks, mathematical achievements in
Africa, and also several links to articles in the MacTutor online resource.
Indian Mathematics. This is a site by an unknown author which deals exclusively with
the history of Indian mathematics. The site features very detailed accounts of
mathematics in the Vedic period, the Indian numeral system, the emergence of calculus
in ancient India and various special ding the value of phi, understanding and finding
prime numbers, developing probability techniques and the binomial theorem, proving
that e is and irrational number, and several more. This is a very interesting and
informative site.
ThinkQuest Challenge site. This site was created by South Korean students of math for
the ThinkQuest Internet Challenge. It contains information on the chronological
development of mathematics, from ancient civilizations through the present, an
alphabetical index of topics, and eight small quizzes on the history of mathematics.
The British Society for the History of Mathematics.
Supplemental Material for Math 218
1/6/03 The map of Mesopotamia mentioned in class is available here.
1/9/03
The map of ancient Greece mentioned in class is available here. The reading assignment
for Monday is to read the text up to (and including Section 2.3). The homework
problems (due next Wednesday) are from the text:
Chapter One: #4, 5, 8, 10, 12, 14, 38
Chapter Two: #1, 4, 6, 10
For those who do not yet have the text, a copy has been placed on close reserve in the
science library.
1/14/03 The URL for the online version of Euclid's Elements can be found here
1/16/03 Homework assignment for this week (due next Wednesday):
Read Chapter 2, Sections 4 and 5
Chapter 2: #11, 12, 13, 14, 15, 16
Most of these problems ask you to reproduce or extend results from the first few books
of Euclid. You are free to use the online version of Euclid linked to above as resources,
but make sure that the results are written in your own words.
Also, as a part of the same assignment, complete the following problems:
Prove that in the geometric figure given below
the area of the lunes I+II+III+ the semicircle on the diameter OD (where O is the
midpoint of the line AD) is equal to the area of the isosceles trapezoid ABCD (where
the length of AD is twice the length of each of AB, BC, and CD).
The figure given below:
can be used to solve the angle trisection problem. (Here, PQ = QR = RS, PTR is a
semicircle with PR as a diameter, and UR is perpendicular to PS.) If <AOB is the tangle
to be trisected, place the tomahawk on the angle so that S lies on OA, the line segment
UR passes through O, and the semicircle with diameter PR is tangent to OB at T. Prove
that the triangles OTQ, ORQ, and ORS are all congruent, whence <ROA is one-third of
<AOB.
(Another proof of the Pythagorean Theorem.) Consider a right triangle ABC with right
angle at C and whose legs have length a and b and whose hypotenuse has length c.
Extend BC to a point D such that BAD is a right angle.
From the similarity of triangles ABC and DBA show that AD = ac/b and DC = a2/b.
Prove that a2 + b
2 = c
2 by relating the areas of triangle ABD to the areas of the two
smaller triangles.
(Another proof of the Pythagorean Theorem.) Consider the two congruent right
triangles EAD and ABC arranged as shown below (so that D, A, and C are colinear) and
draw the line EB so as to form the quadrilateral EBCD.
Prove that a2 + b
2 = c
2 by relating the area of the quadrilateral to the area of the three
triangles ABC, EAD, and EBA.
The results of Book IV of Euclid's elements can be thought of as showing ways to
construct regular n-gons (i.e., n-sided polygons which are equilateral and equiangular).
Show (by performing the actual construction) that using the results of Book IV it is
possible to construct a regular 20-gon.
1/24/03 Here's an assignment which I can't assign to you yet:
Read Chapter Three.
Complete the following problems:
Chapter 2 #24, 26, 27, 28, 37
Chapter 3 #12, 13, 19
We will not be discussing some of the material in Chapter 3 related to Archimedes in
class. I have endeavoured not to assign problems regarding topics we have not
discussed in class, but that should not preclude you from reading those sections
thoroughly.
Also, as a part of this assignment, you should complete the following:
In his work "On Spirals", Archimedes studies spiral curves. He describes how to
construct a spiral as follows:
If a straight line [half-ray] one extremity of which remains fixed be made to revolve at a
uniform rate in the plane until it returns to the position from which it started, and if, at
the same time as the straight line is revolving, a point moves at a uniform rate along the
straight line, starting from the fixed extremity, the point will describe a spiral in the
plane.
These curves have the polar equation r(t) = k t in modern notation. (Try the
Mathematica command PolarPlot[3 t, {t, 0, 2 Pi}] if you don't know what this looks
like.) See Katz, page 114, for more on spirals.
Spirals can be used to trisect angles: Given a spiral, place the angle to be trisected so
that the vertex and the initial side of the angle coincide with the initial point of the spiral
and the initial position OA of the rotation ray. Let the terminal side of the angle
intersect the spiral at P.
Explain how to trisect the segment OP at the points Q and R.
Draw circles with center at O and with OQ and OR as radii. Prove that if these circles
meet the spiral in points U and V, then the lines OU and OV with trisect the angle AOP.
The spiral can also be used to solve the circle quadrature problem. Given a circle with
center at O and radius a, draw the spiral whose equation in polar coordinates is r = a t,
and whose initial point is O.
Prove that when the rotating ray is revolved perpendicular to its inital position OA, the
segment OP will have a length equal to one-fourth the circumference of the circle.
Show that this result can be used to find the quadrature of the circle. (Hint: You do
know how to find the square root of any length which has been handed to you...)
2/4/03 The homework assignment due next Monday:
Read Chapter 4 up to Section 2.1
Chapter 3: #23, 24, 27
Chapter 4: #1, 4 (students with an interest in computers may do problem 5 instead)
Also, complete the following problems:
Complete the unfinished half of the exhaustion proof in Proposition 24 of Archimedes'
Quadrature of the Parabola.
In class, we proved the following result of Ptolemy:
Let there be a circle ABGD on diameter AD and with center Z. From A let there be cut
off in succession two given arcs, AB and BG and join the corresponding chords AB and
BG. Then if we join AG, that chord will also be given.
Show that this result (as proved in class) is equivalent to the angle sum formula for
sin(x). (Note: You will need your class notes for the proof of this result to complete the
problem.)
(Another angle trisection problem) Suppose (in the picture given below) that angle ABC
is to be trisected. Draw AC perpendicular to BC and complete the rectangle ADBC.
Extend DA to the point E which has the property that if BE meets AC in F, then the
segment FE is equal to twice AB.
Let FE be bisected at G. Show that FG = GE = AG = AB.
Show that angle ABG = angle AGB = 2 angle AEG = 2 angle FBC.
Show that angle FBC is 1/3 of angle ABC.
(The crux of using this method to solve the trisection problem is figuring out how to
construct a segment BE which satisfies the conditions of the problem. Conic sections
can be used to do this. See Katz, pages 127-127 for details.)
In Proposition 8 of Book II of the Conics, Apollonius shows that
If a straight line meets an hyperbola in two points, produced both ways it will meet the
asymptotes, and the straight lines cut off on it by the section from the asymptotes will be
equal.
In other words, in the picture below EA = CF.
Give an intuitive argument for why this implies that the two line segments of a tangent
to a hyperbola between the point of tangency and the asymptotes are equal.
Show (without using calculus) that this result implies that the slope of the tangent line to
the curve y = 1/x at the point (x0, 1/x0) equals -1/x02.
As Katz points out (see page 147), nobody really knows how Ptolemy performed his
square root calculations. A later commentator, Theon of Alexandria, proposed the
following:
If we seek the square root of any number, we take first the side of the nearest square
number [beneath it], double it, divide the product into the remainder reduced to
minutes, and subtract the square of the quotient; proceeding in this way, we reduce the
remainder to seconds, divide it by twice the quotient in degrees and minutes, and we
shall have the required approximation the side of the square area.
Here are the essential ideas: Suppose you want to compute the square root of 7200. This
will be of the form x;y,z (in sexagesimal). What is x? "The side of the nearest square
number" beneath 7200. In this case 84, since 842 = 7056 and 85
2 = 7225. So the number
is of the form 84;y,z. How to find y? Double 84 to get 168. Now "divide the product
[i.e., 168] into the remainder reduced to minutes". That is, convert 7200 - 842 = 144 into
units of minutes (i.e., units of 1/60th's) to get 8640 and divide the result by 168 to get 51
(the nearest integer). Then the number is of the form 84;51,z. Now what is z? Do the
same thing: Take 84;51, square it, subtract it from 7200 to get 0;28,39, multiply this by
602 to get 1719. Then divide this by 2 times 84;51 to get 10 (the nearest integer. Hence
the square root of 7200 is 84;51,10.
Why does this work? First, think of the solution as being of the form x + (1/60)y. Then,
as above, x = 84. Next, the object is to find y such that 7200 > (84 + (1/60)y)2. We can
square this right-hand side to get 842 + (2*84*y)/60 + (y/60)
2. This last term is so small
that we can safely ignore it. So we must find the largest y such that 7200 - 842 >
(2*84*y)/60. Written out the procedure for solving for y amounts to multiplying 7200 -
842 by 60 and then dividing by 2*84, which is exactly what you were told to do above.
The closest value of y (from below) will be 51. To find z, repeat the entire procedure
with the inequality 7200 > (84;51 + (1/60)2z)
2. (You'll find that this yields the same
algorithm as described above for finding z.)
Prior to the invention of calculators, this was taught as a standard algorithm (in base 10,
of course) for finding square roots.
Your exercise: Complete Problem 2 of Chapter Four.
2/6/03 Biography Research Topics:
Heron of Alexandria
Nichomachus of Gerasa
Diophantus
Pappus
Liu Hui and other early Chinese Mathematicians
Qin JiuShao
Brahmagupta
Bhaskara
Each group of two or three students is to complete a research project on one of the
above topics. Each individual will be responsible for (1) a three page paper on a specific
area of the mathematician's work and (2) assisting in preparing a poster describing the
mathematician's work.
The research paper should primarily show your understanding of the mathematics you
are presenting. Thus, it should be a formal mathematical presentation. But your paper
should also present a coherent thesis which ties it together. In particular, you should
argue why the results you present are important in the history of mathematics.
The poster should summarize as much as possible the author's work. At minimum it
should include biographical information, references, and examples of the author's
results and methods. The content should be appropriate for a poster-size presentation
that can be browsed by a novice in the field.
I would like each of the topics above to be covered. Consequently, there should be five
groups of two, and one group of three. Students interested in working together or on a
specific topic should e-mail their names and/or desired subject areas to me by Sunday
evening.
The paper and presentation will be due on February 20.
2/11/03 Here is my list of who is working on which projects:
Heron of Alexandria - Trisha Greiner and Susan Vitous
Nichomachus of Gerasa - Chris Enlund and Nathan Wright
Diophantus - Laris Hazners and Ben Stripe
Pappus - Eli Dust and Justin Griffiths
Liu Hui and other early Chinese Mathematicians - Rashid Amid and Martin Winslow
Qin JiuShao - Jason Myers and Bill Stewart
Brahmagupta - Rachel Dahlmeier, Susannah Go, and Nate Totz
Bhaskara - Will Middlecamp and Vanessa Ribeiro
Here some links that might be useful for the biography project:
The MacTutor History of Mathematics Archive
David Joyce's collection of history links (David Joyce created the online version of
Euclid we've been using.)
The Math Forum History of Mathematics Links Page
I've also placed some additional books on reserve in the science library. These may be
helpful for your projects:
Chinese Mathematics: A Concise History by Yen Li
A History of Chinese Mathematics by Jean-Claude Martzloff
An Introduction to the History of Mathematics by Howard Eves
A History of Mathematics by Karl Boyer
2/13/03
Here are the essay questions for tomorrow's exam. For each question you should be
prepared to produced a well-considered essay, complete with thesis and supporting
evidence.
Discuss the influence that mathematicians in the medieval Islamic world had on
subsequent European mathematical developments.
Discuss the influence that Babylonian and Egyptian mathematics had on Greek
mathematics.
Discuss Euclid's success as a model for mathematical rigor.
Discuss the influence of Greek philosophers and the Greek philosophical tradition had
on the development of mathematics in Greece.
Discuss the changes in the roles of geometry and algebra in mathematical thought
between classical times and modern times.
2/14/03 Here's the latest homework batch. These should be turned in by Monday, February 24.
Read Chapters 8 and 9.
Chapter 7: #3, 5, 12, 15
Chapter 8: #17, 26, 38, 39
Chapter 9: #32, 33, 37, 38
2/25/03 The link for the first eleven proposition's from James Gregory's Geometriae Pars
Universalis may be found here.
TRISECTING AN ANGLE
NOTE: all images are schematic and don't represent real values. They are just there to
show the idea.
Problem: Find a general method for trisection of any given angle using the classical
tools.
You start out with an angle:
pic1)
Draw a circle with the angle in its centre and draw to radii's following the sides of the
angle. You can give the radii of the circle any length you want; this is completely
arbitrary. The radii of this circle we call 'r' and the circles diameter we call 'd':
pic2) Our original angle in white. We drew a circle around
it(red) and the circles radii's are the green lines going out and
following the sides of the angle.
You can then erase the circle and connect the tips of the radii lines:
pic3) The green radii's are connected by the red line. Let's call this
red line 'y'
Then split 'y' in two parts by drawing two arbitrarily large circles that intersects each
other. One whose centre is allocated on the tip of one of the green radii lines and
another whose centre lies on the other green radii line. Draw a line that goes through the
two points where the two new circles intersects and you've split 'y' in two:
pic4) The blue circles are the ones that is used for
splitting 'y', they have their centres at the tips of the
radii lines, and the yellow line is the line that goes
through the circle's intersection points and splits 'y.'
We draw a circle whose centre is at the middle of 'y' and that has a radii of 'd'(that is it's
radii is the same length as the diameter of our original circle in pic2). Then draw a line
that goes through the edge of the angle and through 'y's middle point and on to intersect
the newly draw circle. This line may not leave the angle area, we call this line 'l1':
pic5) The blue circle has a radii of 'd' and it's
centre is allocated on the middle of 'y'(the place
where the yellow and red line intersects). The
yellow line is 'l1' and goes from the angles edge,
at the bottom, through 'y' and intersects the blue
circle inside the angle(this will be important
further down).
At the place where this new line intersects the circle with radii 'd' you draw another
circle with the radii 'r'(that is the same radii as the circle in pic2), let's call this circle
'c1':
pic6) The purple circle is 'c1'and has it's centre
where the blue circle and the yellow line
intersects.
Then draw another circle with a diameter 'd' that has it's centre at the angles edge. After
that you draw a straight line that starts from the edge of the angle and continues away
from the interior of the angle and intersects the newly draw circle with the radii 'd', we
call this line 'l2':
pic7) The purple circle is 'c1', which we drew
previously, and the yellow circle is our new
circle with the radii 'd', it has it's centre at
the edge i.e. at the place where the two green
lines meet. The blue line is 'l2'.
See if 'l1' intersects 'y'. If it does you subtract the length from 'y' and to the edge of the
angle from 'd'(not mathematically though, you have to do it graphically by
accommodating your compass to the space between the line 'y' and the intersection point
on the yellow circle) , we call this new length 'z'. If 'l1' doesn't intersect 'y' we're left
with the whole of the diameter(i.e. 'z'='d' and for all angles below 180o 'y' wont be
intersected). Now draw a circle with the centre at the edge of the angle and with the
radii 'z', let's call this circle 'c2':
pic8) The purple circle is 'c1', which we drew
previously, and the blue circle is our new
circle with the radii 'z', the circle we call 'c2'.
Note that since 'l2' didn't intersect 'y' this new,
blue, circle is identical to the yellow circle in
pic7.
We draw two lines from the edge of the angle to both the places where 'c1' and 'c2'
intersects and then we've split the angle in three(though we really haven't):
pic9) The purple circle is 'c1' and the blue
circle is 'c2'. The two yellow lines cut's the
angle in three
Now I'll also show how to do it for angle over 180o:
Draw and split 'y' as described in pic3-4 and draw 'c1' as in pic5-6.
pic10) Note that since the line 'l1' had to go
through both the edge of the angel and the
middle of 'y' and intersect the circle inside the
angle area, the intersection point, and
therefore 'c1', gets much closer to the edge of
the angle:
Then draw the circle with the radii 'd' whose centre is at the edge of the angle and then
draw 'l2', all as described in pic7. Note that 'l2' now intersects 'y'(since it had to move
away from the interior of the angle) and therefore you subtract this length between the
edge of the angle and 'y' from 'd' to get a length 'z':
pic11) The purple line between the edge of the
angle and 'y' is the part we've subtracted from
'l2', the yellow line above is what we're left
with.
Then draw 'c2' as in pic8 and connect the edge of the angle with the intersection points
'c2' and 'c1':
pic12)
Why it doesn't work: Uuumm... It simply doesn't work? I might be a fairly good
approximation.
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Compass and straightedge constructions
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Part of the series on
Mathematics
2+2=4
Axiom
Borel's Law
Complex numbers
Golden ratio
Innumeracy
Paraconsistent logic
v - t - e
This kind of compass
Geometry from the days of the ancient Greeks placed great emphasis on problems of
constructing various geometric figures using only a compass (the circle-drawing
gadget), and a straightedge (like a ruler, but without distance markings). For the most
part, early mathematicians were highly successful in this undertaking: Euclid's
Elements, for example, contains a vast collection of elementary constructions,
including, for example, a way to bisect any angle, or construct a regular pentagon.
A few problems in particular proved resistant to these constructive methods, perhaps
most notoriously that of squaring the circle. Nonetheless they occupy an important role
in the history of mathematics: the first earliest hints of calculus appear in Archimedes'
work on the closely related problem of approximating , and the proof of the
impossibility of these constructions was among the earliest triumphs of new algebraic
methods developed in the 1800s.
The problems also enjoy an important role in the history of pseudomathematics, where
they have long served as a distraction for cranks. In the late 1700s the bogus
suggestions of means for squaring the circle were so numerous that both the French
Academie des Sciences and English Royal Society were forced to pass resolutions
declaring that they would cease to review any manuscripts suggesting solutions to these
problems.[1]
Even though the impossibility of these constructions has been well-
understood for more than a hundred years, circle-squarers persist in their efforts even
today.
Contents
[hide]
1 The problems
2 The attempts
3 The solution
4 The cranks
5 What is possible
6 Some technical stuff
7 Footnotes
[edit] The problems
A few constructions remained that the Greeks were never able to give, and these
remained mysteries until modern times. Three problems in particular attracted the most
attention:
Squaring the circle: given a circle, construct using compass and straightedge a square
having the same area.
Trisecting an angle: given an angle , divide it into three equal angles of measure
.
Doubling the cube: given a cube, construct another cube with twice its volume.
[edit] The attempts
The problem of squaring the circle has a very, very long history. Approximate methods
were known to the Babylonians, Egyptians, and early Indian mathematicians. The
problem was the source of considerable consternation to the Greek geometers, many of
whom struggled with it, perhaps most notably Anaxagoras and Hippocrates of Chios. It
became a famous enough question that it is mentioned in Aristophanes' The Birds, and
the Greeks even found it necessary to coin a world for those who wasted time on the
problem. The other two problems too were studied intensively by the Greeks, though
they appear to have received comparatively little attention in later eras.[2]
The medieval Arabs too invested considerable effort in the quadrature of the circle: Al-
Haytham announced that he had discovered a method, but later failed to produce it.
Various medieval Italians, including Leonardo, also failed to make much progress. Even
non-mathematicians weighed in, and Thomas Hobbes in his later years believed that he
had found the solution.[1]
The first real insight into the problem came from James Gregory, one of the first
mathematicians to come to terms with infinite sequences. He realized that the problem
boiled down to understanding the algebraic properties of . Unfortunately the specifics
of his argument were deeply flawed. In 1761 Lambert made a major advance by proving
that was irrational, but this was not enough to rule out the possibility of circle-
squaring: for example, is possible to construct (it's the length of the diagonal of a
unit square), but it is irrational.
Gauss soon announced that the doubling of the cube and trisection of an angle were
impossible, though his proofs have never been located. The impossibility of trisecting
the angle and doubling the cube, the proofs of which rely on similar methods, were
finally settled in 1837 by Pierre Wantzel, and Sturm soon improved on his methods.
Even with great advances in algebra, the final nail in the coffin of the circle-squarers
was not delivered until 1882 with Lindemann's proof that is transcendental.[3]
[edit] The solution
Animation showing the construction of a heptadecagon with straightedge and
compasses
It's generally easy to prove that it is possible to carry out some geometric construction,
simply by showing how to do it. However, it is not at all clear how one could hope to
show that it's impossible to make other constructions -- how can we hope to rule out the
infinite different techniques that might be applied to the problem?
The key insight in this proof was to translate it from a problem in geometry to a
problem in algebra. Each of the three problems eventually boils down to constructing a
line segment of some certain length , starting with a line segment of length . This
works as follows:
Suppose we are given a circle of radius . Since , this circle has the same
area as a square with side length . Thus the problem of squaring the circle is
equivalent to constructing a line segment of length : if a segment of length
could be constructed, the circle may be squared, and vice versa.
Suppose we are given a cube of side length . If this cube could be duplicated, the new
cube would have side length . So the problem of a doubling the cube is equivalent
to constructing a line segment of length .
To show that it is impossible to trisect the angle, it's enough to show that there is some
specific angle that can be constructed, but one third of which cannot. This would show
that a general method for any angle cannot be created because that general method
should also be applicable to this novel case. One example of such an angle is
; that is, there's no way to construct a angle exactly using just a compass and
straightedge. As illustrated below, trisecting a angle is equivalent to constructing a
line segment of length .
A number is called constructible if a line segment of length can be constructed by
straightedge and compass, starting with a segment of length 1: the proofs of
impossibility eventually come down to an analysis of the algebraic properties of
constructible numbers, and the observation that all three of , , and are not
constructible.
Let's briefly consider which lengths it does seem possible to construct in this way.
We can construct a segment of length for any whole number . Just stick a bunch of
length-1 segments together.
We can construct a segment of length of any rational (i.e. fractional) length.
Construct a segment of length (an integer), and then divide it into pieces (this is
easy to do -- it was known to Euclid).
We can get for any whole .
It's also possible to get things like by similar methods.
For a more exhaustive discussion of which numbers are constructible, see below.
The above examples suggest that the constructible numbers are those which we can get
by adding and repeatedly taking square roots. This intuition turns out to be basically
correct. For example, Gauss showed that it's possible to construct the heptadecagon, a
regular polygon with 17 sides. Such a construction is equivalent to constructing the
number
This number looks complicated, but it's something we get just by taking a bunch of
square roots. So it turns out to be constructible!
It is this basic intuition that makes possible the proofs that , , and are not
constructible:
. Lindemann proved that (and therefore ) is transcendental -- this
implies that it can't be made using roots at all (even allowing cube roots, etc.). But this
means that it isn't constructible.
. This is a cube root, not a square root. But we're not allowed to take cube
roots when writing down a constructible number.
. This one's a little trickier, but a bit of high school trignometry shows
that . Plugging in this shows that
. So
. This is a cubic equation -- such things can be solved in general
by something like the quadratic formula, but it's a mess. The important part is that the
result involves cube roots, and so is some messy expression involving a cube root.
But this means that it isn't constructible!
[edit] The cranks
This was all on very firm footing by 1900, and no serious mathematicians or sane
amateurs believed otherwise. Nevertheless cranks persist in working on these three
problems even today. "The race of circle squarers," wrote Schubert, "will never die out
as long as ignorance and the thirst for glory remain united."[4]
They now have even
bolder goals that the cranks of previous eras: a method of squaring the circle with ruler
and straightedge would not merely settle an old problem and bring fame to its solver,
but would bring down much of the mathematics done in the last 150 years with it.
Needless to say none of them have succeeded, or come close. The exploits of the
modern circle-squarers have been extensively chronicled by the mathematician
Underwood Dudley, in volumes including Mathematical Cranks and The Trisectors.[5]
[edit] What is possible
If tools other than a compass and straightedge are allowed, all of these constructions
become possible. For example, if one allows paper folding (origami-like techniques), it
becomes possible to trisect the angle and double the cube. Similarly, angle trisection is
possible using a marked ruler, which is distinct from the straightedge considered here,
and only requires the meta-step of using the compass to mark off intervals on the
straightedge.
Approximate solutions to these problems are also fairly easy to come by. Ramanujan,
Martin Gardener, and Benjamin Bold have all given methods of squaring the circle
accurate to about 0.0001% based on the approximation
.
[edit] Some technical stuff
A bit of abstract algebra is all that is needed to make the above arguments about
constructible numbers rigorous. The curious reader can find a more careful and accurate
treatment of this material in any reasonable algebra textbook.[6]
A field is, roughly
speaking, a collection of numbers that can be added and multiplied (while staying in the
collection) and for which all not equal to zero, the multiplicative inverse and
the additive inverse are also in the collection. For example, the set of numbers of the
form where and are rational, is a field (denoted ), since:
.
A key theorem is that the constructible numbers are a field. For all three of the above
facts, this is pretty easy to see: if we have two constructible numbers, their sum is
constructible because we can just stick line segments of the two lengths together. The
other important fact is that if is constructible, then is too, so this field is closed
under the operation of taking square roots.
The most important algebraic definition required here is that of the degree of an
algebraic number. A number is said to be algebraic if it satisfies some polynomial with
integer coefficients. The degree of is defined to be the smallest such that there is a
polynomial of degree , with integer coefficients, which has as a root. Thus
Any rational number is of degree 1, since it is a root of .
The square root of two has degree , since it satisfies but does not satisfy
any polynomial of degree one.
This polynomial of minimal degree is called the minimal polynomial of .
For the numbers important to the constructability questions above, we have:
is transcendental, meaning it doesn't satisfy any polynomial at all. Thus it
does not have a degree.
. This satisfies , but no smaller polynomial. Thus it is of
degree .
. The trigonometry recited above shows that this satisfies
. It too is of degree 3.
The intuition above is that a constructible number is an algebraic number built by
repeatedly taking square roots. But any number of this form has a degree which is a
power of 2. Since the numbers above do not have such degrees, we conclude that they
are not constructible!
Once this foundation in Galois theory was in place, problems about straightedge and
compass became fairly routine: in general, it's not terribly difficult to determine whether
a given construction can be done or not, using tools of algebra. For example, there's a
simple criterion to test whether the angle can be trisected, in terms of algebraic
properties of .
[edit] Footnotes
↑ 1.0
1.1
Squaring the circle
↑ [1]
↑ Lindemann's original paper
↑ Beckmann, Petr. A History of Pi (author may be crazy, but nothing wrong with the
book)
↑ Buy it at Amazon!
↑ For example, try Artin's Algebra
Conchoid of Nicomedes
History
Differential Equations, Mechanics, and Computation
According to common modern accounts, the conchoid of Nicomedes was first
conceived around 200 B.C by Nicomedes, to solve the angle trisection problem. The
name conchoid is derived from Greek meaning “shell”, as in the word conch. The curve
is also known as cochloid.
From E. H. Lockwood (1961): The invention of the conchoid (‘mussel-shell shaped’) is
ascribed to Nicomedes (second century B.C.) by Pappus and other classical authors; it
was a favourite with the mathematicians of the seventeenth century as a specimen for
the new method of analytical geometry and calculus. It could be used (as was the
purpose of its invention) to solve the two problems of doubling the cube and of
trisecting a angle; and hence for veery cubic or quartic prblem. For this reason, Newton
suggested that it would be treated as a 'standard' curve.
Did Nicomedes use the curve to double the cube? Who used the name cochloid and in
where?
Description
Conchoid of Nicomedes describes a family of curves of one parameter. They are special
cases of the general conchoid. They are the conchoids of a line.
Step-by-step explaination:
Given a line l, a point O not on l, and a distance k.
Draw a line m passing O and any point P on l.
Mark points Q1 and Q2 on m such that distance[Q1,P]==distance[Q2,p]==k.
The locus of Q1 and Q2 for variable point P on l is the conchoid of Nicomedes.
curve tracing Conchoid of Nicomedes Construction
The point O is called the pole. The given line is called its directrix, and is a asymptote.
Concoid of Nicomedes curve family. Their
pole is at the origin, and directrix y==1. The figure on the left has constants k from -2 to
2. The right has constants k from -100 to 100.
Formulas
Let the distance between pole and line be b, and the given constant be k. The curve only
has one parameter k, because for a given b, all families of the curve can be generated by
varing k (they only differ in scale). (similarly, we can use b as the parameter.) In a
mathematical context, we shoud just use b==1. However, it is convient to have formulas
that has both b and k. Also, for a given k, the curve has two branches. In a math context,
it would be better to define the curve with a signed constant k corresponding to a curve
of only one branch. We will be using this intepretation of k. In this respect, the conchoid
of Nichomedes is then two conchoids of a line with constants k and -k.
The curve with negative offset can be classified into three types: if b < k there is a loop;
if b == k, a cusp; and if b > k, smooth. Curves with positive offsets are always smooth.
The following formulas are for conchoid of a line y==b, pole O at the origin, and offset
k.
Polar: r==b/Sin[θ]+k, -π/2 < θ < π/2. This equation is easily derived. Because a line
x==b in polar equation is r==b/Cos[θ] with -π/2 < θ < π/2.
Some observation of Cosine shows that if θ goes from 0 to 2*π, two conchoids with
offset ±Abs[k] results from a single equation r==b/Cos[θ]+k. To rotate the graph by π/2,
we replace cosine by sine.
Parametric: {t + (k*t)/Sqrt[b^2 + t^2], b + (b*k)/Sqrt[b^2 + t^2]}, -∞ < t < ∞.
This equation follows from the parametric formula for the general conchoid, where the
general point {x,y} is replaced by the curve of a line {t,b}, and pole {a,b} is replaced by
{0,0}.
If we substitute t in the above parametric equation by b*Tan[t], we have this form: (k +
b/Cos[t])*{Sin[t],Cos[t]}, -π/2 < t < 3/2 π. t ≠ π/2. This form is a direct translation from
the polar formula.
For conchoids of a line with positive and negative offsets k and a point at origin, the
Cartesian equation is: (x^2+y^2)*(y-b)^2==k^2*y^2. If k < b, the point at origin is a
isolated point.
if k < 0 and b < Abs[k], the conchoid has a loop with area (b*Sqrt[k^2-b^2]-
2*b*k*Log[(k+Sqrt[k^2-b^2])/b]+k^2*ArcCos[b/k]). The area between any conchoid
of a line and its asymptote is infinite. need proof.
From Lockwood: “The gradient at the node is given by Cos[psi]==+-b/k”. Not sure
what it means.
Properties
Tangent Construction
Look at the conchoid tracing as a mechanical device, where a bar line[O,P] slides on a
line at P and a fixed joint O. The point P on the bar moves along the directrix, and the
point at O moves in the direction of vector[O,P]. We know the direction of motion of
the bar at the two joins O and P at any time. The intersection of normals to these
directions forms the instaneous center of rotation N. Since the tracing points Q1 and Q2
are parts of the apparatus, thus N is also their center of rotation and therefore line[N,Q1]
and line[N,Q2] are the curve's normals.
Angle Trisection
The curve can be used to solve the Greek Angle Trisection problem. Given a acute
angle AOB, we want to construct a angle that is 1/3 of AOB, with the help of conchoid
of Nicomedes. Steps: Draw a line m passing segment[A,O] and perpendicular to it. Let
D:=intersection[m,line[A,O]], L:=intersection[m,line[B,O]]. Suppose we are given a
conchoid of Nicomedes, with pole at O, line m, and offset 2*distance[O,L]. Draw line l
passing L and perpendicular to m. Let C be a intersection of the curve and l, the one on
the opposite side of the pole. angle[A,O,B]==3*angle[A,O,C]. proof in gsp.
Proof: angle[A,O,C]==angle[O,C,L] since line[O,C] cuts parallel lines. Let
q:=line[O,C], N:=intersection[m,q], M:=midpoint[N,C], k:=distance[O,L]. By our
construction, we have distance[N,M]==distance[M,C]==k. Since NLC is a right
triangle, we see that MN, ML, MC, OL all have the same length, thus triangle[M,L,C]
and triangle[M,L,N] are isosceles and it follows that angle[N,M,L]==2*angle[M,C,L].
Since distance[O,L]==distance[M,L], triangle[M,L,O] is also a isosceles, and thus two
of its angles are equal. We have shown that angles equivalent to angle[A,O,C] is 1/3
times angles equivalent to angle[A,O,B]. End of proof.
The essential element where the conchoid made the trisection possible is the
construction of the point C on l such that distance[N,C]==2*distance[O,L], where N is
the intersection of m and line[O,C]. Note that for each angle to trisect, a new conchoid
is needed. This is in contrast to some other trisectrixes such as the quadratrix, where all
angles can be trisected once the curve is given.
Doubling the Cube
The curve can be used to solve the Greek Cube Doubling problem.
How to use this curve to double the cube?
Graphics Gallery
Tangent circles
of a conchoid of a line.
conchoid_conchoid_ani.mov animation code curve tracing
Angle Bisector
The (interior) bisector of an angle, also called the internal angle bisector (Kimberling
1998, pp. 11-12), is the line or line segment that divides the angle into two equal parts.
The angle bisectors meet at the incenter , which has trilinear coordinates 1:1:1.
The length of the bisector of angle in the above triangle is given by
where and .
