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Automated determination of isoptics with dynamic geometry

Authors:

Abstract

We present two approaches to symbolically obtain isoptic curves in GeoGebra in an automated, interactive process. Both methods are based on computing implicit locus equations, by using algebraization of the geometric setup and elimination of the intermediate variables. These methods can be considered as automatic discovery. Our first approach uses pure computer algebra support of GeoGebra, utilizing symbolic differentiation. The second approach hides all details in computer algebra from the user: the input problem is defined by a purely geometric way. In both approaches the output is dynamically changed when using a slider bar or the free points are dragged. Programming the internal GeoGebra computations is an on-going work with various challenges in optimizing computations and to avoiding unnecessary extra curves in the output.
Automated determination of isoptics
with dynamic geometry
Thierry Dana-Picard1Zolt´an Kov´acs2
1Jerusalem College of Technology
2The Private University College of Education of the Diocese of Linz
CICM Hagenberg, Calculemus
August 15, 2018
Abstract
We present two approaches to symbolically obtain isoptic curves
in GeoGebra in an automated, interactive process. Both methods
are based on computing implicit locus equations, by using
algebraization of the geometric setup and elimination of the
intermediate variables. These methods can be considered as
automatic discovery.
Our first approach uses pure computer algebra support of
GeoGebra, utilizing symbolic differentiation.
The second approach hides all details in computer algebra from the
user: the input problem is defined by a purely geometric way.
In both approaches the output is dynamically changed when using
a slider bar or the free points are dragged.
Programming the internal GeoGebra computations is an on-going
work with various challenges in optimizing computations and
to avoiding unnecessary extra curves in the output.
Isoptic curves
Let Cbe a plane curve. For a given angle θsuch that
0θ180, a θ-isoptic curve (or simply a θ-isoptic) of Cis the
geometric locus of points Mthrough which passes a pair of
tangents with an angle of θbetween them.
If θ= 90, i.e. if the tangents are perpendicular, then the isoptic
curve is called an orthoptic curve.
Isoptic curves may either exist or not, depending on the given
curve and on the angle.
Orthoptics of conics
Parabola
The orthoptic curve of
a parabola is its
directrix.
If the parabola has
equation y2= 2px
(for pa non-zero real),
then its directrix has
equation x=p/2.
https://www.geogebra.org/m/pwrWy9dG
Orthoptics of conics
Ellipse
The orthoptic curve of
an ellipse is its director
circle.
If the ellipse is given by
the canonical equation
x2
a2+y2
b2= 1, then the
director circle has the
equation
x2+y2=a2+b2.
https://www.geogebra.org/m/SkQ5qxYr
Orthoptics of conics
Hyperbola
The existence of an
orthoptic curve for a
hyperbola depends on the
eccentricity c/a, where
c2=a2b2.
If it exists, the orthoptic
curve of the hyperbola
with canonical equation
x2
a2y2
b2= 1 (i.e. the focal
axis is the x=axis) is the
circle whose equation is
x2+y2=a2b2, also
called the director circle.https://www.geogebra.org/m/tZcGGrCm
Previous and related work
IDana-Picard, Th., Mann, G. and Zehavi, N.: From conic
intersections to toric intersections: the case of the isoptic curves of
an ellipse, The Montana Mathematical Enthusiast 9 (1), pp. 59–76.
2011.
IDana-Picard, Th.: An automated study of isoptic curves of an
astroid, Preprint, JCT, 2018.
IDana-Picard, Th. and Naiman, A.: Isoptics of Fermat curves,
Preprint, JCT, 2018.
IMiernowski, A. and Mosgawa, W.: Isoptics of Pairs of Nested
Closed Strictly Convex Curves and Crofton-Type Formulas, Beitr¨age
zur Algebra und Geometrie Contributions to Algebra and Geometry
42 (1), pp. 281–288. 2001.
ISza lkowski, D.: Isoptics of open rosettes, Annales Universitatis
Mariae Curie-Sk lodowska, Lublin Polonia LIX, Section A,
pp. 119–128, 2005.
