searching for loci using new technology in teacher's education pech.pdf · teacher’s...

Gordon College of Education, Haifa 2018, November 4

Searching for loci using new technology inteacher’s education

Pavel Pech

University of South Bohemia, Czech Republic


Outline of the talk:

I Introduction

I Command Locus

I Problems 1,2

I Command LocusEquation

I Problems 3,4,5

I Pilot test

I Conclusions


Introduction

I Roughly spoken a locus is a set of points subject to somegeometric constraints.

I Searching for geometric loci belongs to traditional part ofmathematics school curricula all over the world.

I This topic is generally considered to be quite difficult forstudents, despite many well–known loci around us, such aslines, circles or conics.

I Nowadays new computational technologies substantiallyfacilitate investigation of loci, especially in a plane.


Command Locus

I Dynamic geometry software such as Cabri, GeoGebra,Sketchpad and others offer several methods how to describethe locus. The use of this software enables to draw thedesired locus and mostly to obtain its locus equation.

I The tool Locus belongs to one of traditional functions ofdynamic geometry systems. To its application we need twopoints.

I The first point is a mover, the point which usually movesalong a certain object. The second point - a tracer - issomehow dependent on the mover and draws the soughttrajectory.

I The command Locus is very simple and useful, we can use itat all types of schools.


Problem 1Let ABC be a triangle with a base AB and a vertex C on a givenline k. Find the locus of the orthocenter H of ABC when C movesalong the line k.

Using the command Locus, first clicking on the tracer H and thenon the mover C , we get the following curve.


What is it?

I Some students say: It is a parabola.

I Another students say: It is a hyperbola.

I Or it is neither parabola nor hyperbola?

I What is the solution?

I We’ll search for the locus equation.


Introduce a rectangular coordinate system such that A = [0, 0],B = [1, 0], C = [u, v ], H = [p, q]and let k be an arbitrary line k : ax + by + c = 0.


To describe the orthocenter H it holds:

(H − C ) ⊥ (B − A)⇔ h1 := p − u = 0,

(H − A) ⊥ (C − B)⇔ h2 := p(u − 1) + qv = 0.

Further

C ∈ k ⇔ h3 := au + bv + c = 0.


We get the system of three equations h1 = 0, h2 = 0, h3 = 0 invariables u, v , p, q, a, b, c .

To find the locus of H = [p, q] we eliminate variables u, v in thesystem h1 = 0, h2 = 0, h3 = 0 to obtain a relation in p, q whichdepends only on a, b, c . In the program CoCoA1 we enter

Use R::=Q[a,b,c,u,v,p,q];

I:=Ideal(au+bv+c,p-u,(u-1)p+vq);

Elim(u..v,I);

and acquire the equation

κ : bp2 − apq − bp − cq = 0.

1Program CoCoA is freely distributed at http://cocoa.dima.unige.it


I Suppose that (a, b) 6= (0, 0) since in this case the line k is notdefined. Then

bp2 − apq − bp − cq = 0.

is the equation of the conic.

I The cases k ⊥ AB, A ∈ k or B ∈ k lead to singular conicswhich consist of two intersecting lines which are not depicted.

I Considering regular conics we get two cases:


1. If k ‖ AB the locus is a parabola with the vertex(1/2,−b/(4c)) and the parameter |c/(2b)|.


2. If k ∦ AB we obtain a hyperbola centered at(−c/a,−b(a + 2c)/a2) with one asymptote perpendicular to ABand the second asymptote perpendicular to the line k .


Opposite implication (which is often neglected at schools).

Suppose that κ is regular and let H ∈ κ. We are to show that thenC ∈ k or equivalently h3 = 0. If we suppose that q 6= 0, then

Use R::=Q[a,b,c,u,v,p,q];

J:=Ideal(bpp-apq-bp-cq,p-u,(u-1)p+vq);

NF(q(au+bv+c),J);

we get NF = 0. Finally, if q = 0 then H = [0, 0] or H = [1, 0]which both obey the locus conditions.

Conclusion: If k 6⊥ AB, A /∈ k or B /∈ k then

— if k ‖ AB the locus is the parabola,

— if k 6‖ AB the locus is the hyperbola.


