Holy Halved Heaquarters Riddler

Anonymous Philosopher

June 2016

Laser Larry threatens to imminently zap Riddler Headquarters (which is ofregular pentagonal shape with no courtyard or other funny business) with ahigh-powered, vertical planar ray that will slice the building exactly in half byarea, as seen from above. The building is quickly evacuated, but not beforein-house mathematicians move the most sensitive riddling equipment out of theplaces in the building with an extra high risk of getting zapped. Where arethose places, and how much riskier are they than the safest spots?

(It is sufficient to describe those places qualitatively. For extra credit: getquantitative. Seen from above, how many high-risk points are there (where evena point-sized object would be at higher-than-necessary risk)? If infinitely many,what is their total area? What if the shape is an n-sided polygon?)

1 Solution

(Thanks to Daniel Tello for valuable (and productive) blunder-hunting!)Keep away from the center, because area bisectors intersect near that point,

raising the odds that central regions will be zapped. The perimeter is safest(and if Larry randomizes uniformly over perimeter points rather than angle,the middles of the sides are the safest of the safe).

In an even-sided regular polygon, there is not much more to say than thatthe center (point) itself, where all area bisectors intersect, is the most dangerousplace and danger decreases with distance from it for all but point-sized objects(which are at added risk only at the center itself). But for an odd-sided poly-gon, while it remains true that the most dangerous places (even for point-sizedobjects) are where bisectors intersect, it is not true that all bisectors intersectat a single point.

Among the area bisectors of a regular pentagon are the five that join ver-tices to midpoints; call these “medians.” Choose some vertex and neighboringmidpoint, and consider the area bisectors that pass between them. Each passesalso through a non-neighboring side, between another vertex and midpoint, andtogether they comprise a fifth of all bisectors. The area swept out by these linesegments clearly includes the two triangles formed by the intersections of thetwo medians and two sides, and every point not on a median is on a bisectorrunning through its home triangle in this way—its home bisector, we’ll say (and

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Figure 1: An area bisector.

we’ll let a median be the home bisector of the points on it). But the area sweptout includes a bit of terrain on the outside of those triangles.

Figure 2: The curve traced by the midpoints of area bisectors.

The extra terrain forms a curved crotch between the two medians formingthose triangles. It turns out that these curves are segments of hyperbolae, andthe ends of the segments are the midpoints of the medians; indeed the curveis composed of bisector midpoints. Five of these together produce a slender,concave-curvy five-pointed star that occupies a fraction of just 0.001536 of thepentagon’s area. The curves also generate an inner curvy pentagon inside thestar.

How do the crotches affect the number of bisectors a point is on? A pointinside a crotch is on (in addition to its home bisector) two of the bisectors thatform that crotch, namely the two tangents to the hyperbola from that point. Apoint on the crotch curve itself is on only one, the tangent at that point. Anda point on a median bordering a crotch is on (additionally to that median) oneother crotch-forming bisector, nameley the other tangent, besides that median,containing that point. This allows us to tally the number of bisectors through

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Figure 3: The five hyperbolae containing the curves.

Figure 4: The inner star.

every point in the star.On one bisector: points outside the star, plus the cusps of the star. On

two: the perimeter of the star (home bisector plus another due to being on acrotch curve). On three: the interior of the part of the star outside of the innerpentagon (home plus two for being inside a crotch; points along a median are notinside a crotch but border two), and the vertices of the inner pentagon (home

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plus two for being on curves). On four: the sides of the inner pentagon (homeplus one for being on a curve plus two for being inside a crotch; the midpointsare not inside a crotch but border two). On five: the interior of the innerpentagon (the center is on all five medians; points on medians are on medians,on boundaries of two crotches, and inside another; and each other point is onits home bisector and four others thanks to being inside two crotches).

2 Extra Credit: Find the Star’s Area

The extra credit problem is to quantify the most dangerous points, which meansfinding the area of the star, the region of points on multiple area bisectors.

Take a polygon of any odd number (an even-sided polygon has only onepoint on multiple area bisectors) n ≥ 3 of sides. Our figures will feature aheptagon and a pentagon, but the reasoning will be independent of the numberof sides. Take any two sides of this polygon between which there are areabisectors. These bisectors teeter along the crotch curve between the medians tothe two sides. Extend the sides to where the lines they are on intersect, forminga triangle with each of these bisectors. The bisectors, preserving as they do thearea on each side of the polygon, preserve also the area of this triangle. Thisis a defining feature of hyperbolae (https://en.wikipedia.org/wiki/Hyperbola):line segments that yield the same triangular area in that way are tangent, attheir midpoints, to a single hyperbola with the extended sides as asymptotes.

