Searching for loci using new technology in teacher s education

Transcription

1 Searching for loci using new technology in teacher s education Pavel Pech University of South Bohemia, Czech Republic

2 Outline of the talk: Introduction Command Locus Problems 1,2 Command LocusEquation Problems 3,4,5 Pilot test Conclusions

3 Introduction Roughly spoken a locus is a set of points subject to some geometric constraints. Searching for geometric loci belongs to traditional part of mathematics school curricula all over the world. This topic is generally considered to be quite difficult for students, despite many well known loci around us, such as lines, circles or conics. Nowadays new computational technologies substantially facilitate investigation of loci, especially in a plane.

4 Command Locus Dynamic geometry software such as Cabri, GeoGebra, Sketchpad and others offer several methods how to describe the locus. The use of this software enables to draw the desired locus and mostly to obtain its locus equation. The tool Locus belongs to one of traditional functions of dynamic geometry systems. To its application we need two points. The first point is a mover, the point which usually moves along a certain object. The second point - a tracer - is somehow dependent on the mover and draws the sought trajectory. The command Locus is very simple and useful, we can use it at all types of schools.

5 Problem 1 Let ABC be a triangle with a base AB and a vertex C on a given line k. Find the locus of the orthocenter H of ABC when C moves along the line k. Using the command Locus, first clicking on the tracer H and then on the mover C, we get the following curve.

6 What is it? Some students say: It is a parabola. Another students say: It is a hyperbola. Or it is neither parabola nor hyperbola? What is the solution? We ll search for the locus equation.

7 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.

8 To describe the orthocenter H it holds: (H C) (B A) h 1 := p u = 0, (H A) (C B) h 2 := p(u 1) + qv = 0. Further C k h 3 := au + bv + c = 0.

9 We get the system of three equations h 1 = 0, h 2 = 0, h 3 = 0 in variables u, v, p, q, a, b, c. To find the locus of H = [p, q] we eliminate variables u, v in the system h 1 = 0, h 2 = 0, h 3 = 0 to obtain a relation in p, q which depends only on a, b, c. In the program CoCoA 1 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 : bp 2 apq bp cq = 0. 1 Program CoCoA is freely distributed at

10 Suppose that (a, b) (0, 0) since in this case the line k is not defined. Then bp 2 apq bp cq = 0. is the equation of the conic. The cases k AB, A k or B k lead to singular conics which consist of two intersecting lines which are not depicted. Considering regular conics we get two cases:

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

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

13 Opposite implication (which is often neglected at schools). Suppose that κ is regular and let H κ. We are to show that then C k or equivalently h 3 = 0. If we suppose that q 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 AB, A / k or B / k then if k AB the locus is the parabola, if k AB the locus is the hyperbola.

14 Problem 2 Let ABC be a triangle with a side AB and a vertex C on a circle c centered at A and radius AB. Determine the locus of the orthocenter H of ABC when C moves along c. Using the command Locus, first clicking on the tracer H and then on the mover C, we get the curve of the third degree which is a strophoid.

15 How to compute it? Let A = (0, 0), B = (a, 0), C = (u, v) and H = (p, q). Then: (H C) (B A) h 1 : p u = 0, (H A) (C B) h 2 : p(u a) + qv = 0. C c h 3 : u 2 + v 2 a 2 = 0. Elimination of u, v in the system h 1 = 0, h 2 = 0, h 3 = 0 gives Use R::=Q[a,p,q,u,v]; I:=Ideal(p-u,p(u-a)+vq,u 2 + v 2 a 2 ); Elim(u..v,I); the equation of the fourth degree which decomposes into the strophoid and line (p 3 ap 2 + aq 2 + pq 2 )(p a) = 0. (1)

16 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 system h 1 = 0, h 2 = 0, h 3 = 0 transforms into one relation which is the line in (1). p a = 0

17 To avoid this we add the condition B C, i.e. ((u a) 2 + v 2 )t 1 = 0, where t is a slack variable, into the system above. Then we get Use R::=[Q[a,p,q,u,v,t]; J:=Ideal(p-u,p(u-a)+vq,u 2 + v 2 a 2, ((u a) 2 + v 2 )t 1); Elim(u..t,J); the only equation κ : p 3 ap 2 + aq 2 + pq 2 = 0. Using the command LocusEquation we still obtain the equation of the fourth degree. Remark: We are able to detect such extraneous points and get rid of them (e.g. if the point H is not uniquely defined) by the Gröbner Cover algorithm (Abánades 2014).