Problem 2Let ABC be a triangle with a side AB and a vertex C on a circle ccentered at A and radius |AB|. Determine the locus of theorthocenter H of ABC when C moves along c .

Using the command Locus, first clicking on the tracer H and thenon the mover C , we get the curve of the third degree which is astrophoid.


How to compute it?

Let A = (0, 0), B = (a, 0), C = (u, v) and H = (p, q). Then:

(H − C ) ⊥ (B − A)⇔ h1 : p − u = 0,

(H − A) ⊥ (C − B)⇔ h2 : p(u − a) + qv = 0.

C ∈ c ⇔ h3 : u2 + v2 − a2 = 0.

Elimination of u, v in the system h1 = 0, h2 = 0, h3 = 0 gives

Use R::=Q[a,p,q,u,v];

I:=Ideal(p-u,p(u-a)+vq,u2 + v2 − a2);Elim(u..v,I);

the equation of the fourth degree which decomposes into thestrophoid and line

(p3 − ap2 + aq2 + pq2)(p − a) = 0. (1)


How is it possible?

The problem is hidden in the position when C arrives at B, i.e.u = a, v = 0, and the line BC is not defined. Then the systemh1 = 0, h2 = 0, h3 = 0 transforms into one relation

p − a = 0

which is the line in (1).


To avoid this we add the condition B 6= C , i.e.((u − a)2 + v2)t − 1 = 0, where t is a slack variable, into thesystem above. Then we get

Use R::=[Q[a,p,q,u,v,t];

J:=Ideal(p-u,p(u-a)+vq,u2 + v2 − a2, ((u − a)2 + v2)t − 1);Elim(u..t,J);

the only equation

κ : p3 − ap2 + aq2 + pq2 = 0.

Using the command LocusEquation we still obtain the equationof the fourth degree.

Remark: We are able to detect such ”extraneous points” and getrid of them (e.g. if the point H is not uniquely defined) by theGrobner Cover algorithm (Abanades 2014).


Opposite implication.

Let H ∈ κ and suppose that B 6= C , when the orthocenter is notuniquely defined. We ask whether the vertex C lies in thecircumcircle of ABC for a given orthocenter H. Then adding thecondition q 6= 0

Use R::=[Q[a,p,q,u,v,t,s];

L:=Ideal(p-u,p(u-a)+vq,p3 − ap2 + aq2 + pq2,((u − a)2 + v2)t − 1, q(u2 + v2 − a2)s − 1);NF(1,L);

to the respective ideal we get NF (1, L) = 0.

Conclusion:

The locus is the strophoid without the point B = [a, 0].


LocusEquation

I The command Locus cannot be applied to every locus.Problems that we will present further are of this case.

I To solve them we have to use a more advanced toolLocusEquation which has recently been implemented intoGeoGebra version 5.

I This command brings a new approach in searching for loci. Itis based on automated discovery, the part of the theory ofautomated theorem proving.

I This tool uses elimination of variables in a system of algebraicequations describing the locus. It returns an implicit equationof the locus.


LocusEquation

I It is well known that the result is the Zariski closure of aprojection on the space of local coordinates.

I This often leads to the situation that instead of a real locuswe get the smallest variety which contains, besides the locus,also some extraneous objects not pertaining to it.

I Before using the command LocusEquation we have toconstruct in GeoGebra a geometric diagram describing thelocus.

I After constructing the diagram we apply the commandLocusEquation which has two parameters.


LocusEquation

I The first one is the thesis T (which must be a Booleanexpression), the second one is a free point P whose locus weinvestigate.

I The result of LocusEquation [T,P] produces the set V suchthat ”if T is true then P ∈ V ”.

I Several Boolean expressions in the form of commands such asAreCollinear or AreConcyclic are tested in problemswhich results to curves in the plane.

I By searching for loci we apply Grobner bases method usingsoftware CoCoA.


Problem 3Determine the locus of a point P such that its reflections K , L,Min the sides of a given triangle ABC are collinear.