The points , , and have trilinear coordinates , , and ,
respectively, and form the vertices of the incentral triangle.
SEE ALSO: Angle, Angle Bisector Theorem, Angle Trisection, Cyclic Quadrangle,
Exterior Angle Bisector, Incenter, Incentral Triangle, Incircle, Isodynamic Points,
Orthocentric System, Steiner-Lehmus Theorem
REFERENCES:
Altshiller-Court, N. College Geometry: A Second Course in Plane Geometry for
Colleges and Normal Schools, 2nd ed., rev. enl. New York: Barnes and Noble, p. 18,
1952.
Coxeter, H. S. M. and Greitzer, S. L. Geometry Revisited. Washington, DC: Math.
Assoc. Amer., pp. 9-10, 1967.
Dixon, R. Mathographics. New York: Dover, p. 19, 1991.
Kimberling, C. "Triangle Centers and Central Triangles." Congr. Numer. 129, 1-295,
1998.
Mackay, J. S. "Properties Concerned with the Angular Bisectors of a Triangle." Proc.
Edinburgh Math. Soc. 13, 37-102, 1895.
Pedoe, D. Circles: A Mathematical View, rev. ed. Washington, DC: Math. Assoc.
Amer., pp. xiv-xv, 1995.
Referenced on Wolfram|Alpha: Angle Bisector
CITE THIS AS:
Weisstein, Eric W. "Angle Bisector." From MathWorld--A Wolfram Web Resource.
http://mathworld.wolfram.com/AngleBisector.html
HCRP project 2008
The original Conchoid of 280 BC
Conchoid of Nicomedes
by Oliver Knill and Michael Teodorescu
Office: SciCtr 434
Email: [email protected]
Main page
Notes
Exhibits
Second part (Spring 2009)
Conchoids of Nicomedes are planar polar curves with r(t) = 1/cos(t) + c, where c is a
constant called offset. They are the images of a line under the exponential map of the
flat metric. These curves have historical significance since they were used to tackle
classical problems in geometry like angle trisection or cube doubling, problems which
can not be solved by ruler and compass alone. One thinks that Conchoids were first
used by Nicomedes in 200 BC to solve the angle trisection problem.
Nicomedes: 280-210 BC
A page on Conchoids
Questions and comments to [email protected]
Oliver Knill | Department of Mathematics | Harvard University
The Problem of Angle Trisection in Antiquity
A. Jackter
History of Mathematics
Rutgers, Spring 2000
The problem of trisecting an angle was posed by the Greeks in antiquity. For centuries
mathematicians sought a Euclidean construction, using "ruler and compass" methods, as
well as taking a number of other approaches: exact solutions by means of auxiliary
curves, and approximate solutions by Euclidean methods. The most influential
mathematicians to take up the problem were the Greeks Hippias, Archimedes, and
Nicomedes. The early work on this problem exhibits every imaginable grade of skill,
ranging from the most futile attempts, to excellent approximate solutions, as well as
ingenious solutions by the use of "higher" curves [Hobson]. Mathematicians eventually
came to the empirical conclusion that this problem could not be solved via purely
Euclidean constructions, but this raised a deeper problem: the need for a proof of its
impossibility under the stated restriction.
According to a traditional classification used in antiquity there are three types of
problems in geometry: plane, solid, and linear [Thomas]. `Plane' problems are those
which can be solved by "Euclidean" methods, using straight lines and circumferences of
circles. Using one or more of the sections of the cone, called conic surfaces, one solves
'Solid' problems. Problems involving other lines besides those mentioned above, having
a less natural and more complicated origin because they are generated from more
irregular surfaces or by purely mechanical constructions, are called `linear' [Thomas].
The early mathematicians were particularly interested in trisecting angles by `plane'
methods. As it turns out, the trisection of an angle is not a `plane' problem, but a `solid'
one [Heath]. This is why these early mathematicians failed to find a general
construction for angle trisection using Euclidean constructions, though they did find
elegant solutions based on the use of conic sections [Heath] and other more
sophisticated curves.
The trisection of an angle, or, more generally, dividing an angle into any number of
equal parts, is a natural extension of the problem of the bisection of an angle, which was
solved in ancient times. Euclid's solution to the problem of angle bisection, as given in
his Elements, is as follows:
"To bisect a given rectilineal angle: Let the angle BAC be the given rectilineal angle.
Thus it is required to bisect it. Let a point D be taken at random on AB; let AE be cut off
from AC equal to AD; let DE be joined, and on DE let the equilateral triangle DEF be
constructed; let AF be joined. I say that the straight line AF has bisected the angle
BAC. For, since AD is equal to AE, and AF is common, the two sides DA, AF are equal
to the two sides EA, AF respectively. And the base DF is equal to the base EF; therefore
the angle DAF is equal to the angle EAF. Therefore the given rectilineal angle BAC has
been bisected by the straight line AF." [Euclid, p. 264] Note that Euclid's construction
makes use of only lines and circles.
The earliest mathematician whose work bears on the problem of angle trisection was the
Greek Hippias, who was born about 460 BC and died about 400 BC. Hippias' main
contribution to Greek geometry was his invention of a curve called the quadratrix [D.E.
Smith]. The quadratrix was used originally for the problem of squaring the circle, and
later for the trisection of an angle [Smith,Euclid].
As shown the figure above, the construction of the quadratrix is as follows: "Suppose
that ABCD is a square and BED a quadrant of a circle with center A. Suppose
(1) that a radius of the circle moves uniformly about A from the position AB to the
position A D, and
(2) that in the same time the line BC move uniformly, always parallel to itself, and with
its extremity B moving along BA, from the position BC to the position AD"
[Euclid, p. 266]. "Then, in their ultimate positions, the moving straight line and th e
moving radius will both coincide with AD; and at any previous instant during the
motion the moving line and the moving radius will by their intersection determine a
point, as F or L. The locus of these point is the quadratrix" [Heath, p. 227]. "The
property of the curve is that, if F is any point, the arc BED is to the arc ED as AB is to
FH" [Euclid, p. 267]. "In other words, if phi is the angle FAD made by any radius
vector AF with AD, rho the length of AF, and a the length of the side of the square,
"Now the angle EAD can not only be trisected but divided in any given ratio by means
of the quadratrix" [Heath, p. 267]. "For let FH be divided at F' in the given ratio. Draw
F'L parallel to AD to meet the curve in L: join AL, and produce it to meet the circle in
N. Then the angles EAN, NAD are in the ratio of FF' to F'H. Thus the quadratrix lends
itself quite readily to the division of any angle in a given ratio" [Heath, p. 227].
Another mathematician who contributed greatly to the development of ancient geometry
was Archimedes, who was born in 287 BC in Syracuse, Sicily, and died there during the
Roman conquest of Syracuse in 212 BC [Dijksterhuis, pp.9-10]. Archimedes worked in
Sicily, but corresponded heavily with his mathematical colleagues, the Alexandrian
scholars Eratosthenes, Conon, and Dositheus. He would send them statements of the
propositions that he had proved, and waited to see if his colleagues could find a proof
on their own, and would later send detailed proofs to his fellow mathematicians in
Alexandria. [Dijksterhuis, pp.11-12].
Archimedes created a curve that he called the spiral, which was used to investigate the
squaring the circle and the trisection of an angle. He defined the spiral in a letter to
Dositheus as follows: If a straight line drawn in a plane revolves uniformly any num ber
of times about a fixed extremity until it returns to its original position, and if, at the
same time as the line revolves, a point moves uniformly along the straight line
beginning at the fixed extremity; the point will describe a spiral in the plane
[Archimedes, p. 154]. The spiral discovered by Archimedes was used for angle
trisection and more generally to solve the problem of cutting an angle contained by
straight lines in a given ratio. Archimedes used these spirals in most of his works, not
just to attempt to solve the trisection of angles problem [Dijksterhuis, p. 134].
"A proposition leading to the reduction of the trisection of an angle to another `verging'
is included in the collection of Lemmas (Liber Assumptorum) which has come to us
under the name of Archimedes through the Arabic. Though the Lemmas cannot have be
en written by Archimedes in their present form, because his name is quoted in them
more than once, it is probable that some of them are of Archimedean origin, and
especially is this the case with Prop. 8, since the 'verging' suggested by it is of very muc
h the same kind as those the solution of which is assumed in the treatise On Spirals,
Props. 5-8" [Heath, p. 241].
"The proposition is as follows: If AB be any chord of a circle with center O, and AB be
produced to C so that BC is equal to the radius, and if CO meet the circle in C, E, then
the arc AE will be equal to three times the arc BD. Draw the chord EF parallel to AB,
and join OB, OF. Since BO = BC, angle BOC = angle BCO. Now angle COF = 2 angle
OEF = 2 angle BCO, by parallels, = 2 angle BOC.
[Heath [9], p. 241]
Therefore angle BOF = 3 angle BOD, and arc BF = arc AE = 3 arc BD" [Heath, pp.
240-241]. This proposition gives a method of reducing the trisection of any angle.
Suppose that AE is the arc to be trisected, and that ED is the diameter through E of the
circle o f which AE is an arc. In order to find an arc equal to one third of AE, draw a
line ABC through A, meeting the circle again in B and ED produced in C, such that BC
is equal to the radius of the circle [Archimedes,Heath].
Another Greek mathematician to study the trisection of an angle was Nicomedes.
Nicomedes was born about 280 BC and died about 210 BC in Greece [Heath]. There is
not much known about the life of Nicomedes [Thomas]. What is known has been
inferred from mathematical criticisms of Eratosthenes' mathematical efforts and of
Nicomedes' works. Nicomedes is best known for his invention of the curve known as
the conchoid [Heath], which serves as the basis for an angle trisection.
Proclos was a mathematician and Neoplatonist philosopher who lived from 410 to 485
AD [Euclid, p. 29]. In his discussion of the trisection of an angle, he said, "Nicomedes
trisected any rectilineal angle by means of the conchoidal curves, the construction, order
and prope rties of which he handed down, being himself the discoverer of their peculiar
character. Others have done the same thing by means of the quadrtrices of Hippias and
Nicomedes...Others again, starting from the spirals of Archimedes, divided any given
recti lineal angle in any given ratio" [Heath, p. 225]. This curve was originally called
the cochloid by Pappus, a geometer from Alexandria who probably lived in the third
century, and called the conchoid by Proclos [Heath].
"There were varieties of the cochloidal curves; Pappus speaks of the 'first', 'second',
'third', and 'fourth', observing that the 'first' was used for trisecting an angle and
duplicating the cube, while the others were useful for other investigations" [Heath, p.
238]. It is the 'first' curve that is involved with the trisection of an angle. Nicomedes
constructed this curve by means of a mechanical device, as follows:
[Heath, p. 239]
AB is a ruler with a slot in it parallel to its length, FE a second ruler fixed at right angles
to the first, with a peg C fixed in it. A third ruler PC pointed at P has a slot in it parallel
to its length with fits the peg C. D is a fixed peg on PC in a straight line with the slot,
and D can move freely along the slot in AB. If then the ruler PC moves so that the peg
D describes the length of the slot in AB on each side of F, the extremity P of the ruler
describes the curve which is called a conchoid or cochloid. Nicomedes called the
straight line AB the ruler, the fixed point C the pole, and the constant length PD the
distance" [Heath, pp. 238-239].
"The fundamental property of the curve, which in polar coordinates would now be
denoted by the equation r = a + b sec(theta), is that if any radius vector be drawn from C
to the curve, as CP, the length intercepted on the radius vector between the curve and
the straight line AB is constant" [Heath, p. 239]. "Proclos says that Nicomedes gave the
construction, the order, and t he properties of the conchoidal lines; but nothing of his
treatise has come down to us except the construction of the 'first' conchoid, its
fundamental property, and the fact that the curve has the ruler or base as an asymptote
in each direction. The dis tinction, however, drawn by Pappus between the "first',
'second', 'third' and 'fourth' conchoids may well have been taken from the original
treatise, directly or indirectly" [Heath, pp. 239-240].
Carl Friedrich Gauss was born in 1777 and died in 1855 and is considered "the real
founder of modern German mathematics" [Smith, p. 502]. Gauss realized that using
only lines and circles, one could not solve the problems of trisecting an angle, nor that
of duplicating a cube. Unfortunately, Gauss left no proof of either of these statements.
However, he gave the construction of a regular 17-gon using only Euclidean
constructions [Dunham]. Mathematicians were interested in determining whether the
trisection of any angle was possible using only these methods. The first person to prove
its impossibility results was Pierre Laurent Wantzel, who published his proofs in a
paper called "Research on the Means of Knowing If a Problem of Geometry Can Be
Solved with Compass and Straight-edge", in 1837 [Dunham, pp. 245-247]. In Wantzel's
paper he proved the impossibility of the solution under Euclidean restrictions. Wantzel
regarded the magnitudes involved not as geometric segments, but as numerical lengths,
via analytic geometry. This let him use algebra and arithmetic rather than pure geometry
[Dunham, p. 245].
Throughout history the problem of trisecting an angle, or more generally of cutting an
angle into a given ratio, has challenged mathematicians. This quest endured for
centuries. In ancient times Hippias, Archimedes, and Nicomedes all invented curves
which suitable for ad hoc angle trisections, the quadratrix, the spiral, and the conchoid,
but these curves did not provide the desired "Euclidean" construction for the trisection
of angles, using only lines and circles. Nonetheless they were still valuable and
ingenious additions to the field of geometry. The recognition that it was possible to give
a proof of impossibility as precise and rigorous as a conventional mathematical solution
represents a conceptual turning point for mathematics, and a satisfactory solution of this
classical problem was not achieved until the 19th century.
References
Archimedes, The Works of Archimedes, translated and edited by Sir T. L. Heath, Dover
Publications, New York, 1953.
Clagett, M., Archimeds in the Middle Ages, vol. 1, University of Wisconsin Press,
Madison, WI 1954.
Dijksterhuis, E. J., Archimedes, Princeton University Press, Princeton, NJ, 1987.
Dunham, W. The Mathematical Universe, John Wiley & Sons, Inc., New York, 1994.
Euclid, The Thirteen Books of Euclid's Elements, tras. by Sir T. Heath, Dover
Publications, New York, 1956.
Eves, H. W., Mathematical Circles Squared, Prindle, Weber, and Schmidt, Boston, MA,
1972.
Gray, J., The Symbolic Universe, University Press, Oxford, 1999.
Heath, Sir T. L., A History of Greek Mathematics. Vol. 1: From Thales to Euclid, Dover
Publications, New York, 1981.
Hobson, E. W., Squaring the Circle, Cambridge University Press, Cambridge, 1913.
Proclos, A Commentary on the First Book of Euclid's Elements, translated and edited by
Morrow, Princeton University Press, Princeton, NJ 1970.
Roche, J. J., The Mathematics of Measurement: A Critical History, The Athlone Press,
London, 1998.
Smith, D. E., History of Mathematics, vol. 1, Dover Publications, New York, 1923.
Tarrant, D., The Hippias Major, Arno Press, New York, 1976.
Thomas, I., Selections Illustrating the History of Greek Mathematics, William
Heinemann, London, and Harvard University Press, Cambridge, MA, 1939, 1941.
Quadratrix Of Hippias
History
Differential Equations, Mechanics, and Computation
Quadratrix of Hippias is the first named curve other than circle and line. It is conceived
by Hippias of Ellis (≈460 BCE) to trisect the angle thus sometimes called trisectrix of
Hippias. The curve is better known as quadratrix because it is later used to square the
circle.
Description
Step by step description:
Let there be a unit square in the first quardrant with lower-left coner at the origin.
Let there be a line parallel to the x-axis and gradually moving up at a constant speed
from the bottom side of the square until it reaches the upper side of the square.
Let there be a angle in standard position. The angle increase at a constant speed from 0
to π/2. Both the angle and line movement start and finish simultaneously.
The intersections of the line and the angle is quadratrix of Hippias.
Tracing the Curve. Quadratrix of Hippias const
Given the definition, it is easy to derive the parametric equation. Let the parameter t be
the angle, 0 < t < π/2 and the square having side length π/2. So immediately, y == t.
Since we know the angle and one side of the triangle y at any time t, using the identity
Cot[t] = x/y to get the formulas.
Formulas
Polar: r == θ/Sin[θ] , quadratrixOfHippias_polar.gcf
Cartesian: x == y * Cot[y] quadratrixOfHippias_eq.gcf
Properties
Trisecting a Angle
The curve can be used to trisect any acute angle. To trisect angel AOB, one first find
distance OE, take one third of distance OE to get OD (trisection of a segment can be
done with ruler and compass). Let the intersection of the curve and the horizontal line at
D be P, then the angle AOP is one third of angle AOB. This follows because by the
definition of the curve, distance[O,E]/distance[O,C] == angle[A,O,B]/(π/2). If
distance[O,C] == π/2, then distance[O,E] == angle[A,O,B].
Related Web Sites
See: Websites on Plane Curves, Printed References On Plane Curves.
The MacTutor History of Mathematics archive.
Quadratrix.
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Tuesday, February 21, 2012
Surfaces: Catenoid helicoid
Catenoid helicoid
Catenoid-Helicoid is the family of a minimal surface consists of continuous and
isometric deformation of a catenoid to a helicoid such that every member of the
deformation family is minimal.
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Plane Curves: Straight line (polar)
Straight line (polar)
A line is a straight one-dimensional figure having no thickness and extending infinitely
in both directions.
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3D Coordinate Systems: Paracylindrical
Paracylindrical
The Paracylindrical, also called parabolic cylindrical, coordinates are a three-
dimensional orthogonal coordinate system that results from projecting the two-
dimensional parabolic coordinate system in the perpendicular z-direction.
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Plane Curves: Spiral of Archimedes (polar)
Spiral of Archimedes (polar)
The Spiral of Archimedes (polar) is a plane curve.
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Space Curves: Seifferts spherical spiral
Seifferts spherical spiral
The Seifferts spherical spiral is a curve in three-dimensional space.
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Space Curves: Conical helix
Conical helix
A Conical helix is a curve in three-dimensional space. The Conic helix may be defined
as a spiral on a conic surface, with the distance to the apex an exponential function of
the angle indicating direction from the axis.
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Sunday, February 19, 2012
Plane Curves: Tschirnhausen cubic (polar)
Tschirnhausen cubic (polar)
Tschirnhaus's cubic is the negative pedal of a parabola with respect to the focus of the
parabola. The caustic of Tschirnhaus's cubic where the radiant point is the pole is
Neile's parabola.
Get copyable formulas for Tschirnhausen cubic (polar) from Differential Geometry
Library (MathML, TeX, Mathematica input, Maple input)
Автор: Hugo на 2:52 AM 0 коммент.
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Ярлыки: differential geometry library, plane curves
Thursday, February 16, 2012
3D Coordinate Systems: Hyper cylindrical
Hyper cylindrical
The Hyper cylindrical (Hyperbolic cylinder) is three-dimensional coordinate system.
Get copyable formulas for Hyper cylindrical from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Sandra на 4:43 AM 0 коммент.
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Ярлыки: 3D coordinate systems, differential geometry library
Surfaces: Catalan surface
Catalan surface
In geometry, a Catalan surface, named after the Belgian mathematician Eugene Charles
Catalan, is a ruled surface all of whose rulings are parallel to a fixed plane.
Get copyable formulas for Catalan surface from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 4:03 AM 0 коммент.
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Space Curves: Elliptical helix
Elliptical helix
The Elliptical helix is a curve in three-dimensional space.
Get copyable formulas for Elliptical helix from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 4:01 AM 0 коммент.
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Plane Curves: Neile's parabola (polar)
Neile's parabola (polar)
The Neile's parabola, also called semicubical parabola or semi-cubical parabola, is the
curve along which a particle descending under gravity describes equal vertical spacings
within equal times, making it an isochronous curve. The Neile's parabola is the evolute
of a parabola. The Tschirnhausen cubic catacaustic is also a Neile's parabola.
Get copyable formulas for Neile's parabola (polar) from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 3:59 AM 0 коммент.
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3D Coordinate Systems: Tangent sphere
Tangent sphere
The Tangent sphere is three-dimensional coordinate system.
Get copyable formulas for Tangent sphere from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 3:58 AM 0 коммент.
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Plane Curves: Kampyle of Eudoxus (polar)
Kampyle of Eudoxus (polar)
The Kampyle of Eudoxus is a plane curve studied by Eudoxus in relation to the classical
problem of cube duplication.
Get copyable formulas for Kampyle of Eudoxus (polar) from Differential Geometry
Library (MathML, TeX, Mathematica input, Maple input)
Автор: Hugo на 3:56 AM 0 коммент.
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Surfaces: Henneberg
Henneberg
Henneberg's Surface is an immersion of a puctured projective plane. Henneberg's
minimal surface is a nonorientable surface defined over the unit disk and contains a
tractrix as a core curve.
Get copyable formulas for Henneberg from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: Shephard на 3:54 AM 0 коммент.
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Plane Curves: Serpentine
Serpentine
The Serpentine is a plane curve.
Get copyable formulas for Serpentine from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: John на 3:51 AM 0 коммент.
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Plane Curves: Cochleoid
Cochleoid
A Cochleoid is a snail-shaped curve similar to a strophoid. A Cochleoid is the polar
inverse of the Quadratrix of Hippias.
Get copyable formulas for Cochleoid from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: Shephard на 1:24 AM 0 коммент.
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Surfaces: Ellipsoid
Ellipsoid
The general ellipsoid, also called a triaxial ellipsoid, is a quadratic surface. If the lengths
of two axes of an ellipsoid are the same, the figure is called a spheroid, and if all three
are the same, it is a sphere. Tietze calls the general ellipsoid a "triaxial ellipsoid."
Get copyable formulas for Ellipsoid from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: John на 1:21 AM 0 коммент.
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3D Coordinate Systems: Spherical
Spherical
The Spherical coordinates are a three-dimensional coordinate system of curvilinear
coordinates that are natural for describing positions on a sphere (see Sphere surface) or
spheroid.
Get copyable formulas for Spherical from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: Shephard на 1:20 AM 0 коммент.
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Space Curves: Fresnel spiral
Fresnel spiral
The Fresnel spiral is a curve in three-dimensional space.
Get copyable formulas for Fresnel spiral from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: Shephard на 1:18 AM 0 коммент.
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Plane Curves: Cissoid of Diocles
Cissoid of Diocles
The Cissoid of Diocles is a cubic plane curve. It is a member of the Conchoid of de
Sluze family of curves and in form it resembles a Tractrix.
If the cusp of the Cissoid of Diocles is taken as the centre of inversion, the Cissoid of
Diocles inverts to a parabola.
Get copyable formulas for Cissoid of Diocles from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: John на 1:13 AM 0 коммент.
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Plane Curves: Circle curve
Circle curve
A Circle curve is a simple shape of Euclidean geometry consisting of those points in a
plane that are a given distance from a given point, the centre.
Cardioid is the pedal of a circle if the pedal point is taken on the circumference.
If the pedal point is not on the circumference, the pedal of a circle is a Limacon of
Pascal.
Get copyable formulas for Circle curve from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: Shephard на 1:10 AM 0 коммент.
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3D Coordinate Systems: Cassinian cylindrical
Cassinian cylindrical
The Cassinian cylindrical is three-dimensional coordinate system.
Get copyable formulas for Cassinian cylindrical from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Hugo на 1:06 AM 0 коммент.
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Plane Curves: Bifolium
Bifolium
The Bifolium is a quartic plane curve. The Bifolium is a folium with b=0.
Get copyable formulas for Bifolium from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: Hugo на 1:04 AM 0 коммент.
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Plane Curves: Devil's curve (polar)
Devil's curve (polar)
The Devil's curve (polar) is a two-dimensional curve. When a = b the curve changes to a
circle curve.
Get copyable formulas for Devil's curve (polar) from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 1:02 AM 0 коммент.
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Space Curves: Sphere sine wave
Sphere sine wave
The Sphere sine wave is a curve in three-dimensional space.
Get copyable formulas for Sphere sine wave from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 12:57 AM 0 коммент.
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Plane Curves: Parabola
Parabola
In mathematics, the parabola is a conic section, the intersection of a right circular
conical surface and a plane parallel to a generating straight line of that surface. Given a
point (the focus) and a corresponding line (the directrix) on the plane, the locus of
points in that plane that are equidistant from them is a parabola. The line perpendicular
to the directrix and passing through the focus (that is, the line that splits the parabola
through the middle) is called the "axis of symmetry". The point on the axis of symmetry
that intersects the parabola is called the "vertex", and it is the point where the curvature
is greatest. Parabolas can open up, down, left, right, or in some other arbitrary direction.
Any parabola can be repositioned and rescaled to fit exactly on any other parabola - that
is, all parabolas are similar.
Get copyable formulas for Parabola from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: John на 12:20 AM 0 коммент.
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Plane Curves: Freeth's nephroid (polar)
Freeth's nephroid (polar)
The Freeth's nephroid (polar) is a special case of hypocycloid. The curve is a strophoid
of a circle with the pole O at the center of the circle and the fixed point P on the
circumference of the circle.
Get copyable formulas for Freeth's nephroid (polar) from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Hugo на 12:18 AM 0 коммент.
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Wednesday, February 15, 2012
3D Coordinate Systems: Log cylindrical
Log cylindrical
The Log cylindrical is three-dimensional coordinate system.
Get copyable formulas for Log cylindrical from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: John на 1:30 PM 0 коммент.
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Plane Curves: Frequency curve
Frequency curve
The Frequency curve is a smooth curve which corresponds to the limiting case of a
histogram computed for a frequency distribution of a continuous distribution as the
number of data points becomes very large.
Get copyable formulas for Frequency curve from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: John на 1:29 PM 0 коммент.
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Plane Curves: Equiangular spiral (polar)
Equiangular spiral (polar)
The Equiangular spiral (polar), also called logarithmic spiral, growth spiral or Bernoulli
spiral, describes a family of spirals of one parameter. A special case of the Equiangular
spiral (polar) is the circle curve, where the constant angle is 90�.
Get copyable formulas for Equiangular spiral (polar) from Differential Geometry
Library (MathML, TeX, Mathematica input, Maple input)
Автор: John на 1:28 PM 0 коммент.
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3D Coordinate Systems: Maxwell cylindrical
Maxwell cylindrical
The Maxwell cylindrical is three-dimensional coordinate system.
Get copyable formulas for Maxwell cylindrical from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: John на 1:26 PM 0 коммент.
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Surfaces: Paraboloid
Paraboloid
A paraboloid is a quadric surface (any D-dimensional hypersurface in (D + 1)-
dimensional space defined as the locus of zeros of a quadratic polynomial) of special
kind. There are two kinds of paraboloids: elliptic and hyperbolic. The elliptic paraboloid
is shaped like an oval cup and can have a maximum or minimum point. The hyperbolic
paraboloid (not to be confused with a hyperboloid) is a doubly ruled surface shaped like
a saddle.
Get copyable formulas for Paraboloid from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: John на 1:24 PM 0 коммент.
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Plane Curves: Kampyle of Eudoxus
Kampyle of Eudoxus
The Kampyle of Eudoxus is a plane curve studied by Eudoxus in relation to the classical
problem of cube duplication.
Get copyable formulas for Kampyle of Eudoxus from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: John на 1:22 PM 0 коммент.
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Surfaces: Wallis conical edge
Wallis conical edge
Wallis's conical edge is a ruled surface given by the parametric equations. In geometry,
a surface S is ruled if through every point of S there is a straight line that lies on S. The
most familiar examples are the plane and the curved surface of a cylinder or cone.
Wallis's conical edge is also a kind of right conoid. In geometry, a right conoid is a
ruled surface generated by a family of straight lines that all intersect perpendicularly a
fixed straight line, called the axis of the right conoid.
Get copyable formulas for Wallis conical edge from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: John на 1:20 PM 0 коммент.
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Plane Curves: Quadratrix of Hippias (polar)
Quadratrix of Hippias (polar)
The quadratrix was discovered by Hippias of Elias in 430 BC, and later studied by
Dinostratus in 350 BC (MacTutor Archive). It can be used for angle trisection or, more
generally, division of an angle into any integral number of equal parts, and circle
squaring. It has polar equation and Cartesian equation.
Get copyable formulas for Quadratrix of Hippias (polar) from Differential Geometry
Library (MathML, TeX, Mathematica input, Maple input)
Автор: John на 1:18 PM 0 коммент.
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Space Curves: Rotating sine wave
Rotating sine wave
The Rotating sine wave is a curve in three-dimensional space.
Get copyable formulas for Rotating sine wave from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 1:13 PM 0 коммент.
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3D Coordinate Systems: Tangent cylindrical
Tangent cylindrical
The Tangent cylindrical is three-dimensional coordinate system.
Get copyable formulas for Tangent cylindrical from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 11:44 AM 0 коммент.
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Plane Curves: Ellipse
Ellipse
An Ellipse is a plane curve that results from the intersection of a cone by a plane in a
way that produces a closed curve. A circle curve are special cases of an ellipse, obtained
when the cutting plane is orthogonal to the cone's axis. An Ellipse is closed curve and is
the bounded case of the conic sections. The other two cases are parabola and hyperbola.
It is also the simplest Lissajous curve, formed when the horizontal and vertical motions
are sinusoids with the same frequency.
Get copyable formulas for Ellipse from Differential Geometry Library (MathML, TeX,
Mathematica input, Maple input)
Автор: John на 11:42 AM 0 коммент.
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Plane Curves: Freeth's nephroid
Freeth's nephroid
The Freeth's nephroid is a special case of hypocycloid. The curve is a strophoid of a
circle with the pole O at the center of the circle and the fixed point P on the
circumference of the circle.
Get copyable formulas for Freeth's nephroid from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Hugo на 11:41 AM 0 коммент.
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3D Coordinate Systems: Inverse ellcylindrical
Inverse ellcylindrical
The Inverse ellcylindrical (inverse elliptic cylinder) is three-dimensional coordinate
system.
Get copyable formulas for Inverse ellcylindrical from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Hugo на 11:38 AM 0 коммент.
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3D Coordinate Systems: Cardioidal
Cardioidal
The Cardioidal is three-dimensional coordinate system.
Get copyable formulas for Cardioidal from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: Shephard на 11:37 AM 0 коммент.
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Plane Curves: Plateau curve
Plateau curve
The Plateau curves were studied by the Belgian physicist and mathematician Joseph
Plateau. They have Cartesian equation.
Get copyable formulas for Plateau curve from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: John на 11:33 AM 0 коммент.
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Plane Curves: Limacon of Pascal (polar)
Limacon of Pascal (polar)
The Limacon of Pascal (polar) is defined as a roulette formed when a circle rolls around
the outside of a circle of equal radius. The Limacon of Pascal (polar) is a pedal of a
circle cirve and the conchoid of a circle with respect to a point on the circle.
Get copyable formulas for Limacon of Pascal (polar) from Differential Geometry
Library (MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 11:32 AM 0 коммент.
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Surfaces: Right conoid
Right conoid
A right conoid is a ruled surface generated by a family of straight lines that all intersect
perpendicularly a fixed straight line, called the axis of the right conoid. In geometry, a
surface S is ruled if through every point of S there is a straight line that lies on S.
Get copyable formulas for Right conoid from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: Shephard на 11:30 AM 0 коммент.
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Surfaces: Steiner roman
Steiner roman
The Roman surface is a self-intersecting mapping of the real projective plane into three-
dimensional space, with an unusually high degree of symmetry.
Get copyable formulas for Steiner roman from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: John на 11:28 AM 0 коммент.
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Plane Curves: Talbots curve
Talbots curve
A curve investigated by Talbot which is the ellipse negative pedal curve with respect to
the ellipse's center for ellipses with eccentricity
Get copyable formulas for Talbots curve from Differential Geometry Library (MathML,
TeX, Mathematica input, Maple input)
Автор: Shephard на 11:27 AM 0 коммент.
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Space Curves: Spherical helix
Spherical helix
A Spherical helix curve is the tangent indicatrix of a curve of constant precession.
Get copyable formulas for Spherical helix from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: John на 11:25 AM 0 коммент.
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Plane Curves: Sinusoidal spirals (polar)
Sinusoidal spirals (polar)
Sinusoidal spirals are a family of plane curves.
For many values of a the spiral coincides with another curve:
Line (n = −1)
Circle (n = 1)
Hyperbola (n = −2)
Parabola (n = −1/2)
Cardioid (n = 1/2)
Lemniscate (n = 2)
Tschirnhausen cubic (n = −1/3)
Get copyable formulas for Sinusoidal spirals (polar) from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Shephard на 11:20 AM 0 коммент.
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3D Coordinate Systems: Rose cylindrical
Rose cylindrical
The Rose cylindrical is three-dimensional coordinate system.
Get copyable formulas for Rose cylindrical from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: Hugo на 10:36 AM 0 коммент.
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3D Coordinate Systems: Paraboloidal 2
Paraboloidal 2
The Paraboloidal 2 coordinates are a three-dimensional orthogonal coordinate system
(λ,μ,ν) that generalizes the two-dimensional parabolic coordinate system.
Similar to the related ellipsoidal coordinates, the Paraboloidal 2 coordinate system has
orthogonal quadratic coordinate surfaces that are not produced by rotating or projecting
any two-dimensional orthogonal coordinate system.
Se also Paraboloidal coordinate system.