ICsima, G.: Isoptic curves and surfaces . PhD thesis, BUTE,
Math. Institute, Department of Geometry, Budapest, 2017.
Examples of previous work
The orthoptic of a closed Fermat curve, x16 +y16 = 1
Examples of previous work
45-isoptic of an astroid, x2/3+y2/3= 1
Examples of previous work
135-isoptic of an astroid, x2/3+y2/3= 1
Two novel approaches in GeoGebra
An overview
IBoth
Ican be considered as automatic discovery,
Ideliver an algebraic output: a polynomial (with its graphical
representation) via Gr¨obner bases and elimination.
IThe first approach
Iuses pure computer algebra support of GeoGebra:
symbolic differentiation of the input formula,
Iallows the output to be changed dynamically with a slider bar
(dynamic study),
Ican do observations up to quartic curves
(due to computational challenges).
IThe second approach
Ihides all details in computer algebra from the user:
the input problem is given in a a purely geometric way,
Iis a handy method for a new kind of man and machine
communication,
Iworks only for certain conics.
Two novel approaches in GeoGebra
An overview
IBoth
Ican be considered as automatic discovery,
Ideliver an algebraic output: a polynomial (with its graphical
representation) via Gr¨obner bases and elimination.
IThe first approach
Iuses pure computer algebra support of GeoGebra:
symbolic differentiation of the input formula,
Iallows the output to be changed dynamically with a slider bar
(dynamic study),
Ican do observations up to quartic curves
(due to computational challenges).
IThe second approach
Ihides all details in computer algebra from the user:
the input problem is given in a a purely geometric way,
Iis a handy method for a new kind of man and machine
communication,
Iworks only for certain conics.
Two novel approaches in GeoGebra
An overview
IBoth
Ican be considered as automatic discovery,
Ideliver an algebraic output: a polynomial (with its graphical
representation) via Gr¨obner bases and elimination.
IThe first approach
Iuses pure computer algebra support of GeoGebra:
symbolic differentiation of the input formula,
Iallows the output to be changed dynamically with a slider bar
(dynamic study),
Ican do observations up to quartic curves
(due to computational challenges).
IThe second approach
Ihides all details in computer algebra from the user:
the input problem is given in a a purely geometric way,
Iis a handy method for a new kind of man and machine
communication,
Iworks only for certain conics.
The first approach
Let Cbe an algebraic curve given by an implicit equation
F(x,y) = 0.
1. Compute the derivatives dx=F0
xand dy=F0
y.
2. Consider points A(xA,yA) and B(xB,yB) that are assumed to
be points of the curve, that is,
F(xA,yA) = 0 (1)
and
F(xB,yB) = 0 (2)
hold.
3. Compute the partial derivatives px,A=F0
x(xA,yA),
px,B=F0
x(xB,yB), py,A=F0
y(xA,yA) and py,B=F0
y(xB,yB).
The first approach
Let Cbe an algebraic curve given by an implicit equation
F(x,y) = 0.
1. Compute the derivatives dx=F0
xand dy=F0
y.
2. Consider points A(xA,yA) and B(xB,yB) that are assumed to
be points of the curve, that is,
F(xA,yA) = 0 (1)
and
F(xB,yB) = 0 (2)
hold.
3. Compute the partial derivatives px,A=F0
x(xA,yA),
px,B=F0
x(xB,yB), py,A=F0
y(xA,yA) and py,B=F0
y(xB,yB).
The first approach
Let Cbe an algebraic curve given by an implicit equation
F(x,y) = 0.
1. Compute the derivatives dx=F0
xand dy=F0
y.
2. Consider points A(xA,yA) and B(xB,yB) that are assumed to
be points of the curve, that is,
F(xA,yA) = 0 (1)
and
F(xB,yB) = 0 (2)
hold.
3. Compute the partial derivatives px,A=F0
x(xA,yA),
px,B=F0
x(xB,yB), py,A=F0
y(xA,yA) and py,B=F0
y(xB,yB).