It is obvious that in this case the command Locus cannot beapplied. But the command LocusEquation solves the problem.

Procedure determining the locus is following:

1. First construct a geometric diagram

I Construct a triangle ABC .

I Choose an arbitrary point P.

I Construct reflections K , L,M of the point P in the sidesAB,BC and CA.

2. Enter the commandLocusEquation[AreCollinear[K,L,M],P].


Besides the graph of the searched locus

one also gets its equation in given Cartesian coordinates

x2 + y2 − 5x − y = 0.

We see that the locus is the circumcircle of ABC .


This problem is related to the Simson–Wallace theorem, wherereflections of a point P in the sides of ABC are replaced by thefeet of perpendiculars from P onto the triangle sides.

Note that the line passing through K , L,M passes also through theorthocenter H of the triangle (blue) and is parallel to the Simsonline (green).


Which way does the computer proceed?

Let’s show how we arrive at the solution — the circumcircle ofABC — using the theory of automated theorem proving.

Let the coordinates be chosen such that A = [0, 0], B = [a, 0],C = [u, v ], P = [p, q], K = [k1, k2], L = [l1, l2], M = [m1,m2].Suppose that the points K , L,M are collinear.


Then:

PK ⊥ BC ⇔ h1 := (p − k1)(u − a) + (q − k2)v = 0,

K ′ ∈ BC ⇔ h2 := 2av + u(q + k2)− v(p + k1)− a(q + k2) = 0,

PL ⊥ CA⇔ h3 := (p − l1)u + (q − l2)v = 0,

L′ ∈ CA⇔ h4 := (p + l1)v − (q + l2)u = 0,

PM ⊥ AB ⇔ h5 := p −m1 = 0,

M ′ ∈ AB ⇔ h6 := q + m2 = 0,

K , L,M are collinear ⇔

h7 := k1l2 + l1m2 + k2m1 − l2m1 − k1m2 − k2l1 = 0.


After eliminating variables k1, k2, l1, l2,m1,m2 from the systemabove we get

av2 ·M = 0,

where

M = vp2 + vq2 − avp + (au − u2 − v2)q.

If a 6= 0 and v 6= 0, i.e. if A 6= B and A,B,C are not collinear,

then the equation M = 0 represents the circumcircle of ABC .

This can be easily verified by substituting coordinates of thetriangle vertices into M = 0.



Let P ∈ M. We ask whether then K , L,M are collinear, i.e.h7 = 0. Suppose that A 6= C and B 6= C , i.e.(u2 + v2)((u− a)2 + v2)t− 1 = 0, where t is a slack variable. Then

Use R::=Q[a,u,v,k[1..2],l[1..2],m[1..2],p,q,t,s];

J:=Ideal(h1,h2,h3,h4,h5,h6,M,(u2 + v2)((u − a)2 +v2)t-1,h7*s-1);NF(1,J);

NF (1, J) = 0, which means that h7 = 0. Thus every point P of thecircumcircle of ABC satisfies the condition that K , L,M arecollinear.

Conclusion:

If A 6= B, A 6= C , B 6= C and A,B,C are not collinear, then thelocus is the (whole) circumcircle of ABC .


Problem 4Determine the locus of a point P such that feet K , L,M,N ofperpendiculars from P on the sides of a given quadrilateral ABCDare concyclic.




I Draw a quadrilateral ABCD.


I Construct feet K , L,M,N of perpendiculars from P to the linesAB,BC ,CD and DA.

2. Enter the commandLocusEquation[AreConcyclic[K,L,M,N],P].


An algebraic curve of the third degree is displayed.

This curve has many interesting properties.


The ellipse inscribed into ABCD. The focal points F1,F2 lie onlocus curve.The midpoint O of F1 and F2 lies on the Newton-Gauss line g .


If a quadrilateral ABCD is tangetial or extangential then we get acubic curve with the singular point.


Special case:

Determine the locus of a point P such that the feet K , L,M,N ofperpendiculars from P onto the sides of a given parallelogramABCD are concyclic.

After execution the command an equilateral hyperbola

3x2 − 2xy − 3y2 − 15x + 15y = 0

including its equation appears.