Get copyable formulas for Paraboloidal 2 from Differential Geometry Library
(MathML, TeX, Mathematica input, Maple input)
Автор: John на 5:14 AM 0 коммент.
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Blog Archive
▼ 2012 (139)
► October (3)
► June (1)
▼ February (135)
Surfaces: Catenoid helicoid
Plane Curves: Straight line (polar)
3D Coordinate Systems: Paracylindrical
Plane Curves: Spiral of Archimedes (polar)
Space Curves: Seifferts spherical spiral
Space Curves: Conical helix
Plane Curves: Tschirnhausen cubic (polar)
3D Coordinate Systems: Hyper cylindrical
Surfaces: Catalan surface
Space Curves: Elliptical helix
Plane Curves: Neile's parabola (polar)
3D Coordinate Systems: Tangent sphere
Plane Curves: Kampyle of Eudoxus (polar)
Surfaces: Henneberg
Plane Curves: Serpentine
Plane Curves: Cochleoid
Surfaces: Ellipsoid
3D Coordinate Systems: Spherical
Space Curves: Fresnel spiral
Plane Curves: Cissoid of Diocles
Plane Curves: Circle curve
3D Coordinate Systems: Cassinian cylindrical
Plane Curves: Bifolium
Plane Curves: Devil's curve (polar)
Space Curves: Sphere sine wave
Plane Curves: Parabola
Plane Curves: Freeth's nephroid (polar)
3D Coordinate Systems: Log cylindrical
Plane Curves: Frequency curve
Plane Curves: Equiangular spiral (polar)
3D Coordinate Systems: Maxwell cylindrical
Surfaces: Paraboloid
Plane Curves: Kampyle of Eudoxus
Surfaces: Wallis conical edge
Plane Curves: Quadratrix of Hippias (polar)
Space Curves: Rotating sine wave
3D Coordinate Systems: Tangent cylindrical
Plane Curves: Ellipse
Plane Curves: Freeth's nephroid
3D Coordinate Systems: Inverse ellcylindrical
3D Coordinate Systems: Cardioidal
Plane Curves: Plateau curve
Plane Curves: Limacon of Pascal (polar)
Surfaces: Right conoid
Surfaces: Steiner roman
Plane Curves: Talbots curve
Space Curves: Spherical helix
Plane Curves: Sinusoidal spirals (polar)
3D Coordinate Systems: Rose cylindrical
3D Coordinate Systems: Paraboloidal 2
Plane Curves: Eight curve (polar)
Surfaces: Riemann surface
3D Coordinate Systems: Cartesian
Space Curves: Sici spiral
Plane Curves: Tractrix
Plane Curves: Lituus (polar)
Surfaces: Hyperboloid of Two Sheets
2D Coordinate Systems: Tangent
Plane Curves: Parabola (polar)
Surfaces: Pseudosphere
Surfaces: Whitney umbrella
2D Coordinate Systems: Parabolic
3D Coordinate Systems: Ellipsoidal
2D Coordinate Systems: Polar
Plane Curves: Lame curve
Plane Curves: Folium of Descartes (polar)
3D Coordinate Systems: Conical
2D Coordinate Systems: Inverse cassinian
Surfaces: Sphere surface
Surfaces: Monkey saddle
Space Curves: Circular helix
Plane Curves: Hypocycloid
3D Coordinate Systems: Inverse cassinian cylindric...
2D Coordinate Systems: Log cosh
Plane Curves: Tricuspoid
Plane Curves: Conchoid (polar)
3D Coordinate Systems: Oblate spheroidal
Plane Curves: Trifolium (polar)
Plane Curves: Nephroid
Space Curves: Nhelix
Surfaces: Helicoid
3D Coordinate Systems: Confocal parabolical
Plane Curves: Hypotrochoid
Plane Curves: Fermat's spiral
Plane Curves: Eight curve
3D Coordinate Systems: Cylindrical
3D Coordinate Systems: Confocal elliptical
2D Coordinate Systems: Inverse elliptic
Plane Curves: Limacon of Pascal
Plane Curves: Lituus
Plane Curves: Serpentine (polar)
Surfaces: Bohemian dome
Surfaces: Klein bottle
3D Coordinate Systems: Prolate spheroidal
2D Coordinate Systems: Elliptic
Surfaces: Hyperboloid of One Sheet
Surfaces: Torus
Plane Curves: Lissajous curve
Plane Curves: Astroid (polar)
2D Coordinate Systems: Cassinian
Plane Curves: Trisectrix of Maclaurin (polar)
Plane Curves: Trisectrix of Maclaurin
Plane Curves: Lemniscate (polar)
Plane Curves: Hyperbola
3D Coordinate Systems: Bipolar cylindrical
2D Coordinate Systems: Rose
Surfaces: Hyperbolic paraboloid
Space Curves: Viviani
Surfaces: Dini surface
2D Coordinate Systems: Rectangular
Surfaces: Enneper
2D Coordinate Systems: Bipolar
3D Coordinate Systems: Paraboloidal
2D Coordinate Systems: Logarithmic
Plane Curves: Witch of Agnesi
Plane Curves: Folium of Descartes
Plane Curves: Epitrochoid
3D Coordinate Systems: Toroidal
3D Coordinate Systems: Six sphere
2D Coordinate Systems: Cardioid
Surfaces: Sea shell
Surfaces: Boy surface
Plane Curves: Spiral of Archimedes
Plane Curves: Catenary
Surfaces: Pluckers conoid
2D Coordinate Systems: Maxwell
Surfaces: Kuen surface
Surfaces: Scherk
Surfaces: Cross cap
2D Coordinate Systems: Hyperbolic
Plane curve: Astroid
Atlas major update 2.2
3D Coordinate Systems: Bispherical
Surfaces: Moebius strip
Differential Geometry Library is launched!
► 2011 (9)
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Origami Trisection of an angle
How can you trisect an angle? It can be shown it's impossible to do this with ruler and
compass alone, (using Galois theory) - so don't try it!!! But you may be able to find
some good approximations. However, in origami, you can get accurate trisection of an
acute angle.
You can read about this in several places, but since it's so neat, I thought I'd put
instructions up here too - more people should be able to do this for a party trick!
Jim Loy has informed me that this construction is due to to Hisashi Abe in 1980, (see
"Geometric Constructions" by George E. Martin). See Jim Loy's page at
http://www.jimloy.com/geometry/trisect.htm for a description of many other ways to
trisec an angle.
Since we're working with origami, the angle is in a piece of paper:
So what we want is to find how to fold along these dotted lines:
Note, if you don't start with a square, you can always make a square, here's the idea.
We're going to trisect this angle by folding. I'm going to try and describe this in a way
so that you'll remember what to do.
Suppose we could put three congruent triangles in the picture as shown:
These triangles trisect the angle. So we need to know how to get them there.
Choose some height for the lower triangle, any height, and crease a horizontal line at
this height; ie, just crease any horizontal line you want:
We need to get the blue line of the following picture somehow:
We can make a kind of "marked ruler" in the side of the paper, by folding over the
paper again:
Now this "marked ruler" is used to find the bold line we needed:
To do this, fold the paper so point b touches line B, and point d touches line D:
You can check that this really does work out, and that the angles are the same for these
triangles:
(Note, in the above, we don't really have a marked ruler as such, as we can't move the
edge of the paper to any position, as it's attached to the rest of the paper.)
You can find another accounts of this construction at:
Hull, Thomas, A note on "impossible" paper folding, American Mathematical Monthly,
Vol. 103, No. 3 (March 1996), 242-243.
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Challenge Questions
Stop in and exercise your brain. Talk about this month's Challenge from
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Trisection: CR4 Challenge (04/15/08)
Posted April 13, 2008 5:01 PM
This week's CR4 Challenge Question:
It is known that it is not possible exactly to trisect an angle using just compasses and
an unmarked straight edge. However, there is no theoretical limit to the achievable
level of accuracy.
What is the smallest number of predefined quasi-Euclidean 'operations' needed to
produce an angle 1/3 the size of an arbitrary obtuse angle with a systematic error that is
less than 5'? The reduced angle can be wherever is most convenient for the
construction.
For consistency, we need careful definition of both 'rules' and 'operations'
Rules are similar but not quite identical to Euclid's: i.e. the straight-edge can only be
used for drawing straight lines, and has no measurement function; and the compasses
can be adjusted so that the radius passes through a defined point. However, the
compasses are different to Euclid's "awkward compasses" in that they can retain their
setting when the point is removed from the paper. You have only one set of
compasses, and (finally) your draughtsman's estimation of angle size is so poor that
initial guesses are completely pointless.
(In addition, I reserve the right to clarify if ambiguities are identified)
As it would take extreme care in drawing to approach the theoretical precision, the
parts of the drawing process that will take most time are those involving alignment.
Accordingly, the only operations that contribute to the "operations count" are those
requiring careful alignment -i.e.:
the placing of the compass point,
adjustment of the radius of the compasses, and
the alignment of the straight-edge to pass through a single point (alignment of a line to
pass through two points is two operations)
My solution uses 18 operations and has a systematic error of 2.6'. It is unlikely to be
optimal.
Thanks to Physicist? for submitting this question!
(Update: April 22, 8:42 AM EST) And the Answer is...
Conceptually, the simplest approach would be to progressively approach one third of
the angle by bisecting the angle between the latest lines to be placed. That would
create angles of 0.5, 0.25, 0.375, 0.3125, etc. of the original. The systematic error
(hereafter just "error") is always exactly 1/3 of the smallest angle we have drawn. If we
stop after 10 bisections (0.3330 of the original angle) the error will be 1/(3072) of the
original angle, or a maximum of about 3.5 minutes. Assuming we could still see what
we were doing as the angles become so small, this would take a minimum of 32
operations.
Approximating via bisection is a first-order process, in that the error after each
bisection is a fixed proportion (1/2) of the previous error. On the other hand, the
approximate method that uses the angles on a trisected chord is a third order process -
it gives an error that tends (at sufficiently small angles) to 1/(81.θ3) of the angle that is
being divided (angles here in radians). The largest angle we can successfully divide
with an error of 5' or less is just over 29O. As the error using the bisection method
above is 1/3 of the final angle that means we can achieve the desired accuracy by using
this division method after three bisections. In practice, the method of trisecting the
chord requires rather a large number of operations; we therefore use a slightly different
technique (including multiplying the dimensions) instead - using a compact algorithm
that gives an identical error.
The sequence is (with numbers at left used to represent operations):
1. Place the compass point on the angle's vertex, and draw arc #1 to join the rays of the
angle (the compasses should remain at this setting until stated otherwise)
2 & 3. Prepare for bisection1: Draw intersecting arcs of circles centered on intersection
of the rays and circle (arc #3 must include a section for use in bisection2 at operations
6-8).
4 & 5. Draw first angle-bisector through intersection of arcs (#2 & #3) and the vertex
6. Draw arc #4 centered on intersection of bisector line and circle, and intersecting arc
#3 (arc #4 must include a section for bisection3 at operations 9-11).
7 & 8. Draw second angle-bisector through intersection of arcs(#3 & #4) and the
vertex
9. Draw arc #5 centered on intersection of bisector line and circle, and intersecting arc
#4
10 & 11. Draw third angle-bisector through intersection of arcs (#4 & #5) and the
vertex. The length of the section of this line that is inside the original angle must be at
least three times the radius of arc #1.
12. Draw arc #6 centered on intersection of third bisector and arc #1 to cut the third
bisector outside the circle.
13. Center compasses on the intersection of arc #6 and the bisector to produce arc #7.
Arc #7 must cross third bisector and extend at least 1/3 of the way towards the second
bisector
14 & 15. Center compasses on intersection of second bisector and arc #1, and adjust so
arc #8 can be drawn through the intersection of the third bisector and arc #1.
16. Center compasses on intersection of third bisector and arc #7, and draw arc #8
crossing arc #7.
17 & 18. Join vertex of original angle and the intersection of arcs (#8 & #9).
The angle between this final line and the nearest of the rays of the original angle is
approximately 1/3 of the original.
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#89 "Re: Trisection: and the answer should have been.." by Physicist? on 05/04/2008
4:48 PM (score 1)
RickLee
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#1
Re: Trisection: CR4 Challenge (04/15/08)
04/14/2008 2:23 PM
Let me clarify this, Let's say we start with a 45 degree angle.
In the end you want to end up with two lines drawn at 15 and
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30 degrees off one reference side. Is this correct?
Theoretially one could keep bisecting angles until we get close
enough to the target, but I would assume this is not what you
are thinking. In this case one could just keep using the same
compass setting and drawing arcs to get the half point then
draw a line, then repeat.
Also I am not sure what you mean by 5 feet. Could you
express the desired resultant accuracy in terms on precentage
of the orignal angle.
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#2
In reply to #1
Re: Trisection: CR4 Challenge (04/15/08)
04/14/2008 3:45 PM
Hmm. It seems clarification is required.
1) The desired angle is required to be close in value to 1/3 of
the original angle. As the location is unspecified you can place
it wherever you wish (not necessarily sharing any line with the
original).
[Nevertheless, all of my "low-operation-count" procedures to
date have shared an edge and the vertex with the original
angle, but I repeat that this is not a requirement]
2) An obtuse angle is strictly greater than 90O and strictly less
than 180O
3) 5' = 5-arcminutes, or 1/12-degree. Apologies that the
diagonalisation of the single prime got lost in translation.
4) The method of alternating bisections would be legitimate in
principle - but it would take too many operations to attain this
accuracy.
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Kris
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#6
In reply to #1
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 2:56 AM
X = 1/4 + 1/16 + 1/64 + 1/256 +
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= 1/4 + X/4
3X/4 = 1/4
X = 1/3
As you say, bisection would get there but, alas, too late. I just
put in the series above because I like the way it adds up, but
maybe there are other series that work better. It's a good job
Fyz didn't ask for inches ! 1 Degree (o) = 60 minutes ('), 1
minute = 60 seconds (").
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#15
In reply to #6
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 12:44 PM
Edit: dskktb posted his solution while I was working on this.
Here is a sequence I entered into AlgoSim (edited for spacing):
.25+.125 = 0.375
Ans-.0625 = 0.3125
Ans+.03125 = 0.34375
Ans-(.03125/2) = 0.328125
Ans+(.03125/4) = 0.3359375
Ans-(.03125/8) = 0.33203125
Ans+(.03125/16) = 0.333984375
Ans-(.03125/32) = 0.3330078125
Don't know how this applies. Kris got me to thinking, and I
had to follow up on it. I'm beginning to suspect that an
application of the tangent function would prove beneficial. I'll
mull it over later, got other fish to fry right now.
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#3
Re: Trisection: CR4 Challenge (04/15/08)
04/14/2008 9:53 PM
Line AC ........................................... > 1
Cir centre any point on line.................. > 1
Arc through centre ............................ > 1
90° tri line OX, OY, OC........................ > 6
Arc OA ............................................ > 1
First cir R ......................................... > 1
Second cir R .................................... > 1
Line OB ............................................ > 2
Total > 14
Tri BOC = one third of obtuse tri AOC.
Regards JD.
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Re: Trisection: CR4 Challenge (04/15/08)
04/14/2008 10:19 PM
Its wrong. must cut down on the port.
Regards JD.
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#5
Re: Trisection: CR4 Challenge (04/15/08)
04/14/2008 10:50 PM
well, take an angle, triple it, and reverse the process to find the
original angle.
like squaring the circle, it cannot be done as long as you follow
the restrictions on the gear you can use. To get to the minimum
number of moves within that error limit needs some elegance.
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#7
In reply to #5
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 4:17 AM
Aha.
I imagine that you already know this, but "arbitrary" should be
taken to mean imply that you know nothing about the angle
except that it is obtuse - so you haven't created it by
multiplying a known angle by 3.
While writing - I currently have a method that will do it using
15 operations (probably still not optimal) - but anything less
than 18 will still answer the challenge
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Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 4:59 AM
This method should be exact. I don't report all the passage because I'm
busy at moment, but it is sure is possible to use only compass and
straight-edge.
Given the angle BOA:
build the parallelogram BOAC;
Trace the diagonals;
Find the centre point D of OA;
Trace the lines DB and DC;
Find the point H as intersection between EF and GD
The angle HOA is exactly 1/3 of BOA.
This method was found starting from a general method to divide a
segment in any part you desire:
Regards
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#9
In reply to #8
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 5:25 AM
The angle HOA is exactly 1/3 of BOA.
Why ?
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#10
In reply to #8
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 7:57 AM
You have successfully trisected the line - but that is not the
same as trisecting the angle. You could try measuring the
angle starting with an angle of (say) 120O.
I chose 120O because this
a) It is an example of an angle that cannot be trisected exactly
using the methods specified, and
b) Unless you fortuitously have the trisected line in a suitable
position, the error should be large enough to be visible.
[N.B.1 if you choose a position for the trisected line that gives
a relatively small error at 1/3 of the angle, it will increase the
error at 2/3 of the angle, showing that the "exact" angle is pure
coincidence!
N.B.2 Even for bisection this method only works if you start
with an isosceles triangle - and the reason it works here is that
everything is symmetrical]
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RickLee
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#11
In reply to #8
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Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 7:59 AM
Looks like they add up to 18. Interesting, how did you come
up with this little jem?
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#12
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 9:57 AM
I don't know if I'm in the right track, but I've got an idea. I will
elaborate it later if I find some time, but here it is for now:
At first, we draw a circle with its center at O and we get points
A and B. Then we trisect segment AB (that's a simple
Euclidian construction) and get rays Oz and Ow. Needless to
say that Oz and Ow do not trisect the angle, but nevertheless,
they form a good starting point for the iterative process that
will follow.
Now, the iteration idea is that we take the bisector of Oz and
Oy deriving a ray Oz' as well as the bisector of Ox and Ow,
getting ray Ow'. Bisection is again a simple construction. We
then go on and bisect Oz' with Oy to get Oz'', etc.etc. I have a
feeling that this process converges, so if we continue like this
we will be getting closer and closer to the solution. I may be
wrong, but at first it seems as a quick and dirty engineering
approach.
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#13
In reply to #12
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 10:49 AM
I can't reall follow your text. Are you alternately bisecting the
angles:
OBC -> OD'
OAD' -> OC'
OBC' -> OD''
OAD''-> OC''
etc?
If so, it is better to perform the bisections before the trisection,
because the bisections always halve the error, whereas the
quality of the trisection improves when you reduce the initial
angle.
However, I believe that the method using bisections followed
by trisection of a chord requires 21 steps to achieve the
specified accuracy (but it's more than possible I have missed
some useful algorithms).
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#22
In reply to #13
Re: Trisection: CR4 Challenge (04/15/08)
04/16/2008 3:58 AM
You are right to be confused, I made a typing mistake. What I
meant:
"... we take the bisector of Oz and Oy deriving a ray Ow' as
well as the bisector of Ox and Ow, getting ray Oz'..."
BTW, I used some trigonometry to derive the initial angle
φ=zOw as a function of the whole thing, call it θ. I arrived to
the formula: tan(φ/2) = 1/3 tan(θ/2)
For example, if θ=90o then φ=38
o . Not exactly 30
o as it should
be, but it is 4o away the proper value on each side. After the
first step of the iterative process, this error reduces to half, i.e.
2o and after the second goes down to 1
o and so on. Problem is
that for bigger angles, the initial error is higher, so you need
more steps to reach the desired accuracy... I better take back
the "quick and dirty" label I assigned to this methodology and
leave only the "dirty" part!
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#25
In reply to #22
Re: Trisection: CR4 Challenge (04/15/08)
04/16/2008 7:30 AM
I agree the formula for the trisection approximation. It would
be better to use only one of the starting trisections OZ and OY
(OZ), and successively halve the resultant 2/3 angles. The
error halves at each step. However, it is much better to perform
any bisections first and then trisect the final difference
between bisections, as the error in this case falls by a factor of
about eight for each bisection. (Even this requires a large
number of operations if you use the method of trisecting a
chord).
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#14
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 12:12 PM
Well I was able with the help of CAD to prove that the meathod works
and it took a total of 9 circles and 9 lines for a total of 18 marks. The
results below show that it should work for any obtuse angle. The amazing
thing though is that in the work, the accuracy is better than 1/60th of a
degree, about +-0.0141 degrees which is better than the 1/12th that was
asked.
I have numbered the order in which I drew the objects in the two pictures.
Steps 1 - 4: (4 Blue Circles)
1.Given an obtruse angle we draw any size circle with the compass from
the apex of the two lines.
2 and 3.Keeping the compass the same size, draw again at the two points
of intersection between the first circle and the two lines forming the
obtuse angle. (2 circles)
4. Now again with the same size circle draw the center at the new
intersection point between circles 2 and 3.
Steps 5-7: (3 green lines)
5. Draw a line from the apex of the obtuse angle and draw thru the
intersection point of circles 2 and 3 and extend line till the edge of circle
4.
6. Draw a line from the intersection of circles 1 and 4 to the intersection
of circles 3 and 1 and extend to the edge of circle 4.
7. Draw a line from the intersection of circles 4 and 1 to the intersection
of circles 2 and 1 and extend to the dege of circle 4.
Steps 8 - 10: (Yellow circle and two yellow lines)
8. Draw a circle centered at the intersection of circle 4 and line 5 where
the radius meets the intersecting points on circle 4 with lines 6 and 7.
9. Draw a line from the left intersection of circle 8 and circle 4 thru the
intersection of circle 8 and line 5 and extend to line 6.
10. Draw a line from the right intersection of circle 8 and circle 4 thru the
intersection of circle 8 and line 5 and extend to line 7.
Steps 11-13: (Magenta circle and two magenta lines)
11. Draw a circle centered at the intersection of circle 4 and line 5 simalar
to circle 8, but with a larger radius where it intersects at lines (10 and 7)
or (9 and 6).
12. Draw a line from the right intersection of circle 8 and line 10 on circle
4 thru the new intersection point of circle 11 and line 5 and extend to line
7.
13. Draw a line from the left intersection of circle 8 and line 9 on circle 4
thru the new intersection point of circle 11 and line 5 and extend to line 6.
Step 14-16: (3 red circles)
14. Draw a circle centered again a the intersection of circle 4 and line 5
where the radius goes thru the intersection of lines (12 and 6) or (13 and
7).
15. Draw the same size circle as 14 at the intersection of circle 4 and 3 not
on the original obtuse angle.
16. Draw the same size circle as 14 at the intersection of circle 4 and 2 not
on the original obtuse angle.
Steps 17-18: (black lines)
17. Draw a line from the apex of the obtuse angle tangent to the right side
of circle 16, this is one trisection.
18. Draw a line from the apex of the obtuse angle tangent to the left side
of circle 15, this is the other trisection.
I checked the dimensions in CAD for the angles only. The smaller
trisection came out to 52.3193 degrees about. The origional arbitrary
obtuse angle was 157 degrees. The trisect angle should be about 52.3333
repeating degrees. My solution uses 18 steps given the original angle, and
straight ruler with no markings, and a compass where the size can be kept
the same.
In the end, my solution is +-0.0141 degrees or about 1/70th of a degree or
0.846'.
In order to check the answer for a good solution, I drew two aqua
colored lines in CAD at 52.3333 degrees and 104.6666 degrees for
visual comparison.
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#16
In reply to #14
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 1:01 PM
Good work ! - I spent way too long thinking about this today.
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#18
In reply to #14
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 4:36 PM
It is good to see a division by three that is toleranced -
although you haven't said whether it gives similar accuracy at
smaller angles (e.g. 136O, 120
O and 91
O). Also, you could omit
some of the steps, because the challenge only requires a single
angle of 1/3 the original size.
On the other hand, even if all the steps were acceptable (see
next paragraph), the number of operations (as defined in the
challenge) would be 30, not 18 (see final paragraph).
Euclid did not admit of drawing tangents without first defining
the point on the circle through which the tangent passes. I
should have made this clear in the original challenge, but
overlooked it (in mitigation, I would claim that I had not
expected it to be helpful). (N.B. the standard way to define the
point is to draw a line from the centre of the relevant circle to
the other point on the line, and draw an arc from the mid-point
on this line that passes through the centre of the original circle)
Returning to number of operations: Every time you place the
point of the compass, that is an operation (or step if you
prefer). If you then adjust the radius of the compasses to pass
through a specific point, that is another operation. Thus, your
steps ##8,11,14 are each two steps. Similarly, according to the
third bullet of the challenge, each of the lines in steps
5,6,7,9,10,12,13 is two operations. Using the standard
technique, step 17 would have added 9 operations (step 18 is
not necessary for the challenge as written, but could have been
drawn with just a further four operations).
Regards
Fyz
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Mr Gee
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#17
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 3:19 PM
Within the bounds of reasonable precision, it is possible to
trisect any angle. The procedure is as follows.
!. Strike an arc from the vertex of the angle that crosses each
leg. As large as possible is best.
1 placement, 2 strikes
2. Set the compass to a radius slightly more than 1/3 the arc
length.
1 setting
3. Strike a secondary arc from the each of the two intersections
of the primary arc and the angle legs.
2 placements, 2 strikes
4. Without changing the compass strike an arc from the
intersection of the first leg and primary arc and second leg and
primary arc. Without changing the compass strike an arc from
the intersection of the first primary and secondary arcs in the
direction of the other inter-section.
2 placements, 2 strikes
You will now have created a pair of parenthesis that bound the
exact trisection of the angle. The exact trisection is one third of
the distance from the most central arc strike and 2/3 of the way
from the outer mark. If you underestimated the 1/3 radius the
marks still bound the error and a simple estimate of the point
result by eye will give a result so precise extremely good
equipment would be required to measure the error.
5. Draw the line from the vertex to the trisection point.
2 placements, 1 strike.
However, if you need to be more precise simply repeat the
above process on the acute angle of the error. This should
result in a "estimate" that is good to a few arc seconds.
depending on the quality of your equipment and the precision
of your work. (If you are laying this out on flat ground the
whole thing can be done with a string and two stakes.)
Since the exact trisection point is at 1/3 , 2/3 of the new acute
arc and this ratio is exactly the same as the ratio of the first
problem it is possible, in theory, to "prove" that this point can
not be found exactly. However, as a practical matter it can be
found easily and accurately as you mentioned.
Since calculus and the "law of infinitesimals" had not been
formalized in Euclid's time the simple expedient of using the
bounded error marks to more precisely estimate the exact 1/3
radius and rebounding the error by repeating step 4 with the
new compass setting might not have occurred to him. (I would
bet against that myself, Euclid was pretty bright...)
But, the methods of successive approximation, and also of
repeated estimation which in have just outlined certainly were
known to engineers of both past and present times. Euclid's
fascination with minutia challenged peoples thinking for many
centuries and almost certainly contributed to the creation of
calculus and the development of modern computational
algorithm's.
I do not know if my methods fit within your rules and I am
also not certain my operations count is correct. Additionally, it
is not possible to exactly quantify what the arc error of my
method is so I may have missed something along the way.
While it would be possible to approximate the trisection by
successive bisections chosen in the appropriate direction as a
practical matter this is not a method a working engineer would
typically choose. Interestingly enough if the obtuse angle in
question is 360 degrees never changing the first compass
radius produces an exact result.
Thank you for presenting this most interesting challenge.
Sincerely,
Mr. Gee
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#19
In reply to #17
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 5:15 PM
Perhaps this phrase in the challenge is relevant:
"and (finally) your draughtsman's estimation of angle size is so poor
that initial guesses are completely pointless"
Unless you have an enormous piece of paper, visual estimates to 5'
of arc would be quite difficult. But if you can provide bracketing
within about 27 degrees the approximation using a trisected chord
would give the specified theoretical accuracy.
Euclid's geometry was to a large extent a philosophical exercise,
being concerned with the ideas of minimum assumptions and
rigorous proof**, and his construction equipment was quite
deliberately the least capable that could do the job (hence his
collapsing compasses, which I haven't retained for this challenge -
and incidentally consider most unlikely to have been a tool that was
in practical use). But it appears that ideas of limits were already
current; but perhaps problems of definition had not yet been
sufficiently overcome for this to form part of Euclid's thesis.
(Documentation shows that infinite sequences date to at least 4000
BC).
**Though he was by no means as rigorous as is generally believed.
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#20
Re: Trisection: CR4 Challenge (04/15/08)
04/15/2008 6:38 PM
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Sace the wear and tear on your brain and penceils.
With origami, it is possible to trisect an angle without an actual
straight edge, or compass.
Check out this link.
http://www.math.lsu.edu/~verrill/origami/trisect/
HTRN
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#23
In reply to #20
Re: Trisection: CR4 Challenge (04/15/08)
04/16/2008 4:55 AM
You will have to institute rules and a suitable counting
method. Rolling into three should be counted as a sequence of
halving steps (equivalent to successive bisections to the same
precision), as it is multiple successive approximations.
Fyz
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#29
In reply to #20
Re: Trisection: CR4 Challenge (04/15/08)
04/17/2008 7:25 AM
I wasn't going to do this just yet, but how about a tomahawk;
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It's no good as far as the challenge goes, but like the origami
it's fun to know. As you can see, TD and TR do the same as
the single height and double height creases made in the paper
with origami
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#21
Re: Trisection: CR4 Challenge (04/15/08)
04/16/2008 12:27 AM
Second attempt.
Obtuse triangle AOB, scribe two half circles, one radius OA, and the
other at mid point of AB at radius R1, scribe R1 again using centre
point A, join points OC. Scribe radius R2 at intersection and at point B.
Scribe a line through where radius's R2 intersect with line OE. angle
EOB is approx 1/3 of obtuse triangle AOB.
There appears to be a small error of about 27' will have to do some
calculation to check?
Regards JD.
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#24
In reply to #21
Re: Trisection: CR4 Challenge (04/15/08)
04/16/2008 6:57 AM
Is this 17 operations?
I couldn't identify line OD. Is that an "exact" trisection for reference?
(As it doesn't meet the accuracy of the challenge, I haven't checked
how the error varies with angle AOB - did you?)
Regards
Fyz
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#26
In reply to #24
Re: Trisection: CR4 Challenge (04/15/08)
04/16/2008 5:55 PM
Yes OD is a reference, done with Auto-Cad, have done the
maths and it does not meet the required accuracy. Looking at it
with hind sight, R2 only dissects COB, therefore AOC would
have needed to be more accurate. With AOB being 160° the
error for AOC is around 2 1/2° and COB around 1 1/2°. I will
have to check more carefully, third time lucky may be.
Regards JD
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#27
In reply to #26
Re: Trisection: CR4 Challenge (04/15/08)
04/17/2008 5:02 AM
Even though 15 steps (maybe fewer) is possible, it's still a
"hard" problem to do it in 18...
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#28
Re: Trisection: CR4 Challenge (04/15/08)
04/17/2008 5:28 AM
Above are obtuse triangles AOB. Top left 174°, top right 96°, and bottom
135°. The object being to find the worst case error of angle X. being 1/3
of angle AOB. E1 being the first error. Dissection of angle YOB being
error 2 (E2). Then a dissection of AOZ, followed by dissection BOY etc.
FIND X.
(1) Find OY, by cosine rule, side R1=AW, and side AO and angle A+60°.
(2) Find X, by sine rule, (sineX)/R1 = (sineA+60)/YO.
(3) Error 1 = (E1) = AOB/3 - X.
(4) Error 2 = (AOB - E1)/2
(5) Error 3 = (AOB - E2)/2
(6) Etc.
Worst case, AOB = 135° = error 1 = 1.3°
Error 2 = 0.65
Error 3 = 0.325
Error 4 = 0.1625
Error 5 = 0.08125 (E5 x 60' = 4.8' minimum error. )
---------------------------------------------
Find X.
Line .......... AB ->1
Semi cir A to B -> 1
Find W ..........-> 3
Find Y ..........- > 3
.................Total 8
Error 2 ...........-> 3
Error 3 ..........-> 3
Error 4 ..........-> 3
Error 5 ..........-> 3
Total ...............20
Best I can do is twenty unless I'm counting wrong
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#30
In reply to #28
Re: Trisection: CR4 Challenge (04/15/08)
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04/17/2008 7:27 AM
Assuming that I've understood correctly, I have probably made a mistake
with my trig, as I get a slightly larger error than yours: E2 = 1.44-degrees
at 135-degrees, and 1.48-degrees at 155-degrees. That's not too
significant, but the process seems to take rather more steps than you give:
1 Arc centre O to define A, B
2,3 Chord AB (all your lines have one alignment at each end)
4 Arc centre A
5 Arc centre B crosses previous arc at W
6,7 Line OW bisects O at C
8,9 Arc centre C (place point on C, and align scribe to pass through A)
10 Arc centre A to cross previous at Y
11,12 First approximate trisection line OY (crosses first arc at Y1)
13,14 Arcs centred at Y1 and B respectively to define Z
15,16 Trisector through OZ
I agree that subsequent bisections can each reuse one existing arc, so each
takes three steps.
That makes 25 steps to reach E5, I think (one more if you make an
approximate reset of the compasses from R1 to R2)
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#31
Re: Trisection: CR4 Challenge (04/15/08)
04/17/2008 2:27 PM
Good Answers: 2
Here are the steps:
Draw the arc PQ
Construct the perpendicular MO to OQ at O
Set the compass to distance OQ and draw the arc ON up to the PQ
arc. This produces a 60 degree angle NOQ at O, trisecting the
right angle MOQ.
Bisect the angle MON, yielding a 15 degree angle MOX.
