Образователни технологии
COMPUTER DISCOVERED MATHEMATICS: ORTHOLOGYCENTERS OFTHE EULERTRIANGLES
Резюме. The paper studies the orthologic triangles of the Euler triangles and their orthology centers. The results are discovered by the computer program “Discoverer”.
Ключови думи: Euler triangles, orthologic triangles, orthology center, computer-generated mathematics, Discoverer.
1. Introduction
In the recent years many theorems are discovered by experimenting with dynamical geometry software. See e.g. (Baralić et al., 2014). But the possibilities of this approach are limited. The next step is the use of computers-discoverers. In this paper we present the computer-program “Discoverer”, created by the authors. The computer program “Discoverer” is the first computer program, able easily to discover new theorems in mathematics, and possibly, the first computer program, able easily to discover new knowledge in science. The computer program discovers theorems in Euclidean geometry. The “Discoverer” was able to discover an essential improvement of the classical Steiner’s solution of the construction of the Malfatti circles (see (Grozdev & Dekov, 2015)), as well as a number of theorems in Euclidean geometry see e.g. (Grozdev & Dekov, 2014a,b).
The orthologic triangles are studied since 1827 when Jacob Steiner discovered some basic facts about them. In the same year August Ferdinand Möbius discovered the barycentric coordinates. In this paper we present some results discovered by the “Discoverer” and related to the orthologic triangles and orthology centers. All theorems and examples, as well as their proofs, are produced by the computer program “Discoverer”.
2. Preliminaries
In this section we review some basic facts about barycentric coordinates. We refer the reader to (Grozdev & Nenkov, 2012a,b), (Stanilov, 1979), (Yiu, 2013).
Given a triangle \(A B C\) with side lengths \(B C=a, C A=b\) a and \(A B=c\). The barycentric coordinates with respect to \(\triangle A B C\) are used to define points, lines, circles, triangles, etc. The reference triangle \(A B C\) has vertices \(A=(1,0,0)\), \(B=(0,1,0)\) and \(C=(0,0,1)\). A point is an element of \(\mathbb{R}^{3}\), defined up to a proportionality factor, that is:
For all \(k \in \mathbb{R}-\{0\}: P=(u, v, w)\) means that \(P \cong(u, v, w) \cong(k u, k v, k w)\).
We use the Conway triangle notation:
\[ S_{A}=\tfrac{1}{2}\left(b^{2}+c^{2}-a^{2}\right), S_{B}=\tfrac{1}{2}\left(c^{2}+a^{2}-b^{2}\right), S_{C}=\tfrac{1}{2}\left(a^{2}+b^{2}-c^{2}\right), \] \(S=2 \Delta\),
where \(\Delta\) is the area of the reference triangle.
The equation of the line joining two points with coordinates \(\left(u_{1}, v_{1}, w_{1}\right)\) and \(\left(u_{2}, v_{2}, w_{2}\right)\) is
(1) \[ \left|\begin{array}{ccc} u_{1} & v_{1} & w_{1} \\ u_{2} & v_{2} & w_{2} \\ x & y & z \end{array}\right|=0 . \]
The infinite point of the line \(L: p x+q y+r z=0\) is the point \((f, g, h)\) \(=(q-r, r-p, p-q)\). Denote \(F=S_{B} g-S_{C} h, \quad G=S_{C} h-S_{A} f \quad\) and \(H=S_{A} f-S_{B} g\). Then the line through the point \(P=(u, v, w)\) and perpendicular to the line \(L: p x+q y+r z=0\) has equation:
(2) \[ \left|\begin{array}{lll} F & G & H \\ u & v & w \\ x & y & z \end{array}\right|=0 \]
Three lines \(p_{i} x+q_{i} y+r_{i} z=0, i=1,2,3\) are concurrent iff
(3) \[ \left|\begin{array}{lll} p_{1} & q_{1} & r_{1} \\ p_{2} & q_{2} & r_{2} \\ p_{3} & q_{3} & r_{3} \end{array}\right|=0 \]
The intersection of the two lines \(L_{1}: p_{1} x+q_{1} y+r_{1} z=0\) and \(L_{2}:\) \(p_{2} x+q_{2} y+r_{2} z=0\) is the point
(4) \[ \left(q_{1} r_{2}-q_{2} r_{1}, r_{1} p_{2}-r_{2} p_{1}, p_{1} q_{2}-p_{2} q_{1}\right) \]
If the first barycentric coordinate of a triangle center \(R=(u R, v R, w R)\) is \(u R=f(a, b, c, u, v, w)\), then \(v R=f(b, c, a, v, w, u)\) and \(w R=f(c, a, b, w, u, v)\). Hence, in order to define the triangle center \(R\) it is enough to define the first barycentric coordinate \(u R\).
