The Problems of Ukrainian Authors

Question

Given acute-angled triangle ABC. O is the circumcenter, H is the orthocenter and AH_A, BH_B, CH_C are the altitudes of ABC. Denote by A_1, B_1, C_1 the circumcenters of triangles BOC, COA and AOB respectively. Prove that the lines A_1H_A, B_1H_B, C_1H_C meet at a point, which lies on the Euler line of ABC. (Euler line of ABC is the line passing through the circumcenter and centroid of a triangle).

Accepted Answer

It is easy to see that point O is the incenter of A_1B_1C_1, since OC is perpendicular to the tangent of the circumcircle of ABC at the point C and is parallel to H_AH_B. Therefore OC H_A H_B A_1 B_1 H_A H_B. And so A_1 B_1 C_1 H_A H_B H_C (Fig.29). Hence A_1 H_A B_1 H_B C_1 H_C = X, where point X is the homothety center of these triangles. Since O is the incenter of A_1 B_1 C_1, and H is the incenter of H_A H_B H_C, then X OH.