Set the compass to distance MX, then mark-off the arc segments
MB, PC and CA.
Construct the straight line AB.
Tri-sect the line AB. (*)
Construct line OD.
The arc labeled ST will be approximately 1/3rd
the arc PQ, and
thus the angle DON will be the tri-section of POQ.
(*) For small angles, the tangent of DOB is equivalent to the angle
DOB in radians. For a 2 degree angle the difference amounts to
2.9 arc-seconds.
Note that if angle POM were less than 45 degrees, an extra step
would be needed to tri-sect angle MOX to yield a 7.5 degree
angle, which would then be used to mark-off the arc segments PC,
CA and MB. Similarly, if angle POM were more than 60 degrees
the 30-degree arc segment could be used to generate the arc
segments PC, CA and MB.
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#32
In reply to #31
Re: Trisection: CR4 Challenge (04/15/08)
04/17/2008 4:15 PM
Agree the precision - possibly overkill, as the spec is 5 arc-minute.
But if we take just the first 6 stages of the construction shown on
your drawing we have taken 17 operations - and we still have to
trisect AOB.
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#58
In reply to #32
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 9:33 AM
Yeah,
When I looked at my steps and counted them your way, I knew
it was too many steps. But I decided to post my solution
anyway since I thought it was an interesting challenge.
It was proved years ago, of course, that angle trisection is
impossible except for a few special cases like a right triangle.
But trisecting an angle to a desired degree of accuracy is a
very different problem and, I decided, worth spending some
time thinking about.
At least I was able to prove a correlary of the problem (though
this too was probably proved a long time ago) -- that the
trisection of an obtuse angle can be reduced to the trisection of
an accute angle, since the right-angle portion can itself be
exactly trisected.
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#59
In reply to #58
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 10:27 AM
Sorry I misunderstood the intent.
In fact, the 'official' solution is similar to yours in that it comes
down to creating an angle that is small enough that it can be
trisected accurately enough. The final trisection uses a method
that is equivalent to trisecting the arc - but in a way that
requires far fewer steps.
In the end, I found two completely different solutions that take
just 15 operations - there are probably more - and there may
even be some that require even fewer operations.
One of these is a refinement of the 'official' solution. For larger
angles it uses a slightly more accurate pair of first
"approximations" than bisection, and then uses a slightly more
accurate approximation to trisection than the one given
(although it is very similar in most of the detail).
The other is based on Mark Stark's method. It uses a simple
first approximation that is within 20-degrees of the actual
trisection, and then a compact version of the Mark Stark
refinement technique to provide the required angle.
Either of these methods would easily include acute angles. The
restriction to obtuse was intended to allow the possibility of
methods that didn't work for smaller angles (JDretired showed
an interesting one, though it too is a lot of steps).
Personally, I prefer the 'refinement' of the official solution -
even though the theoretical accuracy is not as good. The
reason is that (for the same size of total drawing)
constructional errors in Mark Stark's method give worst-case
angular error that is amplified by about a factor of six relative
to their effects in the 'refinement'.
I propose to delay posting these solutions for a few weeks to
allow space for others to find better methods.
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Re: Trisection: CR4 Challenge (04/15/08)
04/18/2008 2:14 AM
I'm not sure if my operations count is right, but I might have done this
in 17 steps:
1. Draw line AB
2. Draw line AC
3. Draw line CB
4. Set compass to length of AC
5. Strike arc from C intersecting CB to find point a
6. Set compass to Ba
7. Strike intersecting arcs from B and a to find point b
8. Draw line Ab to find vertex D
9. Set compass to Da
10. Strike intersecting arcs from D and a to find point c
11. Draw line Ac to find vertex E
12. Set compass to DE
13. Strike intersecting arcs from D and E to find point d
14. Draw line Ad to find vertex F
15. Set compass to DF
16. Strike intersecting arcs from D and F to find point e
17. Draw line Ae to find vertex G
Angle of GAB measured by AutoCAD to 3 decimal places came to
40.399°.
If I'm off in my count or the margin of error, let me know. It's too late at
night for me to convert from decimal to degree-minutes-seconds format.
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#34
In reply to #33
Re: Trisection: CR4 Challenge (04/15/08)
04/18/2008 6:12 AM
Assuming that I have understood the construction: I think it only works
for a very small range of angles - if my quick algebra is correct, this
would be slightly larger (but not including) 144O. On that basis, if you
try this with 90O I think you will find the error is greater than 7.5
O.
Regarding the steps that will be required:
As the initial angle is provided, you don't have to count construction of
AB or AC. On the other hand, if you want AB=AC you have to draw a
circle centre O.
Regarding counting: every point you align to counts as one, so:
. drawing a line through two points is two operations,
. placing the compass point is one operation, and
. adjusting the scribe to pass accurately through a point is another.
So - stages with number of operations in braces:
1 (0); 2 (0); 3 (3); 4 (already done); 5 (2); 6+7 (3); 8 (2), etc.
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#35
Re: Trisection: CR4 Challenge (04/15/08)
04/19/2008 12:42 AM
Haven't done my maths or count yet, but on the Auto-CAD it looks exact?
scribe a line from the centre of a equilateral triangle to the inter section,
and where it intersects the arc is one third of the obtuse angle?
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Re: Trisection: CR4 Challenge (04/15/08)
04/19/2008 4:58 AM
Count
Line AB ……………………………….. > 1
Cir C ………………………...………… > 2
Line OD ………….…………………….> 3
Arc E ……………….……………………> 4
Line WG ……….……………………….> 5
Cir F ……………………………………..> 6
Intersection H ………………………. >7
Cir J centre I ……….………………. > 8
Cir L centre K …….………………… > 9
Line OM …………….…………………. > 10
Cir N centre P .…………………….. > 11
Cir Q centre O …………………….. > 12
Line RS ……………..…………………. > 13
Line TH ………………………………... >14 Intersection U = 1/3 of arc
E?
Maths to follow ( getting late, tomorrow). Sorry about the drawings got the
wrong ink cartage.
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#37
In reply to #36
Re: Trisection: CR4 Challenge (04/15/08)
04/19/2008 10:00 AM
The basic construction is new to me, so it's of interest even though the
construction takes rather more than 18 operations**. It is impossible for it
to be exact (mathematically provable), but it may be very accurate. I'll
come back to you on the size of the error if/when I can get my head
around it.
**Your lines are constructed by aligning both position and angle (or we
could say you are defining two points on each line). So they are each two
operations.
Similarly, most of your circles require both the centre and the position of
the scribe to be aligned - that too is two operations.
The sole exception ( I think) is circle L,, which looks like a single
operation.
That would make this 27 operations - though I may have misinterpreted
some part of the procedure.
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#38
Re: Trisection: CR4 Challenge (04/15/08)
04/19/2008 9:23 PM
This is the actual construct, I make it 12 but could be wrong?
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Calculation references.
MATHS
Given. Obtuse angle AOB = 120°, Length OA = 100, Length OF = 50,
Angle FAO = 30°.
(1) Find OK. Angle FOK = 90+60 = 150. FK = radius, 100 x Cos30 =
86.66
By Sine rule find angle FKO.
Sine150/FK = Sine(K)/50. Angle K = 16.7787°
Therefore angle KFH = 180-(150+16.7787) = 13.2213°
By Sine rule find length KO.
OK/ Sine13.2213 = 50/Sine16.7787. Length OK = 39.6141.
(2) Find ON.
Triangle height = OK x Cos30. ON = 2/3 of height.
ON = (39.6141x0.866x2)/3 = 22.8712.
(3) Find angle WND.
WND = Tan WD/( WF+FO+ON) = 43.3013/ (75+50+22.8712) =
16.3217°
(4) Find angle OPN
By Sine rule Sine(P)/ON = Sine(N)/AO.
OPN = (Sine16.3217x22.8712)/100 = 3.6852°
(5) Find angle POB.
Angle POH = 180-( 3.6852+16.3217) = 159.993°
Angle FOP = 180 - 159.993 = 20.007°
Angle POB = FOB - FOP. 60 - 20.007 = 39.993°
Error trisecting angle AOB = 40 - 39.993 = 0.007°. 0.4' = 25".
Accurate to within 25 seconds.
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#39
In reply to #38
Re: Trisection: CR4 Challenge (04/15/08)
04/20/2008 4:53 AM
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Different construct which I believe to be more accurate and a
smaller count? I hope. This is more accurate on the larger
angles, which where getting close to the limit.
Regards JD.
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#40
In reply to #39
Re: Trisection: CR4 Challenge (04/15/08)
04/20/2008 9:54 AM
Your claimed accuracy looks impressive; unlike Mark Stark's
method** (which can be part of a sequence that reduces to 15
steps) you use no constructions that magnify small errors,
which is a great improvement. I'll restart independent
calculations of the first error because I mis-transcribed the
original version
On the other hand, I think this one is 22 operations - it may be
fewer if I have misinterpreted, so I'd appreciate your
confirmation. (Did you see my previous attempts at
clarification?)
Arc BCO ->2 (centre O to find centre for OCB, then draw
OCB)
Arc AEB -> 2 (Centre O and adjust span to OB)
Line AB -> 2 (align at A and B)
Line FG -> 2 (align at O and F)
Arcs centre F, radius AF -> 2 (centre and adjust span to AF)
Arc centre B -> 1 (defines D)
Arc IJ -> 2 ((centre H and adjust span to OH)
Arc JH -> 1 (recentre on I)
Line JO -> 2 (align at J and O)
Line KB -> 2 (align at K and B)
Line ND -> 2 (align at N and D)
Line OP -> 2 (align at O and P)
**Kris (many thanks) provided a link that includes Mark
Stark's basic method in earlier discussions.
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#41
In reply to #40
Re: Trisection: CR4 Challenge (04/15/08)
04/20/2008 11:05 AM
Here's a more direct link to Mark Starks method, a nice applet that can be
tinkered with to see how it works ;
http://www.math.umbc.edu/~rouben/Geometry/trisect-stark.html
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#46
In reply to #40
Re: Trisection: CR4 Challenge (04/15/08)
04/20/2008 8:23 PM
Yes I do confirm your number of operations as 22, I now see my mistake
of adding only 1 operation for a line that uses two points. Thank you, and
Kris, for the link, have put same into favorites and will look closer at it. I
had not seen the Mark Stark's solution prior to the above, when looking at
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it, it does seem have some similarities.
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#49
In reply to #46
Re: Trisection: CR4 Challenge (04/15/08)
04/21/2008 7:00 AM
Hi JD
Thank you for sharing this remarkably precise "single-step"
approach - it's really interesting, even if it can't beat the
number of operations that can be achieved using more obvious
approaches.
After numerous false entries into the spreadsheet, I finally
managed to confirm the accuracy at 120O. So far as I can see,
this family of methods is exact at 90O and 180
O, and the
version in post #38 give a theoretical peak error of just under
0.8' (at close to 143.1O); that's more than a factor of 5 more
precise than required by the challenge**. So I'm not quite
certain what you mean in post #39 where you say "the larger
angles, which where getting close to the limit".
Regards
Fyz
**Even near 53.13-degrees (i.e. before the method fails
altogether), the error is just marginally over 0.5O (or 30').
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#51
In reply to #49
Re: Trisection: CR4 Challenge (04/15/08)
04/21/2008 8:54 AM
Yes your correct the approach in #39 is not correct, and my
statement is in error, use of a hand calculator and confusion
does not give good results. Thank you for going to the trouble
to check the accuracy, much appreciated. There is a small
floating error in the method which I was trying to eliminate,
plus reduce the count,but my efforts did not work.
Regards JD.
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#52
In reply to #51
Re: Trisection: CR4 Challenge (04/15/08)
04/21/2008 10:27 AM
It's a pleasure to look at something that works this well.
Although I found a few alternative approaches that would need
the same number of operations, I haven't managed to reduce it
below your 23 (I think) operations.
BTW, there are a couple of free spreadsheets on the web
which should easily handle what is needed - "gnumerics" and
OpenOffice "Calc" both seem quite good. Perhaps they are not
quite as 'pretty' as EXCEL, but the interface is basically quite
similar [including working in radians rather than degrees, so
you would need to know that pi needs to be written as pi() ].
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Anonymous Poster #42
Re: Trisection: CR4 Challenge (04/15/08)
04/20/2008 11:56 AM
We obviously live in an irrational universe.There is no way to
escape the irrationality of Pi in any construct, using current
mathematical processes.Perhaps we need a different kind of
math to measure circles,spheres, etc.? Obviously, a bubble is
the most powerful computer in the universe.It can calculate Pi
to the last place instantly when it forms.That makes it apparent
that factors other than radius determine circumference.I do not
claim to know what the factors are are, but they must exist, or
a bubble could never completely form, it would always be
ALMOST formed, chasing the elusive Pi out to infnity.An
older sphere would be more complete than a younger one,
etc.Perhaps our "straight" lines are actually slightly curved and
we cannot detect it? Can a straight line not be defined as a
segment of a circle with an Infinite Radius?
Beam me up Scotty.
Our work is finished here.!
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Anonymous Poster #43
In reply to #42
Re: Trisection: CR4 Challenge (04/15/08)
04/20/2008 12:41 PM
Physicist? #44
In reply to #42
Re: Trisection: CR4 Challenge (04/15/08)
04/20/2008 12:48 PM
Anonymous Poster #45
In reply to #44
Re: Trisection: CR4 Challenge (04/15/08)
04/20/2008 2:02 PM
Physicist? #50
In reply to #45
Re: Trisection: CR4 Challenge (04/15/08)
04/21/2008 7:58 AM
3Doug
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#47
Re: Trisection: CR4 Challenge (04/15/08)
04/21/2008 1:33 AM
This method uses 28 operations, but the accuracy is impressive (at least
to me).
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Given line AB:
1. Set compass to AB
2. Draw circle centered on A with radius of AB
3, 4. Draw AC
5. Draw line perpendicular to AB from A to circle to find point D (top
quadrant).
6. Draw circle centered on D with radius of AB
7,8. Draw horizontal line through the 2 points where the circles
intersect.
Bisect the right half of this line:
9. Set compass to half of horizontal line.
10. Draw circle at midpoint of line. (This is the left endpoint of this
section. The reason for noting this will be seen in next paragraph.)
11. Draw circle centered on right endpoint.
12, 13. Draw vertical line from upper intersection point to horizontal
line.
Repeat the bisection 3 times on right, then left, then right halves of
previous bisections.
Angle from A to last bisection point is 40.023° or 40° 1' 22.8".
Notes:
I counted drawing the first vertical line to find point D as one operation
because I would be aligning the squared end of my straightedge with
AB, unless I'm allowed use of a drafting triangle as a straightedge. I am
aligning with only one point to draw the line.
I counted the drawing of the other vertical lines as 2 operations because
I have to align my straightedge with both intersection points because
they are known points.
I'm not really concerned with the number of operations in light of the
accuracy I've achieved.
Apparently, my thinking in post #15 paid off.
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#48
In reply to #47
Re: Trisection: CR4 Challenge (04/15/08)
04/21/2008 4:27 AM
Hi Doug
I think you demonstrated the accuracy at 120O. As the process is hard to
analyse, can you indicate the errors at 90, 150 and 180?
Am I right that you used the right-angle at the tip of your straight edge?
Unfortunately, Euclidian straight-edges do not have calibrated right-
angles at their ends, so you need to use a couple of arcs* and a line to
create a right-angle. (Even if you have a reference angle, I think that you
would have to count two alignments - one for the point you take the line
through, and a second to give you the direction)
Regards
Fyz
*One of which may often be a reuse
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In reply to #48
Re: Trisection: CR4 Challenge (04/15/08)
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04/22/2008 12:10 AM
Well, as far as the vertical line from A to the top quadrant, to stay
within the constraints, I'd have to do these steps:
1. With compass set to AB (already done), strike arcs on circle from B.
2. Extend AB to left quadrant.
3. With compass still set to AB, strike arcs on circle from left quadrant.
4. Draw line from left lower intersection to right lower intersection.
5. Set compass to length of this line.
6, 7. Draw circles centered on the endpoints of the line.
8, 9. Draw vertical line through the intersections of these circles.
As for other angles, that will take more drawing time. I'm sure the basic
method is sound, the key issue is the placement of the horizontal line.
I have a feeling this thread is going to be one of those that lives on long
past the posting of the official answer.
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#55
In reply to #54
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 4:42 AM
I'm probably being thick, but can you mark up which points and angles
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#73
In reply to #55
Re: Trisection: CR4 Challenge (04/15/08)
04/23/2008 9:55 PM
You ask, Fyz, and I deliver.
I have numbered the items according to the steps in which I created
them. I'm also reposting the steps so you don't have to scroll back and
forth.
1. With compass set to AB (already done), strike arcs on circle from B.
2. Extend AB to left quadrant.
3. With compass still set to AB, strike arcs on circle from left quadrant.
4. Draw line from left lower intersection to right lower intersection.
5. Set compass to length of this line.
6, 7. Draw circles centered on the endpoints of the line.
8, 9. Draw vertical line through the intersections of these circles to
locate top quadrant.
Notes:
The horizontal line at the bottom of the main circle isn't entirely needed.
I decided it would help in setting the compass and locating the center
points of the circles. As it turns out, the line equals AB, but I had no
way of knowing this beforehand.
The 2 arcs at the top of main circle are really needed either, but again, I
had no way of knowing beforehand if I would use them.
Edit: I just noticed that I could have extended AC to intersect the circle
at the right end of the horizontal line. A diagonal through A and the
intersection of the upper right arc would have given the left end.
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#74
In reply to #73
Re: Trisection: CR4 Challenge (04/15/08)
04/24/2008 4:54 AM
I still don't understand.
I assume you intend to trisecting the angle CAB - but I can't identify the
angle that should be 1/3 of CAB. There are plenty of 60O angles
available (though the lines are not constructed), plus the right angle and
CAB-90O.
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#76
In reply to #74
Re: Trisection: CR4 Challenge (04/15/08)
04/24/2008 2:58 PM
The illustration in post #73 is a clarification of the illustration
in post #54, where I presented a Euclidian approach to locating
the top quadrant point on the main circle, which is what I
thought you wanted. Apparently, you want a clarification of
the illustration in post #47. OK, I'll do that too.
BTW, I am working on an illustration of trisecting a 60° angle.
I chose that angle because it was either the link in post #40 or
some other comment somewhere said that 60° is particularly
hard to trisect.
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#77
In reply to #76
Re: Trisection: CR4 Challenge (04/15/08)
04/24/2008 4:45 PM
Thanks, you've clarified that. The reason I didn't understand
was that I didn't follow back through the thread . And my
original question was how you apparently counted the
production of a right angle as a single operation.
Purely communications, I think. On which topic - realigning a
straight-edge to pass through a point is normally counted as an
operation, even if it was immediately previously aligned
through that point; I think you could argue (with considerable
justification) that this was not a separate alignment if you used
pins for each placement and didn't need to move the pin for
drawing purposes. But accepting that would save operations
using other methods as well (typically at least two), so the
goalposts would move along with the scoring method**.
Regarding specific angles. You are right that 60O is often
given as an example of an angle that is impossible to trisect
using Euclidean methods. So far as I can make out the reason
it is singled out is that it is regarded as a "commonplace"
angle, and the measure (number of degrees) is so obviously
divisible by three. Most trisection methods give a more
accurate result for 60-degrees than (say) 120-degrees - though
naturally it is possible to tailor a method for obtuse angles that
will be less accurate at 60-degrees (the method presented by
jdretired is one such).
**That's not a mixed metaphor - it's scrambled
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#53
Re: Trisection: CR4 Challenge (04/15/08)
04/21/2008 9:25 PM
I think I saw the angle trisected in Popular Science ages ago.
All I remember was three equal circles on a common
centerline, with the trisected angle within. The vertex may
have been on the perimiter of the circle on the left, on the
Good Answers: 46 common centerline, and the trisected angles going to the point
where a vertical centerline hit the circles. Just did a rough test
on a piece of paper with some desk stuff, and it looked close.
Very simple, too.
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#56
In reply to #53
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 6:07 AM
You might have seen an approximation, but not the real deal.
It's a bit like trying to work back from Cos(3ß) = 4 cos3(ß)-
3cos(ß). As Fyz mentioned, you end up trying to find a cube
root which can't be done (in all cases) with euclidean tools.
< I shall hide until Fyz gives a better explanation !>
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In reply to #56
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 7:22 AM
Kris #65
In reply to #57
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 12:01 PM
dskktb
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#60
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 10:29 AM
Physicist,
I guess I am having a hard time following your answer. If you
would post a visual representation of the solution with
references to each arc and line that would be great. I have a
bunch of geometry teachers awaiting an answer. Thanks.
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#61
In reply to #60
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 11:25 AM
My tools won't talk to CR4, unfortunately. If you can
recommend a free web-based drawing tool that I can learn
quickly and will easily upload onto CR4, I'll have a go.
Regards
Fyz
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#62
In reply to #61
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 11:33 AM
I am not sure of an online tool, but I used AutoCAD to draw
in, then printed to PDF which is free, then opened the PDF,
took a snapshot, copied into a new bitmat image using paint
and it worked well for me. I was able to use colors also, which
help quite a bit and distinguishing different arcs. As far as a
web-based drawing program, I do not know of any.
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Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 11:37 AM
Sorry, incorrect description. I meant a package that is free to
download. (It's just that I don't have an use for this type of
drawing in my day-to-day work).
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#64
In reply to #63
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 11:57 AM
http://sketchup.google.com/download/
Physicist? try this link, I think it may work for you, it is a free
sketch tool through google. What will they come up with next.
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In reply to #64
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 12:02 PM
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#67
In reply to #62
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 6:12 PM
I also did my drawings in AutoCAD, then highlighted the parts
I wanted to post, clipboarded into Paint, then clipboarded from
Paint into Serif Photo Plus (free version), cropped to the size I
wanted and exported to .jpg. When I tried clipboarding directly
into Photo Plus, the lines came out too faint, so I would go into
the drawing and convert everything to polylines and widen the
lines. Clipboarding into and from Paint gives me a bitmap I
can edit and the lines remain dark enough to see. Also I don't
have to save the bitmaps or the .spp format image that Photo
Plus creates.
As for free drawing programs, Kris had a thread in the General
Discussion a few months ago about that.
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#69
In reply to #67
Re: Trisection: CR4 Challenge (04/15/08)
04/23/2008 3:44 AM
...much to my shame, the drawing Fyz just posted at #68 (after
some rapid learning of a drawing package) is far better than
anything I've yet produced ! What did you use Fyz ?
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#70
In reply to #69
Re: Trisection: CR4 Challenge (04/15/08)
04/23/2008 4:26 AM
Pure luck (I mistyped this as "puer" first time - equally
appropriate).
I used the free version of CadStd, transferred the 'raster' to
OpenOffice draw, and saved it as a .bmp file (ouch)
N.B. I tried with Google SketchUp initially - not my sort of
thing at all.
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Re: Trisection: CR4 Challenge (04/15/08)
04/23/2008 1:14 PM
A bit of advice, not just to you Fyz, but to anyone who hasn't
yet posted a drawing in CR4, but is thinking of doing so: Try
to set your background color to white to make things more
readable in this forum. You can probably find it under Display
Settings or Visual Styles under Options, Tools, or Format. I
know that while drawing, a dark background sometimes works
better, but a light background with dark lines makes for a
better presentation in CR4.
I do intend to add some drawings later in answer to earlier
requests by Fyz. It might take a while, especially if some
storms mosey through my neighborhood and ask for my
attention as a storm spotter.
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#72
In reply to #71
Re: Trisection: CR4 Challenge (04/15/08)
04/23/2008 4:10 PM
Thanks - I agree entirely, but it took a while to find the invert
background command in a new package - but I've got it now.
But I haven't found out how to hide the paper edges yet...
(I don't expect I'll have time to fully learn this tool...)
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In reply to #72
Re: Trisection: CR4 Challenge (04/15/08)
04/24/2008 5:55 AM
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In reply to #60
Re: Trisection: CR4 Challenge (04/15/08)
04/22/2008 6:22 PM
Apologies that this is not pretty - but I hope this is sufficient to work
with. I have numbered the constructions, but have not marked all the
intersections, as these would be hard to see on this scale. Vertex names
are in the text in case this helps. It's not identical to the official answer,
but it is exactly equivalent. The reason for the difference is partly that
the drawing would be even more cluttered, and also because this is
slightly less prone to accumulating errors from the bisections.
There may be mistypes - please ask about specifics if still unclear.
The original angle is AOB.
First construction: Arc1, centre O, (arbitrary radius R1 to mark AB on
rays of original angle at equal distances from O. (1 operation)
Second construction: Arc2, centre B, radius R1 (1 operation)
Third construction: Arc3, centre A, radius R1 (1 operation)
Arc2 and Arc3 cross at Z
Fourth construction: Line OZ (2 operations)
OZ is "First bisector". It bisects AOB, and crosses Arc1 at C
Fifth construction: Arc4, centre C, radius R1 (1 operation)
Arc2 and Arc4 cross at Y
Sixth construction: Line OY (2 operations)
OY is "Second Bisector". It bisects ZOB, and crosses Arc1 at D
Seventh construction: Arc5, centre D, radius R1 (1 operation)
Arc3 and Arc5 cross at X
Eighth construction: Line OX (2 operations)
OX is "Third Bisector". It bisects YOA, and crosses Arc1 at E
We now mark of three times the radius of Arc1 from O along the line
OE.
Ninth construction: Arc6 centre E, radius R1 (1 operation)
Arc 6 crosses OE at W
Tenth construction: Arc7 centre W, radius R1 (1 operation)
Arc 6 crosses OE at V
We now measure CE
Eleventh construction: Arc8, centre C, to pass through E (2 operations)
[Radius R2]
We now draw an arc this radius three times as far from O to create an
angle that is approximately 1/3 of angle OCE. We use the same arc we
used to mark point V to define the final position. This is not ideal, but it
avoids extra operations. The precision is identical to using the method
of trisecting an arc.
Twelfth construction: Arc9 centre V, radius R2 (1 operation)
Arc nine crosses Arc7 at T
Final operation: Draw OT
The angle AOT is approximately 1/3 of AOB.
AOT is built from AOE - COE/3** = 3.AOB/8 - AOB/8/3**
**This represents that the division by 3 is not exact.
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#78
Re: Trisection: CR4 Challenge (04/15/08)
04/25/2008 1:07 AM
OK, this will seem cumbersome and tedious to get through, but here's
my attempt at trisecting a 60° angle. I broke the procedure down into
discrete steps that may or may not equate to Euclidian operations. Part
of this will be redundant, but it might help people understand my
approach. It does involve a major modification of my procedure in post
#47.
Phase 1 - Preparing for Bisections.
Draw AB.
Set compass to AB.
Draw circle 1 centered on A.
Strike arc on circle 1 above AB and centered on B.
Label this intersection as point C.
Draw AC. Angle CAB is 60°.
Extend AC to diameter of circle 1.
Label this end as point D.
Strike arc on circle 1 below AB centered on B.
Label this intersection as point E.
Draw line DE.
Set compass to DE if needed. (DE should = AB, so check first to see
they are the same.)
Draw circle 2 centered on D.
Draw circle3 centered on E.
Draw vertical line through the intersections of circles 2 and 3 up to the
top quadrant of circle 1.
Label the quadrant as point F.
Phase 2 - Bisecting to (almost) Trisect
Set compass to AF if needed. (AF should = DE or AB.)
Draw circle 4 centered on F.
Draw horizontal bisector 1 through the intersections of circles 1 and 4.
Label the intersection of bisector 1 with AF as point G.
Set compass to AG.
Draw circle 5 centered on A.
Draw circle 6 centered on G.
Draw bisector 2 through the intersections of circles 5 and 6 and over to
circle 1.
Label the intersection of bisector 2 with AF as point H.
Set compass to AH.
Draw circle 7 centered on A.
Draw circle 8 centered on H.
Draw bisector 3 through the intersections of circles 7 and 8 and over to
circle 1.
Label the intersection of bisector 3 with AF as point J.
Set compass to HJ.
Draw circle 9 centered on H.
Draw circle 10 centered on J.
Draw bisector 4 through intersections of circles 9 and 10 over to circle
1.
Label the intersection of bisector 4 and AF as point K.
Set the compass to JK.
Draw circle 11 centered on J.
Draw circle 12 centered on K.
Draw bisector 5 through intersections of circles 11 and 12 over to circle
1.
Label the intersection of circle 1 and bisector 5 as point L.
Draw AL.
Angle LAB = 20.106 degrees or 20` 6' 21.6".
Of course the specific procedure has changed, but the basic approach of
alternating bisections appears to be sound. This will require me to
experiment more, but not right away. I need a break from AutoCAD for
a couple of days.
FWIW, this page has good info on these types of constructions beyond
trisecting angles.
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#80
In reply to #78
Re: Trisection: CR4 Challenge (04/15/08)
04/25/2008 7:45 AM
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There's no question that alternating bisections halves the error
each time. Choosing the ray closest to the trisection as the
initial estimate gives an error of θ/(3.2N), where N is the
number of bisections. My preferred bisection method is to
make the initial bisection by scribing circle AB, placing the
compasses unadjusted centred on A and B successively to
mark the bisection, and scribing the bisector to cross the arc
AB at C. I would then adjust the compasses to AC, scribe new
arcs centred on A and B. For each succeeding trisection I
would use one of these arcs and an arc centred on the
intersection of the preceding trisection with the arc AB. The
only problem with this is that the reference arcs get rather
crowded, so it can be worth adjusting the compass again - but
in the end the crowding of the estimates will inevitably
become problematical...
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#86
In reply to #80
Re: Trisection: CR4 Challenge (04/15/08)
04/28/2008 10:32 PM
In ruminating on this, and trying different trisections in
AutoCAD, a question has come to my mind: how can you
know which angle you have drawn using Euclidian methods?
From what I can see, you can summarize Euclidian
construction as drawing from at least one known point using
either compass or straightedge, except for drawing a simple
line of any length. Using compass and straightedge, you can
locate points to draw right angles, plus angles of 60, 30 and 45
degrees without too much thought. You can bisect 30° to get
15°, and you can do the same the same to 45° to find 22.5°,
etc. But let's say you want to draw an angle of 10°? or 20°?
How would you know you have drawn that angle from the
method you used to construct it? If you drew an angle at
random, how would you know which angle you have drawn?
Of course, using CAD, a protractor, or distances and
trigonometry will tell you the angle. But remember, we are
talking Euclidian construction, using only the designated tools,
and methods conforming to the given restrictions.
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#87
In reply to #86
Re: Trisection: CR4 Challenge (04/15/08)
04/29/2008 3:59 AM
For whole numbers, it's possible to construct 3o, and therefor any multiple
of it. As far as the challenge goes, it's not necessary to know the size or
the initial, or trisected angle ; the error in trisection can be worked out as a
maximum value. If you wanted to draw 10o, you'd have to draw 30
o and
approximate a trisection (or use some other approximating method).
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#79
Re: Trisection: CR4 Challenge (04/15/08)
04/25/2008 3:13 AM
Obtuse triangle AOB
Arc BCO ->2 (centre to find centre for OCB,(X) draw OCB (W).
Arc AYB >2 (centre O and adjust span to OB. full circle)
Line AB ->2 (align A and B).
Line FG ->2 ( align O and F).
Arcs centre F, radius AF ->2 ( centre and adjust span to AF full circle).
Arc centre B ->1 (defines D).
Line AG ->2 (align A and G).
Line KM ->2 (align K and M).
Line ND ->2 (align N and D).
Line OP ->2 (align O and P).
Count 19? I thought it was less.
Regards JD
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#81
In reply to #79
Re: Trisection: CR4 Challenge (04/15/08)
04/25/2008 9:53 AM
Interesting (and I agree the count). As the method is not immediately
obvious to me, I'll ask rather than heading down blind alleys: - does this
place point N identically to your previous method?
If so. it's economical for the accuracy - methods I now know can give:
Your method: 0.26'
19 operations: 0.14' **
18 operations: 0.55' **
17 operations: 0.77' **
16 operations: 1.3'
15 operations: 4.4'
14 operations: 6.2' (fails challenge on accuracy)
**These are based on the 16, 15 and 14 operation methods, with the first
estimates improved by an additional bisection.
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#82
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Re: Trisection: CR4 Challenge (04/15/08)
04/25/2008 6:16 PM
Will analyse it further, N is pretty close, but the theory is, can line AKG
be curved to eliminate the error?
Regards JD.
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#83
In reply to #82
Re: Trisection: CR4 Challenge (04/15/08)
04/26/2008 1:47 AM
I have come to the conclusion that the above does not in any
way improve on what I have already done, the end.
Regards JD.
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#84
Trisection: in 13 operations
04/28/2008 4:50 PM
When I first proposed this challenge, I thought CR4 would do
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well to manage it in 21 operations. By the time it was
accepted, 18 seemed more sensible. In the end, it seems it is
possible to solve using 13 operations - until someone finds an
even more efficient method (I doubt it will be me, however).
The proposed solution uses the method due to Mark Stark. It is
by no means the most robust method (in terms of its sensitivity
to drawing errors). But it is easily the most efficient in terms of
operations - provided the operations are performed in the
optimum order.
N.B. The next most effecient method I found requires 15
operations - but it does have the advantage of significantly
lower sensitivity to drawing error - and doesn't present
additional problems with acute angles.