Given \(\triangle A B C\) and an arbitrary point \(P\), the Euler triangle of \(P\) is the triangle whose vertices \(E_{1}, E_{2}\), and \(E_{3}\) are the midpoints of the segments \(A P, B P\) and \(C P\), respectively. The Euler triangle of the orthocenter is the ordinary Euler Triangle (See (Weisstein \({ }^{3)}\), Euler Triangle). The vertices of the Euler triangle \(E_{1} E_{2} E_{3}\) of point \(P=(u, v, w)\) have barycentric coordinates, See (Grozdev \& Dekov, 2014b):
\(E_{1}=(2 u+v+w, v, w), E_{2}=(u, u+2 v+w, w)\) and \(E_{3}=(u, v, u+v+2 w)\).
The \(\Delta A_{1} B_{1} C_{1}\) is orthologic wrt \(\Delta A_{2} B_{2} C_{2}\) if the perpendiculars from the vertices \(A_{1}\), \(B_{1}, C_{1}\) to the sides \(B_{2} C_{2}, C_{2} A_{2}, A_{2} B_{2}\) are concurrent. If this is the case, the point of intersection of the perpendiculars is known to be the orthology center of \(\Delta A_{1} B_{1} C_{1}\) wrt \(\Delta A_{2} B_{2} C_{2}\). In 1827 Steiner showed that if \(\Delta A_{1} B_{1} C_{1}\) is othologic wrt \(\Delta A_{2} B_{2} C_{2}\), then the converse is true, that is, \(\Delta A_{2} B_{2} C_{2}\) is orthologic wrt \(\Delta A_{1} B_{1} C_{1}\). For examples of orthologic triangles see e.g. (Danneels \& Dergiades, 2004), (Gibert¹), Table 7).
In this paper we denote by \(E(T, P)\) the orthology center of the Euler Triangle of point \(P\) wrt triangle \(T\), and by \(C(T, P)\) the orthology center of triangle \(T\) wrt the Euler Triangle of point \(P\).
The labeling of triangle centers follows ETC (Kimberling \({ }^{2)}\) ). Hence, X(1) denotes the Incenter, X(2) denotes the Centroid, etc. For the definition of various triangles and the barycentric coordinates of their vertices we refer the reader to (Weisstein \({ }^{3}\) ). Note that the computer program "Discoverer" uses the term "Inner (Outer) Yff triangle" instead of "First (Second) Yff circles triangle".