How does computer arrive at this result?

Let A = [0, 0], B = [a, 0], C = [u, v ], D = [u − a, v ]. Furtherdenote K = [k1, k2], L = [l1, l2], M = [m1,m2], N = [n1, n2],P = [p, q].

Suppose that K , L,M,N are concyclic.


Then:

K ∈ AB ⇔ h1 := k2 = 0,

L ∈ BC ⇔ h2 := vl1 + al2 − av − ul2 = 0,

M ∈ CD ⇔ h3 := vm1+uv +(u−a)m2−v(u−a)−vm1−um2 = 0,

N ∈ DA⇔ h4 := vn1 − (u − a)n2 = 0,

PK ⊥ AB ⇔ h5 := p − k1 = 0,

PL ⊥ BC ⇔ h6 := (p − l1)(u − a) + (q − l2)v = 0,

PM ⊥ CD ⇔ h7 := p −m1 = 0,

PN ⊥ DA⇔ h8 := (p − n1)(u − a)w + (q − n2)v = 0,


K , L,M,N are concyclic ⇔

h9 :=

∣∣∣∣∣∣∣∣k21 , k1, 0, 1

l21 + l22 , l1, l2, 1m2

1 + m22, m1, m2, 1

n21 + n22, n1, n2, 1

∣∣∣∣∣∣∣∣ = 0.

Eliminating variables k1, k2, l1, l2,m1,m2, n1, n2 in the systemh1 = 0, h2 = 0, . . . , h9 = 0, gives

H := vp2 + 2(a− u)pq − vq2 − avp + (u2 + v2 − au)q = 0. (2)

Ifu2 − 2au + v2 6= 0

then (2) is the equilateral hyperbola.

Ifu2 − 2au + v2 = 0, (3)

then (2) decomposes into two mutually orthogonal lines.


The condition (3), together with z = v and w = u − a lead to|AB| = |BC | = |CD| = |DA|, and ABCD is a rhombus.

This case is also easy to prove straightforward. For a point P onthe diagonal

|PK | · |PM| = |PL| · |PN|

holds. Then from the theorem on the power of a point withrespect to a circle concyclicity of the points K , L,M,N follows.


Opposite implication. Suppose that P ∈ H. We want to knowwhether h9 = 0. If we add a 6= 0 to the related ideal then

Use

R::=Q[a,u,v,w,z,k[1..2],l[1..2],m[1..2],n[1..2],t,p,q];

J:=Ideal(h1,h2,h3,h4,h5,h6,h7,h8,H,a*h9*t-1);

NF(1,J);

We get NF (1, J) = 0, i.e., the points K , L,M,N are concyclic.

Conclusion:

— If ABCD is a parallelogram with distinct side lengths, then thelocus is the equilateral hyperbola.

— If ABCD is a rhombus, then the locus is formed by its diagonals.


Problem 5Let ABCD be a quadrilateral and K , L,M,N feet of perpendicularsfrom a point P to the lines AB, BC , CD, DA. Determine the locusof P such that the lines KN and LM are parallel.

It is obvious that in this case the command Locus cannot beapplied. But the command LocusEquation solves the problem.




I Draw a quadrilateral ABCD.


I Construct feet K , L,M,N of perpendiculars from P to the linesAB,BC ,CD and DA.

I Denote m = KN and n = LM.

2. Enter the commandLocusEquation[AreParallel[m,n],P].


Besides the graph of the searched locus

one also gets its equation in given rectangular coordinates

(x − 1)2 + (y − 3)2 = 10. (4)

By (4) we see that the locus is a circle.


Following questions may arise:

Is the solution really the entire circle?

Is the solution always a circle (and not another curve)?

If the solution is not a circle, for which positions of the verticesA,B,C ,D does it happen?


Which way does the computer proceed?

Let’s show how we arrive at the solution — the circle — using thetheory of automated theorem proving.

Let the coordinates be chosen such that A = [0, 0], B = [a, 0],C = [u, v ], D = [w , z ] P = [p, q], K = [k , 0], L = [l1, l2],M = [m1,m2] N = [n1, n2].