In order to leave opportunity for any other obsessives to
struggle, I don't plan to post this solution for a few weeks yet.
But I can send it to anyone who sends a request via my CR4
mailbox.
N.B. If you provide Mark Stark's estimation method with an
initial solution with an error ε (radians), the method will
produce a new estimate with an error of about ε3/48 - you will
find the full formula here (although, as I did, you may prefer to
rewrite the formula in terms of the error rather than arbitrary
intermediate stages). I used a construction that gave an initial
estimate of the trisection that was accurate to within 20-
degrees - that gives a next-stage error of 3.046 arc-minute.
I was also struck by the efficacy of the method at the following
complexities (all using Mark Stark's method) interesting:
11 operations: 10.28 arc-minute (initial estimate is exactly
60O)
16 operations: 0.381 arc-minute (initial estimate within 10O)
17 operations: 0.000115 arc-seconds (starts from the 10.28
arc-minute error)
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#85
Re: Trisection: CR4 Challenge (04/15/08)
04/28/2008 8:08 PM
Hello Fyz,
Trisecting angles is not quite my cup of tea, but I went through
your procedure as a matter of curiosity, and satisfied myself
that the practical procedure works (for an angle which I
measured as 104°).
I think a little re-wording is called for in the description of
steps 14 onwards. Adjusting the compass in step 15 is only a
radius setting (and no arc #8 need be drawn at that point?),
while the last step should refer to arc #7 rather than arc #9 (?).
Perhaps this is mere quibbling, unless I have misunderstood
something. Hope I'm not repeating something that others may
have pointed out, as I haven't read all the posts.
I have not tried to investigate whether the method works for
acute angles, and if not why not. Maybe you have a ready
answer to that.
I presume you would have employed suitable analytical or
numerical methods to establish the accuracy of the method or
maximum possible error over the whole range of obtuse
angles. I am not however interested in that aspect, and it may
be beyond my mathematical capabilities anyway.
My interest as an engineer is more oriented towards "accuracy
of construction", whereas the values mentioned in your post 81
are no doubt based on "theoretically perfect" construction.
I have often had to resort to graphical procedures to deal with
problems relating to beams, trusses, theory of machines, etc.
Difficulties would arise whenever intersecting arcs cut each
other at a shallow angle, because there could be an error in
radius setting or compass point placement. A repeat
construction could yield a substantially different result at
times! When I could choose a radius freely, as in simple angle
bisection, I would naturally select one which would allow the
arcs to intersect close to a right angle.
In the trisection procedure, as the original angle approaches
180°, the very first practical bisection becomes increasingly
error-prone. Suppose each point placement could be 'out' by
say 0.2 mm absolute (or 0.2% of the initially selected radius)
the cumulative error can become enormous. I certainly don't
expect you to address this kind of problem in any detail, but if
you have any qualitative insights or suggested references
regarding constructional inaccuracies in general, I'd be grateful
for your comments. I've never come across anything suitable,
or never knew where to look, so I just tried to do my
construction work each time with eyes open for possible
errors. I don't think I've tried doing any accurate graphical
work in the last two decades though.
Thanks, =TeeSquare=
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#88
In reply to #85
Re: Trisection: CR4 Challenge (04/15/08)
04/29/2008 5:06 AM
Agreed. The 13-operation approach doesn't suffer from this,
but is not brilliant at any angle, and fails completely at 60-
degrees. The 15-operation approach is sound at all angles - and
is reasonable for constructional sensitivity at all angles.
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Physicist?
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#89
Re: Trisection: and the answer should have been..
05/04/2008 4:48 PM
And the answer should have been... 10 - giving a maximum error of 2.6
arc-minutes. That is most unexpected (at least to me). The reason is that a
minor modification to Mark Stark's method can increase its accuracy by a
factor of approximately four. The construction is:
Provided with angle, vertex O
Operation 1: Draw circle to intersect the rays of the angle at A and B
respectively
. all subsequent arcs will use the same radius as the circle
Operation 2: Draw arc passing through O to intersect the circle at E inside
the angle AOB
. BOE = 60O (not constructed) is used as the first estimate for the
trisection.
. the error ε of this first estimate grows from zero for AOC=180O to 30
O at
AOC=90O
Operations 3,4: Draw Chord AB, to intersect arc at C.
Operations 5,6: Draw line EC, cutting the circle again at G and extending
outside the circle by at least one diameter.
Operation 7: Draw arc, centre G, intersecting EG at H outside EG
Operation 8: Draw arc, centre H, intersecting EH at K outside EH
Operation 9: Draw line KO, intersecting the circle inside the angle AOB at
T
. TOB is the improved estimate of the trisection. Its maximum error for
obtuse angles is 2.6 arc-minutes
Error calculation:
Following the provision of the (rather poor) initial estimate, the initial
construction is the same as for Mark Stark's original, and provides an
angle CEO (EO not drawn) that is 3/4 of the error ε in the initial estimate
. i.e. CEO = 3.ε/4 = 3.(AOB/3-EOB)/4.
. So we need a line that is ε/4 rotated w.r.t. EC. An approximation will be
provided by KO.
Now consider the triangle GKO. Clearly, GOK = 180O - CEO
So, by the cosine rule:
. OK = OG2 + GK
2 - 2.OG.GK.cos(GOK) = r
2(1
2 + 2
2 + 2.1.2.cos(CEO))
. = r2(5 + 4.cos(3.ε/4))
Now, by the sine rule:
. sin(GKO)/GO = sin(OGK)/OK, i.e.
. GKO = arcsin(sin(OGK)/(5 + 4.cos(3.ε/4)))
Which gives the error for the second estimate angle TOB as:
. arcsin(sin(OGK)/(5 + 4.cos(3.ε/4))) - 3.ε/4
Comment:
The improved accuracy provided by the slight reduction in the length OK
from the 3.r value provided by Mark Stark's construction allows the
challenge accuracy to be achieved with an initial estimate that has an error
of up to 37.2-degrees. This makes it possible to use a fixed non-zero angle
anywhere between -7.2 and 67.2 for the initial estimate. Keeping the
compass radius fixed throughout the construction keeps the number of
operations to a minimum.
. Note1: that, if a secoond stage of improvement is needed, the original
Mark Stark method can be more economical than this modified - and
given the extreme theoretical accuracy this would already provide, the
factor of 256 accuracy improvement that this "improved" method would
give is probably irrelevant.
. Note2: that the constructional error in the angle CEO becomes
progressively more sensitive to alignment to the points C and E as the
angle for trisection reduces. This particular method also fails completely at
AOB=60O. Thus, while I would a-priori have expected trisection of a
small angle (to any given accuracy) to be less costly than a larger one, it
appears that no comparably economical method exists that will cover the
full range of acute angles - unless someone knows different.
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#90
In reply to #89
Re: Trisection: and the answer should have been..
05/07/2008 5:09 AM
What no other comments yet!
This is incredible. I checked the construction for 100o and 170
o with
results shown. 2 decimal places doesn't quite do them justice sorry.
33.36o
56.67o
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#91
In reply to #90
Re: Trisection: and the answer should have been..
05/07/2008 6:07 AM
I actually chose the obtuse angles because I assumed they would need
the more refined construction - and I feared the descriptions "acute or
obtuse", "non-reflex" or "internal angle" would further overcomplicate
the reading of the problem.
So you can imagine why I still find it surprising that such economy is
possible for obtuse angles - but not apparently for all acute ones.
Plus, I should have mentioned:
The third-order approximation
. ε' ≈ ε3/192 (angles in radians) is remarkably close throughout this
range.
As the error in the initial estimate is ∏/3 - AOB/3**, that gives:
. Final error ≈(∏/3 - OAB/3)3/192 ≈ (∏ - AOB)
3/5184**
i.e. maximum error in the initial estimate is 30O (at AOB=90
O)
[Any chance of an improved symbol for pi?]
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#92
Re: Trisection: CR4 Challenge (04/15/08)
05/11/2008 12:49 AM
I must be an engineer by nature; I can't leave well enough alone. I took
another stab (several, actually) at trisecting a 60° angle. Here's what I
came up with as my best shot:
Draw AB
Set compass to AB
Draw circle 1 centered on A
Strike arcs on circle 1 from B
Label these intersections as C and D
Draw AC
Angle CAB = 60°
Draw CD
Draw circle 2 centered on C
Draw circle 3 centered on intersection of CD with circle 2
Label the intersections of circles of 2 & 3 as E and F
Draw EF
Draw AF
Label intersection of AF and circle 1 as G
Set compass to AG
Draw circle 4 centered on F
Label intersection of circle 1 and EF as H
Set compass to FH
Draw circle 5 centered on H
Draw vertical line through intersections of circles 4 & 5
Label the intersection of this line and AF as I
Set compass to FI
Draw circle 6 centered on I
Draw circle 7 centered on G
Draw line through intersections of lines 6 & 7
Label the intersection of this line with EF as J
Label the intersection of this line with AF as K
Set compass to JK
Draw circle 8 centered on J
Draw circle 9 centered on the left intersection of circle 8 and EF
Draw line through the intersections of circles 8 & 9
Label the intersection of this line with EF as L
Draw AL
Angle of LAB = 20°
Is this a perfect trisection? NO!
I did it in AutoCAD, but tried to stay within the constraints of drawing
with compass and staightedge. Except for drawing AB, I only used
points created (or, more accurately, marked) by something already
drawn. Of course, AutoCAD positions elements of a drawing precisely
on the those points. You can not count on that happening when drawing
manually. The amount of error then depends on the skill of the person,
and the quality of the instruments.
I am sure that Tee Square and some others here on CR4 could probably
come within an error on 1' or less. I'm sorry I can't think of other names
just now
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#93
Re: Trisection: CR4 Challenge (04/15/08)
05/13/2008 11:26 AM
I tried to follow the steps a few times but got a little (or a lot) confused
along the way. I wasn't always sure which bisector or circle or arc was
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the right one. It would help me if each bisector, circle, or arc had a
specific designation (most have already) and then these designations
were used in following steps.
Perhaps I'll be able to work through this and post a graphical
representation.
Thanks
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#94
Re: Trisection: CR4 Challenge (04/15/08)
04/03/2009 8:39 AM
If you have compass you can always mark the straight edge. And you
do not need to mark too many divisions for this solution. After that
the link below gives this answer. So easy with the internet but I
hunted down this solution in High school, 40 years back after I was
provoked by an uncle who was a very good Mathematician. It can
also be done by the conchoid of Nicomedes.
http://www-groups.dcs.st-
and.ac.uk/~history/HistTopics/Trisecting_an_angle.html#s31
One of these days I will put in a problem, (the answer for which I do
not know) he gave but expired before I could get his solution.
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#95
Re: Trisection: CR4 Challenge (04/15/08)
01/15/2012 11:12 AM
I wondered if the TRiSeK V3 (http://www.zolkorp/TRiSeK.html)
could be simplified to be competitive, but the best I can do is 6
circles and 7 lines. I'm not sure how many steps that would count
as, but it gets to within a hundredth of a degree for a 45 degree
angle. Adding another 2 circles and 1 line gets to less than a
thousandth. (see tha SIMPL8 on that page). Obviously, more steps
would be added to do an obtuse angle.
A few questions about this thread.
Who's Mark Stark?
Why bother with minutes and radians?
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Z man
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#96
Re: Trisection: CR4 Challenge (04/15/08)
02/10/2012 9:16 AM
I ran Mark Starks trisector with the refinement step.
Thats incredible!
It works all the way up to 180 and is accurate to 13 decimal
places over most of the range.
Definitely has my trisectors beat for simplicity, and they only
work up to 104. My latest one may only have a slite accuracy
advantage. http://www.zolkorp.com/TRiSeK.html
I can't really tell though, since the computer measures the
angle to be below perfect from one side and above from the
other. But it measures Mark's consistently.
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Exploring possible human knowledge
Paul Vjecsner
PRESUMED IMPOSSIBILITIES
(18 November 2004. To the reader: This page has become the most visited one on this website, and I therefore feel I should call to the attention of the visitor that I treat other what may be viewed as thought-provoking subjects in the book mentioned on the home page. On that page I now added at the start a parenthesized paragraph like the present one, giving more information about the contents of the book. Perhaps the visitor of this page will find those contents likewise stimulating.)
(16 January 2006. After further consideration of the present issues, I thought I should bring out some matters I see as quite important. These pages are mainly confined to what I see as falsely presumed impossibilities. However, I also find many other false conclusions reached in highly respected fields, as I explore them in my book. What can be especially startling is that these falsehoods abound particularly in mathematics, a science considered most exact. What is more, the mathematical community fails to recognize a single thing wrong in the field. Physical sciences at least admit, if reluctantly, that its accepted views are revisable theories, but in mathematics the closest thing to this uncertainty appear to be what are known as conjectures. Writing more fully on this, as indicated, elsewhere, I feel compelled to at least mention here areas containing the falsities. Here noted ones are in non-Euclidean geometry, which consistently redefines terms to so commit the fallacy of equivocation; others concern set theory [a fitting name perhaps], regarding which infinite magnitudes are contradictorily considered and paradoxes wrongly interpreted; and still others are arbitrary postulates on which reasoning is built. The questioning of this and more is expectably resented, but it should instead be looked at with at least somewhat open minds.)
In the past—going back often to antiquity—thinkers have tackled many problems pronounced more recently as insoluble. Most have occurred in mathematics, and especially in geometry, owing to its renowned Greek legacy. It is my sense that the continued futility in solving the problems, and the eagerness to prove one's mettle, led many to prematurely decide that, rather than finding the answer sought, they found it to be impossible. Maybe the most famous of these problems is a proof of what is known as Euclid's 5th, or the parallel, postulate. My own consideration of it is included in the book referred to on the home page, so I will not consider it directly at present. But I might note, shocking as the idea is today, that I find the claim of that postulate's unprovability false and, yet more shocking,
offer a proof.
At this time I would like to turn to some other ancient problems in geometry. A well known threesome of them is the trisection of an arbitrary angle, the duplication of the cube, and the squaring of the circle. Related is the construction of a regular heptagon. These are of interest to me at this moment because I just recently wrote about some to Professor Robin Hartshorne of the University of California at Berkeley, whose book, Geometry: Euclid and Beyond, I previously obtained.
At issue in the above mentioned constructions is performing them by use of merely compass and unmarked ruler, in the Euclidean tradition. They were only found instead possible by using added devices such as marks on the ruler, and also involving a "sliding" or "insertion" to make the tool fit certain points and lines, rather than depending on points determined. Whatever action was used, however, the constructions were never accomplished by compass and unmarked ruler alone.
This is here my main interest, without considering any proof presented that the constructions are in some manner impossible. With regard to only the past inability of using compass and unmarked ruler alone however applied, I indeed found ways of using these tools for such purposes. In my letter to Professor Hartshorne I turned specifically to the trisection and the heptagon, which he dealt with in detail in his book and are treated in the below drawings I sent him.
Both of them concern what was previously an insertion of a line by means of a marked ruler. The first drawing is of a simplified trisection that can correspondingly not be constructed with such a ruler, which relies on a length already given. The second drawing will serve to illustrate this reliance.
It is about a heptagon construction shown on page 265 of the book by Professor Hartshorne. In it the distance CF is marked on a ruler by which that distance is then transferred to segment GH by "sliding" the ruler so that the mark for G falls
on arc CG and the mark for H falls on line HF while line HGC goes through C. What was not recognized is that the same is accomplished if a compass with a center G on arc GC is positioned to describe an arc from F to a point H lying on lines HF and HGC which goes through C. In this drawing an unmarked ruler pivots on C till H is on the horizontal. (To briefly describe what in the construction is done before, with the right-side intersection of the circle and its extended horizontal diameter as center, and the radius of the circle, an arc is marked on the circle at C and the point vertically below it; with the distance between these two points as radius, and the same center as of the circle, the bottom arc on the vertical is marked; from that point the line to C is drawn, and with its intersection F on the horizontal as center the arc CG is described.)
Similarly to the preceding, in the first drawing a ruler for line ADC pivots on A till from B, on arc BD with center A, an arc BC with center D is drawn so that C is also on horizontal BC. Unlike before and in other cases, there is no preexisting radius like DC that could be marked on a ruler, making this a simpler construction than for instance on page 262 of the above book.
How this construction trisects an angle, here ABC, how for that matter the other drawing leads to a heptagon, will presently be omitted, there being opportunity for these elsewhere. (It may be helpful though to note that in the trisection the angle DBC is a third of angle ABC, and for the heptagon its initial sides are to be found by, with center H and radius of the circle, marking arcs on it above and below the horizontal—the distance then between the intersections on the circle of either of them and of the horizontal makes a side of the heptagon, with the other sides easily following.) What I wish to include is the following response I received by e-mail from Professor Hartshorne.
From: Robin Hartshorne [mailto:[email protected]] Sent: Friday, November 15, 2002 6:40 PM To: [email protected] Cc: [email protected] Subject: constructions
Dear Mr. Vjecsner Thank you for your letter about constructions. In spite of your protestations, I find nothing controversial in your letter. You have given a mathematically correct construction of the trisection and the heptagon neusis step. These constructions are not classical "ruler and compass constructions" as required in the classical problems, because they make use of the ruler and compass in ways not allowed by the traditional constructions. (See explanation on p. 21 of my book). I would say you have invented a new tool, that is simultaneous use of the ruler and compass in a sliding manner, and with this tool you have solved these problems. I wonder if your tool is equivalent to the use of the marked ruler--that is can it solve all and only those problems that can be solved with marked ruler? Sincerely yours,
Robin Hartshorne
The in this e-mail cited p.21, as I mentioned in my reply, allows that "at any time one may choose a point at random, or [a point] subject to conditions such as that it should lie on a given line or circle", which is what my use of "neusis" or "sliding" consists in. And as I indicated, previous attempts at making any of the constructions with only unmarked ruler and compass were unsuccessful regardless of whether those movements are allowed or not. I am reiterating this to bring to attention that the present solutions mean more than the various ones produced with the aid of additional devices.
If this sounds like boasting, I might lessen the effect by saying that an intention of mine is to bring my capacities to people's awareness, so they will give a listen to other things I offer, which I feel are of benefit. The preceding puzzles, again, are among many others I confronted, and are not included in the book of mine spoken of on the home page. This does not mean I do not like to carry the preceding explorations farther, and I have in fact done so and will try to describe them subsequently.
14 December 2002
Viewing the first above construction once more, I can add that the pivoting of the ruler, too, can be avoided here, as well as in the other cases. One can start by drawing arcs from B with a center D on arc BD, till a straight line ADC can be drawn with the ruler.
At this point it may be useful to say that the proof of this trisection follows from Euclid's Proposition I.32. By it angle ADB is equal to angles DBC+DCB, which are equal to each other, because triangle DBC is isosceles; hence angle ADB is twice angle DBC, and since triangle ABD is likewise isosceles, angle ABD, also, is twice angle DBC.
In light of such needed complication in even this trisection, probably the simplest compared to those known, which largely depend on the same proposition of Euclid, it may seem implausible if I offer a trisection simpler by far, whose rightness is so obvious that the impulse may be that the process is not legitimate.
As to be seen, however, the process is none other than the "insertion" used on the many other constructions. What is more, the only tool for it in this case is a compass.
To trisect angle AOB, with center O draw an arc AB; with a center close to C at estimated third of the angle draw the arc close to AD beginning at A; with reached center close to D at two thirds of the angle draw the arc close to CB beginning at last used center. Then "slide" the compass at C for a more accurate radius till the second arc ends at B.
The job is made still easier than the previous ones by knowing that because thirds are concerned, of the reached difference from the goal of B can be taken an again estimated but much finer third by which the radius is rightly changed, since each small angle must change by that third to attain the correct three angles.
To almost end this session let me return to the heptagon. As was seen, the described construction of it, notwithstanding that the drawing left out parts unrelated to the present issue, is quite elaborate and had to avail itself of insertion regardless, with an added device to boot. Other past attempts, equally requiring insertion and additional tools, were similarly intricate. As the preceding drawing may suggest, however, a much simpler process is possible here, too, as the following drawing will illustrate.
In likeness to the last case, the vertices of the heptagon can be determined by progressive arcs, to end at a specific point. Only 3 arcs are needed; upon estimating the length of a side close to the distance between A and B, draw with center near B an arc from A to a point near C on the circle; with center at that point, draw from A an arc to a point near D on the circle; and with that point as center draw from the point at B an arc toward A to complete the cycle. If that arc ends at A, distance AB becomes a side of the heptagon. As before, of the difference on the circle from that goal of A can be taken an estimated finer seventh, by which the initial radius for BA can be changed, till the third arc ends at A.
One could, to be sure, keep repeating the first arc, with each successive one centered at the endpoint of the former, till completion. But reducing instead the number of arcs not only shortens the job of "insertion", but the fewer the steps the smaller the chance of physical inaccuracy. The added dashed arcs in the drawing suggest a way to locate the remaining vertices.
It can be observed that the method in the last two examples can be employed to divide an angle into any number of equal parts, and to construct any regular polygon. The steps can, further, be drastically reduced by the likes of doubling each successive arc as in the preceding. Thus a 100-gon is constructible with 7 arcs.
Estimating the length of its side as radius of the smallest arc in the present drawing, draw the arc from a point A; draw succeeding arcs with centers at preceding endpoints and starting from points shown, until an endpoint near C is reached; and with that point as center, draw an arc from the point at B toward A to complete the cycle. Then, since distance AB amounts to eight times the concerned first radius, take of an eighth of the difference—found by halving the difference 3 times—on the circle from the goal of A an estimated finer hundredth by which to change that radius, till the last arc ends at A. The first radius equals then a side of this polygon. The dashed arc with center A suggests a beginning for correspondingly locating the other vertices.
The viewer should not have too much trouble calculating that the length of the first radius is contained 100 times in the measure starting and ending at A. Counting from A, the first endpoint reached after B can be seen to be 14 times that radius, the next endpoint 27 times, C 54 times, and since B is from C less by 8, namely 46, so is on the opposite side A; and 46+54=100.
Where and to what extent arcs can in this manner be increased (which they needn't be) for a polygon, or for a divided angle, may not be always easy to detect, although there are normally several relations possible for such arcs. My objective was rather, again, to demonstrate that not only can these methods be economical, but that the process of "sliding" or "insertion" can often be performed with simplified structures and compass alone, let alone with an unmarked ruler besides.
It may be added that the above method of constructing any regular polygon or dividing an angle into any number of equal parts is more significant in geometry than it may seem. Geometers may be at a complete loss if asked to construct for example a regular polygon of some arbitrary number of sides. An aid like a protractor may be of assistance, if deficiently so, especially the larger the polygon. The same is the case when approximations for certain polygons exist. In contrast, the above method allows for more precision the larger the polygon, because the
relative difference in accuracy can continually be reduced by any practicable amount.
26 December 2002
The next drawing utilizes once more the increasing arcs, this time without a need for "insertion", but the drawing is only an expected approximation in regard to "squaring the circle", one of the threesome of ancient problems mentioned. Its aim has been to construct a square of the same area as a circle, a task hindered by the irrational nature of connected pi, the measure of a circle's circumference in relation to the diameter.
In those constructions, pi, which is 3.14159... to infinity, has been approximated to half a dozen and more correct decimal places, and it is not the intention here to trump these, but rather to present a good approximation of exceptionally simple construction, by again only an unmarked ruler and compass. The approximation is to almost five decimal places, 3.14158..., which will be seen to go very far.
Past such approximations relied on rather involved constructions which in the end usually yielded the side of the square of area near that of the circle, and separate from it. One may be at a loss to instead find a construction of the square centered over the circle. From ancient Egypt it may be one that is of some fame, known as
of the Rhind Papyrus, and in which the side of the square is 8/9 of the diameter of the circle. Its accuracy as to pi is only somewhere near two decimal places, 3.16..., which nonetheless can in practice be quite adequate. (A substantial reference book on the subject is Pi: A Source Book, published by Springer.) (See a somewhat better approximation with an extremely simple construction on the next page.)
How the Egyptian square was realized appears uncertain, but evidently it was not by use of straightedge and compass. There is such a way, but now to the present drawing. The measurements resulting in its above approximation are 23 units for AB, and 44 for AC, with A and C respectively at half the side and corner of the square, whose intersection with the circle is at B, the point at issue.
Here, too, it may be asked how this can be realized when only the circle is given. The steps used are as follows. With center O of the circle draw prospective diagonals of the square; bisect as shown; on the bisecting line measure off chosen units of 21 (44 minus 23) and 44, at the end of which draw the perpendicular to that bisecting line; with their meeting point as center draw the arc from end of 21 units to that perpendicular; from the point meeting it draw the line to O. At the line's intersection B of the circle and perpendicular to the above bisector draw the side of the square, etc.
The preceding description may sound more difficult than the drawing obviously is, and I think the steps are self-explanatory and need not be justified. I could remark that the counting process, here shown with arcs on the left, is as a rule not depicted in the approximations, with only the resulting structure shown. As noted, further, the measurements are here made by means of increasing and few arcs, with the first chosen bottom radius the unit, and the numbering, which designates distance from center O, should aid in seeing the placement of those arcs.
It remains to give the calculation of the degree of accuracy in this construction as regards pi, known by the Greek letter π. The area of a circle is pi times radius squared, or A = πr 2. Therefore A/r2 = π; that is, the area of the above square divided by the square of the circle's radius will yield the approximation to pi. The area of the square is 882 = 7744, and the square of the radius is found by way of the Pythagorean theorem, through the sum of AB squared, 232 = 529, and AO squared, 442 = 1936, for the square of BO, 2465. And 7744/2465 = 3.14158...
If the square roots of pi and of this approximation, 1.772453... and 1.772450... respectively, are considered as sides of corresponding squares, and either of them is divided by the difference, about 0.000003, then 1 mile for either is found to differ from the other by less than 1/8 of an inch.
21 January 2003
9 June 2005 As done elsewhere, I am sandwiching in another item here, of a new approximation that occurred to me of a circle squared. The approximation is fairly good considering that this is the simplest way I found of performing it with unmarked ruler and compass, utilizing the above "sliding" or "insertion" and my circular arcs for division.
On a circle with a suitably vertical radius OA (=1) and A as center draw an arc BC such that on drawing with center B arc AD, with center C arc AE, and with center E arc BF, the arc drawn with center F from B meets D. Any part of the circle spanned by radius AB is then 1/13 of the circle, with angle AOB 27.692...°. The cosine of that angle is 0.88545..., and, with a side of the square drawn through B and C, this cosine (=the vertical from O to that side) is half that side. The full side then is 1.7709..., and the square 3.1360..., compared to pi, 3.1415..., the area of the circle (pi times OA² or 1). The difference between that side, 1.7709... again, and the square root of the circle, 1.7724... (the side of the corresponding square), is less than a millimeter for a meter.
It may be noted again that the circular arcs I employ to find the ultimate point to be reached are far more accurate than many methods not depending on found points but on predetermined ones, as in various other constructions leading to squares approximating the area of a circle. In those cases, though the steps arrive mathematically at a good approximation, there is no way of checking—by straightedge and compass—that the steps are physically accurate enough. In contrast, the very act of adjusting my arcs for proper "insertion" assures as close an accuracy as physically enabled (mainly the first arc needs adjustment, the others depending on it, which is why I sometimes show it alone drawn twice).
Someone may argue that what is of account in the other cases is their mentioned mathematical reliability, not the physical limitations. However, the methods of "sliding" or "insertion" are equally mathematically correct. As in the preceding, if distances or other requirements are met by the "inserted" elements, then the accuracy at issue pertains. (The paragraph following here and the rest belong to previous matter on this page.)
7 January 2008. I am again squeezing in another item, and the preceding last sentence, in parentheses, refers now to the following matter that starts on a yellow panel. At this time I am depicting an approximation of a "rectified" circle, for an additional method of squaring the circle with straightedge and compass to a fairly good accuracy. The number 3 17/120 yields 3.14166... as an approximation of pi accurate to almost four decimal places, only by one less than the first of the two constructions above and with the present construction simpler by far, at least in comprehension if not in execution.
As should be easy to follow from the image, one way to measure off 3 17/120, after first drawing a line 4 times the length of the circle's diameter, is to continually halve the last of these lengths until 15 units are reached, divide the second 15 by three for 5 units, and then count 2 more fifths for 17/120. The square root of the total (the dashed line discounted), by which to construct the square, is known to be found by a simple method utilizing continued proportion similar to one shown below at right. (In practice, it is only necessary to extend 3 17/120 at left to 4 17/120, use this total length as a diameter for a semicircle, and construct a perpendicular from the start of 3 17/120 to that semicircle. This perpendicular is then the square root sought.)
1 February 2008. I am replacing below a former image, because this new one represents a more general approximation of pi as 3.1416, which is better than the preceding 3.14166..., since the 6 alone is much closer to the actual 59... in pi than is the repeating 66.... In fact it is also closer than the 58... in the above first of the four cases depicted. The present approximation is given by 3 177/1250. Accordingly, pictured is the pertinent part of the "rectification" of this pi, similar to the preceding.
As seen, the denominator of the fraction, 1250, standing for the diameter of the circle, or a unit, is the main dimension shown, and above it is measured the numerator, 177, the part of that unit belonging to the length of the pi. Of the 3 units, the whole numbers, that are to precede that part a portion only is shown at left as if a panhandle. Explaining the measuring, 1250 is divided into fifths of 250, the first of which is divided into fifths of 50, with the fourth divided into fifths of 10; from this is obtained 45, divided into fifths of 9, to obtain its multiple 27. This plus the 150 shown then equal the needed 177 out of 1250, the dashed line again signifying the discounted portion.
11 February 2008. I thought of a way to "rectify" approximations of pi, of a circle's circumference, similarly to the preceding two but much easier and as usable for any fractional form of such approximations. Since this page is getting quite cluttered, I am illustrating the method on the next page.
So as to treat of each of the triad referred to, a look is here taken at the problem of duplicating the cube. The issue is doubling the volume of a given cube, by again use of only straightedge and compass. Ironically, doing so, by way of “sliding” or “insertion”, was in antiquity virtually accomplished, failing only to put the proverbial “two and two together”.
As described in the book The Ancient Tradition of Geometric Problems by W.R. Knorr, the solution was in different periods attempted both by means of a pivoting ruler similar to the one in the first two drawings here but without recognition that the compass can be of aid (pp.188-9, Dover edition, 1993), and by means of a compass while overlooking the role of a ruler (p.305). Complying with the present drawing (it may be noted that rectangle ABCD is twice as wide as high, the rest to become evident), by the first of these cases a ruler “able to pivot about [B]” would “be manipulated until it assume the position such that [EF] = [EG]”; by the second of the cases, “instead of a sliding ruler [FBG], there is conceived a circle centered at [E] such that the vertex [B] and the two intercepts [F], [G] lie in a straight line”.
It is really the second case where a sliding ruler, other than a pivoting one, can literally be employed. As depicted, the radius of an arc with center E can be adjusted until a line connecting the arc’s endpoints F, G and point B can with a sliding ruler be drawn.
There should be no need now to demonstrate the resulting duplication, the information if of interest obtainable elsewhere. It may be of use though to note that if BC is the edge, viz. the cube root, of the first cube, then CG is the edge of the cube of double the volume.
28 January 2003
6 August 2004 Recently I became aware of another duplication of the cube offered (included on page 270 of the book by Robin Hartshorne referred to on top of this page) which uses a marked ruler. In accordance with the construction below, segment AB makes a right angle with BC, which makes a 30° angle with BD. With distance AB marked on a ruler, it is by "sliding" transferred as distance EF on line AF such that the line lies on A, E is on BC, and F on BD. I again found the following way of achieving this with only an unmarked ruler and compass, as held impossible.
With center B and radius AB draw circle; by "sliding" the compass, find a center G on the circle for a circular arc of radius BG, intersecting BC at a point E, with which as center and radius GE an arc intersects BD at a point F, such that A, E and F lie on a straight line, as determined by the ruler.
The critical first arc is once again shown in two tries, indicating that the appropriate center G must be located. Since the radii are constant, AB and EF are of equal lengths. In this construction, with AB the cube root, edge, of the first cube, AE is to be the cube root of the cube twice the size.
16 September 2004 It should be no surprise if I continue with these constructions, thought to require some device other than compass and unmarked straightedge alone, whether or not performed by way of "sliding" or "insertion", by the indirect way of finding unspecified points through which exact requirements are reached.
For instance, I have been able to so construct with straightedge and compass all forms of cube duplication and angle trisection I have encountered and which often are believed to need involved mechanical devices or geometric shapes ranging from special curves to combinations of solids.
First let me return to a previous cube duplication for which "insertion" is considered used, but the nature of which is not specified. As I noted there, the process can be of greatest simplicity with straightedge and compass alone. One need merely adjust the compass for the radius of the arc until F, B and G lie in a straight line as determined by the straightedge.
Nonetheless, the mentioned book by W. R. Knorr, which contains these, presents also (p.306) a more elaborate version, depicted below first and said to be simpler to execute by "neusis" ("insertion"), without giving an explanation. I am including it here not only because I find it doable with straightedge and compass, though much more involved than the other, but because the result has a certain symmetric beauty.