18 Opposite implication. Let H κ and suppose that B C, when the orthocenter is not uniquely defined. We ask whether the vertex C lies in the circumcircle of ABC for a given orthocenter H. Then adding the condition q 0 Use R::=[Q[a,p,q,u,v,t,s]; L:=Ideal(p-u,p(u-a)+vq,p 3 ap 2 + aq 2 + pq 2, ((u a) 2 + v 2 )t 1, q(u 2 + v 2 a 2 )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].

19 LocusEquation The command Locus cannot be applied to every locus. Problems that we will present further are of this case. To solve them we have to use a more advanced tool LocusEquation which has recently been implemented into GeoGebra version 5. This command brings a new approach in searching for loci. It is based on automated discovery, the part of the theory of automated theorem proving. This tool uses elimination of variables in a system of algebraic equations describing the locus. It returns an implicit equation of the locus.

20 LocusEquation It is well known that the result is the Zariski closure of a projection on the space of local coordinates. This often leads to the situation that instead of a real locus we get the smallest variety which contains, besides the locus, also some extraneous objects not pertaining to it. Before using the command LocusEquation we have to construct in GeoGebra a geometric diagram describing the locus. After constructing the diagram we apply the command LocusEquation which has two parameters.

21 LocusEquation The first one is the thesis T (which must be a Boolean expression), the second one is a free point P whose locus we investigate. The result of LocusEquation [T,P] produces the set V such that if T is true then P V. Several Boolean expressions in the form of commands such as AreCollinear or AreConcyclic are tested in problems which results to curves in the plane. By searching for loci we apply Gröbner bases method using software CoCoA.

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

23 It is obvious that in this case the command Locus cannot be applied. But the command LocusEquation solves the problem. Procedure determining the locus is following: 1. First construct a geometric diagram Construct a triangle ABC. Choose an arbitrary point P. Construct reflections K, L, M of the point P in the sides AB, BC and CA. 2. Enter the command LocusEquation[AreCollinear[K,L,M],P].

24 Besides the graph of the searched locus one also gets its equation in given Cartesian coordinates x 2 + y 2 5x y = 0. We see that the locus is the circumcircle of ABC.

25 This problem is related to the Simson Wallace theorem, where reflections of a point P in the sides of ABC are replaced by the feet of perpendiculars from P onto the triangle sides. Note that the line passing through K, L, M passes also through the orthocenter H of the triangle (blue) and is parallel to the Simson line (green).

26 Which way does the computer proceed? Let s show how we arrive at the solution the circumcircle of ABC 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 = [k 1, k 2 ], L = [l 1, l 2 ], M = [m 1, m 2 ]. Suppose that the points K, L, M are collinear.

27 Then: PK BC h 1 := (p k 1 )(u a) + (q k 2 )v = 0, K BC h 2 := 2av + u(q + k 2 ) v(p + k 1 ) a(q + k 2 ) = 0, PL CA h 3 := (p l 1 )u + (q l 2 )v = 0, L CA h 4 := (p + l 1 )v (q + l 2 )u = 0, PM AB h 5 := p m 1 = 0, M AB h 6 := q + m 2 = 0, K, L, M are collinear h 7 := k 1 l 2 + l 1 m 2 + k 2 m 1 l 2 m 1 k 1 m 2 k 2 l 1 = 0.

28 After eliminating variables k 1, k 2, l 1, l 2, m 1, m 2 from the system above we get av 2 M = 0, where M = vp 2 + vq 2 avp + (au u 2 v 2 )q. If a 0 and v 0, i.e. if A 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 the triangle vertices into M = 0.

29 Opposite implication. Let P M. We ask whether then K, L, M are collinear, i.e. h 7 = 0. Suppose that A C and B C, i.e. (u 2 + v 2 )((u a) 2 + v 2 )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,(u 2 + v 2 )((u a) 2 + v 2 )t-1,h7*s-1); NF(1,J); NF (1, J) = 0, which means that h 7 = 0. Thus every point P of the circumcircle of ABC satisfies the condition that K, L, M are collinear. Conclusion: If A B, A C, B C and A, B, C are not collinear, then the locus is the (whole) circumcircle of ABC.

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

31 Procedure determining the locus is following: 1. First construct a geometric diagram Draw a quadrilateral ABCD. Choose an arbitrary point P. Construct feet K, L, M, N of perpendiculars from P to the lines AB, BC, CD and DA. 2. Enter the command LocusEquation[AreConcyclic[K,L,M,N],P].