So we know the shape of our curvy star: its sides are segments of hyperbo-lae whose asymptotes extend the polygon’s opposing sides. There are n suchhyperbolae, each tangent to two medians, the points of tangency forming thecusps of the star. The curves are the locus of the bisector midpoints, and thecusps occur exactly halfway along the medians, which is very close to the cen-ter of the polygon even for n = 5 (.0955 of the distance to a vertex). I havederived the star’s area (for any odd n ≥ 3) in two ways, a la Euclid (withoutspecifying axes or coordinates) and a la Descartes. My preferred version is theEuclidean, because at least in my hands it has produced a tidier expression forthe area (and it also makes it easy to show that the result applies also to affinetransformations of regular polygons; open question: does it apply to an evenlarger class of polygons?). I did a second derivation to check the results, andI include the Cartesian version here because the math will be more familiar tomany readers.

2.1 Euclidean Approach

Consulting the figure, and assuming an out-radius (distance from the center Pof our polygon ABCDE to its vertices) of 1:

6 EOC =π

n

The apothem d = PF = cosπ

n

4

Figure 5: A polygon with opposing sides extended.

a = OE =d+ 1

sin πn

= cotπ

2n

b = OF =d+ 1

tan πn

The half-side c = FE = sinπ

n

Let the vectors vE and vc be (E −O) and (C −O), respectively.Note that d = b

a . EH bisects the polygon and the area of 4EOH is12ab sin π

n . Let GJ be an area bisector of the polygon with endpoints on EF andCH. Then, by constancy of area:

OG ·OJ = ab

We will parameterize the area bisectors that have endpoints on EF and CHas a function of t for t ∈ [d, 1]:

Gt = O + tvE

Jt = O +d

tvC

5

The midpoint of GJ is:

Mt =Gt + Jt

2= O +

t

2vE +

d

2tvC

Because H = O + bavC , the center P of the polygon is:

P =d

d+ 1E +

1

d+ 1H = O +

d

d+ 1vE +

d

d+ 1vC

Their difference is the vector vM from P to Mt:

vM =

(t

2− d

d+ 1

)vE +

(d

2t− d

d+ 1

)vC

We will integrate the area of the section of 12n of the central star consisting

of triangles with area 12 |vM ×

ddtvM |dt (the magnitude of the cross-product of

two vectors is the area of the parallelogram of which they form two sides).

Figure 6: Visualizing the integral.

d

dtvM =

1

2vE −

d

2t2vC

1

2|vM ×

d

dtvM | =

1

2|4EOC| ·

(d

2t2

(t

2− d

d+ 1

)+

1

2

(d

2t− d

(d+ 1)

))

=1

4a2 sin

π

n· d ·

(1

t− d

(d+ 1)t2− 1

(d+ 1)

)

6

The area of the star is 2n times the integral of this for d ∈ [t, t′], where t′ isthe t such that vM = k(P −A) for some k (i.e., Mt′ is on the median containingAP ). To find t′ we first find an expression for A:

A = O +a− 2c

avE = O + (2d− 1)vE

(That last step follows from trigonometric identities.)

P −A =1− 2d2

d+ 1vE +

d

d+ 1vC

Mt is on the median containing AP when P −A is a scalar multiple of vM :

d

d+ 1·(t

2− d

d+ 1

)=

(1− 2d2

d+ 1

)·(d

2t− d

d+ 1

)This simplifies algebraically to:

t2 + 2(1− 2d)t+ (2d2 − 1) = 0

The quadratic formula gives:

t = 2d− 1±√

2(1− d)

Only the plus solution falls between d and 1, and so:

t′ = (2−√

2)d+√

2− 1

The area of the star is:

1

2na2 sin

π

n· d ·

∫ t′

t=d

(1

t− d

(d+ 1)t2− 1

(d+ 1)

)dt

=1

2na2 sin

π

n· d ·

(ln

(t′

d

)+

d

(d+ 1)t′− d

(d+ 1)d− t′ − dd+ 1

)

=(n sin

π

ncos

π

n

)·

cot2 π2n

2·(

ln

(t′

d

)− (t′ + 1)(t′ − d)

(d+ 1)t′

)Since n sin π

n cos πn is the area of the polygon, the ratio of the star’s area tothat of the polygon is:

cot2 π2n

2·(

ln

(t′

d

)− (t′ + 1)(t′ − d)

(d+ 1)t′

)

=cot2 π

2n

2·

[ln

(2−√

2 +

√2− 1

cos πn

)−

(4− 3√

2) cos2 πn + (4√

2− 6) cos πn + 2−√

2

(2−√

2) cos2 πn + cos πn +√

2− 1

]Here are some values of this ratio:

7

n Ratio

3 0.01986038542

5 0.00153624226

7 0.0003544589389

9 0.0001236877431

11 0.00005413173298

13 0.0000273775531

15 0.00001531615375

17 0.00000923179205

19 0.000005893391885

In the case of n = 3 (where the inner shape is not so much a star as a curvytriangle) this reduces to a tidy:

3

4ln(2)− 1

2

2.2 Cartesian Approach

Figure 7: Heptagon.