The idea in this version is that if rectangle ABCD is, with center E, circumscribed by a circle, then upon its intersection of correct line FG at points I and B the segments FI and BG are equal. My way of attaining this with straightedge and compass is as follows. Rotate the straightedge on B so that when with B as center an arc with a radius BG is drawn, and from the midpoint of line FG or its part IB is drawn the perpendicular to meet that arc at a point H, and with H as center and same radius, now HB, is drawn an arc BI, and with I as center and same radius, IH, is drawn an arc HF, the arc will meet line FG at line DA extended. (A much shorter way occurred to me, but I leave the depicted, as perhaps interesting. Rotate line FG on B until a perpendicular through the midpoint of FG or its part IB meets E beneath. The perpendicular is usually drawn in locating that midpoint.)
The second preceding drawing is likewise for doubling the cube and is accepted as requiring a device like a marked ruler. A somewhat differing form of it is in the above mentioned book by Robin Hartshorne (p.263), but I found the present one more frequent, and it is credited to Newton.
In it an equilateral triangle ABC is constructed; AB is then extended for BD of equal length and line DC is drawn and prolonged, as is done to BC. Thereafter a ruler is marked with distance AB and applied so that one of the marks falls on line DC at a point E, the other mark falls on line BC at a point F, and the ruler goes through A. With AB the edge of the smaller cube, AE is then the edge of the doubled one.
The way I used unmarked ruler and compass instead is by first drawing an arc with center B and radius BA (one could draw a different arc, but this one nicely preserves a circle used to get the equilateral triangle by Euclid's Proposition 1). Then a center G on the arc is found such that with radius GB an arc BE is drawn and with center E an arc GF, with AEF a straight line.
Following further are two depictions of ancient angle trisections, equally included in the book by Robin Hartshorne (pp.262, 269), and that have likewise held to require devices like marked rulers, or curves like the conchoid of Nicomedes (used also for above cube duplications and dealt with in the books by Hartshorne (pp.263-264) and Knorr (pp.219-226). They are similar to an above trisection of mine, which, as noted, is simpler by disposing of the need for marking distances.
First here depicted is a trisection, of angle ABC, attributed to Archimedes. Drawn is semicircle CAE on CB extended; distance AB is then marked on a ruler, which is thereafter so positioned that one of the marks falls on line DC (point D), the other on the semicircle (point E), and the ruler goes through A.
Keeping the arrangement, I dispense with the marked ruler by drawing an arc with a center E so positioned on the semicircle that the arc, drawn from B, meets line DC at a point D that forms a straight line with E and A.
It may have been bothersome to some in antiquity that in that arrangement angle ABC does not contain its sought after third, angle EDB. In any event, the second preceding depiction is of another, similar, trisection that had been given. There seem to have been two basic versions, describable here. In both, to the to be trisected angle ABC is added line AF parallel to BC. In one case, then, a perpendicular is dropped from A to BC; distance AB is then marked twice successively on a ruler, which is then so positioned that it lies on B, with the first mark on AC (point D) and the last mark on AF (point F); from central mark, E, a line is then drawn to A for elucidation. The result is that angle FBC, contained in angle ABC, is one third of it. In the second case the perpendicular AC is omitted; an arc with center A and radius AB is drawn; distance AB is marked on a ruler, which is positioned then so that the farther mark is on AF (point F), the nearer mark is on arc BE (point E), and the ruler passes through B.
My own solution, without marks, is, on drawing arc BE, position a center E on it for another arc, AF, so that B, E and F are aligned.
Let me in the first following depiction add another trisection, similar to an above one. One difference is that instead of moving the center of the initial arc, its radius alone is adjusted. To trisect angle ABC, with B as center draw arbitrary arc CA; draw with center A an arc DE with a radius of an estimated third of angle ABD, and adjust the radius until with center D an arc CE meets on extended arc CA both arc DE (at E) and line BC (at C).
17 January 2008. This late insertion is about the figure directly below. I felt it would be good to give a simple, non-algebraic, proof of this cube duplication, having to do with observed continued proportionality and also applying to the other cases.
Noting again that in the three triangles ADF, AFE and AEC, AD is to AF as AF is to AE and AE to AC, it follows that with AD length 1, the length (for n x 1 or n) by which AF exceeds AD is the length (for n
2) by which AE
exceeds AF, and the length (for n
3) by
which AC exceeds AE. That is to say, AF (n) is the cube root of AC (n
3), which, as
2, stands for twice the cube 1, their edges as cube roots respectively AF and AD.
The second preceding drawing returns to cube duplication. I thought it of much interest because it illustrates a basic concept involved, called mean proportionality. In the drawing, for instance, line AE is the mean proportional between AF and AC, because in the right triangles ACE, AEF and AFD, AC is to AE as AE is to AF. It was found that if in a semicircle like the present one the radius AB, namely half the diameter AC, is the edge of the smaller cube, then regarding the same triangles the continued mean proportional AF between AD (=AB) and AE is the edge of the double cube.
Making the construction proved to be a problem, however. The issue is to find point E on the semicircle such that all perpendiculars meet as required (angle AEC is always right by Euclid's Proposition III.31). A famed ancient solution, described in the above book by Knorr (pp.50-54) was given by Archytas, an associate of Plato. He used elaborate intersections of a cylinder, a cone, and a torus, and has been hailed for his ingenuity. Another recent book, Geometry: Our cultural Heritage by Audun Holme (Springer), even utilizes his construction as cover. A much simpler solution should nevertheless be more welcome. The same book also gives a solution (pp.54-55) where two T-squares are used, one of them marking yet distance AD on the crossbar at top from its right inside corner toward the left; that corner is then on arc BD here placed to coincide with a point D, with the other mark on A, the crossbar along a line AE, and the long bar along a line DF; the second T-square has the crossbar at bottom with the inside edge aligned with diameter AC, and the long bar upward. The two T-squares are then so adjusted that the right edge of the long bar of the first meets the left inside corner of the second on AC, while the bottom right edge of the crossbar of the first meets the left edge of the long bar of the second on semicircle AEC.
This would be better visualized with a picture, as in the book, but let me turn to my construction of the same with straightedge and compass. To make it easier to follow the text and picture, I am using the top of the column to the right for the description.
5 October 2004 Presently I felt it of interest to sandwich between earlier items this further consideration of the ancient answer by Archytas to the preceding cube duplication. As noted, the answer availed itself of the intersection of three solids, a cylinder, a torus, and a cone. These were really used only partially, as to be seen. Viewing the drawing below, in correspondence to the last drawn semicircle's diameter and radius as lengths for which mean proportionals are
This text comes after the first three paragraphs at left, and is about the construction immediately above. It is done with straightedge and compass by rotating a line AE on A so that on dropping a perpendicular from E to AC, and from meeting point F to AE, the meeting point D falls on arc BD.
I may remark that for these perpendiculars it suffices mathematically to with center E draw an arc tangent to AC, and with center F an arc tangent to AE. But unlike the standard way, after Euclid's Proposition 12 or 11, this is physically not very practical (as is not at all practical the
sought the bottom oval below, depicting a circle in three dimensions, has its diameter AB given again as one length, and the chord AC as half that length. The chord is extended to D, meeting line BD, tangent to circle ACB. I may remark that some postulations, like that tangency, are not required. Triangle ABD is to be rotated on axis AB (to describe part of the cone) toward position ABD', for which BD needn't be tangent as said. Also, with arc CEF described by the rotation, line CF is needlessly posited as parallel to BD or at right angles to AB. It will be so by the rotation. To continue, a right cylinder or half of it is constructed on circle ACB, and a vertical semicircle on diameter AB. This semicircle is rotated about A horizontally (to describe part of the torus) toward position AGB', intersecting, like triangle ABD', the cylinder in the process. The same semicircle and triangle will meet on the surface of the cylinder at a point G. Through a series of deductions triangle AGB' is, as above, found to yield the sought after result.
intersecting of three solids as noted at left).
28 August 2008. Regarding the figure directly above, I wrote an explanation above it of how the cube root is determined in it, and not being sure that I made myself clear enough, on the next page I did the like for drawing a square root, maybe helping in the above.
What I want to observe is that this elaborate structure relies like others on "sliding" or "insertion", though it may not be at once evident. The "cone" and "torus" are "loci", signifying the continuous motions of the respective triangle and semicircle by tracing their routes. Accordingly one is by moving them searching for the point, here G, at which they will appropriately meet. In other words, one has to by, as described, rotating both the triangle and semicircle find the point at
which they meet on the cylinder. In the drawing preceding this one the point at which instead the semicircle and vertex of the inscribed triangle appropriately meet, is E. As seen, furthermore, the last pictured method is hardly achievable in practice. It is difficult to imagine how the concerned semicircle and triangle are physically rotated so as to meet on the cylinder as required, not to mention a desired solution using only straightedge and compass.
28 September 2004 Possibly more interesting than the construction of double mean proportionality in the last two drawings is the next one. Because of the adjacency of the triangles in their clockwise arrangement, the mean proportionals are easily seen. AB is to BD as BD is to BE and as BE is to BC. A difference is that in the last case the proportionality can continue in principle indefinitely (in the present case this too is possible if supposing overlapping spirals), not necessary for present cube duplication. With AB here twice as long as perpendicular BC, which is the edge of the small cube, BE is the edge of the double cube.
Given ABC, to draw this construction with only straight edge and compass may be viewed as my coup de grâce to the presumed impossibility of the like. The drawing is well known to have been approached since antiquity with a fairly elaborate mechanical device, used in a "sliding" manner of equal involvement. The method is described in the above book by Knorr (pp.57-60) and also in the often referenced books Science awakening by B.L.van der Waerden (pp.163-165) and, especially, History of Greek mathematics (vol.1, pp.255-258) by the most relied on English author in the field, T.L. Heath. He writes:
"If now the inner regular point between the strut KL and the leg FG does not lie on [CB] produced, the machine has to be turned again and the strut moved until the said point does lie on [CB] produced, as [D], care being taken that during the whole of the motion the inner edges of KL and FG pass through [A], [C] respectively and the inner angular point at G moves along [AB] produced.¶That it is possible for the machine to take up the desired position is clear from the figure of Menaechmus, in which [DB], [EB] are the means between [AB] and [CB] and the angles [ADE], [DEC] are right angles, although to get it into the required position is perhaps not quite easy."
Instead, my use of straightedge and compass only, in likewise a "sliding" manner, can be seen to be child's play. Rotate the straightedge on A so that on drawing at meeting point D of CB extended the perpendicular to AD toward AB extended, the perpendicular to DE at meeting point E meets C.
The second preceding drawing is once again a version of a previous cube duplication. It is an admired one by a contemporary of Archimedes, Nicomedes, known also for his invention of the conchoid, a special curve. He used the curve for this construction, and created it with also a mechanical device he built for it. Moreover, the curve had to be specially created for any construction at hand. The mechanical device, the construction, etc., are again found in the books by Heath, van der Waerden, Knorr, or Holme. In their figure, given rectangle ABCD draw line from B through midpoint E of DC to extension of AD at G; from midpoint F of AD draw perpendicular FH of such length that AH equals DE; join GH and draw AI parallel to it; from H draw HIJ meeting AI and DA extended at J so that IJ equals AH (=DE); draw line JBK to meet DC extended in K.
For my construction with straightedge and compass, with center D draw arc EHL for distance DH to get equal AH; on that arc find center L for an arc from D to a point I on AI so that with center I an arc from L meets DA extended at a point J to make straight line HIJ.
5 September 2007 Again, I am sandwiching in with this text and the next image a cube duplication that will likely be the last one by me, inasmuch as I seemed to have managed to draw with straightedge and compass, using the method of "sliding", all the known variations of the figures (excepting the squaring of the circle) thought impossible to construct with these tools.
The present duplication is a corresponding solution of one that was likewise done with a mechanical device, by Eratosthenes, a contemporary of Archimedes and mentioned previously. He, as also described by T.L. Heath in History of Greek mathematics (vol.1, pp.258-260), built a frame represented here by the top (dashed) and bottom horizontals and the far-left vertical (AE); the horizontal parts had interior grooves in which panes like the leftmost rectangle (which remained fixed) could slide and overlap; the two other rectangles here represent the sliding panes, which were the same as the first but partly hidden; each pane also had the same diagonal (like AF); there was a rod probably attached, to rotate, at A. Now, the edge of the small cube is marked on the last pane as distance DH, which is half of AE; the doubled cube's edge, which is CG, is again found through mean proportionals; to have AE, BF, CG and DH in required continued proportion, the two right panes were made to slide until B and C, the meeting
points of the corresponding verticals and diagonals, formed a straight line with A and D.
Here then is my construction of the preceding with straightedge and compass only. With only the solid lines considered, draw lines EH, AE and, at a random distance from the last, BF; draw diagonal AF and auxiliary horizontal at height D. Now rotate a line AD until on drawing a diagonal BG parallel to AF, a vertical GC, and a parallel CH, the vertical from H meets D.
22 October 2004 Let me now return to one of the first problems on this page, the heptagon, drawn for a communication with professor Hartshorne, by whom an e-mail is reproduced. That drawing is about a solution by the sixteenth-century mathematician Viète, and it involves the use of a marked ruler. As was seen, I was able to do the same with an unmarked ruler and also offered a much simpler method, applicable to any regular polygon.
It may be of interest to include more here on the pentagon, in particular a discussion of an answer known as given by Archimedes. It is considered in most of the books cited above, in Van der Waerden (pp.226-7), Knorr (pp.178-182), Hartshorne (p.270), Holme (pp.90-93).
Viewing the first below figure, Archimedes was to observe that in the heptagon if line AB and triangle CDE are drawn, then, with intersections F and G, AG x AF equals GB squared, and FB x FG equals AF squared. On dividing a line AB into these segments then, a heptagon can be constructed. To attain this division, Archimedes, in accord with the second figure below, is said to draw a square ABCD with diagonal AC, and, with AB extended, rotate a straightedge on D for a line DE so that, with intersections F and H, triangles CDF and BEH are equal in area; by dropping the perpendicular FG to AB, the segments AG, GB, and BE correspond to AF, FG, and GB in the previous figure.
To add a comment before proceeding, there is interestingly no indication of how on so sliding the straightedge the equal areas of those triangles are determined. What is more, it appears to me that one could instead similarly adjust in the first figure the three concerned segments until the required areas of the rectangles and squares are attained, dispensing with the second figure. In fact, while I have no knowledge how areas of dissimilar triangles, as in that figure, are with Euclidean tools found equal (in this case, if reversely the segments of AE are known, the triangles can be found), there is a way of finding those segments (of AB in the first figure) by Euclid III.36. If starting with e.g. a segment AF squared, one can find rectangles like FG x FB of the same area, and use "sliding" until GB squared equals in area rectangle AF x AG. (I found later that in the second figure the rectangle formed by the height and base of either triangle in question can be made into a square by Euclid II.14, which then can be tried to be made by preceding III.36 into the like rectangle for the other triangle, having to of course rotate DE again to find the required equal areas in this elaborate process.)
There is, however, an easier way. As noted in Knorr (pp.181-2, 204), the roots of Archimedes' procedure may be compared with those of Euclid's pentagon (IV.11), dealing with properties of isosceles triangles. It has been correspondingly known that, as in triangle ABC in the first figure below, the triangles can be divided into adjacent isosceles triangles with accordingly equal sides, here AB, BD, DE, and EC, as also discussed in Hartshorne (pp.266-7) in connection with the above heptagon by Viète. Now, I want to demonstrate how presently, too, I use only straightedge and compass to construct the triangle and with it the heptagon.
At the midpoint of chosen line AB for a side of the heptagon erect perpendicular FC; with B as center draw circular arc AD; rotate on A a line AC through a point D on arc AD toward FC so that, on completing from the meeting point on FC a triangle ABC, if with D as center an arc BE is drawn, then with center E on the right side of that triangle an arc DC will meet its vertex. (Dashed transversals BD and DE are not required for the construction. They were merely used here for preceding information.)
The next figure, above, replicates the first one in the preceding pair, but for
the addition of some dashed lines. It will be recalled that a special relationship of
areas had to be found regarding the segments of line AB for their length.
Those lengths, were they still needed for the purpose, are given by the figure at left, without search for the areas. In it
AD, BE, and EC correspond above
Triangle ABC can then be circumscribed with a circle (Euclid IV.5), and the dashed arcs with center C and through E and D suggest a way to complete the heptagon. It can also be completed without the circle, with an arc from B centered at A and meeting the lower dashed line at left, and so forth.
respectively to FG, AG, and GB. To demonstrate, draw HB; then line AFGB is duplicated in DKGC; let triangle EDB in the figure at left be replicated above
by GID, for parallelogram GIDB since the sides are of equal length (note the
parallelogram above it for length CG and GB); draw IJ parallel to HB and ED. Then
IE = JD = KG = FG, and of course DG = AG and CG = GB.
26 October 2007. This is again squeezing in an item, after the text following on the yellow panel. At the start of that text I mentioned solutions to well-known problems with some constructions by straightedge and compass. Less known is similar use of those tools for other purposes. Archimedes used for such "neusis", an indirect connection of points, further aids on preliminary constructions in his work On Spirals. Here I am showing one of these as related to that treated in T.L. Heath's A History of Greek Mathematics, Vol. II, pp.386-8. He speaks of the use of special curves, taking it that straightedge and compass fail (Archimedes may have used a marked ruler, as above). One can, however, again do the job without the aids, as I describe below the diagram at left.
It is stated, "Given a circle, a chord [AB] in it less than the diameter, and a point [C] on the circle the perpendicular from which to [AB] cuts [AB] in a point [D] such that [AD is greater than DB] and meets the circle again in [E], it is possible to draw...a straight line [CHG] cutting [AB] in [H] and the circle in [G] in such a way that [HG is] equal to [DE]."
My use of straightedge and compass only is as follows. On drawing the circle and perpendiculars AB and CDE, with E as center draw arc FD, and with an F on it as center draw an arc EG such that with G as center for arc FH, GHC make a straight line.
31 October 2007. It seems I never stop adding to these pages. "Neusis", the indirect connections on this page, was also employed by Apollonius of Perga, regarded as second to Archimedes among the great Greek geometers, and he in fact wrote two books, now lost, on the subject. Their content was described in later work, likewise dealt with by Heath in the volume mentioned in the preceding. On pages 190-2 he speaks about a "construction by Apollonius [that] can be restored with certainty". It contains a rhombus (a parallelogram with four equal sides, like a square), here depicted as ABCD. The object of this "neusis" is that given a length EF, a straight line DHI be drawn such that segment HI from BC to AB extended equal EF.
(5 November. I am indicating in parentheses here the complex way this "neusis" was to be accomplished. The diagonal DB, not shown here, was extended upward to locate a line of the same ratio to EF as AB is to DB, the diagonal; for that line, its square had to equal a rectangle formed by that diagonal extended to a point and the distance from that point to B; "then with [that point] as centre and radius equal to [the line sought, drawn is] a circle cutting [AB] produced in [I] and [BC] in [H]. [HI] is then equal to [EF] and...verges towards [D]". I may note that this equality, etc., may have as well been reached by the known method utilizing marking of a ruler.)
I show now how the neusis can again be simply performed by straightedge and compass. With E as center draw arc FG, and with a G on that arc as center draw an arc EH such that with H as center for arc GI, DHI make a straight line.
My preceding solutions to trisection, cube duplication, and the heptagon, by using only straightedge and compass are difficult for the profession to acknowledge, because of the reluctance to recognize that any of its widely held convictions could be revealed wrong. I consider myself unique in being able to resolve many difficult issues, and therefore the views by many others that a number of general mathematical or scientific claims is false are, although they may instinctively be justified, insufficiently supported and correspondingly dismissed by professionals en masse as coming from the ignorant.
That the above problems could be solved by straightedge and compass alone, even if—in what is called "neusis", "sliding", "insertion", or "verging"—one has to, in order to arrive at the desired construction, find points not given in advance, is, as seen, not held possible, in believing that only the described marked rulers, special curves, mechanical instruments, or other tools can do the task. In the above reproduced e-mail by professor Hartshorne he twice speaks of my method as a tool, in apparent unwillingness to admit that what it consists in is a bypassing of the tools. (14 September 2007. A while ago someone wrote on the web page http://en.wikipedia.org/wiki/Talk:Heptagon, a discussion page for the heptagon entry in Wikipedia, that the heptagon can be constructed with straightedge and compass only, the person giving a link to this page. The discussion continues there, with someone else answering, and me responding. If the reader is
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interested, I explain there that Euclid did not place a limit on the use of straightedge and compass, such as what points can be connected, unlike contended in trying to prove the concerned constructions impossible, a proof not accomplished otherwise.)
There are in the fields, as I indicated, other assertions—quite a few of them—that I dispute, beside claiming that I can answer many further questions. The preceding is intended to contribute to any confidence I may engender in what I do.
30 November 2004 Some recent findings of mine made me decide to add a little more geometry, but I do it on the next page for which click the succeeding link, my not wanting to overload the present page. To propose to prove Euclid's parallel postulate, the 5th, is ridiculed almost as much as to propose to prove the existence of God. Consequently, if I enter this arena I can only remind the reader that the depictions on this page are in the main of solutions (constructions with only Euclidean tools) not achieved in over 2,000 years by the best of minds. The reader may recognize by this that I am competent and conscientious in my work and therefore do not contend a result lightly. The impossibility alleged of proving the 5th postulate is based on fallacious reasoning I consider elsewhere, and so I encourage the reader to click the following for the next page.
MORE ON PRESUMED IMPOSSIBILITIES
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Trisectrix: An Angle Trisection Curve
The Trisectrix is the curve, in green, with polar equation r = 1 + 2 cos(θ)
Angle PCD is trisected by the curve, because the measure of angle OPC is one third that
of PCD.
Proof
We will denote the measure of angle OPC by α, and show that PCD has a measure of
3α. Begin by drawing OP, intersecting the red circle at Q. Note that OCQ is isosceles,
because CO and CQ are radii of the red circle. Note, too, that CQP is isosceles, because
the rays OQ and OP differ in length by 1 (note their equations to see why), and so
QP=1.
Now in triangle CQP, we see if QCP=α, then QPC=α as well, so the external angle
OQC=2α.
In triangle OCQ, we see since OQC=2α, QOC=2α, and so its external angle QCD is 4α,
so the measure of PCD is 3α.
Internet references
Xah Lee's Visual Dictionary of Famous Plane Curves: Trisectrix
Wikipedia: Trisectrix
Related pages in this website:
Back to the Geometry and Trig home, or Triangles and Polygons
Law of Sines - Given triangle ABC with opposite sides a, b, and c, a/(sin A) = b/(sin B)
= c/(sin C) = the diameter of the circumscribed circle.
Circumscribed Circle - The radius of a circle circumscribed around a triangle is R =
abc/(4K), where K is the area of the triangle.
Inscribed Angle -- proof that an angle inscribed in a circle is half the central angle that
is subtended by the same arc
Triangle Trisection -- If a point, P, on the median of triangle ABC is the isogonal
conjugate of point Q, on the altitude of ABC, then ABC is a right triangle.
Why Trisecting the Angle is Impossible
Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
First-time Visitors: Please visit Site Map and Disclaimer. Use "Back" to return here.
A Note to Visitors
I will respond to questions and comments as time permits, but if you want to take issue
with any position expressed here, you first have to answer this question:
What evidence would it take to prove your beliefs wrong?
I simply will not reply to challenges that do not address this question. Refutability is
one of the classic determinants of whether a theory can be called scientific. Moreover, I
have found it to be a great general-purpose cut-through-the-crap question to determine
whether somebody is interested in serious intellectual inquiry or just playing mind
games. It's easy to criticize science for being "closed-minded". Are you open-minded
enough to consider whether your ideas might be wrong?
The ancient Greeks founded Western mathematics, but as ingenious as they were, they
could not solve three problems:
Trisect an angle using only a straightedge and compass
Construct a cube with twice the volume of a given cube
Construct a square with the same area as a given circle
It was not until the 19th century that mathematicians showed that these problems could
not be solved using the methods specified by the Greeks. Any good draftsman can do all
these constructions accurate to any desired limits of accuracy - but not to absolute
accuracy. The Greeks themselves invented ways to solve the first two exactly, using
tools other than a straightedge and compass. But under the conditions the Greeks
specified, the problems are impossible.
Since we can do these tasks to any desired accuracy already, there is no practical use
whatever for an exact geometrical solution of these problems. So if you think you'll get
headlines, endorsement contracts and dates with supermodels for doing so, it is my sad
duty to tell you otherwise.
Also, there are a few trivial special cases, like a right angle, where it is possible to
"trisect" the angle. More specifically, it is possible to construct an angle one-third the
given angle. For example, if you draw a diameter of a circle and mark off 60 degree
intervals on the circle, you "trisect" the straight angle. This isn't trisection in any
meaningful sense because it doesn't generalize to other angles.
"Impossible" is a welcome challenge to a lot of people. The problems are so easy to
understand, but the impossibility proofs are so advanced, that many people flatly refuse
to accept the problems are impossible. I am not out to persuade these people.
Mathematicians have spent years corresponding with some of them, and many are
absolutely immune to persuasion (The trisectors, I mean. The mathematicians are too,
but they have reason to be). But if you want a (hopefully) intelligible explanation of
why mathematicians regard the problems as impossible - as proven to be impossible -
then this site might help you.
Warning: you need trigonometry and an understanding of polynomials - that is, the
equivalent of a good high school math education - to follow this discussion.
What is a Proof?
Enough people have asked me to look at their constructions that one thing is very
obvious. Most people do not have any idea what constitutes a proof in mathematics.
Step by step instructions for doing a construction are not a proof. A traditional
geometric proof for a construction would require:
A justification for every step in the construction. Most of the time this is simple:
you can always bisect an angle, draw a line through two points, and so on. But if
you say "draw a line through points A, B and C," you have to be able to prove
that points A, B, and C actually do lie on a straight line. If you say draw a line
through a particular point, you have to be able to show that the point actually
exists and can be constructed using the classical rules of geometry. There are a
lot of interesting geometrical fallacy puzzles that rely on assuming a step is
possible when it is not. Sometimes it's subtle. If you draw a construction that
passes through a circle, for example, you have to be prepared to show what
happens if the line misses the circle. Does it matter? It may or may not, but you
have to be prepared to deal with it.
Second, you have to prove that the construction does what it claims to do. If the
proof isn't inherent in the construction you have to come up with a way to prove
it.
Detailed instructions and diagrams are not proof.
Measurement is not proof.
If you have any questions, get a (preferably old-school) geometry textbook and study
their proofs.
Proving Things Impossible
One of the major problems people have with angle trisection is the very idea that
something can be proven impossible. Many people flatly deny that anything can be
proven to be impossible. But isn't that a contradiction? If nothing can be proven to be
impossible, and you can prove it, then you've proven something to be impossible, and
contradicted yourself. In fact, showing that something entails a contradiction is a
powerful means of showing that some things are impossible. So before tackling the
trisection problem, let's spend some time proving a few things impossible just to show
that it can be done.
The Domino Problem
Imagine you have a checkerboard and
start covering it with dominoes so that
each domino covers two squares.
Obviously there are a vast number of
ways to do it.
Now imagine you remove two opposite
corner squares. Can you still cover the
checkerboard with dominoes?
No. Rather than try every possible way
to place dominoes on the board,
consider this: each domino covers both
a red and a white square. If you remove
the two corners shown, there will be 32
white squares but only 30 red squares.
There's no way to cover the board with
dominoes without leaving two unpaired
white squares. So it can't be done. We
have proven that something is
impossible.
I expect some die-hard to ask what about coloring a white square red so there are 31 of
each color. It doesn't matter. (Actually, it's easy to see that coloring a white square red
will create a situation where you have to cover two red squares with a domino. Once
you cover them, you have an unequal number of red and white squares remaining, and
then we're back to square one, literally.) You can paint the board psychedelic if you like
and it still can't be done. The traditional checkerboard coloring makes it easy to prove it,
but if it can't be done on a traditional checkerboard pattern, it can't be done, period. In
fact, a lot of proofs depend on marking, labeling, or grouping items in a certain way to
show that some particular arrangement either is, or is not, possible.
An Oldie but Goodie: The Largest Prime Number
Prime numbers like 2, 3, 5, 7, 11 .... are divisible only by themselves and one. As
numbers get bigger, primes get more rare. Is there a largest one?
The ancient Greeks showed there is not. Imagine there is a largest prime p. Now
calculate the number q, which is 2 x 3 x 5 x 7 x 11 x...(all the primes less than p) x p. It
will be a huge number. Now consider q+1. It's not divisible by 2, because q is divisible
by 2. Likewise, it's not divisible by 3, 5, 7, or any other prime up to p, because q is
divisible by all those numbers. So there are only two possibilities. Either q+1 is prime,
or it's divisible by primes bigger than p (q+1 will be vastly bigger than p - if p is only 19
than q is 9,699,690 - there's lots of room for bigger primes). Either way the initial
assumption leads to a contradiction. Hence it must be wrong. There is no largest prime.
It is impossible to find a largest prime. It's not that people have tried, failed, and given
up. It's impossible because the idea itself leads to a contradiction. This method of proof
- making an assumption and then showing that it leads to a contradiction - is called
reductio ad absurdum.
It is impossible to find a largest prime, but it is possible to find arbitrarily long stretches
of numbers without them. Our number q is composite - it is the product of smaller
numbers. It's easy to see that q+2 must be composite, as well as q+3, q+4 (divisible by
2) q+5, q+6 (divisible by 2 and 3) and so on up to q+p. We can find so-called "prime
deserts" of any desired length, but there are always primes after them. Indeed, we keep
finding pairs of primes, like 5 and 7, or 101 and 103, however high we go, though
nobody has yet shown there is an infinite number of them.
Another Golden Oldie: Square Root of Two
Can you represent the square root of two as a fraction? The ancient Greeks also found
out that this is impossible. Imagine that there is a fraction p/q, where p and q are whole
numbers and p/q is in lowest terms (that qualifier is important), that equals the square
root of two. We can conclude:
p2/q
2 = 2 and therefore:
p2 = 2q
2, therefore,
p is even, since p2 is even and only even numbers have even squares. Also,
p2 is divisible by 4, since all even squares are divisible by 4. Thus 2q
2 is
divisible by 4 which means q2 must be divisible by 2. Therefore q must also be
even.
Thus p and q must both be even. Since we assumed they were in lowest terms,
and have arrived at a contradiction, the assumption must be false. We have
another reductio ad absurdum. It is impossible to represent the square root of
two as a fraction of whole numbers.
The sect called the Pythagoreans believed that everything was ultimately based on
whole numbers. They were horrified by this discovery. Tradition has it that they decreed
death to any member who divulged the secret. They called numbers that could not be
represented as ratios as irrational, and to this day irrational carries a negative
connotation. Actually this proof is related to the proofs that trisecting the angle,
doubling the cube and squaring the circle are impossible.
Uniform Polyhedra
Reductio ad absurdum is a powerful way of showing that some things are impossible,
but not the only way. If we can succeed in showing all the things that are possible, then
anything else must be impossible.
In simple cases we can list the possibilities by brute force. For example, what kinds of
three dimensional shapes have regular polygon faces, with all faces and vertices
identical? Clearly you have to have at least three faces meeting at a vertex, and the sum
of all the angles meeting at a vertex can't equal or exceed 360 degrees. For triangles, we
can have three, four or five meeting at a vertex (six would equal 360 degrees). For
squares we can have only three, same for pentagons, and for hexagons and above, it
can't be done (three hexagons add up to 360 degrees). So that's it. There are five shapes,
and no others, shown below. (If you allow faces and edges to cross through each other,
it gets a bit more interesting.)
What Kinds of Plane Patterns are Possible?
When there are a potentially infinite number of possibilities, we have to use the
general properties of the problem to devise a proof. Here's a fairly simple example.
What kinds of repeating patterns can we have on a plane, for example, wallpaper? We
say something has n-fold symmetry if, when you rotate it 360 degrees, there are n
positions where it looks the same. For example, a honeycomb has six-fold symmetry -
you can rotate it 360 degrees and there will be six positions where the pattern looks the
same. A sheet of square graph paper has four-fold symmetry. So does a checkerboard if
we ignore the colors, but if we include the colors, then it only has two-fold symmetry.
Something with no symmetry at all has one-fold symmetry.
Now a
single
object
like a
flower
or
propelle
r can
have
any
kind of
symmet
ry at all.
But
what
about a
pattern
that
repeats?
Look at
the
pattern
at right.
The pattern consists of points, only some of which are shown here. All the points are
identical (that's what a repeating pattern means). So if we can rotate the pattern around
one point and see symmetry, we can rotate it around every identical point. If we rotate
the pattern by angle 360/n so it still looks the same, then a point originally at on the
bottom row will be rotated to the same position as some other point on the top row (or
on the bottom row if it rotates 180 or 360 degrees). Note that we can rotate in either
direction. The pattern shown here is the familiar honeycomb pattern, but notice we have
not made any assumptions about the angles in the pattern at all. The pattern shown is
just for illustration.
Let's assume each point is distance one from its neighbor (ao = ob = ov = ou) The
distance between the two points u and v in the upper row has to be a whole number, but
it need not be 1. If a = 90 degrees (4-fold symmetry), it could be zero. But it has to be a
whole number, because it's the distance between two points on a row of the pattern.
From elementary trigonometry, we can also see that uv = 2 cos (360/n).