3. Orthology Centers of the Euler Triangles wrt Triangle \(\boldsymbol{A} \boldsymbol{B} \boldsymbol{C}\)
Theorem 3.1. For any point \(P=(u, v, w)\), v, w) , the Euler triangle of \(P\) is orthologic wrt \(\triangle A B C\). The first barycentric coordinate of the orthology center \(R=(u R, v R, w R)\) is
\[ \begin{aligned} u R & =v a^{4}+w a^{4}-v b^{4}-w b^{4}-v c^{4}-w c^{4}-2 u b^{4}-2 u c^{4} \\ & +4 u c^{2} b^{2}+2 u a^{2} c^{2}+2 u a^{2} b^{2}+2 v b^{2} c^{2}+2 w b^{2} c^{2} \end{aligned} \]
Proof. The equation of the line \(B C\) is \(x=0\). By using (1) we easily obtain this equation. The infinite point of the line \(B C\) is \((0,-1,1)\), -1, 1) , so that the infinite point of any line, perpendicular to \(B C\), is \(\left(2 a^{2}, c^{2}-a^{2}-b^{2}, b^{2}-c^{2}-a^{2}\right)\). Next we calculate the equation of the line \(L_{1}\) through the point \(E_{1}=(2 u+v+w, v, w)\) and perpendicular to the line \(B A\). By using (2), we obtain \(L_{1}: p_{1} x+q_{1} y+r_{1} z=0\), where
\[ \begin{aligned} & p_{1}=w a^{2}-v a^{2}+v b^{2}+w b^{2}-v c^{2}-w c^{2} \\ & q_{1}=3 w a^{2}-2 u b^{2}+2 u c^{2}+2 u a^{2}-v b^{2}+v c^{2}+v a^{2}-w b^{2}+w c^{2} \\ & r_{1}=-3 v a^{2}-2 u a^{2}+2 u c^{2}-2 u b^{2}-v b^{2}+v c^{2}-w a^{2}-w b^{2}+w c^{3} \end{aligned} \]
The equations of the line \(L_{2}\) through the point \(E_{2}\) and perpendicular to the line \(C A\), is \(L_{2}: p_{2} x+q_{2} y+r_{2} z=0\), where
\[ \begin{aligned} & p=-3 w b^{2}+2 v a^{2}-2 v b^{2}-2 v c^{2}+u a^{2}+w a^{2}-u b^{2}-u c^{2}-w c^{2} \\ & q_{2}=-w a^{2}-u a^{2}+u b^{2}-w b^{2}+u c^{2}+w c^{2} \\ & r_{2}=3 u b^{2}-2 v c^{2}+2 v b^{2}+2 v a^{2}+u a^{2}-u c^{2}-w c^{2}+w a^{2}+w b^{2} \end{aligned} \]
The equations of the line \(L_{3}\) through the point \(E_{3}\) and perpendicular to the line \(A B\), is \(L_{3}: p_{3} x+q_{3} y+r_{3} z=0\), where
\[ \begin{aligned} & p_{3}=3 v c^{2}-2 w a^{2}+2 w b^{2}+2 w c^{2}-u a^{2}-v a^{2}+u b^{2}+v b^{2}+u c^{2} \\ & q_{3}=-3 u c^{2}-2 w a^{2}+2 w b^{2}-2 w c^{2}+u b^{2}-u a^{2}+v b^{2}-v c^{2}-v a^{2} \\ & r_{3}=-u c^{2}+v a^{2}+u a^{2}-u b^{2}+v c^{2}-v b^{2} \end{aligned} \]
We use (3) in order to prove that the three lines \(L_{1}, L_{2}\) and \(L_{3}\) concur in a point. Finally, we use (4) to find the barycentric coordinates of the point of concurrence \(R\), that is, the orthology center, as the point of concurrence of the lines \(L_{1}\) and \(L_{2}\). The barycentric coordinates of the point \(R\) are given in the statement of the theorem.
Theorem 3.1 defines a transform \(E\) (Triangle \(A B C, P\) ) \(=R\) of the plane of the Triangle \(A B C\). If \(T\) is any triangle homothetic with the triangle \(A B C\), then for any point \(P, E(T, P)=E\) (Triangle \(A B C, P\) ). We remind that the following triangles are homothetic with the triangle \(A B C\) : Medial triangle, Antimedial triangle, Johnson triangle, Inner Yff triangle, Outer Yff triangle, Half-Median triangle, etc. Hence, \(E\) (Triangle \(A B C, P)=E(\) Medial triangle, \(P)\), etc. Note that if we want directly to calculate e.g. the \(E\) (Inner Yff triangle, \(P\) ), we have to perform much complicated calculation, but finally we will obtain \(E(\) Inner Yff triangle, \(P)=E(\) Triangle \(A B C, P)\).