Suppose that the lines KN and LM are parallel.


Then:

PK ⊥ AB ⇔ h1 := p − k = 0,

L ∈ BC ⇔ h2 := ul2 + av − al2 − vl1 = 0,

PL ⊥ BC ⇔ h3 := (p − l1)(u − a) + (q − l2)v = 0,

M ∈ CD ⇔ h4 := um2 + zm1 + vw − wm2 − uz − vm1 = 0,

PM ⊥ CD ⇔ h5 := (p −m1)(w − u) + (q −m2)(z − v) = 0,

N ∈ DA⇔ h6 := wn2 − zn1 = 0,

PN ⊥ DA⇔ h7 := (p − n1)w + (q −m2)z = 0,

KN ‖ LM ⇔ h8 := (l1 −m1)n2 − (l2 −m2)(n1 − k) = 0.


After eliminating variables k, l1, l2,m1,m2, n1, n2 from the systemabove we get

z(av − vw − az + uz)S = 0, (5)

where

S = (p2 + q2)(avw − 2uvw + vw2 − auz + u2z − v2z + vz2)+

p(u2vw + v3w − avw2 + au2z − u3z + av2z − uv2z − avz2)−

q(au2w − u3w + av2w − uv2w − auw2 + u2w2 + v2w2 − u2vz−

v3z − auz2 + u2z2 + v2z2).

In (5) we can suppose that z 6= 0, av − vw − az + uz 6= 0,otherwise ABCD degenerates.

Thus (5) implies the locus equation of P = [p, q]

S = 0.


Denote the coefficient at p2 + q2 by

T = avw − 2uvw + vw2 − auz + u2z − v2z + vz2.

If T 6= 0, then (5) is the circle passing through the points A,C andMiguel point H.


Now suppose that T = 0.

Then S = 0 is the line passing through the vertices A,C .

For P ∈ AC the angles by B and D are equal.



Does every point P = [p, q] satisfying S = 0 have the requiredproperty KN ‖ LM?

Suppose that P obeys S = 0. We want to show that thenKN ‖ LM, i.e. h8 = 0.

Using the command Normal Form NF in CoCoA we enter

Use R::=Q[a,u,v,w,z,k,l[1..2],m[1..2],n[1..2],p,q,t];

J:=Ideal(h1,h2,h3,h4,h5,h6,h7,S,z(av-vw-az+uz)h8*t-1);

NF(1,J);

and get NF 6= 0. The answer is negative.


Adding the conditions P 6= A and P 6= C , i.e.(p2 + q2)((p − u)2 + (q − v)2)s − 1 = 0 to the ideal J, then

Use

R::=Q[a,u,v,w,z,k,l[1..2],m[1..2],n[1..2],p,q,t,s];

K:=Ideal(h1,h2,h3,h4,h5,h6,h7,S,

(p2 + q2)((p − u)2 + (q − v)2)s − 1,z(av-vw-az+uz)h8*t-1);NF(1,K);

we get NF(1,K)=0. This implies that h8 = 0.

Realize that if P = A or P = C , then K = N or L = M and thelines KN or LM are not defined.


Conclusion:

— If the angles by B and D of a quadrilateral ABCD are distinct,then the locus is the circle through the points A,C and the Miquelpoint H, without A and C .

— If the angles by B and D are equal, then the locus is the lineAC without the points A and C .


Pilot test

I How efficient is the help of DGS by searching for the locus?

I The aim of the test is to find out how to increase theavailability of a synthetic solution if students use DGSsoftware.

I What are the most frequent causes of failure.

I Together with my doctor student Jirı Blazek, we haveformulated, the following questions:


I What is the effect of the software on the process of findingthe solution?

I How significant is the help of the software to find relevantproperties?

I What are the causes of failure in solving problems? In moredetail:

I Are students capable of assembling all the important factsrelated to the solution?

I Are they able to use these facts for a mathematical proof?


The experiment was as follows:

I Seven students from the University of South Bohemia duringthe lesson of geometry solved two geometric problems.

I For half an hour they solved these problems with paper andpencil only.