Now the cosine can only have values between -1 and 1, so 2 cos a, which must be a
whole number, can only take on values -2, -1, 0, 1 and 2. Thus cos a can only be -1, -
1/2, 0, 1/2 and 1, or a = 180, 120, 90, 60 and 0 (or 360), respectively. Since the rotations
are symmetry rotations, a = 360/n, so n = 2, 3, 4, 6, or 1.
Thus, repeating patterns in the plane can only have 1, 2, 3, 4 or 6-fold symmetry. In
particular, repeating patterns in the plane cannot have five-fold symmetry.
What about the stars in the flag? The stars have five-fold symmetry, but the
overall symmetry of the pattern does not. Rotate a flag by 72 degrees and see.
In recent years some fascinating materials have turned up with five-fold
symmetry. These materials, termed quasicrystals, do not have repeating patterns.
These are related to beautiful patterns called Penrose Tilings, which also have 5-
fold symmetry but do not repeat regularly.
Notice, this is not a matter of people trying to find five-fold repeating patterns and
failing, then concluding it must be impossible. We devised a method to find all the
patterns that are possible. We showed that only certain patterns are possible; therefore
all the others are impossible. Many proofs of impossibility are based on the fact that we
show what is possible, thus ruling out everything else.
More (much more) on symmetry and patterns
What's With the Ruler and Compass Anyway?
Plato's famous analogy of the Cave gives a lot of insight into the Greek mathematical
mind. In this analogy, a prisoner in a cave knows the outside world only from shadows
cast on the wall of the cave. Needless to say, his knowledge of the world is pretty
imperfect. To many Greeks, this world, or at least our sensory picture of it, were crude
images of a perfect, real world. Thus, to Greek mathematicians, the lines we drew were
crude approximations of real lines, which were infinitely long, infinitely sharp,
infinitesimally narrow, and perfectly straight. In principle, lines intersected in infinitely
tiny, precise points. In geometry, "the truth is out there", the answer already exists, and
they sought constructions that would find the answer infallibly and with infinite
precision. Thus the Greeks rejected any technique that smacked of approximation or
trial and error. The problem with trial and error is that, regardless how closely the
construction seems to fit, we can never be sure there isn't some microscopic error below
the limits of our visibility. The Greeks wanted to find the correct point directly, the first
try, with absolute precision.
It is possible to build
tools that will solve
the problem, using a
ruler and compass.
One of the simplest,
at right, was dubbed
the "hatchet." It
consists of a right-
angled T as shown
with a semicircle on
one side. Points A, B,
C and D are all
equally spaced. Place
A on one side of the
angle with the long
arm of the T passing
through O, and slide
the hatchet so that the
semicircle just
touches the other side
of the angle at E.
Lines OB and OC
trisect the angle. The
upper diagram shows
how it works, the
lower diagram shows
how one might be
constructed from
cardboard.
Proof: Since AB=BC and angles ABO and CBO are right angles, triangles AOB and
BOC are identical and angles AOB and BOC must be equal. Also BC=CE and angle
CEO is a right angle, so triangles BOC and COE are also identical and angles BOC and
COE are identical. Thus angle AOB = BOC = COE.
So what's wrong with this device? It can be laid out with ruler and compass, but not
physically made. A real hatchet, however perfect, would have imperfections. Even if it
were absolutely perfect, the process of lining the device up would be one of trial and
error. So this device was unsatisfactory to the Greeks. It's tempting to try to use a
straightedge and compass to lay out the hatchet right on the desired angle, but it also
can't be done without trial and error.
It is possible to trisect the angle using a marked straightedge, but that's not allowed by
the ancient Greek rules, since a mark can't be lined up against another point, line or arc
without trial and error, and without some inherent error of alignment. Likewise, you can
solve the problem with two carpenters' squares, but that's also not allowed since it also
involves trial and error and since no carpenters' square will have an absolutely perfect
right angle.
There are subterfuges people have developed for seeming to get around these
restrictions. You can hold the compass against the ruler to measure distance; technically
you haven't "marked" the ruler but the effect is the same. There are a variety of
constructions that involve sliding the pivot point of the compass along a line or arc;
these all involve some measure of trial and error as well (though the people who devise
these constructions sometimes hotly deny it). Using the compass for anything but
drawing an arc around a fixed center is forbidden.
I have modified this page several times as people come up with evasions of the classical
restrictions. So let's clarify the general principle: all loopholes violate the rules.
This is perhaps the simplest
trisection using a marked
straightedge. It was discovered by
Archimedes. Given the angle
AOX, draw a circle of arbitrary
radius centered at O. Extend one
side of the angle through the
opposite side of the circle at D
(top).
Mark off interval BC on the
straightedge. BC = OX = radius
of the circle.
Slide the straightedge so that B
lies on line DOX, C lies on the
circle and the straightedge passes
through A. Angle CBD is one-
third of AOX. Note the element
of trial and error inherent in
positioning the straightedge.
Proof:
Since BC=OC, angles CBD and COD are equal and angle BCO = 180-2CBD.
And since OC = AO, angles OCA and OAC are equal.
OCA + COD = 180 so OCA = 2CBD.
The angles in triangle ABO sum to 180, so we have CBD + DOA + OAC = 180
= CBD + DOA + 2CBD.
Rearranging, we get DOA = 180 - 3CBD, and since DOA + AOX = 180, AOX =
3CBD.
You can eliminate the marking by setting a compass with radius BC and sliding the
pivot along the line until B, C, and A are on a straight line. However, if you do, there
are two additional points to consider: where the radius of the compass hits the circle
(call it C') and where the radius hits the ruler (call it C"). You can never be certain that
C' = C" = C, regardless of how well the points seem to agree. Even if the same pencil
line covers all three points, if they are different in the most microscopic degree, the
construction is not exact.
It also doesn't do to indulge in arm waving and say "slide the compass and ruler until B,
C, and A are on a straight line." We know point C exists somewhere. You have to prove
that you have actually found it. You have to be able to prove that C' = C" = C.
There are other curves besides circles
that can be drawn, and can be used to
trisect the angle. The simplest, shown at
right, is the Archimedean Spiral. The
radius of the spiral is proportional to
azimuth. Obviously an angle can be
trisected by drawing an Archimedean
Spiral over the angle, finding the radius
where the other side of the angle
intersects the spiral, then trisecting the
radius. Draw arcs from the trisection
points of the radius to the spiral, then
draw radii through those points on the
spiral. In fact, an angle can be divided
into any number of equal parts using
this method.
The problem with the Archimedean Spiral, or any other curve used to trisect the angle,
is that although you can construct an infinite number of points on the curve using a
straightedge and compass, you can't construct every point. However closely spaced the
points are, there will always be tiny gaps between them, so that any curve we draw will
always be approximate, not exact.
If you think about it, this whole business is actually pretty artificial. The Greek
compass, unlike modern ones, had a spring so that it snapped shut once it was lifted off
the page. Anyone who has ever had a compass change radius while drawing a circle can
picture the potential for error in such a device. Of course, we can do all the
constructions the Greeks did, in many cases a lot more simply, by using a compass that
stays fixed. To use something as error-prone as a spring-loaded compass, then worry
about possible imperfections in constructing a tool like the hatchet, or positioning a
marked straightedge, is completely arbitrary.
Also, if you think about it, the business of lining a straightedge up through two points to
draw a line also has a lot of trial and error about it - as much as the hatchet tool. You
line the straightedge up against one point, then position it against the other, then go back
and correct any shifting at the first point, and so on. Then, actually to draw the line, you
need to take into account the fact that any marking tool has finite width, so that as often
as not the drawn line doesn't pass exactly through the two points.
Renaissance instrument makers soon discovered this problem. They found out that
markings plotted using only compasses were more accurate than those made using
straightedges. They began devising alternative constructions that eliminated use of the
straightedge. Although the constructions were often more complex, they were still more
accurate than those that required straightedges. Mathematicians finally showed that
every construction that can be done with a compass and straightedge can be done with a
compass alone. The only qualification is that we define constructing two points on a
given line as equivalent to constructing the line itself.
Proving It Can't Be Done
Follow the Rules
Many people who "solve" the angle trisection problem inadvertently violate the rules:
You can only use a compass and a plain straightedge.
You cannot use the straightedge for measuring, or put marks on it.
You can only use the compass for drawing arcs around a fixed center. You
cannot slide the pivot.
You cannot use a straightedge and compass to construct some other tool.
You cannot use a straightedge and compass to construct some other curve.
All loopholes and evasions of these restrictions violate the rules.
Why It's Impossible
The triple angle formula in trigonometry for the sine is: sin 3a = 3 sin a - 4 sin3a.
We can rewrite this to 4 sin3a - 3 sin a + sin 3a = 0
In other words, trisecting an angle amounts to solving a cubic equation. That's why
nothing has been said about doubling the cube. Doubling the cube amounts to finding
the cube root of two, that is, also solving a cubic equation. So algebraically, the two
constructions are equivalent. Squaring the circle is a bit more complicated.
Recall how the Pythagoreans, to their horror, found out that there are other kinds of
numbers than integers (whole numbers) and rational fractions. The process of
discovering all the types of numbers that exist turns out to be directly related to the
proof that the three classic problems are unsolvable. Since the time of Pythagoras,
mathematicians have discovered that there are many types of numbers:
Integers (whole numbers), like 1,2 and 3
Rational Numbers, numbers that are fractions involving integers, like 2/3, 1/2,
and 19/10. Integers are also rational numbers, since they can be written as
fractions like 1/1, 2/1, 3/1, and so on.
Irrational Numbers, those that cannot be written as fractions involving integers.
Almost all roots, trigonometric functions and logarithms are irrational. So is pi.
Positive and Negative Numbers. Also zero, which was unknown to the ancient
Greeks.
Imaginary Numbers, square roots of negative numbers. All other numbers are
called Real Numbers
Complex Numbers, combinations of Imaginary and Real Numbers.
The final solution to the
three famous problems of
antiquity came from
studying classes and
properties of numbers.
Geometrical constructions
can be considered the
equivalent of mathematical
operations. For example,
using only a straightedge,
you can add, subtract,
multiply and divide:
If you can draw circles, you
can also construct square
roots.
Proof: Triangle ACD is a
right triangle since all
angles inscribed in a
semicircle are right angles.
So if one angle is p, q = 90
- p and the remaining
angles have the values
shown. So triangles ABD
and DBC are similar. Thus
we can write BD/1 = X/BD,
or BD squared = X. Hence
BD equals the square root
of X.
One of the first fruits of these studies was the discovery by the young Karl Friedrich
Gauss that it was possible, using a ruler and straightedge, to construct a polygon of 17
sides. This was something completely unsuspected. He also found that polygons of 257
and 65537 sides could be constructed. (Strictly speaking, Gauss only discovered it was
possible; other mathematicians devised constructions and showed that no other
polygons were possible, but Gauss made the pivotal discovery.) A number of books
give constructions for the 17-sided polygon; constructions for the other two have been
devised but are hardly worth the effort.
Numbers that can be expressed as combinations of rational numbers and square roots,
however complicated the combination, are called Constructible Numbers. Only
Constructible Numbers can be constructed using a compass and straightedge. We can
now pose (and answer without proof) the following questions:
Are all numbers Constructible Numbers? (No - there are other kinds of numbers,
just as the Pythagoreans found there were other kinds of numbers than integers
and rational fractions.)
Are cube roots Constructible Numbers? (No - hence we cannot trisect the angle
or duplicate the cube using straightedge and compass. In fact higher roots like
fifth roots, and so on, are also not surds. Of course we ignore special cases like
the cube root of 27 or the fifth root of 32, and so on.)
Is pi a Constructible Number? (No - in fact pi belongs to yet another class of
numbers called transcendental numbers that cannot be obtained as the solution
of any finite-sized polynomial. Obviously, if pi is not a Constructible Number,
then we can't square the circle, either.)
Gauss did more than just find new polygons. Building on his results, other
mathematicians showed that polygons of 2, 3, 5, 17, 257 ... sides are the only ones with
prime numbers of sides that can be constructed. (Of course you can also construct
polygons by repeatedly bisecting the angles of these polygons to construct polygons
with 4, 6, 8, 10, 12, 16, etc. sides. You can also construct polygons of 15 sides by
combining the constructions for 3 and 5 sides, etc.) By enumerating what was possible,
he ruled out many other things as impossible. In particular, 7 and 9-sided polygons
cannot be constructed using straightedge and compass. Constructing a 9-sided polygon
requires trisecting a 120-degree angle. Since this can't be done, obviously trisecting any
desired angle is impossible.
Note, by the way, that 2=1+1, 3=21+1, 5=2
2+1, 17=2
4+1, 257=2
8+1, 65537=2
16+1. The
numbers are all primes, and equal to some power of 2 plus one, and the exponents are
all powers of 2. 65537 is the largest prime of this type known, although extensive
searches for larger ones have been made.
"But You Have to Solve the Problem Geometrically"
Most people who still send solutions to the three classic problems to mathematics
departments don't have a clue how the problems were finally solved. Many seriously
think mathematicians just gave up and decreed the problems unsolvable.
The problems actually can't be solved because they require properties that a straightedge
and compass simply do not have. You can't draw an ellipse with a straightedge and
compass (although you can construct as many points on the ellipse as you like), so why
is it a shock that you can't trisect the angle, duplicate the cube, or square the circle? You
can't tighten nuts with a saw or cut a board with a wrench, and expecting a straightedge
and compass to do something beyond their capabilities is equally futile.
Here's another analogy: if you spend your entire life driving a truck, you might
eventually be lulled into thinking you can see the entire country from the highway. It's
only if you get out and walk or fly over the landscape that you discover there are a lot of
other places as well. Geometry with straightedge and compass creates a similar illusion;
eventually we believe the points we can construct are all the points that exist. It was
only when mathematicians began studying the properties of numbers that they found out
it wasn't so. Just as you have to get off the highway to see that other places exist, to find
the limitations of geometry you have to get outside of geometry.
More mathematically literate angle trisectors are sometimes aware of the number-theory
approach, but reject it because they think a geometrical problem can only be properly
solved geometrically. But if the problem is that the solution requires capabilities beyond
those of a straightedge and compass, how in the world can that be discovered from
within geometry? Anyway, who says geometrical problems can only be properly solved
geometrically? The only thing that would justify that rule is some demonstration that the
geometrical solution to a problem and the algebraic solution yielded different results -
and then you'd have to prove the geometrical approach was the correct one. But there
are no cases where this has ever happened, so there is no justification for rejecting the
algebraic solution to the three classic problems. Indeed, it has been shown that if there
is an inconsistency anywhere in mathematics, it is possible to prove any proposition
whatsoever.
In fact, one of the most famous mathematical proofs of all times, Kurt Godel's
Incompleteness Theorem, showed that there is always an "outside" to mathematics.
Once a rule system gets complex enough (and Euclidean geometry is plenty complex
enough) it is always possible to make true statements that cannot be proven using only
the rules of that system. Thus is possible to make true geometrical statements (like
"angles cannot be trisected using ruler and compass") that cannot be proven using the
rules of geometry alone.
Reference Angles
I've seen several constructions that start off by generating an arbitrary angle, then
trisecting some angle derived from it. In all the cases I've seen, the construction
amounts to constructing an angle A, then trisecting 180-3A. Since one third of 180-3A
= 60-A, of course it's possible to construct 60-A. Note that I didn't say "trisect 180-3A,"
since constructing a desired angle isn't the same thing as trisecting three times that
angle.
If your method involves a "reference angle," you can bet that it's not a true
trisection.
If you ever construct one angle three times the size of another, or any derivative
of it, it's not going to be a trisection.
One Final Note
You can construct a 20-sided polygon with central angles of 18 degrees. You can
construct a 24-sided polygon with central angles of 15 degrees. Obviously you can
superimpose the constructions, and the difference between the two angles is three
degrees. That's the smallest integer value we can get using straightedge and compass
constructions. Since an arbitrary angle can't be trisected, and we can't find any integer
angles except multiples of three degrees,
Therefore you cannot construct an exact one-degree angle with ruler and compass!
I had one correspondent who got rather worked up over this. But what's so special about
360 degrees? It's divisible by a lot of numbers, but so is 240. If we defined a circle as
having 240 degrees, each degree would be 1-1/2 of our degrees. And we can construct
1-1/2 degrees exactly.
But so what?
We can't construct a one-degree angle exactly using a ruler and compass, but we can
subdivide line segments into any number of intervals. In principle, we could construct a
right triangle with one edge 100 million kilometers in length and another edge
1,745,506.492821758576 kilometers long at right angles to it. The hypotenuse and the
long edge define an angle of one degree to an accuracy of 10-8
cm, the diameter of an
atom. You'll need a very big sheet of paper, a lot of patience, and a very sharp pencil,
but in Greek terms it's "possible". So we can construct a one-degree angle to precision
so fine it surpasses any possible practical need.
Reference
Lots of books cover the mathematics behind the impossibility proofs. A really good
recent one is Robin Hartshorne, Geometry: Euclid and Beyond, Springer, 2000.
Warning: this is not fireside reading. The book is very clear but it will take serious study
to master it.
Return to Pseudoscience Index
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Z man
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Trisecting an Angle
10/31/2011 3:57 PM
Has Joe Keating done the impossible?
http://users.tpg.com.au/musodata/trisection/trisecting_any_a
ngle.htm
I tried it in Photoshop & it appeared to be perfect. The
ancient Greeks started trying to find a way. A
mathematician named Pierre Wantzel proved it to be
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impossible several hundred years ago. But, as far as I can
tell, this method works.
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#11 "Re: Trisecting an Angle" by Randall on 11/01/2011 8:11 AM
(score 2)
#25 "Re: Trisecting an Angle" by Randall on 11/02/2011 7:12 AM
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#5 "Re: Trisecting an Angle" by Usbport on 10/31/2011 10:27 PM
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#27 "Re: Trisecting an Angle" by Randall on 11/02/2011 8:21 AM
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#60 "Re: Trisecting an Angle" by Tornado on 11/05/2011 3:10
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#87 "Re: Trisecting an Angle" by Tornado on 11/11/2011 10:14
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#113 "Re: Trisecting an Angle" by cwarner7_11 on 11/12/2011
11:50 PM (score 1)
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#1
Re: Trisecting an Angle
10/31/2011 9:21 PM
The gotcha (at least one of them) is that the
problem is defined as an arbitrary triangle and
not a right triangle. A right triangle is easily
solvable, but if the triangle is not a right triangle
that link's solution will not work.
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Z man
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#2
Re: Trisecting an Angle
10/31/2011 9:59 PM
The right triangle is only part of the method of
trisecting an arbitrary angle.
You will see there are 3 diagrams for 3
situations.
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#3
In reply to #2
Re: Trisecting an Angle
10/31/2011 10:07 PM
But all are right angle triangles.
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#4
Re: Trisecting an Angle
10/31/2011 10:16 PM
This is total baloney. In each case a smaller angle
is produced first, and then a 3x larger angle is
constructed. This triples the original angle, but
does not trisect it. Joe Keating has completely
wasted his time.
The "I tried it in Photoshop & it appeared to be
perfect" is entirely irrelevant.
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Joe Keating
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#7
In reply to #4
Re: Trisecting an Angle
11/01/2011 5:10 AM
Not quite true: In Diagram-1, angles IHJ and ICJ
are created simultaneously.
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Z man
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#8
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Join Date: Jan 2006
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Re: Trisecting an Angle
11/01/2011 5:38 AM
Welcome to the forum Joe.
About the comment 'Looked perfect in
Photoshop is irrelevant', I suppose doing the trig
will be the final word, but the Photoshop test
essentially proves that it can be done with a
compass and streiaght edge way beyond the
accuracy of pencil marks on paper to show any
error.
I don't have a proper CAD program on my comp,
so I laboriously draw the test with a big 1000 dpi
picture using 3 pixel lines; virtually impossible
to do with any manual writing instrument and
you can zoom way in for microscopic views.
I was bungling around with this problem myself
earlier this year and got pretty good at using
Photoshop for the task.
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#10
In reply to #7
Re: Trisecting an Angle
11/01/2011 7:34 AM
In fact I do not understand why you make simple things
complicated.
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If you want to CONSTRUCT the angle which is 3 x BAC
then you do not need the triangle. As above picture shows
you only have to draw a series of triangles using same
circle and yo obtain not only the 3x but also the 2x and if
you go further other multiples. What you have done in a
complicated way is NOT what is considered as an
unsolvable problem since you did NOT cut (secate) in 3 a
given angle but ONLY build up the 3 fold of a given angle.
O2O1O3 = 2x BAC and O3O2P = 2x O2O1O3 = 4x BAC
but PO2O4=BAC thus O3O2O4=(4-1)xBAC = 3xBAC
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#26
In reply to #10
Re: Trisecting an Angle
11/02/2011 7:37 AM
Surely this is the easy way to create multiples of an angle.
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Anyway Joe is not trying to triple an angle: he starts with
any random angle constructs a 90/60/30 triangle at the
vertex then proceeds to recreate the original angle and an
angle which is 1/3rd of it.
The trouble is he does not explain/or prove his step 12 in
the third diagram:-
12. From the above the angle OMN = the angle HBJ
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#13
In reply to #7
Re: Trisecting an Angle
11/01/2011 4:22 PM
Hello Joe Keating.
Can you tell us more about yourself? As in, who is Joe
Keating? I find a few of you on the interworld information
web... are you one of these?
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#23
In reply to #7
Re: Trisecting an Angle
11/02/2011 12:31 AM
Not so. Angle ICJ was created first. Only later
(by at least two steps) was angle IHC
constructed. Then, by two applications of
(exterior angle = sum of alternate interior angles)
[which was not actually cited], it could be shown
that IHJ = 3 x ICJ.
That was actually backward. A correct
construction would need to start with the larger
angle IHJ and then produce the smaller angle
ICJ.
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Joe Keating
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#24
In reply to #23
Re: Trisecting an Angle
11/02/2011 3:36 AM
"Not so. Angle ICJ was created first. Only later
(by at least two steps) was angle IHC
constructed. Then, by two applications of
(exterior angle = sum of alternate interior angles)
[which was not actually cited], it could be shown
that IHJ = 3 x ICJ.That was actually backward. A
correct construction would need to start with the
larger angle IHJ and then produce the smaller
angle ICJ."
Your Message:
You have got it entirely wrong: You can start
from J if you want.
I have been advised by an honorary maths
professor that it is geometrically correct.
It is so elementary and serves merely as an
introduction to Diagram-2.
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#33
In reply to #24
Re: Trisecting an Angle
11/02/2011 10:01 PM
Wrong.
If you start with J, and then proceed backward to
construct HD, D may not bisect AC, and DC
may not equal DH.
The classical problem requires trisecting any
general angle, not tripling some angle. While
true that ICJ = 1/3 IHJ, that's the wrong order of
events. Your construction may be accurate, but it
does not do what you claim. Shame on your math
prof for ratifying this (it might be interesting to
see the exact wording).
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Joe Keating
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#42
In reply to #33
Re: Trisecting an Angle
11/03/2011 8:30 PM
Quote: "If you start with J, and then proceed
backward to construct HD, D may not bisect AC,
and DC may not equal DH."
It is a given of the construction that JH = HD =
DC for ALL angles IHJ.
Quote: "The classical problem requires trisecting
any general angle, not tripling some angle. While
true that ICJ = 1/3 IHJ, that's the wrong order of
events."
There is no correct order of events. Start at J, I,
H, or C. It does not matter.
Quote: "Your construction may be accurate, but
it does not do what you claim. Shame on your
math prof for ratifying this (it might be
interesting to see the exact wording)."
Are you serious? He told me what I have
already stated here: Diagram-1 simultaneously
constructs and trisects any random angle IHJ on
the base IHC without addressing the trisection of
a 'given' angle. Unlike some, he can distinguish
between random and given angles. He did not
say Diagrams 2 and 3 are correct.
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#43
In reply to #42
Re: Trisecting an Angle
11/03/2011 9:43 PM
I am absolutely as dead serious as it is possible to
be. You have completely blown it. More later.
It might be interesting to know the name and
institution of your math professor, who should
have seen right through all of this.
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Joe Keating
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#44
In reply to #43
Join Date: Nov 2011
Posts: 62 Re: Trisecting an Angle
11/03/2011 10:28 PM
There is NOTHING blown.
There is NOTHING for anyone to see through.
It is but an elementary geometrical random angle
trisecting-tripling model.
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#45
In reply to #44
Re: Trisecting an Angle
11/03/2011 10:59 PM
Wrong again, as usual. Your method first triples
an angle, making the larger angle specified--not
random. What you have done is tripling-
"trisecting," not trisecting-tripling. The order of
events is absolutely critical.
To ask again, what is your professor's name and
institution? Have you shown this thread's
comments thus far to your professor? If so, how
has he modified his opinion, if at all?
When all is said and done, I intend to give him,
as well as you, a failing grade. Just be
forewarned.
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Joe Keating
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#46
In reply to #45
Re: Trisecting an Angle
11/03/2011 11:19 PM
You have already established that you are control
freak.
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#47
In reply to #46
Re: Trisecting an Angle
11/03/2011 11:47 PM
Given it bears no resemblance to how it's
actually done, and does not stand up to scrutiny,
I'd say you are still "approaching impossible".
It's a pity you connived to bend the question to
your solution and didn't explore the accuracy, but
go instead for internet glory. And therein is a life
lesson.
Now you are famous as an internet idiot, with
apparently an idiot, or dupe, for a professor.
Perhaps this is why 'real science' involves 'peer
review'.
Mind you, even so; "control freak" is unlikely to
be viewed as a valid, or substantive, rebuttal.
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Joe Keating
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#49
In reply to #47
Re: Trisecting an Angle
11/04/2011 12:17 AM
It was put here for feedback.
I do not mind the questioning, criticism, or
rejection.
To me it is no more than a curious puzzle.
I have opened one door, and will see if I can
open another.
It is interesting to me that so many have a serious
hang-up about it.
If I progress some more that will be nice; if I
don't I could not care less.
Randall is very helpful.
Joe.
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#50
In reply to #49
Re: Trisecting an Angle
11/04/2011 12:51 AM
"It was put here for feedback"
Really?
Given it was posted on the net signed and dated
24 September and linked by Z man here on 31
October, isn't it a case of 'publish first' then seek
'publicity' - rather than 'feedback'?
"Randall"? is that Z man or the professor? Either
way, neither appear to have sheet-metal
knowledge. I.e. I do find it humorous that
academia and apparently Wiki folk, still haven't
caught up with age old dividers and straight edge
'marking out' techniques.
So no, it's nothing to "have a serious hang-up
about"
I'd be very surprised if I was the first to explain
how to do it.
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#51
In reply to #49
Re: Trisecting an Angle
11/04/2011 12:58 AM
Oh, really; you don't much care about this one
way or the other? Then why the goofy business
about signing and dating the three diagrams?
And now, on to diagram 3.
Angle HBJ is supposedly random. It looks like
about 30°, so that is what I used to follow out
this construction. When this is done, angle HCJ
comes out 10.89° (to 2 decimal places). Angle
OMN turns out to be 30.61°, and angle OCN to
be 10.20°. (There was some slight rounding;
actually OMN does equal 3 x OCN.) However,
the lower group of angles differ significantly
from the upper group.
I recommend not only that you study geometry
better, but also invest in a CAD program and
learn how to use it. This advice also applies to
your professor. I can understand your reticence
in furnishing any information about him; if word
got back around, the repercussions could be
embarrassing.
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Z man
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#52
In reply to #51
Re: Trisecting an Angle
11/04/2011 3:13 AM
You are a very easily riled bunch.
I deserve the blame/credit for bringing this here,
Posts: 292 Joe just happened across the same subject in
another forum. One with a substantially more
mannerly membership.
I admit to being enthusiastic about the possibilty,
and since I'm busy with several projects and
endless chores, the easy way to verify the
method was to show it to you guys, since you
have the expertese and software to do it properly.
The idea of proportionally translating a known
trisected angle occured to me before, but I never
came up with a way to do it. Joe's method seems
to be a start.
So, instead of escalating your attacks, take a
Zanax, gulp some Pepto Bismol, apply some
Preperation H and think 'co-operation' instead of
'confrontation'.
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#53
In reply to #52
Re: Trisecting an Angle
11/04/2011 7:17 AM
Unfortunately,the peer review process can
sometimes be a rough one and engineers know
that all too well.
Everyone has an ego and it doesn't take much to
get bruised. I have learned, long ago, that the
best way to do this is to do all your home work
as best you can before the presentation, and then
be humble as your peers stab at with the sharp
steely knives. ;-)
In the end you get a better "product" out of it and
your peers still respect you.
However, if the presenter gets defensive things
will quickly slide down hill.
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#48
In reply to #46
Re: Trisecting an Angle
11/04/2011 12:09 AM
And just what is accurately geometric about that
irrelevant remark?
In diagram 2, the angle IHJ that you claim to be
60° is only 59.4º (actually a bit less).
If you extend the 0, 15, 30, 45, 60, 75, and 90
lines to the right to intersect arc AHB, the angles
you get are 0°, 15.8°, 30.6°, 45°, 59.4°, 74.2°,
and 90°. The 0, 45, and 90 ones are exact and
correct; the others are not.
Instead of dividing line segment AB into six
equal intervals, you needed to divide arc AHB
into six equal angles. But that entails two
trisections that are not yet established, because
you are still trying to establish them. Vicious
circle.
(Next up will be diagram 3.)
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#54
In reply to #44
Re: Trisecting an Angle
11/04/2011 9:17 AM
Only according to your knowledge and opinion.
Of course as many INVENTORS you do not
accept critics and the fact you made an erroneous
reasoning. You are convinced that you are right
because some prof said it to you either he wanted
to get rid of you or he is since such a long time
out of profession that he forgot his basics too.
As for your remark the angle would have been
the 3 fold if the PERPENDICULAR would have
been the other way. You do not want to look in a
critical way what people say you consider it as
an attack and on the contrary it is a help they
give to you.
I am sure that in your way of thinking you all of
us are so stupid that we cannot understand your
high skills and your possibility to solve a
problem considered as impossible. Be pleased
with your feelings I do NOT comment any more.
As I several times wrote in similar situations :
"There is no blinder than the one who does not
want to see" !
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Joe Keating #56
In reply to #54
Re: Trisecting an Angle
11/04/2011 6:20 PM
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#57
In reply to #56
Re: Trisecting an Angle
11/04/2011 10:49 PM
Arrogant and wrong - a fatal combination
Have another OT
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In reply to #56
Re: Trisecting an Angle
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11/04/2011 11:55 PM
If you anticipated everything, then you must
have anticipated that someone would say that
one of your 60° angles was less than 59.4°. Have
you yet anticipated a cogent reply to this and
other difficulties?
Maybe you should cut your teeth on a simpler
problem: Find two even integers that can be
added to make an odd integer. It may surprise
you that the reason a general angle cannot be
trisected is the same (though more complicated)
as why two evens cannot add to an odd. It has to
do with constructible vs. nonconstructible
numbers.
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#55
In reply to #44
Re: Trisecting an Angle
11/04/2011 10:12 AM
Hi,
I already mentioned that you CONSTRUCTED and did
NOT CUT the angle. Your approach is too complicated for
what you did (not for what you had the intention or
claimed to do) and today's CAD have routines to CUT an
angle as in the following picture.
118
If you had the intention to make a geometry demonstration
you are wrong, if you wanted to give a help there is no
need. As you see the error depends of the number of digits
behind the point. As for the way you reacted at Tornado's
comment it is something I consider as ABSOLUTELY not
compatible with CR4's mentality.
Revise you basic geometry especially the properties of
triangles.
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Rose
#66
In reply to #33
Re: Trisecting an Angle
11/05/2011 3:32 PM
Tornado #67
In reply to #66
Re: Trisecting an Angle
11/05/2011 4:09 PM
English Rose #78
In reply to #67
Re: Trisecting an Angle
11/08/2011 4:49 AM
Tornado #79
In reply to #78
Re: Trisecting an Angle
11/09/2011 12:58 AM
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#28
In reply to #7
Re: Trisecting an Angle
11/02/2011 11:55 AM
I think I found a flaw in your construction, of
course I am not an honorary math prof so that I
can be wrong. But if we look at you last sketch
we can say that you are right if via a rotation
with center C line CJ covers CE and line CH
covers CO. This can happen if the angles BCK =
ACM. For simplification I wrote the
trigonometric function tangens for both angles if
values are equal angles are equal as well:
tan (BCK)= BK/(AC*cos 30°) since your
triangle has AB= AD=DC then ABD is
equilateral and Angle BAC is 60° thus BCA=
30°
tan(MCA)= AM/MC Since AM not equal to KB
and ACM< ACB the 2 values cannot be equal so
that after a rotation the angles will not cover and
the angle NMO is not equal to JBH.
There is an other error. Since AN is not equal to
AD the angle NMO is NOT 3x DCM.
I am sorry but the whole construction is -
according to my limited knowledge - not what
you claim to be.
Of course if you find an error in my
demonstration I shall be glad to have your
correction or anybody else's.
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Joe Keating
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#37
In reply to #28
Re: Trisecting an Angle
11/03/2011 6:37 AM
Quote...."There is an other error. Since AN is not
equal to AD the angle NMO is NOT 3x DCM."
AN does not have to be equal to AD for NMO to
be 3DCM.
