APMO

Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Asia Pacific Mathematics Olympiad (APMO)

Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles intersecting at $P$ and $Q$ . The common tangent, closer to $P$ , of $\Gamma_{1}$ and $\Gamma_{2}$ touches $\Gamma_{1}$ at $A$ and $\Gamma_{2}$ at $B$ . The tangent of $\Gamma_{1}$ at $P$ meets $\Gamma_{2}$ at $C$ , which is different from $P$ and the extension of $A P$ meets $B C$ at $R$ . Prove that the circumcircle of triangle $P Q R$ is tangent to $B P$ and $B R$ .

Solution

Let $\alpha=\angle P A B$ , $\beta=\angle A B P$ and $\gamma=\angle Q A P$ . Then, since $P C$ is tangent to $\Gamma_{1}$ , we have $\angle Q P C= \angle Q B C=\gamma$ . Thus $A, B, R, Q$ are concyclic.

Since $A B$ is a common tangent to $\Gamma_{1}$ and $\Gamma_{2}$ then $\angle A Q P=\alpha$ and $\angle P Q B=\angle P C B=\beta$ . Therefore, since $A, B, R, Q$ are concyclic, $\angle A R B=\angle A Q B=\alpha+\beta$ and $\angle B Q R=\alpha$ . Thus $\angle P Q R=\angle P Q B+ \angle B Q R=\alpha+\beta$ .

Since $\angle B P R$ is an exterior angle of triangle $A B P$ , $\angle B P R=\alpha+\beta$ . We have
$\angle P Q R=\angle B P R=\angle B R P$

So circumcircle of $P Q R$ is tangent to $B P$ and $B R$ .

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