I If they were not successful, they wrote down all thehypotheses and ideas they thought to be relevant to thesolution and for another half an hour repeated the wholeprocess again, now with the help of GeoGebra.

I The problems were chosen so that the key empirical factswere not difficult to observe. Let us first outline the two tasksand their idealized solution.


1st Problem (Holfeld 1773)

Given two mutually orthogonal lines a, b which intersect at thepoint O. Let points A and M lie on a and b. Circle c with diameter|MO| intersects the line AM at the point P. What is the locus ofP when M moves along the line b?


Conventional solution (without aid of software):

We focus our attention on the angle MPO and recall Thales’theorem: All angles above the diameter of a circle are right.

Statement 1: The angle MPO is right.


At first, it is not clear whether Statement 1 has a relationship tothe solution. The question is: Does this claim have anyconsequences?

Answer: The angle OPA is right. Now one should again apply theThales’ theorem, this time in its opposite formulation:

Statement 2: The locus of P from which the segment OA is seenat the right angle is a circle with diameter OA.


One concludes:

Statement 3: The locus of P is a circle with diameter |OA|,without the point A.


In the previous solution several not quite obvious steps andassociations require some experience and intuition.

Why, for instance, should students focus their attention on theangle OPA when they have no idea that this angle is the key to thesolution?

And even when they know that this angle is right, they need notrecall the Thales’ theorem.

In these cases the experiments in DGS can help and inspire.


Solution with GeoGebra support:

We first define a set of points P using the Locus command. Thelocus is a circle with the diameter |AO| (to be sure that this is acircle and not a curve similar to a circle, we should rather use theLocusEquation command).

Notice that the angles MPO and OPA are right and that theyremain right even if we change the position of the point M.


Let us summarize the experimental facts:

The angles MPO and OPA are right. The locus of P is the circlewith diameter |OA|.

It is now up to the student to utilize these facts. However, ifhe/she does not recall the Thales’ theorem at this stage, thesoftware will not help him/her longer.


2nd ProblemTwo mutually orthogonal lines a, b intersecting at the point O, anda right triangle APB with the angle ∠PBA = 30◦ are given. Thepoints A and B lie on the lines a and b, the length |AB| is fixed.Determine the locus of P when A moves along the line a.


Conventional solution (without software):

By experiments we find out that points A,P,B and O lie on acircle (Thales’ theorem or the characteristics of a cyclicquadrilateral).

Statement 1: Points A,P,B,O lie on the circle.


As an implication of the previous statement and according to angleinscribed theorem the angles PBA and POA are equal. In otherwords, the angle POA is constant and is equal to 30◦.

Statement 2: The angle POA = 30◦. The point P lies on astraight line forming the angle 30◦ with the line a.


Since |PO| ≤ |AB|, another requirement is that the distance of thepoint P to the point O does not exceed |AB|.

Statement 3: The locus of P is the line segment XY of the length2|AB| with center at O and forming the angle 30◦ with the line a.


This example is much more difficult than the first one.

For a beginner, it is difficult to imagine that the fact that thequadrilateral APBO is cyclic leads to the solution.

GeoGebra can significantly help in the search for the locus.


Solution with GeoGebra support

Using the Locus command, we find that the locus is a line segment.

By measuring we find out that the angle AOP is 30◦. Itimmediately follows that the points A,O,B and P are concyclic,which can be easily verified.


With the help of DGS we have found the following empirical facts:

The points A,P,B,O are concyclic. The angle POA is equal tothe angle PBA. The locus is a segment.

I It is difficult for an inexperienced student to infer the solutionfrom these facts. The solution process is complicated by thefact that the hierarchy of experimentally obtained findings isnot clear in advance.

I For example, the equality ∠POA = ∠PBA leads toconcyclicity of the points, or vice versa. In any case thestudent has to be equipped with the theory. In this casehe/she has to know the condition for four points to lie on thesame circle and the inscribed angle theorem.


I Let us recall again that the acquisition of crucial facts isdependent not only on visual perception, but also on thesubject’s knowledge, experience and thinking.


Pilot test — results

The first question:

How big is the increase of right solutions if students can use thesoftware?