The first approach
Let Cbe an algebraic curve given by an implicit equation
F(x,y) = 0.
1. Compute the derivatives dx=F0
xand dy=F0
y.
2. Consider points A(xA,yA) and B(xB,yB) that are assumed to
be points of the curve,
that is,
F(xA,yA) = 0 (1)
and
F(xB,yB) = 0 (2)
hold.
3. Compute the partial derivatives px,A=F0
x(xA,yA),
px,B=F0
x(xB,yB), py,A=F0
y(xA,yA) and py,B=F0
y(xB,yB).
The first approach
Let Cbe an algebraic curve given by an implicit equation
F(x,y) = 0.
1. Compute the derivatives dx=F0
xand dy=F0
y.
2. Consider points A(xA,yA) and B(xB,yB) that are assumed to
be points of the curve, that is,
F(xA,yA) = 0 (1)
and
F(xB,yB) = 0 (2)
hold.
3. Compute the partial derivatives px,A=F0
x(xA,yA),
px,B=F0
x(xB,yB), py,A=F0
y(xA,yA) and py,B=F0
y(xB,yB).
The first approach
Let Cbe an algebraic curve given by an implicit equation
F(x,y) = 0.
1. Compute the derivatives dx=F0
xand dy=F0
y.
2. Consider points A(xA,yA) and B(xB,yB) that are assumed to
be points of the curve, that is,
F(xA,yA) = 0 (1)
and
F(xB,yB) = 0 (2)
hold.
3. Compute the partial derivatives px,A=F0
x(xA,yA),
px,B=F0
x(xB,yB), py,A=F0
y(xA,yA) and py,B=F0
y(xB,yB).
The first approach
(cont’d)
4. Now, when speaking about orthoptic curves, we can assume
that
px,A·px,B+py,A·py,B= 0,(3)
otherwise, when speaking about θ-isoptics, the following
equation holds:
(px,A·px,B+py,A·py,B)2= cos2θ·(p2
x,A+p2
y,A)·(p2
x,B+p2
y,B).(3’)
The first approach
(cont’’d)
5. When defining a point P(x,y) that is an element of both
tangents t1and t2to c, the points
A,A0= (xA+py,A,yApx,A) and Pmust be collinear;
for the same reason, also
B,B0= (xB+py,B,yBpx,B) and Pare collinear.
So the following equations hold:
xAyA1
xA+py,AyApx,A1
x y 1
= 0,(4)
xByB1
xB+py,ByBpx,B1
x y 1
= 0.(5)
The first approach
(cont’’d)
5. When defining a point P(x,y) that is an element of both
tangents t1and t2to c, the points
A,A0= (xA+py,A,yApx,A) and Pmust be collinear;
for the same reason, also
B,B0= (xB+py,B,yBpx,B) and Pare collinear.
So the following equations hold:
xAyA1
xA+py,AyApx,A1
x y 1
= 0,(4)
xByB1
xB+py,ByBpx,B1
x y 1
= 0.(5)
The first approach
(cont’’’d)
6. Now we have 5 equations.
By eliminating all variables
but xand ywe obtain an implicit equation whose graphical
representation is, at least partly, the θ-isoptic curve.
This technique (“elimination theory”, “automated geometry
theorem proving”, “automated discovery”) is discussed
in detail in:
ICox, D., Little, J. and O’Shea, D.: Ideals, varieties and
algorithms. Third edition. Springer, 2007.
IChou, S.-C.: Mechanical Geometry Theorem Proving,
Reidel Dordrecht, 1987.
IAb´anades, M. A., Botana, F., Kov´acs, Z., Recio, T. and
olyom-Gecse, C.: Development of automatic reasoning tools
in GeoGebra. Software Demonstration at the ISSAC 2016
Conf. ACM Comm. in Comp. Alg. 50 (3), pp. 85–88. 2016.
Theoretically, the obtained implicit equation is a multiple of
the algebraic closure of the geometrically expected set.
