Autumn Tournament

Geometry Difficulty 5.1 AIME, harder Prove it Bulgaria

The points $A_1$ , $B_1$ , $C_1$ are chosen on the sides $BC$ , $CA$ , $AB$ of a triangle $ABC$ so that $BA_1 = BC_1$ and $CA_1 = CB_1$ . The lines $C_1A_1$ and $A_1B_1$ meet the line through $A$ , parallel to $BC$ , at $P$ , $Q$ . Let the circumcircles of the triangles $APC_1$ and $AQB_1$ meet at $R$ . Given that $R$ lies on $AA_1$ , show that $R$ lies on the incircle of $ABC$ .
(Emil Kolev)

Solution

Observe that $(A_1, C_1, R, B_1)$ are concyclic. Let $I$ be the incenter of triangle $ABC$ . Then $BI$ and $CI$ perpendicularly bisect $C_1A_1$ and $A_1B_1$ respectively. Hence, $I$ is the circumcenter of $\triangle A_1B_1C_1$ . Let $\angle ACB = 2\gamma$ . $\angle A_1IB_1 = 2\angle B_1C_1A_1 = 2\angle A_1RB_1 = 2\angle AQB_1 = 2\angle B_1A_1C = 180^\circ - 2\gamma$ . Therefore $(I, A_1, B_1, C)$ concyclic $\rightarrow IA_1 \perp BC$ , $IB_1 \perp AC$ . Similarly $IC_1 \perp AB$ , so circle $(A_1, C_1, R, B_1)$ is the incircle of $ABC$ . Thus, $R$ lies on the incircle of $ABC$ . $\square$

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