Without computer, none of the seven students solved any task,with the support of computer the first task was solved by fourstudents, the second one by one:

Without computer: 0/14, With the help of DGS: (4 + 1) / 14.

Software helped significantly in these cases. It can be assumedthat this was also due to the appropriate selection of tasks.


The second question:

How significant is the help of the software in finding relevantproperties?

Without computer

I In the case of the first task without computer, only onestudent guessed that the locus is a circle. The hypotheses ofall the others were wrong, most often the students thoughtthe locus was a parabola.

I In the second task without computer the students were moresuccessful: five students gave a presumption that provedcorrect. Four guessed that the locus of P is in a straight lineand one student guessed ∠POA = 30◦.


With the help of DGS

I In the first task with the aid of computer, the locus wasdetermined by everyone, however, only four students came outwith the idea to measure the magnitude of the angles MPOand OPB, and they ultimately solved the task.

I In the second task with computer, several students measuredthe magnitude of the angle POA, and when asked toexperiment, they also found that a circle can by circumscribedaround the quadrilateral.

We can say that software helps not just with verifying hypotheses,but mainly with discovering completely new facts.


The third question:

What are the causes of failure in problem solving?

The causes of the failure can be divided into two types:

I Type 1) Insufficient mathematical knowledge - unfamiliarity orinability to apply mathematical theory.

In this case, the student will not solve the problem, eventhough he gets all data needed to solve the problem.

I Type 2) Insufficient software help

The student has sufficient knowledge, but lacks the rightintuition or sometimes good luck, and can not find all thefacts needed for the solution although he/she has the softwaresupport. But if we give him/her a clue, the student is usuallyable to make a deductive proof himself/herself.


I The first problem: Three students, who did not solve thisproblem, did not even realize that the angles MPO and OPBcould be significant. They did not measure these angles at all.

It can be assumed that their failure can be attributed tounfamiliarity or inability to apply mathematical theory(Thales’ theorem).

I The second problem: Out of the six students who did notsolve the problem, four found all the empirical facts crucial forsolution. Their solution was incomplete because they did notexplain why the APBO is a cyclic quadrilateral.

It is not easy to decide whether it was because of theirunfamiliarity with the mathematical theory or whether theydid not realize that they should justify this step precisely.


Due to the difficulty of the problem, it is possible that a similarphenomenon will occur frequently: students do not solve theproblem because there are too many facts that they cannotlogically connect.

In other words, they cannot produce deductive evidence becausethey are not able to link isolated mathematical knowledge.

The remaining two students were unable to find all the factscrucial to the solution. Just as in the previous case, the lack ofknowledge of mathematical theory could play its role here, it is alsonot possible to rule out the wrong intuition in estimating whichfacts could be used to find the solution.


Conclusions

I Realize that the locus equation generated by the commandLocusEquation is only the necessary condition for validity ofthe condition given by a Boolean expression in the command.

I We strongly recommend, mainly for educational reasons, toimplement the verification of the opposite implication —which part of the found set of points belongs to the locus —into the program in future.

I This question is often omitted at schools.

I Otherwise students will consider the procedure given by thecommand LocusEquation as correct.


References

M. A. Abanades, F. Botana, A. Montes and T. Recio, Analgebraic taxonomy for locus computation in dynamicgeometry, Computer-Aided Design 56, 22-33 (2014).

A. Capani, G. Niesi and L. Robbiano, CoCoA, a System forDoing Computations in Commutative Algebra,http://cocoa.dima.unige.it

S. C. Chou, Mechanical Geometry Theorem Proving, D. ReidelPublishing Company, Dordrecht (1987).

I. Holfeld: Exercitationes Geometricae, Jesuit College of St.Clement in Prague, 1773.

E. Roanes-Lozano and E. Roanes-Macıas, AutomaticDetermination of Geometric Loci. 3D-Extension ofSimson–Steiner Theorem, Lecture Notes in ArtificialIntelligence, 1930, AISC 2000, pp. 157-173 (2000).

E. V. Shikin, Handbook and Atlas of Curves, CRC Press, BocaRaton (1995).


Thank you for your attention

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