In Table 1 below \(T\) is the \(\triangle A B C\) or any triangle homothetic with \(\triangle A B C\).
Table 1
Remark 3.2. For any point \(P\) :
(1) \(C(\) Triangle \(A B C, P)=\mathrm{X}(4)\) Orthocenter;
(2) \(C(\) Medial triangle, \(P)=\mathrm{X}(3)\) C(Medial triangle, P) = X(3) Circumcenter;
(3) \(C\) (Antimedial triangle, \(P\) ) \(=\mathrm{X}\) (20) de Longchamps P) = X(20) de Longchamps Point;
(4) \(C(\) Johnson triangle, \(P)=\mathrm{X}(3)\) C(Johnson triangle, P) = X(3) Circumcenter;
(5) \(C\) (Inner Yff triangle, P ) \(=\mathrm{X}(1478)\) Center of the Inner Johnson-Yff Circle;
(6) \(C\) (Outer Yff triangle, \(P\) ) \(=\mathrm{X}(1479)\) Center of the Outer Johnson-Yff Circle.
4. Orthology Centers of the Euler Triangles wrt the Excentral Triangle
Theorem 4.1. For any point \(P=(u, v, w)\), , the Euler triangle of \(P\) is orthologic wrt the Excentral triangle. The first barycentric coordinate of the orthology center \(R=(u R, v R, w R)\) is \(u R=(2 a+b+c) u+a v+a w\).
Proof. We use the algorithm of the proof of theorem 3.1. The vertices of the Excentral triangle \(J_{a} J_{b} L_{c}\) have barycentric coordinates \(J_{a}=(-a, b, c)\), \(J_{b}=(a,-b, c)\) and \(J_{c}=(a, b,-c)\). The equation of the line \(J_{b} J_{c}\) is \(c y+b z=0\), the infinite point of the line \(J_{b} J_{c}\) is ( \(c-b, b,-c\) ), so that the infinite point of any line, perpendicular to \(J_{b} J_{c}\) is \((b+c,-b,-c)\). The equation of the line \(L_{1}\) through the point \(E_{1}-(2 u \quad v \quad w, v, w)\) and perpendicular to the line \(J_{b} J_{c}\) is
\[ L_{1}:(-b w+c v) x-(b w+c v+2 c u+2 c w) y+(b w+c v+2 b u+2 b v) z=0 \]
Similarly, we calculate the equations of the line \(L_{2}\) through the point \(E_{2}\) and perpendicular to the line \(J_{c} J_{a}\), and the line \(L_{3}\) through the point \(E_{3}\) and perpendicular to the line \(J_{a} J_{b}\) :
\[ \begin{aligned} & L_{2}:-(u c+w a+2 c v+2 c w) x+(u c-w a) y+(u c+w a+2 a u+2 a v) z=0, \\ & L_{3}:(a v+u b+2 b v+2 b w) x-(a v+u b+2 a u+2 w a) y+(a v-u b) z=0 . \end{aligned} \] By using (3) we prove that these three lines concur in a point. Finally, we use (4) to find the barycentric coordinates of the point of concurrence \(R\) as the point of concurrence of the lines \(L_{1}\) and \(L_{2}\). The barycentric coordinates are given in the statement of the theorem.
In Table 2 below, \(T\) is the Excentral triangle or any triangle homothetic with it. Recall that the following triangles are homothetic with the Excentral triangle: Intouch triangle, Hexyl triangle, Yff Central triangle, etc.
Table 2
Remark 4.2. For any point \(P\) :
(1) \(C(\) Excentral triangle, \(P)=\mathrm{X}(40)\) Bevan P) = X(40) Bevan Point;
(2) \(C(\) Intouch triangle, \(P)=\mathrm{X}(1)\) Incenter;
(3) \(C(\) Hexyl triangle, \(P)=\mathrm{X}(1)\) Incenter;
(4) \(C(\) Yff C(Yff Central Triangle, \(P)=\) Circumcenter of the Yff Central Triangle. This point is not available in (Kimberling \({ }^{2)}\) ).