32 An algebraic curve of the third degree is displayed. This curve has many interesting properties.

33 The ellipse inscribed into ABCD. The focal points F 1, F 2 lie on locus curve. The midpoint O of F 1 and F 2 lies on the Newton-Gauss line g.

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

35 Special case: Determine the locus of a point P such that the feet K, L, M, N of perpendiculars from P onto the sides of a given parallelogram ABCD are concyclic. After execution the command an equilateral hyperbola 3x 2 2xy 3y 2 15x + 15y = 0 including its equation appears.

36 How does computer arrive at this result? Let A = [0, 0], B = [a, 0], C = [u, v], D = [u a, v]. Further denote K = [k 1, k 2 ], L = [l 1, l 2 ], M = [m 1, m 2 ], N = [n 1, n 2 ], P = [p, q]. Suppose that K, L, M, N are concyclic.

37 Then: K AB h 1 := k 2 = 0, L BC h 2 := vl 1 + al 2 av ul 2 = 0, M CD h 3 := vm 1 +uv +(u a)m 2 v(u a) vm 1 um 2 = 0, N DA h 4 := vn 1 (u a)n 2 = 0, PK AB h 5 := p k 1 = 0, PL BC h 6 := (p l 1 )(u a) + (q l 2 )v = 0, PM CD h 7 := p m 1 = 0, PN DA h 8 := (p n 1 )(u a)w + (q n 2 )v = 0,

38 K, L, M, N are concyclic h 9 := k1 2, k 1, 0, 1 l1 2 + l 2 2, l 1, l 2, 1 m1 2 + m2 2, m 1, m 2, 1 n1 2 + n2 2, n 1, n 2, 1 = 0. Eliminating variables k 1, k 2, l 1, l 2, m 1, m 2, n 1, n 2 in the system h 1 = 0, h 2 = 0,..., h 9 = 0, gives If H := vp 2 + 2(a u)pq vq 2 avp + (u 2 + v 2 au)q = 0. (2) u 2 2au + v 2 0 then (2) is the equilateral hyperbola. If u 2 2au + v 2 = 0, (3) then (2) decomposes into two mutually orthogonal lines.

39 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 on the diagonal PK PM = PL PN holds. Then from the theorem on the power of a point with respect to a circle concyclicity of the points K, L, M, N follows.

40 Opposite implication. Suppose that P H. We want to know whether h 9 = 0. If we add a 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 the locus is the equilateral hyperbola. If ABCD is a rhombus, then the locus is formed by its diagonals.

41 Problem 5 Let ABCD be a quadrilateral and K, L, M, N feet of perpendiculars from a point P to the lines AB, BC, CD, DA. Determine the locus of P such that the lines KN and LM are parallel. It is obvious that in this case the command Locus cannot be applied. But the command LocusEquation solves the problem.

42 Procedure determining the locus is following: 1. First construct a geometric diagram Draw a quadrilateral ABCD. Choose an arbitrary point P. Construct feet K, L, M, N of perpendiculars from P to the lines AB, BC, CD and DA. Denote m = KN and n = LM. 2. Enter the command LocusEquation[AreParallel[m,n],P].

43 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.

44 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 vertices A, B, C, D does it happen?

45 Which way does the computer proceed? Let s show how we arrive at the solution the circle using the theory 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 = [l 1, l 2 ], M = [m 1, m 2 ] N = [n 1, n 2 ]. Suppose that the lines KN and LM are parallel.

46 Then: PK AB h 1 := p k = 0, L BC h 2 := ul 2 + av al 2 vl 1 = 0, PL BC h 3 := (p l 1 )(u a) + (q l 2 )v = 0, M CD h 4 := um 2 + zm 1 + vw wm 2 uz vm 1 = 0, PM CD h 5 := (p m 1 )(w u) + (q m 2 )(z v) = 0, N DA h 6 := wn 2 zn 1 = 0, PN DA h 7 := (p n 1 )w + (q m 2 )z = 0, KN LM h 8 := (l 1 m 1 )n 2 (l 2 m 2 )(n 1 k) = 0.

