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Problems of Ukrainian Authors

Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

Given acute-angled triangle $ABC$ . $O$ is the circumcenter, $H$ is the orthocenter and $AH_A$ , $BH_B$ , $CH_C$ are the altitudes of $\triangle ABC$ . 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 $\triangle ABC$ . (Euler line of $\triangle ABC$ is the line passing through the circumcenter and centroid of a triangle).

Solution

It is easy to see that point $O$ is the incenter of $\triangle A_1B_1C_1$ , since $OC$ is perpendicular to the tangent of the circumcircle of $\triangle ABC$ at the point $C$ and is parallel to $H_AH_B$ .

Therefore $OC \perp H_A H_B \Rightarrow A_1 B_1 \parallel H_A H_B$ . And so $\Delta A_1 B_1 C_1 \sim \Delta H_A H_B H_C$ (Fig.29).

Hence $A_1 H_A \cap B_1 H_B \cap C_1 H_C = X$ , where point $X$ is the homothety center of these triangles. Since $O$ is the incenter of $\Delta A_1 B_1 C_1$ , and $H$ is the incenter of $\Delta H_A H_B H_C$ , then $X \in OH$ .

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