Резюме. By using the computer program “Discoverer”, we propose theorems about antipedal corner products.

Ключови думи: antipedal corner product, triangle geometry, remarkable point, computerdiscovered mathematics, Discoverer.

1. Introduction

The computer program “Discoverer”, created by the authors, 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. See (Grozdev & Dekov, 2014, 2015). In this paper, by using the “Discoverer”, we investigate the antipedal corner products. The paper contains more than 30 theorems about antipedal corner products. We expect that the majority of these theorems are new theorems, discovered by a computer.

The labeling of triangle centers follows (Kimberling). Hence, \(X(1)\) denotes the Incenter, \(\mathrm{X}(2)\) denotes the Centroid, \(\mathrm{X}(37)\) is the Grinberg Point, etc. We refer the reader to (Kimberling, Glossary) for the definition of a triangle center.

Given triangle \(A B C\). Let \(P\) and \(Q\) be finite triangle centers of \(\triangle A B C\) and let \(\triangle P a P b P c\) be the antipedal triangle of \(P\). We say that \(\triangle P a C B, \triangle C P b A\) and \(\triangle B A P c\) are the antipedal corner triangles of \(P\). Denote by \(H a\) the \(Q\)-triangle center of \(\triangle P a C B\), by \(H b\) the \(Q\)-triangle center of \(\triangle C P b A\), and by \(H c\) the \(Q\)-triangle center of \(\triangle B A P c\). If the lines \(A H a, B H b\) and \(C H c\) concur in a point, we say that the antipedal corner product of \(P\) and \(Q\) exists, and we call the point of concurrence of the lines the antipedal corner product of \(P\) and \(Q\).

The computer program “Discoverer” has discovered the following theorems:

Theorem 1. The Antipedal Corner Product of a finite triangle center \(P\) and the Orthocenter is the Complement of point \(P\).

Figure 1 illustrates theorem 1. In Fig.1, \(P a P b P c\) is the antipedal triangle of point \(P, H a\) is the Orthocenter of \(\triangle P a C B, H b\) is the Orthocenter of \(\triangle C P b A\), and \(H c\) is the Orthocenter of \(\triangle B A P c\). Then the lines \(A H a, B H b\) and \(C H c\) concur in point \(c P\), the complement of point \(P\).

Theorem 2. The Antipedal Corner Product of the Orthocenter and a finite triangle center \(P\) is the Complement of point \(P\).

Figure 2 illustrates theorem 2. In Fig.2, \(M a M b M c\) is the antipedal triangle of the Orthocenter, that is, the Antimedial triangle, \(P a\) is the \(P\)-triangle center of \(\triangle M a C B, P b\) is the \(P\)-triangle center of \(\triangle C M b A\), and \(P c\) is the \(P\)-triangle center of \(\triangle B A M c\). Then the lines \(A P a, B P b\) and \(C P c\) concur in point \(c P\), the complement of point \(P\).

Form the proofs of theorems 1 and 2, given below, it follows:

Theorem 3. The Triangle of the Orthocenters of the Antipedal Corner Triangles of a finite triangle center \(P\) is the Triangle of \(P\)-triangle centers of the Antipedal Corner Triangles of the Orthocenter.

Figure 3 illustrates theorem 3 for the special case when \(P\) is the Circumcenter. In Fig.3, \(O\) is the Circumcenter, \(M a M b M c\) is the Antipedal triangle of the Orthocenter, that is, the Antimedial triangle, \(P a P b P c\) is the Antipedal triangle of the Circumcenter, \(H a\) is the Orthocenter of \(\triangle P a C B\) and the Circumcenter of \(\triangle M a C B, H b\) is the Orthocenter of \(\triangle C P b A\) and the Circumcenter of the \(\triangle C M b A\), and \(H c\) is the Orthocenter of \(\triangle B A P c\) and the Circumcenter of the \(\triangle B A M c\).

In this paper we give proofs of theorems 1 and 2 by using barycentric coordinates. Also, we give examples of antipedal corner products, discovered by the “Discoverer”. The enclosed files contain 37 examples of antipedal corner products. Of these, 24 are included in Kimberling and the rest 13 examples are not included in Kimberling. We recommend to the reader to prove the examples of the antipedal corner products, given in the enclosed files.

2. Preliminaries

In this section we review some basic facts about barycentric coordinates. We refer the reader to (Grozdev and Nenkov, 2012a,b), (Paskalev & Tchobanov, 1985), (Yiu, 2001, edition of 2013), (Douillet, 2012).

We use barycentric coordinates. The reference triangle \(A B C\) has vertices \(A=(1,0,0)\), \(B=(0,1,0)\) and \(C=(0,0,1)\). The side lengths of \(\triangle A B C\) are denoted by \(a=B C, b=C A\)

and \(c=A B\). A point is an element of \(\mathbb{R}^{3}\), defined up to a proportionality factor, that is, for \(\forall k \in \mathbb{R} \backslash\{0\}: P=(u, v, w)\) means that \(P \simeq(u, v, w) \simeq(k u, k v, k w)\).