47 After eliminating variables k, l 1, l 2, m 1, m 2, n 1, n 2 from the system above we get z(av vw az + uz)s = 0, (5) where S = (p 2 + q 2 )(avw 2uvw + vw 2 auz + u 2 z v 2 z + vz 2 )+ p(u 2 vw + v 3 w avw 2 + au 2 z u 3 z + av 2 z uv 2 z avz 2 ) q(au 2 w u 3 w + av 2 w uv 2 w auw 2 + u 2 w 2 + v 2 w 2 u 2 vz v 3 z auz 2 + u 2 z 2 + v 2 z 2 ). In (5) we can suppose that z 0, av vw az + uz 0, otherwise ABCD degenerates. Thus (5) implies the locus equation of P = [p, q] S = 0.

48 Denote the coefficient at p 2 + q 2 by T = avw 2uvw + vw 2 auz + u 2 z v 2 z + vz 2. If T 0, then (5) is the circle passing through the points A, C and Miguel point H.

49 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.

50 Opposite implication. Does every point P = [p, q] satisfying S = 0 have the required property KN LM? Suppose that P obeys S = 0. We want to show that then KN LM, i.e. h 8 = 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 0. The answer is negative.

51 Adding the conditions P A and P C, i.e. (p 2 + q 2 )((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, (p 2 + q 2 )((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 h 8 = 0. Realize that if P = A or P = C, then K = N or L = M and the lines KN or LM are not defined.

52 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 Miquel point H, without A and C. If the angles by B and D are equal, then the locus is the line AC without the points A and C.

53 Pilot test How efficient is the help of DGS by searching for the locus? The aim of the test is to find out how to increase the availability of a synthetic solution if students use DGS software. What are the most frequent causes of failure. Together with my doctor student Jiří Blažek, we have formulated, the following questions:

54 What is the effect of the software on the process of finding the solution? How significant is the help of the software to find relevant properties? What are the causes of failure in solving problems? In more detail: Are students capable of assembling all the important facts related to the solution? Are they able to use these facts for a mathematical proof?

55 The experiment was as follows: Seven students from the University of South Bohemia during the lesson of geometry solved two geometric problems. For half an hour they solved these problems with paper and pencil only. If they were not successful, they wrote down all the hypotheses and ideas they thought to be relevant to the solution and for another half an hour repeated the whole process again, now with the help of GeoGebra. The problems were chosen so that the key empirical facts were not difficult to observe. Let us first outline the two tasks and their idealized solution.

56 1st Problem (Holfeld 1773) Given two mutually orthogonal lines a, b which intersect at the point 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 of P when M moves along the line b?

57 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.

58 At first, it is not clear whether Statement 1 has a relationship to the solution. The question is: Does this claim have any consequences? Answer: The angle OPA is right. Now one should again apply the Thales theorem, this time in its opposite formulation: Statement 2: The locus of P from which the segment OA is seen at the right angle is a circle with diameter OA.

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

60 In the previous solution several not quite obvious steps and associations require some experience and intuition. Why, for instance, should students focus their attention on the angle OPA when they have no idea that this angle is the key to the solution? And even when they know that this angle is right, they need not recall the Thales theorem. In these cases the experiments in DGS can help and inspire.

61 Solution with GeoGebra support: We first define a set of points P using the Locus command. The locus is a circle with the diameter AO (to be sure that this is a circle and not a curve similar to a circle, we should rather use the LocusEquation command). Notice that the angles MPO and OPA are right and that they remain right even if we change the position of the point M.

62 Let us summarize the experimental facts: The angles MPO and OPA are right. The locus of P is the circle with diameter OA. It is now up to the student to utilize these facts. However, if he/she does not recall the Thales theorem at this stage, the software will not help him/her longer.

63 2nd Problem Two mutually orthogonal lines a, b intersecting at the point O, and a right triangle APB with the angle PBA = 30 are given. The points 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.

64 Conventional solution (without software): By experiments we find out that points A, P, B and O lie on a circle (Thales theorem or the characteristics of a cyclic quadrilateral). Statement 1: Points A, P, B, O lie on the circle.

65 As an implication of the previous statement and according to angle inscribed theorem the angles PBA and POA are equal. In other words, the angle POA is constant and is equal to 30. Statement 2: The angle POA = 30. The point P lies on a straight line forming the angle 30 with the line a.

66 Since PO AB, another requirement is that the distance of the point P to the point O does not exceed AB. Statement 3: The locus of P is the line segment XY of the length 2 AB with center at O and forming the angle 30 with the line a.

67 This example is much more difficult than the first one. For a beginner, it is difficult to imagine that the fact that the quadrilateral APBO is cyclic leads to the solution. GeoGebra can significantly help in the search for the locus.