That is, some factors of the obtained implicit equation will
contain the expected curve.
The first approach
(cont’’’d)
6. Now we have 5 equations. By eliminating all variables
but xand ywe obtain an implicit equation whose graphical
representation is, at least partly, the θ-isoptic curve.
This technique (“elimination theory”, “automated geometry
theorem proving”, “automated discovery”) is discussed
in detail in:
ICox, D., Little, J. and O’Shea, D.: Ideals, varieties and
algorithms. Third edition. Springer, 2007.
IChou, S.-C.: Mechanical Geometry Theorem Proving,
Reidel Dordrecht, 1987.
IAb´anades, M. A., Botana, F., Kov´acs, Z., Recio, T. and
olyom-Gecse, C.: Development of automatic reasoning tools
in GeoGebra. Software Demonstration at the ISSAC 2016
Conf. ACM Comm. in Comp. Alg. 50 (3), pp. 85–88. 2016.
Theoretically, the obtained implicit equation is a multiple of
the algebraic closure of the geometrically expected set.
That is, some factors of the obtained implicit equation will
contain the expected curve.
The first approach
(cont’’’d)
6. Now we have 5 equations. By eliminating all variables
but xand ywe obtain an implicit equation whose graphical
representation is, at least partly, the θ-isoptic curve.
This technique (“elimination theory”, “automated geometry
theorem proving”, “automated discovery”) is discussed
in detail in:
ICox, D., Little, J. and O’Shea, D.: Ideals, varieties and
algorithms. Third edition. Springer, 2007.
IChou, S.-C.: Mechanical Geometry Theorem Proving,
Reidel Dordrecht, 1987.
IAb´anades, M. A., Botana, F., Kov´acs, Z., Recio, T. and
olyom-Gecse, C.: Development of automatic reasoning tools
in GeoGebra. Software Demonstration at the ISSAC 2016
Conf. ACM Comm. in Comp. Alg. 50 (3), pp. 85–88. 2016.
Theoretically, the obtained implicit equation is a multiple of
the algebraic closure of the geometrically expected set.
That is, some factors of the obtained implicit equation will
contain the expected curve.
The first approach
(cont’’’d)
6. Now we have 5 equations. By eliminating all variables
but xand ywe obtain an implicit equation whose graphical
representation is, at least partly, the θ-isoptic curve.
This technique (“elimination theory”, “automated geometry
theorem proving”, “automated discovery”) is discussed
in detail in:
ICox, D., Little, J. and O’Shea, D.: Ideals, varieties and
algorithms. Third edition. Springer, 2007.
IChou, S.-C.: Mechanical Geometry Theorem Proving,
Reidel Dordrecht, 1987.
IAb´anades, M. A., Botana, F., Kov´acs, Z., Recio, T. and
olyom-Gecse, C.: Development of automatic reasoning tools
in GeoGebra. Software Demonstration at the ISSAC 2016
Conf. ACM Comm. in Comp. Alg. 50 (3), pp. 85–88. 2016.
Theoretically, the obtained implicit equation is a multiple of
the algebraic closure of the geometrically expected set.
That is, some factors of the obtained implicit equation will
contain the expected curve.
Examples
The orthoptic of y=x4
The equations to consider are as follows:
x4
AyA= 0,(1)
x4
ByB= 0,(2)
4x3
A·4x3
B+ 1 = 0,(3)
4x4
A+ 4x3
Ax+yAy= 0,(4)
4x4
B+ 4x3
Bx+yBy= 0.(5)
Examples
The orthoptic of y=x4
The equations to consider are as follows:
x4
AyA= 0,(1)
x4
ByB= 0,(2)
4x3
A·4x3
B+ 1 = 0,(3)
4x4
A+ 4x3
Ax+yAy= 0,(4)
4x4
B+ 4x3
Bx+yBy= 0.(5)
Examples
The orthoptic of y=x4(cont’d)
After eliminating all variables but xand yfrom this system by
using a CAS, we obtain the equation
65536x6+ 196608x4y2+ 196608x2y4
41472x2y+ 65536y6+ 13824y3+ 729·
(16777216x6y3+ 50331648x4y5+ 5308416x4y2+ 50331648x2y7+
5308416x2y4+ 559872x2y+ 16777216y9
1769472y6
186624y3+ 19683) = 0.