A point \(P=(u, v, w)\) is finite, if \(u+v+w \neq 0\). A point \(P=(u, v, w)\) is normalized, if \(u+v+w=1\). A finite point cou + v + w = 1 . A finite point could be put in a normalized form by \(P=\left(\tfrac{u}{s}, \tfrac{v}{s}, \tfrac{w}{s}\right)\), where \(s=u+v+w\).

The barycentric coordinates of the antipedal triangle \(P a P b P c, P a=(u P a, v P a, w P a)\), \(P b=(u P b, v P b, w P b), P c=(u P c, v P c, w P c)\) of point \(P=(u, v, w)\) are as it follows:

\[ \begin{gathered} u P a=-\left(2 v a^{2}+u a^{2}+u b^{2}-u c^{2}\right)\left(2 w a^{2}+u c^{2}+u a^{2}-u b^{2}\right), \\ v P a=\left(2 w a^{2}+u c^{2}+u a^{2}-u b^{2}\right)\left(2 u b^{2}+v a^{2}+v b^{2}-v c^{2}\right), \\ w P a=\left(2 v a^{2}+u a^{2}+u b^{2}-u c^{2}\right)\left(2 u c^{2}+w c^{2}+w a^{2}-w b^{2}\right), \\ u P b=\left(2 w b^{2}+v b^{2}+v c^{2}-v a^{2}\right)\left(2 v a^{2}+u a^{2}+u b^{2}-u c^{2}\right), \\ v P b=-\left(2 w b^{2}+v b^{2}+v c^{2}-v a^{2}\right)\left(2 u b^{2}+v a^{2}+v b^{2}-v c^{2}\right), \\ w P b=\left(2 u b^{2}+v a^{2}+v b^{2}-v c^{2}\right)\left(2 v c^{2}+w b^{2}+w c^{2}-w a^{2}\right), \\ u P c=\left(2 v c^{2}+w b^{2}+w c^{2}-w a^{2}\right)\left(2 w a^{2}+u c^{2}+u a^{2}-u b^{2}\right), \\ v P c=\left(2 u c^{2}+w c^{2}+w a^{2}-w b^{2}\right)\left(2 w b^{2}+v b^{2}+v c^{2}-v a^{2}\right), \\ w P c=-\left(2 u c^{2}+w c^{2}+w a^{2}-w b^{2}\right)\left(2 v c^{2}+w b^{2}+w c^{2}-w a^{2}\right), \end{gathered} \]

Given two normalized points \(P=\left(u_{1}, v_{1}, w_{1}\right)\) and \(Q=\left(u_{2}, v_{2}, w_{2}\right)\), v2, w2) , then (Paskalev & Tchobanov, 1985, § 15, Proposition 1):

(1) \[ |P Q|^{2}=-a^{2} v w-b^{2} w u-c^{2} u v, \]

where \(u=u_{1}-u_{2}, v=v_{1}-v_{2}\) and \(w=w_{1}-w_{2}\).

Let \(D E F\) be a triangle whose vertices have normalized barycentric coordinates wrt \(\triangle A B C\) as it follows: \(D=\left(p_{1}, q_{1}, r_{1}\right), E=\left(p_{2}, q_{2}, r_{2}\right)\) and \(F=\left(p_{3}, q_{3}, r_{3}\right)\). Let \(P\) be a point with normalized barycentric coordinates \(P=(p, q, r)\) wrt \(\triangle D E F\). Then the barycentric coordinates of \(P=(u, v, w)\) wrt \(\triangle A B C\) are as it follows (Paskalev & Tchobanov, 1985, § 30):

(2) \[ \begin{aligned} & u=p_{1} p+p_{2} q+p_{3} r \\ & v=q_{1} p+q_{2} q+q_{3} r \\ & w=r_{1} p+r_{2} q+r_{3} r \end{aligned} \]

The equation of the line joining two points with coordinates (\(u_{1}, v_{1}, w_{1}\) ) and (\(u_{2}, v_{2}, w_{2}\) ) is

(3) \[ \left|\begin{array}{ccc} u_{1} & v_{1} & w_{1} \\ u_{2} & v_{2} & w_{2} \\ x & y & z \end{array}\right|=0 \]

Three lines \(p_{i} x+q_{i} y+r_{i} z=0, i=1,2,3\) are concurrent if and only if

(4) \[ \left|\begin{array}{lll} p_{1} & q_{1} & r_{1} \\ p_{2} & q_{2} & r_{2} \\ p_{3} & q_{3} & r_{3} \end{array}\right|=0 \]

The intersection of 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

(5) \[ \left(q_{1} r_{2}-q_{2} r_{1}, r_{1} p_{2}-r_{2} p_{1}, p_{1} q_{2}-p_{2} q_{1}\right) \]

Given a point \(P=(u, v, w)\), v, w) , then the complement of P is the point \((v+w, w+u, u+v)\).

