ONE PROPERTY OF THE JERABEK HYPERBOLA AND ITS COROLLARIES ALEXEY A. ZASLAVSKY Abstract. We study the locus of the points P having the following property: if A 1 B 1 C 1 is the circumcevian triangle of P with respect to the given triangle ABC, and A 2 , B 2 , C 2 are the reflections of A 1 , B 1 , C 1 in BC, CA, AB, respectively, then the triangles ABC and A 2 B 2 C 2 are perspective. We show that this locus consists of the infinite line and the Jerabek hyperbola of ABC. This fact yields some interesting corollaries. We start with the following well-known fact [1, p. 4.4.5]. Statement 1. Let the tangents to the circumcircle of ABC at A and B meet in C 0 . The line CC 0 meets the circumcircle of ABC for the second time in C 1 , and C 2 is the reflection of C 1 in AB. Then CC 2 is a median in ABC. Proof. Let C  be the common point of CC 1 and AB, and A  , B  be the common points of AC 2 , BC 2 with BC, AC, respectively. Since ∠C  CB = ∠C 1 AB = ∠BAA  , the triangles BCC  and BAA  are similar; therefore, BA  = AB BC · BC  . But CC  is a symedian in ABC, so BC  = BC 2 BC 2 +AC 2 · AB. Therefore, BA  BC = AB 2 AC 2 +BC 2 . Analogously, the ratio AB  AC has the same value, yielding the claim.  A B C C 0 C 1 C 2 A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  Fig. 1. This fact yields the following corollary: let A 1 B 1 C 1 be the circumcevian tri- angle of the Lemoine point, L, and let A 2 , B 2 , C 2 be the reflections of A 1 , B 1 , 53 54 ALEXEY A. ZASLAVSKY C 1 in BC, CA, AB, respectively. Then the triangles ABC and A 2 B 2 C 2 are perspective. Our goal is to find the locus of the points sharing this property. To this end, first we formulate the following assertion. Lemma 1. Let CC 1 divide AB in ratio x : y. Then CC 2 divides AB in ratio x(b 2 (x + y) − c 2 x) : y(a 2 (x + y) − c 2 y). In order to prove this, it suffices to repeat the argument by which we demon- strated the previous assertion and then to apply Ceva’s theorem. Now let P be the point with barycentric coordinates (x : y : z). Using Lemma 1 and Ceva’s theorem, we see that a point P has the property in question iff it lies on some cubic c. From the following assertion, we infer that c is degenerated. Statement 2. Let three parallel lines passing through the vertices of ABC meet its circumcircle in A 1 , B 1 , C 1 . The points A 2 , B 2 , C 2 are the reflections of A 1 , B 1 , C 1 in BC, CA, AB, respectively. Then the lines AA 2 , BB 2 , CC 2 are concurrent. Proof. Consider the three lines which pass through A 1 , B 1 , C 1 and are parallel to BC, CA, AB, respectively. It is easy to see that they meet at a point on the circumcircle of ABC. The points A 2 , B 2 , C 2 are the reflections of this point in the midpoints of the sides of ABC. Therefore, the triangles ABC and A 2 B 2 C 2 are centrosymmetric.  A B C A 1 B 1 C 1 A 2 B 2 C 2 Fig. 2. Note also that the center of perspective in this claim lies on the Euler circle. And so, c consists of the infinite line and some conic k. In order to determine k completely, it suffices to indicate five point lying on it. We already know that k contains the Lemoine point, L. Furthermore, k contains the vertices of ABC as well as its orthocenter H (in this case, all of A 2 , B 2 , C 2 coincide with H). Therefore, k is the Jerabek hyperbola. Here follow some corollaries of this fact. ONE PROPERTY OF THE JERABEK HYPERBOLA AND ITS COROLLARIES 55 Statement 3. Let P be a point on the Euler line of ABC, A 1 B 1 C 1 be the cir- cumcevian triangle of P , and A 2 , B 2 , C 2 be the reflections of A 1 , B 1 , C 1 in the midpoints of BC, CA, AB, respectively. Then the lines AA 2 , BB 2 , CC 2 are concurrent. Proof. The isogonal conjugate, Q, of P lies on the Jerabek hyperbola. Let CQ meet the circumcircle of ABC for the second time in C 3 . Then C 2 and C 3 are symmetric with respect to AB.  A B C A 1 B 1 C 1 A 2 B 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 C 2 P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P Fig. 3. Statement 4. Let P be a point on the Euler line of ABC, A 0 , B 0 , C 0 be the midpoints of BC, CA, AB, respectively, and A 1 , B 1 , C 1 be the projections of the circumcenter O of ABC onto AP , BP , CP , respectively. Then the lines A 0 A 1 , B 0 B 1 , C 0 C 1 are concurrent. A B C A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 B 1 C 1 A 0 B 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O Fig. 4. Proof. It suffices to apply homothety of center the centroid of ABC and coeffi- cient − 1 2 to the configuration of the previous claim.  Statement 5. Let O, I be the circumcenter and incenter of ABC. An arbitrary line perpendicular to OI meets BC, CA, AB in A 1 , B 1 , C 1 , respectively. Then the circumcenters of the triangles IAA 1 , IBB 1 , ICC 1 are collinear. Proof. Applying an inversion of center I and the previous assertion we see that the circumcircles of the three triangles in question have a common point other than I.  56 ALEXEY A. ZASLAVSKY A B C A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 B 1 C 1 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I O P Fig. 5. References [1] A. V. Akopyan. Geometry in figures. Createspace, 2011. Central Economic and Mathematical Institute RAS E-mail address: zaslavsky@mccme.ru . B C A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 B 1 C 1 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I O P Fig respectively. Then the lines A 0 A 1 , B 0 B 1 , C 0 C 1 are concurrent. A B C A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 A 1 B 1 C 1 A 0 B 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 C 0 P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P P O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O Fig B C C 0 C 1 C 2 A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  A  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  C  Fig.