Let the origin be the center of the hyperbola whose asymptotes extend twoopposing sides the polygon, for odd n ≥ 3 (for n = 3, the origin is a vertex of

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the triangle), with out-radius (center to vertex) r = 1. The asymptotes meet ata π/n angle to each other.

We are going to find equations for two medians and a hyperbola that bounda region (too small to be seen in the figure above, so there is a close-up below)that forms 1

2n of our star. Because CAO is a right triangle and 6 COA is π2n

(where d is the apothem length cos πn ):

C =

(0,

d

sin(π2n

))AB is at a slope of m = tan

(π2n

), and so:

A =(−d cos

( π2n

), yC − d sin

( π2n

))B =

(cos( π

2n

), yC + sin

( π2n

))CD is at a slope of m′ = tan

(3 π2n

), and so:

D =

(− cos

(3π

2n

), yC − sin

(3π

2n

))

Figure 8: Close-up of 1/(2n) of the inner star.

The cusp of our region is at the midpoint M of AB:

M =

(xA + xB

2,yA + yB

2

)H has asymptotes with slopes ± 1

tan( π2n )

and passes through M, which de-

termines an equation for H by settling the values of a and b in the standardformat for an upright hyperbola (where the slope of the asymptotes is alwaysab ):

H :y2

a2− x2

b2= 1

a

b=

1

tan(π2n

)9

a =

√y2M −

x2Mtan2

(π2n

)b = a tan

( π2n

)We will divide R into the part where it is bounded above and below by

DE and AB, and the rest, where it is bounded by H and AB. Where I is theintersection of DE and H, we want to divide at the line x = xI , which we willfind by substituting m′x + c (where c = yC) for y in the equation for H. Weobtain a quadratic equation and so use the quadratic formula:

x2I(b2m′2 − a2) + xI(2cm

′b2) + (c2b2 − a2b2) = 0

xI =−2cm′b2 +

√(2cm′b2)2 − 4(b2m′2 − a2)(c2b2 − a2b2)

2(b2m′2 − a2)

First we find the area of the small triangle bounded by AB, DE and x = XI :

A1 =1

2xI(m

′xI −mxI)

Then the remaining area:

A2 =

∫ xM

x=xI

a

√1 +

x2

b2− (mx+ yc)dx

=

[xMx=xI

1

2a

(x

√1 +

x2

b2+ b sinh−1

(xb

))−(m

2x2 + ycx

)]The area of the polygon is:

A =n

4 tan πn

and so the ratio of the area of the star to that of the polygon is:

Rn =2n(A1 +A2)

A

3 Hypocycloid?

A natural hunch for Spirograph aficionadi is that the star’s perimeter is a self-intersecting hypocycloid, a curve traced by a point on a smaller circle rollingwithin a larger circle. There is a single five-point self-intersecting hypocycloid,which can be formed identically (an interesting fact) with circles whose radiiare in a ratio of 5:2 or 5:3. It looks promising, doesn’t it?

The hypocycloid’s radius (center to cusp) is that of the larger circle. Ifour star is a 5-2 hypocycloid, then it is formed from circles of radii .0955 and2/5 times that, or .0382. Then the distance from the pentagon’s center to the

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Figure 9: 5-2 hypocycloid.

nearest points on the curves, which are where the bisectors parallel to sides aretangent to them, is the larger radius minus twice the smaller, or .0191. But infact that distance is .0147, so the hunch is incorrect; our star is significantlymore svelte.

To reach .0147 from the center, the inner circle would have to have a radiusof (.0955− .0147)/2 = .0404, which of course is not an integral divisor of .0955and so would not make a five-pointed star. A natural revised hunch is that ourstar’s shape can be formed with an elliptical inner “wheel” of semi-major axis.0382 and circumference 2/5 of the outer circle. However, despite being closethere is no cigar to be had; the resulting shape is clearly not hyperbolic.

Figure 10: A 5-2 elliptical hypocycloid.

There is, however, an inner wheel that would produce our hyperbolic star,and so the star is hypocyclic in a broad sense. Here’s a thought-experiment toshow that. Start with an inner circle just large enough to reach from the outercircle to the points on the hyperbolae nearest the center. It’s too big, so weneed to shave a bit off the sides. To see where we need to cut we measure fromeach point on the outer circle to the nearest point on the hyperbola (on a lineperpendicular to its tangent). That’s how far the “drawing point” of the innerwheel needs to be from this point on the outer circle when the inner wheel is

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tangent to the outer at that point, because the tangent to a hypocyclic curveis perpendicular to the line from the drawing point to the point of tangencybetween the wheels. Relying on these measurements from several points, wecan construct an inner wheel as in the figure below. It will draw a pretty closeapproximation, composed of circular arcs, of the hyperbola segment. As weincrease the number of points, we have our desired shape at the limit.

Figure 11: A custom hyperbolic hypocycloid.

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