3. Proofs of Theorem 1 and 2

Proof of theorem 1. Let \(P\) be a finite triangle center, \(P=(u, v, w)\). By using (1), we find the side lengths \(a_{1}=|B C|=a, b_{1}=|P a B|\), a and \(c_{1}=|P a C|\) of \(\triangle P a C B\). The barycentric coordinates of the orthocenter \(H a\) of \(\triangle P a C B\) wrt \(\triangle P a C B\) are as it follows:

\(H a=\left(\tfrac{1}{b_{1}^{2}+c_{1}^{2}-a_{1}^{2}}, \tfrac{1}{c_{1}^{2}+a_{1}^{2}-b_{1}^{2}}, \tfrac{1}{a_{1}^{2}+b_{1}^{2}-c_{1}^{2}}\right)\). By using (2), we find the barycentric coordinates of \(H a\) wrt \(\triangle A B C\) as it follows: \(H a=(-u, u+w, u+v)\). Similarly, we find the barycentric coordinates of \(H b\) wrt \(\triangle A B C\) as it follows: \(H b=(v+w,-v, u+v)\), , and the barycentric coordinates of \(H c\) wrt \(\triangle A B C\) as it follows: \(H c=(v+w, u+w,-w)\).

Now by using (3) we find the barycentric equations of the lines \(A H a, B H b\) and \(C H c\) as it follows:

\[ \begin{aligned} & A H a:(u+v) y-(u+w) z=0 \\ & B H b:(u+v) x-(v+w) z=0 \\ & C H c:(u+w) x-(v+w) y=0 \end{aligned} \]

By using (4), we prove that these lines concur in a point. Then, by using (5), we find the point of intersection of the lines \(A H a, B H b\) and \(C H c\) as the point of intersection \(R\) of the lines \(A H a\) and \(B H b\) as it follows: \(R=(v+w, w+u, u+v)\). Point \(R\) is the antipedal corner product of point \(P\) and the Orthocenter. It is easy to see that \(R\) is the complement of point \(P\). This completes the proof.

Proof of theorem 2. Let \(P\) be a point with barycentric coordinates \(P=(u, v, w)\) w wrt \(\triangle A B C\). The antipedal triangle of the Orthocenter is the antimedial triangle \(M a M b\)\(M c\). Hence, the side lengths of \(\triangle M a C B\) are as it follows: \(a_{1}=|B C|=a, b_{1}=|M a B|=b\), and \(c_{1}=|M a C|=c\). The barycentric coordinates of \(P a\) wrt \(\triangle M a C B\) are as it follows: \(P a=(u, v, w)\). By using ( (2), we find the barycentric coordinates of \(P a\) wrt \(\triangle A B C\) as it follows: Pa = (-u, u + ws: \(P a=(-u, u+w, u+v)\). Similarly, we find the barycentric coordinates of \(P b\) and of \(P c\) wrt \(\triangle A B C\) as it follows: \(P b=(v+w,-v, u+v)\) and \(P c=(v+w, u+w,-w)\). Points \(P a, P b\) and \(P c\) coincide with points \(H a, H b\) and \(H c\) in the proof of theorem 1. Hence the lines \(A P a, B P b\) and \(C P c\) are the same as the lines \(A H a, B H b\) and \(C H c\) in the proof of theorem 1. This completes the proof.

Supplementary material

The enclosed file “2015-apcp.zip” contains the files quoted in this paper. The reader may download it from the web page of the journal.

RЕFERENCES

Douillet, P. (2012). Translation of the Kimberling’s Glossary into barycentrics, http:// eg-enc.webege.com/htm/links/glossary.pdf

Grozdev, S. & Dekov, D. (2014). Computer-generated mathematics: Points on the Kiepert hyperbola, The Mathmatical Gazette, vol. 98, no. 543, 509 – 511.

Grozdev, S. & Dekov, D. (2015). A Computer Improves the Steiner’s Construction of the Malfatti Circles, Mathematics and Infomatics, vol. 58, no.1, 40 – 51.

Grozdev, S. & Nenkov, V. (2012a). Three Remarkable Points on the Medians of a Triangle (in Bulgarian), Sofia, Archimedes.

Grozdev, S. & Nenkov, V. (2012b). On the Orthocenter in the Plane and in the Space (in Bulgarian), Sofia, Archimedes.

Kimberling, C., Encyclopedia of Triangle Centers, http://faculty.evansville.edu/ck6/ encyclopedia/ETC.html

Paskalev, G. & Tchobanov, I. (1985). Remarkable Points in the Triangle (in Bulgarian), Sofa, Narodna Prosveta.

Yiu, P. (2001, version of 2013). Introduction to the Geometry of the Triangle, http:// math.fau.edu/Yiu/YIUIntroductionToTriangleGeometry130411.pdf