This can be written as f1·f2= 0. Both f1and f2are reducible
over C. After numerical and visual experiments, it turns out that f1
has no real geometrical meaning, but f2has. Also, f2has a divisor
f=x2y+y33/8y23
29y3
4
64 +27
256.
According to GeoGebra’s numerical precision the cubic f= 0 is
indeed the orthoptic of y=x4.
Examples
The orthoptic of y=x4(cont’d)
After eliminating all variables but xand yfrom this system by
using a CAS, we obtain the equation
65536x6+ 196608x4y2+ 196608x2y4
41472x2y+ 65536y6+ 13824y3+ 729·
(16777216x6y3+ 50331648x4y5+ 5308416x4y2+ 50331648x2y7+
5308416x2y4+ 559872x2y+ 16777216y9
1769472y6
186624y3+ 19683) = 0.
This can be written as f1·f2= 0.
Both f1and f2are reducible
over C. After numerical and visual experiments, it turns out that f1
has no real geometrical meaning, but f2has. Also, f2has a divisor
f=x2y+y33/8y23
29y3
4
64 +27
256.
According to GeoGebra’s numerical precision the cubic f= 0 is
indeed the orthoptic of y=x4.
Examples
The orthoptic of y=x4(cont’d)
After eliminating all variables but xand yfrom this system by
using a CAS, we obtain the equation
65536x6+ 196608x4y2+ 196608x2y4
41472x2y+ 65536y6+ 13824y3+ 729·
(16777216x6y3+ 50331648x4y5+ 5308416x4y2+ 50331648x2y7+
5308416x2y4+ 559872x2y+ 16777216y9
1769472y6
186624y3+ 19683) = 0.
This can be written as f1·f2= 0. Both f1and f2are reducible
over C.
After numerical and visual experiments, it turns out that f1
has no real geometrical meaning, but f2has. Also, f2has a divisor
f=x2y+y33/8y23
29y3
4
64 +27
256.
According to GeoGebra’s numerical precision the cubic f= 0 is
indeed the orthoptic of y=x4.
Examples
The orthoptic of y=x4(cont’d)
After eliminating all variables but xand yfrom this system by
using a CAS, we obtain the equation
65536x6+ 196608x4y2+ 196608x2y4
41472x2y+ 65536y6+ 13824y3+ 729·
(16777216x6y3+ 50331648x4y5+ 5308416x4y2+ 50331648x2y7+
5308416x2y4+ 559872x2y+ 16777216y9
1769472y6
186624y3+ 19683) = 0.
This can be written as f1·f2= 0. Both f1and f2are reducible
over C. After numerical and visual experiments, it turns out that f1
has no real geometrical meaning, but f2has.
Also, f2has a divisor
f=x2y+y33/8y23
29y3
4
64 +27
256.
According to GeoGebra’s numerical precision the cubic f= 0 is
indeed the orthoptic of y=x4.
Examples
The orthoptic of y=x4(cont’d)
After eliminating all variables but xand yfrom this system by
using a CAS, we obtain the equation
65536x6+ 196608x4y2+ 196608x2y4
41472x2y+ 65536y6+ 13824y3+ 729·
(16777216x6y3+ 50331648x4y5+ 5308416x4y2+ 50331648x2y7+
5308416x2y4+ 559872x2y+ 16777216y9
1769472y6
186624y3+ 19683) = 0.
This can be written as f1·f2= 0. Both f1and f2are reducible
over C. After numerical and visual experiments, it turns out that f1
has no real geometrical meaning, but f2has. Also, f2has a divisor
f=x2y+y33/8y23
29y3
4
64 +27
256.
According to GeoGebra’s numerical precision the cubic f= 0 is
indeed the orthoptic of y=x4.