5. Orthology Centers of the Euler Triangles wrt the Tangential Triangle
Theorem 5.1. For any point \(P=(u, v, w)\), , the Euler triangle of \(P\) is orthologic wrt the Tangential triangle. The first barycentric coordinate of the orthology center \(R=(u R, v R, w R)\) is
\[ u R=a^{4}(2 u+v+w)+\left(b^{2}-c^{2}\right)^{2} u-\left(a^{2} b^{2}+c^{2} a^{2}\right)(3 u+v+w) \]
To prove theorem 5.1 we use the same algorithm as in the proofs of the previous theorems. Recall that the vertices of the Tangential triangle \(T_{a} T_{b} T_{c}\) a are \(T_{a}=\left(-a^{2}, b^{2}, c^{2}\right), T_{b}=\left(a^{2},-b^{2}, c^{2}\right)\) a and \(T_{c}=\left(a^{2}, b^{2},-c^{2}\right)\).
In Table 3 below, \(T\) is the Tangential triangle or any triangle homothetic with it. Recall that the following triangles are homothetic with the Tangential triangle: Orthic triangle, Intangents triangle, Extangents triangle, etc.
Table 3
Remark 5.2. For any point \(P\) :
(1) \(C(\) Tangential triangle, \(P)=\mathrm{X}(3)\) Circumcenter;
(2) \(C\) (Orthic triangle, \(P\) ) \(=\mathrm{X}(4)\) Orthocenter;
(3) \(C(\) Intangents triangle, \(P)=\mathrm{X}(1)\) Incenter;
(4) \(C(\) Extangents triangle, \(P)=\mathrm{X}(40)\) Bevan P) = X(40) Bevan Point.
6. Orthology Centers of the Euler Triangles wrt the Malfatti Squares Triangle
Theorem 6.1. For any point \(P=(u, v, w)\), , the Euler triangle of \(P\) is orthologic wrt the Malfatti squares triangle. The first barycentric coordinate of the orthology center \(R=(u R, v R, w R)\) is
\(\begin{aligned} u R= & 2(a+b+c)^{2}(a-b-c)^{2}(a-b+c)^{2}(a+b-c)^{2} \\ & \left(-5 a^{4}-5 b^{4}-5 c^{4}+26 a^{2} b^{2}+26 b^{2} c^{2}+26 c^{2} a^{2}\right)(4 u+v+w) \\ & +12\left(a^{2}+b^{2}+c^{2}\right) \sqrt{-(a+b+c)(a-b-c)(a-b+c)(a+b-c)} \end{aligned}\) .
To prove theorem 6.1 we use the same algorithm as in the proofs of the previous theorems. Recall that the barycentric coordinates of the vertices of the Malfatti squares triangle \(M_{a} M_{b} M_{c}\) are \(M_{a}=\left(S, S+S_{A}+2 S_{C}, S+S_{A}+2 S_{B}\right), \quad M_{b}=\left(S+S_{B}+2 S_{C}, S, S+S_{B}+2 S_{A}\right)\) and \(M_{c}=\left(S+S_{C}+2 S_{B}, S+S_{C}+2 S_{A}, S\right)\).
In Table 4 below, \(T\) is the Malfatti squares triangle or any triangle homothetic with it.
Table 4
Remark 6.2. For any point \(P, C(\) Malfatti squares triangle, \(P)=\mathrm{X}(3068)\) Malfatti-Moses Point.
2010 Mathematics Subject Classification. Primary 51-04, 68T01, 68T99
NOTES
1. Gibert, B., Cubics in the Triangle Plane, http://bernard.gibert.pagesperso-orange. fr/index.html
2. Kimberling, Encyclopedia of Triangle Centers, http://faculty.evansville.edu/ck6/encyclopedia/ETC.html
3. Weisstein, E.W., MathWorld – A Wolfram Web Resource. http://mathworld.wolfram.com/
http://math.fau.edu/Yiu/YIUIntroductionToTriangleGeometry130411.pdf.
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