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#38
In reply to #37
Re: Trisecting an Angle
11/03/2011 8:50 AM
I think you should review your triangle geometry
basics!
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Joe Keating
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#41
In reply to #38
Re: Trisecting an Angle
11/03/2011 5:30 PM
Quote "I think you should review your triangle
geometry basics!"
Are you sure about this?
The ONE-ONLY angle trisection where AN =
AD is 0 degrees.
Further, using this model, when trisecting 90
degrees N will coincide with A.
For all angles between O and 90 degrees N is
variously located between A and E.
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#5
Re: Trisecting an Angle
10/31/2011 10:27 PM
One problem I see is that the angles are not truly
arbitrary; they are arbitrary within an already
generated diagram. Even if it works (see my next
comment) I'm not sure that this system would
work given an arbitrary angle by itself drawn on
a sheet of paper. Conceivably, one could
[stealthily] generate angle A in some non-
obvious manner and then construct angle B that
is 3 times the size of angle A; then reveal angle
A and show, since A is 1/3rd the size of B, that
you have trisected B.
I also am not sure this system really works. In all
cases the 'verification' relies on the statement that
3 segments are equal in length. I may of course
be missing something, but I don't see how that is
a proof that one angle is precisely 1/3rd the size
of another angle.
It's a problem that is more complicated than it
looks, so when I get some time I'll study it in
more depth and work through the proof carefully.
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cwarner7_11
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Re: Trisecting an Angle
11/01/2011 1:09 AM
I believe the key here is that the specification
that Line AC=2AB says that angle ACB is
arcsine of 0.5, which is not very general. From
the images provided, it appears that Point D is
the center point of AC, although this is not
specified.
The question is, can I construct the 30 degree
right triangle and arc in relation to H such that I
can complete this analysis? Unfortunately, this
evening, Dionysus rules, so my head is not clear
enough to follow through...
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velisj
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#9
In reply to #6
Re: Trisecting an Angle
11/01/2011 7:31 AM
You are correct when you note that Step 2 states
that Line (should be "Segment") AC = 2AB. This
limitation now generates a 30°/60°/90° triangle
which is indeed a very special case with Angle A
= 30° and Angle B = 60°. If you follow the
constructions, you will find many 30° and 60°
angles throughout.
There are any number of angles that can be
trisected, but it has been show in the late 18th
century that the general proof is impossible,
much like squaring the circle, where the number
pi must be constructed using only a "compass"
and straight edge.
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Randall
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Re: Trisecting an Angle
11/01/2011 8:11 AM
To Joseph rather than the original poster.
12. From the above the angle OMN = the angle HBJ
No it doesn't. How did you prove that?
I started with the random 15.95o angle then constructed the
rest of the picture as per your instructions. I didn't measure
any angle 'til the end.
Welcome to CR4.
I suggest that you get yourself a free CAD package.
Or Geogebra http://www.geogebra.org/cms/
It makes trying out these sort of things very easy.
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#12
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Re: Trisecting an Angle
11/01/2011 2:34 PM
Very nice simple program thanks for giving the
link.
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#14
In reply to #11
Re: Trisecting an Angle
11/01/2011 5:37 PM
I tried to get that program, Randall, but the
webpage sent my browzer into a tizzy. Probably
some malware, so anybody who downloads it
should do a complete security sweep.
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#18
In reply to #14
Re: Trisecting an Angle
11/01/2011 8:08 PM
Z-man- I could not get the Linux version
GeoGebra from the web site, but was able to get
the package here. Older versions are also
available here. These links include the packages
for Windows, MAC, etc. as well as the Linux
version. The Windows version comes as an *exe
installer, and the Linux version comes in a
*.tar.gz that only requires unpacking- no make or
other build options. Apparently, no external
dependencies (except, of course, Java).
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Joe Keating
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#15
Join Date: Nov 2011
Posts: 62
In reply to #11
Re: Trisecting an Angle
11/01/2011 5:58 PM
One query: Are we here dealing with a question
of draftsmanship or geometry?
Visually on my computer screen, and on a
printout, the 'trisecting' line to the right hand
corner angle does not appear to be going directly
to that angle, but slightly beyond it to the right,
which if true will increase the size of the
'trisected' angle at the top left.
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#16
In reply to #15
Re: Trisecting an Angle
11/01/2011 7:06 PM
I would like to see how your method works on triangles like
these:
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Re: Trisecting an Angle
11/01/2011 9:39 PM
GA, because it is the right question.
The classical problem is about pure mathematics/geometry,
not about draftsmanship or CAD or approximations, no
matter to how many decimal places, nor how good they
"look." The classical "tools" are an unmarked straightedge
and a compass. The compass is also supposed to be
collapsible, meaning you can't pick it up and simply transfer
the radius elsewhere. Many attempts at angle trisection
founder by not sticking to these rules (such as using a marked
straightedge or a device known as a "tomahawk").
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#20
In reply to #19
Re: Trisecting an Angle
11/01/2011 10:20 PM
Thanks for the link, cwarner.
I wonder how the drafting software works, concerning
accuracy. At some level it has to be deciding where to put
a pixel, how many decimal places to calculate at each step,
and many other little 'decisions' for the actual geometry
and how to present it.
This is helpful:
From PlanetMath:
"Compass and straightedge constructions are of historical
significance. The ancient Greeks are the most well-known
civilization for investigating these constructions on an
elementary level. It should be pointed out that the
compasses that they used were collapsible. That is, you
could open the compass and draw an arc, but immediately
after you removed a point of the compass from the plane
where you drew the arc, the compass would close
completely. It turns out that whether a collapsible compass
or a modern-day compass is used to perform these
constructions makes no difference. This statement is
justified by the fact that one can use a collapsible compass
to construct a circle with a given radius at any point as
shown by this entry."
http://planetmath.org/encyclopedia/CompassAndRulerCon
struction.html
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#21
In reply to #20
Re: Trisecting an Angle
11/01/2011 10:36 PM
Further to this observation (GA), compass alone can
construct all the relevant points of intersection of
compass and straightedge together. Also, if a circular arc
is given, a straightedge alone will suffice.
AutoCAD is accurate to something like 16 decimal
places (I forget exactly how many), but the pixel screen
is not as accurate.
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#22
In reply to #20
Re: Trisecting an Angle
11/01/2011 10:40 PM
My experience with CAD is that, unless one is
using a plotter, one does not get a true circle (and
there is a good chance that if one looked at a
plotted drawing at high enough magnification, it
would not be a true circle either). Typically, a
circle is represented by a number of points
equidistant from the center, connected by straight
line segments- a polygon of very high side count.
Of course, the larger the number of sides, the
closer one gets to a true circle. A circle drawn
with a compass would be a continuous line of
constant curvature, containing EVERY point at a
given distance from the center- essentially, an
infinite number, which would be rather difficult
to store (or even generate) on a digital
computer...
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#31
In reply to #19
Re: Trisecting an Angle
11/02/2011 3:55 PM
You've reminded of me of this coming up before
on CR4.
As I recall I demonstrated, principally to
Canadian Chris, how it's done (within the rules).
Unfortunately I'm away from my CAD system,
so can't access the file, and don't have time to do
it all again. But I'm fairly sure I posted it back
then.
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#359
In reply to #31
Re: Trisecting an Angle
02/10/2012 9:12 PM
You annoying hound ! It was on forum before,
and I, that is I, I, I, I, offered up one of the
mathematical proofs that it cannot be done.
No way am I going to fetch (go on, boy fetch ,
fetch ) at this time of night, but it boils down
to the impossibilility of constructing a cube root.
The proof was not of my own devising, but very
elegent (purists would go loopy, but even Phyz
gave it a nudging 'OK').
If anybody wants me to track where this was, I
will. Conditional upon me recieving 5 PM's titled
'Bad dawg'.
Look, it can't be done. This makes it one of lifes
great pleasures. If you succeed, don't tell me, tell
the unwife/hubby .
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#361
In reply to #359
Re: Trisecting an Angle
02/13/2012 2:02 AM
Nice article, Kris.
I like your line trisector, your attempted angle
trisection thru projection and especially your
perspective of trisection as an addiction! It's
totally true!
I started reading Dudley's book yesterday and
can't stop chuckling about how he would be
disgusted with me. I am like the perfect trisector
crank! If he were to write a fictional story about
this, he would not go so far with a character to
avoid cartoonishness!
Anyway, I think I'm done. Or cured.
http://www.zolkorp.com/TRiSeK.html
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#362
In reply to #361
Re: Trisecting an Angle
02/13/2012 4:58 AM
Brilliant link - until you trisect, you've never
lived !
I'm still trying to understand the Squaw on a
Hippopotamus.
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#363
In reply to #359
Re: Trisecting an Angle
04/04/2012 6:09 AM
Alright, I'll fetch
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#364
In reply to #363
Re: Trisecting an Angle
04/04/2012 9:45 AM
Wu? Duz it work????!!!!
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#365
In reply to #364
Re: Trisecting an Angle
04/04/2012 9:51 AM
Oh, yeah. Might be able to dig up a bit of a
clearer explanation, but it all boils down to 'can
you construct an arbitrary cube root'. Nope, you
can't. This has never stopped me trying, though
.
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#25
In reply to #15
Re: Trisecting an Angle
11/02/2011 7:12 AM
How far do you want me to zoom in:-
A CAD program "snaps" ends of lines to the points you
choose: grid points, vertex, mid point etc.
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#17
In reply to #11
Re: Trisecting an Angle
11/01/2011 7:58 PM
Randall- excellent link!
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#27
Re: Trisecting an Angle
11/02/2011 8:21 AM
By the way you all might find this challenge
question interesting:-
http://cr4.globalspec.com/blogentry/5532/Trisection-
CR4-Challenge-04-15-08
When you submit a challenge question, you have to
submit a solution, and the solution given i.e 18
operations appears at the top of the thread. However
make sure you look at fyz's post #89 where he beat
his original challenge with a solution of 10
operations: quite incredible!!
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#29
Re: Trisecting an Angle
11/02/2011 3:14 PM
I still haven't got my head around Diagram 3 and your
point 12 yet, but looking at diagram 2: you are totally
doomed to failure trying to map a linear scale onto an
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angular scale (or vice versa) by a construction.
AB=AD=DH=HJ and let this distance = 90 units length
let angle DCH=alpha
now drop a perpendicular from D to CH and it becomes
obvious that
CH=2*90*cos(alpha)
Sin(30-alpha) = h/CH (where h is the height of H above
KC)
so h = 180*Cos(alpha)*Sin(30-alpha)
pop this formula into a spread sheet:-
And you can see that although 3*alpha is very close to
90-h it's not quite exact.
EDIT Sorry Z man and Joe: I keep replying to the
original post when I should be directing my observations
at Joe.
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Joe Keating
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#32
In reply to #29
Re: Trisecting an Angle
11/02/2011 6:25 PM
Wherever H is the geometry of the model is
standard for any acute angle IHJ trisection, and I
will not presume to you why.
The linear scale could sit on AB with a
corresponding 'point' on AB at the level of H.
Using this procedure: If the angle IHJ is actually
60 degrees using the linear scale, then the angle
ICJ is 20 degrees.
Whatever the angle IHJ is it is trisected.
Is the angle IHJ 60 degrees?
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#36
In reply to #32
Re: Trisecting an Angle
11/03/2011 6:36 AM
If IHJ is 60 then H is not 1/3 of the "height" of
AB.
If H is 1/3 the height of AB then IHJ is not 60°.
Just do the calculation using AB as 90 units long
h=180*cos(20)*sin(10)
you get 29.37166401 not 30
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#40
In reply to #36
Re: Trisecting an Angle
11/03/2011 5:11 PM
Quote "If IHJ is 60 then H is not 1/3 of the
"height" of AB.
If H is 1/3 the height of AB then IHJ is not 60°.
Just do the calculation using AB as 90 units long
h=180*cos(20)*sin(10)
you get 29.37166401 not 30."
Thanks.
From this I read that IHJ in Diagram-2 is 58.74+
degrees.
With respect to Diagram-3, if BK as the correct
height for the angle HBJ, as distinct from a 60-30
type presumption, is geometrically as distinct
from draftsmanship located as AL, does this
render the angle OMN equal to HBJ?
Joe.
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#39
In reply to #32
Re: Trisecting an Angle
11/03/2011 4:32 PM
No, as constructed, angle IHJ is not 60°; it is
about 59.4°.
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#34
In reply to #29
Re: Trisecting an Angle
11/03/2011 2:13 AM
Quote: "....I still haven't got my head around
Diagram 3 and your point 12 yet...."
In Diagram-C the process is essentially to 'lift'
the 'given' angle from the bottom of AB to its
top.
If, for example, to be cheeky in the context of
impossibilty, the given angle HBJ is 60 degrees,
you will notice that the line JC crosses AB at
point K, which for 60 degrees lies 2/3rds up AB.
To locate the given angle at the top of AB the
point L is positioned 2/3rds of the way down AB
and the point B becomes the point M on the arc
at the same level as the point L, for the angle 60
degrees to be constructed and automatically
trisected.
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#35
In reply to #34
Re: Trisecting an Angle
11/03/2011 4:54 AM
I see no reaction from your side for the comment
I lately made. Do you consider that it is wrong?
Then please give me the errors I made. Or do
you think it is not interesting to see your errors?
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#30
Re: Trisecting an Angle
11/02/2011 3:39 PM
Thanks again for the drafting software; doing
this in Photoshop is very tedious. Your example
of how far in you can zoom convinces me theres
no problem with accuracy.
I did my last Photoshop test last night. Diagram 3
does fail to duplicate the angle in the new
position. Its off by a 3 or 4 degrees. But the
bogus angle OMN does appear to be trisected by
OCE.
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Z man
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#59
Re: Trisecting an Angle
11/05/2011 2:08 AM
Just so I'm clear, are the angles produced by the
1st diagram 3 to 1 exactly or just close?
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#60
In reply to #59
Re: Trisecting an Angle
11/05/2011 3:10 AM
In Diagram 1, / IHJ = 3 / ICJ exactly.
The problem is that / ICJ happens first, so that /
IHJ is not a given initial angle.
In other words, this construction triples / ICJ,
rather than trisecting / IHJ.
By the way, in the sad history of angle trisection
crackpots, this is one of the common errors:
tripling an angle and then just finding the
original angle again.
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#61
Re: Trisecting an Angle
11/05/2011 6:14 AM
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Thats the answer I wanted to hear!
While you geniuses were thinking up snide
remarks, I figured out a way to make Joe's
creation work properly.
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#62
In reply to #61
Re: Trisecting an Angle
11/05/2011 6:41 AM
You had your answer already in comment #10. I
am sorry that my demonstration was not
understood by you. Read it again and if
something unclear it will be a pleasure togive
you further explanations.
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#64
In reply to #62
Re: Trisecting an Angle
11/05/2011 3:23 PM
I understood it before, nick. But you are
multiplying the angle, which iz easy.
So you're 100% sure we haven't done it, Randall?
Not only 99.99999%?
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#65
In reply to #64
Re: Trisecting an Angle
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11/05/2011 3:31 PM
IT IS EXACTLY WHAT HE DID in a more
complex way but THERE is NO difference since
in my sketch I pictured ONLY what was of
importance for his method!
Please compare what he presents in the st
drawing and what I sketched and you will see it.
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#68
In reply to #64
Re: Trisecting an Angle
11/05/2011 4:14 PM
Even if Randall is only 99.99999% sure, I am
fully 100% sure. So would you be, if you were to
read and understand the Wikipedia article on it,
to say nothing of copious other literature.
This is why ignoramuses are so offensive. They
expect other people to waste time finding the
inevitable mistakes in their assertions.
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#63
In reply to #61
Re: Trisecting an Angle
11/05/2011 2:26 PM
You have done nothing of the kind. The task is
mathematically impossible.
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#69
Re: Trisecting an Angle
11/05/2011 8:24 PM
OOPS! Sorry Randall! I meant Tornado. The
layout here isn't very good; can't see the whole
topic while youre typing a response, plus I have a
narrow screen.
I'll let Joe show you why your certainty should
not have been 100%.
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#70
In reply to #69
Re: Trisecting an Angle
11/05/2011 8:42 PM
That'll be the day. Joe hasn't been doing too well
thus far, but maybe his anonymous math
professor might show up to complete this folie à
trois.
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Joe Keating
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#71
Re: Trisecting an Angle
11/06/2011 3:24 PM
A dummies edited random trisection model for friends
and freaks.
http://users.tpg.com.au/musodata/trisection/trisecting_any
_angle.htm
Joe.
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#72
In reply to #71
Re: Trisecting an Angle
11/06/2011 5:02 PM
/ HIJ is NOT randomly constructed, and it is NOT ANY
(i.e, general) angle.
Instead, it is specially constructed (and predetermined)
from the already established / DCI.
This does NOT solve the classical problem of starting
with the larger angle and constructing an angle of 1/3 the
size. All you have done is triple an angle.
Calling knowledgeable people "freaks" is spectacularly
offensive on your part.
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#73
In reply to #72
Re: Trisecting an Angle
11/07/2011 5:26 AM
It has no reason to continue this discussion with
somebody who does not either want or can
understand explanations.
If you look at my first sketch you find his
method explained.
What he all time does is to hang on his triangle
which is not at all required for the construction.
This is the best proof that he does not understand
what he does.
His last demonstration is the same approach but
backwards and the angle he wants to cut is not
given it is the result in a way.
I consider that defending his own ideas is good
but not when flaws are clear and one has to
accept the defeat. It is not any more a fight it is
to be stubborn (I try to stay polite what he is not).
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#80
In reply to #72
Re: Trisecting an Angle
11/11/2011 2:10 AM
" / HIJ is NOT randomly constructed,.. "
Incorrect: The line segment IC is not needed to
construct any random angle HIJ.
Any random angle HIJ is constructable in the
diagram, by an unmarked straight edge and
compass, without being trisected.
The frequently repeated claim that the diagram is
the tripling of an angle is misleading.
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#81
In reply to #80
Re: Trisecting an Angle
11/11/2011 4:32 AM
I'm sorry, but that is completely dead false.
I still would like to know your professor's name
and location. You still have not answered how an
angle you claimed to be 60° was only a bit less
thn 59.4°, and how the upper group of angles in
diagram 3 differed from the lower group.
Major flunk on all counts.
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#82
In reply to #81
Re: Trisecting an Angle
11/11/2011 5:04 AM
Selecting a point on the arc effectively creates
both angles simultaniously. Joe is right.
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#83
In reply to #82
Re: Trisecting an Angle
11/11/2011 5:25 AM
No, it does not. And, even if it did, that is not
enough. The larger angle must be given first.
It is abundantly clear that neither you nor Joe,
nor maybe even his professor, is familiar with
the literature on this problem. Instead, your
collective misunderstandings have been rampant,
and riddled with inaccuracies. (As you yourself
have noted in one of Joe's constructions.)
By the way, it should be mentioned that CAD
programs cannot prove the absolute accuracy of
geometric constructions. They may be accurate
to say 10 decimal places, but off in the 11th and
subsequent places. However, they are useful in
proving that constructions are inaccurate,
providing that the inaccuracy is at or before the
10th decimal place.
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#85
In reply to #83
Re: Trisecting an Angle
11/11/2011 5:49 PM
"...The larger angle must be given first..."
The random angle model will easily randomly
construct any angle HIC on the arc, first, without
in any way referring to the point C.
The random angle so constructed will not be
trisected until the point I is connected to the
point C, whereupon the random angle HIC will
automatically be perfectly trisected.
Should one wish to leave a random angle HIC
constructed but not trisected, so be it.
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#86
In reply to #83
Re: Trisecting an Angle
11/11/2011 7:17 PM
"..The larger angle must be given first.."
Correcting angle naming error in post 85: 'angle
HIC' should read 'angle HIJ'.
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#87
In reply to #86
Re: Trisecting an Angle
11/11/2011 10:14 PM
That is precisely your problem: /HIJ is
constructed; it is not random. In particular, HI
was picked to lie on line IC, with /ICJ already
preexisting. The issue is not about whether /HIJ
= 3 /ICJ (which is trivial); it is that /HIJ is not
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given first.
To repeat, what you have really done is to triple
/ICJ, not trisect /HIJ. These are not the same
thing; rather, they are opposites.
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#89
In reply to #87
Re: Trisecting an Angle
11/11/2011 11:21 PM
You have it totally wrong.
HIJ can be constructed from H and be there
without any connection to C.
You do not know how to do it, do you?
And you said you are the knowledgeable one!!!
I now see your responses are merely deliberate
reversals of what others post so that you can
control the forum by confusing the direction of
peoples statements and thinking.
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#90
In reply to #89
Re: Trisecting an Angle
11/12/2011 12:01 AM
Once again, you understand nothing. /HIJ must
not be constructed; it must be given.
Not to mince words here, you are the most
horrifyingly incompetent "student" I have ever
encountered.
And I still want to know who your "professor" is,
because he too needs some sharp words. No
wonder you are hiding this information. I would
suggest thet your professor read this entire thread
and then offer some better evalution/advice than
so far.
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#91
In reply to #90
Re: Trisecting an Angle
11/12/2011 12:14 AM
(/HIJ must not be constructed; it must be given)
No, you again have it wrong: The random angle
construction model is not for 'given' angles.
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#93
In reply to #91
Re: Trisecting an Angle
11/12/2011 12:23 AM
No, I am not wrong. Your /HIJ is NOT
RANDOM. The classical problem requires the
angle to be trisected to be given initially. When
you say your "model is not for 'given' angles,"
you fail utterly--because that is the key
requirement.
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#84
In reply to #81
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Re: Trisecting an Angle
11/11/2011 5:33 PM
"...I'm sorry, but that is completely dead false..."
I am staggered that you say that. Have another look at the
elementary geometry of the random angle model.
"...I still would like to know your professor's name and
location..."
My math academic friend commented ONLY on the
random angle model. One does not need to be a scholar to
know it is geometrically correct.
"...You still have not answered how an angle you claimed
to be 60° was only a bit less than 59.4°, and how the
upper group of angles in diagram 3 differed from the
lower group..."
Diagrams 2 and 3 are no longer under consideration: I
accepted the analysis by Randall explaining why they are
not quite right.
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#74
In reply to #71
Re: Trisecting an Angle
11/07/2011 6:12 AM
The above sketch shows that you do NOT need any
TRIANGLE to make the CONSTRUCTION you
developed. It is enough to proceed as follows:- draw a line
CE (I kept your notations)- set a point D and draw a circle
with radius R = DC- choose a point H on the line CE and
draw a circle with same radius R- circles will cut in point
I - draw the lines HI, CI and DI - the angle HI and CI will
ALWAYS be 3x angle ECI.This is TRUE and for ALL
positions of point H as long as HD<2RBUT how do you
define H when you ONLY know the angle between HI
and CI ? HERE is THE problem. If you solved it then you
can claim you solved the angle trisection problem!In fact
you DISCOVERED a BASIC property NOTHING new, I
am sincerely sorry for you but you did all you could to
deserve it.
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#75
Re: Trisecting an Angle
11/08/2011 1:28 AM
There is a book about this by Underwood Dudley, The
Trisectors. (Also one or more older Martin Gardner
articles from the 1960s/1970s, and lots of other
literature.) I will be reading Dudley's book when it
arrives. It would surely be good for others to read.
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#76
Re: Trisecting an Angle
11/08/2011 1:48 AM
On another forum, skepticforum.com, our beloved
Z_man has the identity JO753 (same avatar as here).
This is his contribution on a similar thread over there (in
his inimitable txtspik style ):
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CR4 Moderator - Resized image to fit the screen.
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#77
Re: Trisecting an Angle
11/08/2011 4:07 AM
Ye. That one's another that almost works, damit!
Too close to tell on paper, but
I tried it with Geogebra (thanks again, Randall &
cwarner) and came up about 1/4 degree short. So,
I was gloating too soon. Tornado scores again.
Reality has not been altered yet.
(could you ask admin to shrink the pik down? It
stretches the text in the entire topic way off my
screen.)
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#88
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Re: Trisecting an Angle
11/11/2011 10:37 PM
You are actually quite close to the right method
there.
Hint: Try again with the mid point of the
constructed line coincident with the circle.
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#92
Re: Trisecting an Angle
11/12/2011 12:21 AM
NOW I think I got it!
I've been trying to test the new method with
Geogebra, but it's not as user freindly as I first
thought. Probably a learning curve thing, but it
also doesn't seem to be working properly on my
computer. My exasperation level was rising
toward Hulk green level, so I gave up for now.
I have the Photoshop version at
zolkorp.com/aNGL_TRiSeKsN_WeB.gif if you
want to try it out. Its rotated so you can print it
without any rigamarole.
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#94
In reply to #92
Re: Trisecting an Angle
11/12/2011 12:32 AM
No points are labeled in this alleged construction,
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which is poorly described and guaranteed not to
work, anyway.
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#11 "Re: Trisecting an Angle" by Randall on 11/01/2011 8:11 AM
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Quadratrix: Trisect Any Angle
Arielle Alford
While creating and representing the quadrix proved to be an interesting task,
the true utility of this curve lies in its ability to help trisect angles (something
not possible using simply the compass and straightedge). Let's take a look at
how to accomplish this.
By trisecting the radius of our circle, we were able to use the quadratrix to
trisect the associated 90 degree angle as shown below
Utilizing the quadratrix curve to trisect any angle is accomplished in a very
similar manner. The trick is to align one side of the angle to be bisected with
the base of the quadratrix (segment AO in our diagram). Here's an example.
Once the angle is lined up with the quadratrix, one must locate the intersection
of the curve and the second leg of the angle. By constructing a perpendicular
throught that point, one can easily mark a point , M, at that height on the
radius CO. The MO becomes the segment that we must trisect. In doing so,
we find the points X and Y along CO.
By constructing perpendicular lines to CO through X and Y, we can locate the
points on the quadratrix curve through which the angle "trisectors" must pass.
One must simply connect those intersection points to the vertex of the angle to
complete the trisection. Check it out below.
What Angles Can Now be Constructed?
The ability to bisect angles, combined with the ability to perform angle
trisection means that any angle can be divided in half or in third. By
combining these procedures, an angle can be divided may ways, including
into fourths, sixths, and twelfths.
One question that remains to be answered is what integer valued angles can
now be constructed with our bisection and trisection tools. To answer this
question, one must consider iterations of bisection and trisection.
In particular, any possible integer-valued angle would be created by
conducting a a series of bisection and trisections on a 360 degree angle. The
resulting degree measure could be represented as follows:
The following table offers a glimpse of different degree measure that can be
created through bisection and trisection:
m
\n 0 1 2 3 4 5
0 360 180 90 45 22.5 11.25
1 120 60 30 15 7.5 3.75
2 40 20 10 5 2.5 1.25
3 13.3333333
333333
6.666666666
66667
3.33333333
333333
1.666666666
66667
0.8333333333
33333
0.4166666666
66667
4 4.44444444
444444
2.222222222
22222
1.11111111
111111
0.555555555
555556
0.2777777777
77778
0.1388888888
88889
5 1.48148148
148148
0.740740740
740741
0.37037037
037037
0.185185185
185185
0.0925925925
925926
0.0462962962
962963
We see that angles with integer measure occur with 2 or fewer trisections and
3 or fewer bisections. This table shows that angles with the following integer
measures are possible to construct:
Now a question - is it possible to add or subtract any of these 12 angle
measures to create other angles? Notice that the smallest integer angle listed
has a measure of 5 degrees. Certainly it is possible to add multiple 5 degree
angles together to form new angles. Consequently, we can form angles having
any measure equal to a multiple of 5. These angles are shown in the chart
below.
Thus far, we've only found about 1/5 of the total number of angles with
integer measure can be created through bisection and trisection. Are there
more? Perhaps we can add or subtract some of these angles (shown in blue) to
find more angles . . .
Notice that every possible angle is a multiple of 5. What would happen if we
added or subtracted combinations of these angles. In general we could
imagine adding 5g to 5f, where 5g and 5f represent one of the possible angles.
Note f and g must be positive integer values less than 72 (that is, 360/5).
Through adding or subtracting we find the following occurs:
And so addition or subtraction of possible angles always yields a multiple of
5. Since we have already included all multiples of 5 that are less than or equal
to 360, no new angles are formed.
So it seems that with bisection and trisection we are equipt to create about 1/5
of the total number of angles having integer measure.
Back to Quadratrix
Home
Creating the Quadratrix
Arielle Alford
The Problem: Trisect an Angle
While the statement of this problem is absurdly simple, finding a solution has
proven challenging (impossible?). Any high school geometry student is
familiar with the procedure to bisect an angle. However, dividing an angle
into three congruent pieces is another matter . . .
One solution is proposed through the use of the Quadratrix Curve [insert
history of the quadratrix].
What is a Quadratrix?
The Quadratrix is built within a circle (or quarter of a circle) much like the
one shown below.
The idea is to move a ray through O from point I to point A and
simultaneously move a line perpendicular to IO from point I to point O. These
two lines will each move at a constant rate complete their travel at precisely
the same moment (i.e. the ray reaches point A when the perpendicular line
reaches point O).
Getting Started
Accomplishing this task relies on determining how fast the ray and line must
move in relation to one another.
Selecting a rate of movement for the line is a relatively simple task - we can
simply choose any value of linear motion (e.g. 3cm/s).
However, the rate of movement of the ray can be considered in either of the
following two ways: (1) how many radians are swept out per unit time (e.g. 2
radians/s) or (2) what the linear distance that a point on the circumference of
the circle traveled in a unit of time (e.g. 1.5cm/s)
In using Geometer's Sketchpad softward to create our curve, it proves easiest
to consider the motion of the ray as a linear distance per unit time.
It turns out that what we want is for point K (where the line intersects the
radius of the circle) to move a distance equal to the radius in the same amount
of time that it takes for point M to move 1/4(2 r).
So, what is the ratio of rate for the movement of these two points? Check it
out below.
Therefore for a given rate 'a' of the movement of the line, the rate of the
movement of the point M must be
In this Geometer's Sketchpad file, the movement of the line and the movement
of M have been set to have a ratio of .
So, what does the Quadratrix curve have to do with this?
The curve is created by the intersection point of the ray and the line (point C
on the figure below).
A trace of the intersection point yields the following curve:
And this curve is lovingly referred to as the quadratrix!
Using the curve to trisect an angle
Essentially, this curve serves to correlate linear distance to angular distance.
For example, the bisector of angle AOI can be found by constructing the
midpoint of segment OH. The point where the line perpendicular to HO
passing through this midpoint intersects with the curve lies on the angle
bisector. Why? By construction, we know that when the line (see previous
diagrams) has travelled halfway down the radius, the ray has moved halfway
around the arc (and consequently swept out half of the the total measure of
angle AOH).
Trisecting the angle is accomplished by trisecting the line segment OH! Check
it out below.
By construction, the ray swept out the area of the sector in the time it took for
the point to move the length of the radius. Correspondingly, when the point
had moved 1/3 of the length of segement OH, the ray must have swept out 1/3
of the sector (or 1/3 of the 90 degree angle). While this may not be impressive
given the fact that one can easily construct a 60 degree angle which can be
bisected to form a 30 degree angle, an explanation of how this curve can be
used to trisect any angle can be found here.
Also, you can click here to take a look at an algebraic representation of the
quadratrix.
An application of the quadratrix is the problem of Squaring a Circle. Find out
more about that here.
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The Quadratrix: An Algebraic
Representation
Arielle Alford
Having created a geometric representation of the quadratrix, a next challenge
might be to create an algebraic representation of this curve. To accomplish
this task we turn to a little trigonometry.
Getting started
We'll begin by examining a particular instance in the sketching of the
quadratrix curve. Check it out below.
We have (-x,y), a point on our curve and theta, the complement of the angle of
rotation. To determine an algebraic representation, we will assume that the
radius of our circle is 1.
Relating x, y and theta is a relatively simple task which can be accomplished
by incorporating the tangent ratio as follows:
However, our algebraic represnetation should only be a function of x and y.
Consequently, we must find a way to represent theta in terms of x and y. To
accomplish this goal, we turn to our knowledge of how the quadratrix is
constructed.
A key feature of this curve is that the proportion of the right angle that has
been swept out over a period of time equals the proportion of the segment
[0,1] that has been covered. Consequently, the following proportion is true:
And so we discover that
Using this information we can rewrite our equatio for the quadratrix as
Now that we have an equation candidate, let's take a look at a graph of this
equation to see if it resembles the quadratrix curve we've seen before. Check it
out below.
Hmmm . . . is this the graph we were expecting? There certainly seems to be a
lot more happening in this graph than the rather simple curve we saw in our
geometric representation. So, what's going on?
Domain and Range
In developing an algebraic representation, we've left out a very important
consideration of the range for our variables. Looking back at our original
sketches, we see that the minimum value for y is zero. Additionally, the
maximum value for y is the radius of the circle, in this case 1. Consequently, x
ranges from -1 to 0. Thus we should actually only be examining the graph
above on the following range: -1<x<0 and 0<y<1. Let's take another look at a
graph of our function.
By zooming in on our graph we see a curve that looks very familiar - its the
quadratrix!
Not surprisingly, the y-intercept of this curve is (0,1). Perhaps more
interesting is the x-intercept of the quadratrix. What exactly is the point where
this curve crosses the x-axis? Click here to find out more.
Back to Quadratrix
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