Examples
The orthoptic of y=x4(cont’d)
After eliminating all variables but xand yfrom this system by
using a CAS, we obtain the equation
65536x6+ 196608x4y2+ 196608x2y4
41472x2y+ 65536y6+ 13824y3+ 729·
(16777216x6y3+ 50331648x4y5+ 5308416x4y2+ 50331648x2y7+
5308416x2y4+ 559872x2y+ 16777216y9
1769472y6
186624y3+ 19683) = 0.
This can be written as f1·f2= 0. Both f1and f2are reducible
over C. After numerical and visual experiments, it turns out that f1
has no real geometrical meaning, but f2has. Also, f2has a divisor
f=x2y+y33/8y23
29y3
4
64 +27
256.
According to GeoGebra’s numerical precision the cubic f= 0 is
indeed the orthoptic of y=x4.
Examples
The orthoptic of y=x4(cont’’d)
https://www.geogebra.org/m/JvhNwAzF
Examples
The orthoptic of y=x4x
https://www.geogebra.org/m/mfrwfGNc
Examples
35-isoptic of a hyperbola
Examples
35-isoptic of a hyperbola (cont’d)
Algebraically, after elimination, GeoGebra obtains
2x14 2y14 c2x12 c2y12 10x2y12 18x4y10 10x6y8+ 10x8y6+ 18x10y4
+ 10x12y26c2x2y10 15c2x4y820c2x6y615c2x8y46c2x10 y223x12
23y12 + 12c2x10 12c2y10 58x2y10 25x4y8+ 20x6y625x8y458x10y2
36c2x2y824c2x4y6+ 24c2x6y4+ 36c2x8y2+ 112x10 112y10 60c2x8
60c2y880x2y8+ 32x4y632x6y4+ 80x8y248c2x2y6+ 24c2x4y448c2x6y2
300x8300y8+ 160c2x6160c2y6+ 144x2y6136x4y4+ 144x6y2+ 96c2x2y4
96c2x4y2+ 480x6480y6240c2x4240c2y4+ 544x2y4544x4y2+ 288c2x2y2
464x4464y4+ 192c2x2192c2y2+ 608x2y264c2+ 256x2256y264 = 0,
where c= cos27
36 π.
After factorization this can be simplified to
cx4+ 2cx2y2+cy 4x42x2y24cx 2y4+ 4cy 2+ 4c= 0,
that is, the isoptic curve is a quartic (containing also the set of
points for the 145-isoptic).
Examples
35-isoptic of a hyperbola (cont’d)
Algebraically, after elimination, GeoGebra obtains
2x14 2y14 c2x12 c2y12 10x2y12 18x4y10 10x6y8+ 10x8y6+ 18x10y4
+ 10x12y26c2x2y10 15c2x4y820c2x6y615c2x8y46c2x10 y223x12
23y12 + 12c2x10 12c2y10 58x2y10 25x4y8+ 20x6y625x8y458x10y2
36c2x2y824c2x4y6+ 24c2x6y4+ 36c2x8y2+ 112x10 112y10 60c2x8
60c2y880x2y8+ 32x4y632x6y4+ 80x8y248c2x2y6+ 24c2x4y448c2x6y2
300x8300y8+ 160c2x6160c2y6+ 144x2y6136x4y4+ 144x6y2+ 96c2x2y4
96c2x4y2+ 480x6480y6240c2x4240c2y4+ 544x2y4544x4y2+ 288c2x2y2
464x4464y4+ 192c2x2192c2y2+ 608x2y264c2+ 256x2256y264 = 0,
where c= cos27
36 π. After factorization this can be simplified to
cx4+ 2cx2y2+cy 4x42x2y24cx 2y4+ 4cy 2+ 4c= 0,
that is, the isoptic curve is a quartic (containing also the set of
points for the 145-isoptic).