68 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. It immediately follows that the points A, O, B and P are concyclic, which can be easily verified.

69 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 to the angle PBA. The locus is a segment. It is difficult for an inexperienced student to infer the solution from these facts. The solution process is complicated by the fact that the hierarchy of experimentally obtained findings is not clear in advance. For example, the equality POA = PBA leads to concyclicity of the points, or vice versa. In any case the student has to be equipped with the theory. In this case he/she has to know the condition for four points to lie on the same circle and the inscribed angle theorem.

70 Let us recall again that the acquisition of crucial facts is dependent not only on visual perception, but also on the subject s knowledge, experience and thinking.

71 Pilot test results The first question: How big is the increase of right solutions if students can use the software? Without computer, none of the seven students solved any task, with the support of computer the first task was solved by four students, 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 assumed that this was also due to the appropriate selection of tasks.

72 The second question: How significant is the help of the software in finding relevant properties? Without computer In the case of the first task without computer, only one student guessed that the locus is a circle. The hypotheses of all the others were wrong, most often the students thought the locus was a parabola. In the second task without computer the students were more successful: five students gave a presumption that proved correct. Four guessed that the locus of P is in a straight line and one student guessed POA = 30.

73 With the help of DGS In the first task with the aid of computer, the locus was determined by everyone, however, only four students came out with the idea to measure the magnitude of the angles MPO and OPB, and they ultimately solved the task. In the second task with computer, several students measured the magnitude of the angle POA, and when asked to experiment, they also found that a circle can by circumscribed around the quadrilateral. We can say that software helps not just with verifying hypotheses, but mainly with discovering completely new facts.

74 The third question: What are the causes of failure in problem solving? The causes of the failure can be divided into two types: Type 1) Insufficient mathematical knowledge - unfamiliarity or inability to apply mathematical theory. In this case, the student will not solve the problem, even though he gets all data needed to solve the problem. Type 2) Insufficient software help The student has sufficient knowledge, but lacks the right intuition or sometimes good luck, and can not find all the facts needed for the solution although he/she has the software support. But if we give him/her a clue, the student is usually able to make a deductive proof himself/herself.

75 The first problem: Three students, who did not solve this problem, did not even realize that the angles MPO and OPB could be significant. They did not measure these angles at all. It can be assumed that their failure can be attributed to unfamiliarity or inability to apply mathematical theory (Thales theorem). The second problem: Out of the six students who did not solve the problem, four found all the empirical facts crucial for solution. Their solution was incomplete because they did not explain why the APBO is a cyclic quadrilateral. It is not easy to decide whether it was because of their unfamiliarity with the mathematical theory or whether they did not realize that they should justify this step precisely.

76 Due to the difficulty of the problem, it is possible that a similar phenomenon will occur frequently: students do not solve the problem because there are too many facts that they cannot logically connect. In other words, they cannot produce deductive evidence because they are not able to link isolated mathematical knowledge. The remaining two students were unable to find all the facts crucial to the solution. Just as in the previous case, the lack of knowledge of mathematical theory could play its role here, it is also not possible to rule out the wrong intuition in estimating which facts could be used to find the solution.

77 Conclusions Realize that the locus equation generated by the command LocusEquation is only the necessary condition for validity of the condition given by a Boolean expression in the command. We strongly recommend, mainly for educational reasons, to implement the verification of the opposite implication which part of the found set of points belongs to the locus into the program in future. This question is often omitted at schools. Otherwise students will consider the procedure given by the command LocusEquation as correct.

78 References M. A. Abánades, F. Botana, A. Montes and T. Recio, An algebraic taxonomy for locus computation in dynamic geometry, Computer-Aided Design 56, (2014). A. Capani, G. Niesi and L. Robbiano, CoCoA, a System for Doing Computations in Commutative Algebra, S. C. Chou, Mechanical Geometry Theorem Proving, D. Reidel Publishing Company, Dordrecht (1987). I. Holfeld: Exercitationes Geometricae, Jesuit College of St. Clement in Prague, E. Roanes-Lozano and E. Roanes-Macías, Automatic Determination of Geometric Loci. 3D-Extension of Simson Steiner Theorem, Lecture Notes in Artificial Intelligence, 1930, AISC 2000, pp (2000). E. V. Shikin, Handbook and Atlas of Curves, CRC Press, Boca

79 Thank you for your attention