Computational features of the first approach
IFast computations for conics (dragging of θis possible)
IFeasible (but slow) computations for certain quartics
IInfeasible computations for most quartics
and other higher degree polynomials
IGeoGebra’s CAS View is involved
IIn most cases, the output contains additional factors that
have no geometrical meaning (“extended output”)
IGeoGebra’s Graphics View correctly plots the extended output
IFactorization of the extended output may be incomplete
in GeoGebra (Maple or Singular can be used for absolute
factorization): the minimal algebraic form of the curve is
difficult to determine
The second approach
A “dynamic geometry” approach
Let Aand Bbe arbitrary points in the plane.
(Aand Bare called free points).
Where to put point Pin the plane
to have the lengths AP and BP equal?
https://www.geogebra.org/classic
LocusEquation(f== g,P)
The second approach
A “dynamic geometry” approach
Let Aand Bbe arbitrary points in the plane.
(Aand Bare called free points).
Where to put point Pin the plane
to have the lengths AP and BP equal?
https://www.geogebra.org/classic
LocusEquation(f== g,P)
The second approach
A “dynamic geometry” approach
Let Aand Bbe arbitrary points in the plane.
(Aand Bare called free points).
Where to put point Pin the plane
to have the lengths AP and BP equal?
https://www.geogebra.org/classic
LocusEquation(f== g,P)
The second approach
A “dynamic geometry” approach
Let Aand Bbe arbitrary points in the plane.
(Aand Bare called free points).
Where to put point Pin the plane
to have the lengths AP and BP equal?
https://www.geogebra.org/classic
LocusEquation(f== g,P)
The second approach
A “dynamic geometry” approach
Let Aand Bbe arbitrary points in the plane.
(Aand Bare called free points).
Where to put point Pin the plane
to have the lengths AP and BP equal?
https://www.geogebra.org/classic
LocusEquation(f== g,P)
The second approach
A “dynamic geometry” approach
Let Aand Bbe arbitrary points in the plane.
(Aand Bare called free points).
Where to put point Pin the plane
to have the lengths AP and BP equal?
https://www.geogebra.org/classic
LocusEquation(f== g,P)
The second approach
Orthoptic of a circle
https://www.geogebra.org/m/z2uNpHCU
LocusEquation(fg,P)
Some features of the second approach
IGeoGebra’s CAS View is not involved
IEach type of input (circle, parabola, . . .) must be separately
implemented (=programmed) internally in GeoGebra
IComputations are feasible for orthoptics of circle and parabola
(moderately slow dragging of θis possible)
ITo obtain isoptics, the AreCongruent command must be used
IComputations are slow for isoptics of circle and parabola
IIsoptic curves may contain extra linear components
due to algebraic issues
IOther curves (ellipse, hyperbola and non-conics)
are not yet implemented
IThe output may contain additional factors that have no
geometrical meaning (“extended output”)
IFinding the “best” equation system describing the geometric
setup can be tricky
Creating the equation system programmatically
GeoGebra’s source code is at https://github.com/geogebra/geogebra
Examples
Orthoptic of a parabola
https://www.geogebra.org/m/dtgzjzcj
Examples
45-isoptic of the circle
LocusEquation(AreCongruent(α,β),P)
Examples
135-isoptic of the parabola
LocusEquation(AreCongruent(α,β),P)
Conclusion
INo longer a researchers-only topic? Students can be involved!
IAnother application of Gr¨obner bases and elimination
(for polynomial input)
IExperiments exploiting (computer) algebraic and
(dynamic geometric) graphical representations
IFurther studies may involve more efficient computations
and further tricks
Thank you for your kind attention!
Thank you for your kind attention!
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Chapter
For a long time, the traditional way to convey mathematics was structured according to the definition-theorem-proof-example scheme. A few decades ago, digital tools appeared: calculators working numerically, then graphical features (also numerically based) were added. With the developments of algorithms for symbolic computations, things changed profoundly. Computer Algebra Systems (CAS) are now important systems involving symbolic and numerical features, together with efficient graphical abilities. The affordances of symbolic computations had a great influence on the proof of new theoretical results.
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