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Problem 2442

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.4 Prove it Team Selection Test for IMO · Turkey · 2011

Let $K$ be a point in the interior of an acute triangle $ABC$ and $ARBPCQ$ be a convex hexagon whose vertices lie on the circumcircle $\Gamma$ of the triangle $ABC$ . Let $A_1$ be the second point where the circle passing through $K$ and tangent to $\Gamma$ at $A$ intersects the line $AP$ . The points $B_1$ and $C_1$ are defined similarly. Prove that
$\min \left\{ \frac{PA_1}{AA_1}, \frac{QB_1}{BB_1}, \frac{RC_1}{CC_1} \right\} \le 1.$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $O$ be the center of $\Gamma$ . Since $ABC$ is an acute triangle $O$ lies inside $ABC$ . Assume that $K$ lies on the same side of the lines $AO$ and $BO$ as $C$ , and on the same side of the bisector of the line segment $AB$ as $B$ . Then $KA \ge OA$ .

Let $\omega$ be the circle passing through $K$ and tangent to $\Gamma$ at $A$ . Then $\Gamma$ and $\omega$ are homothetic with center $A$ and ratio $PA/A_1A$ . Since $KA \ge OA$ , $O$ lies inside $\omega$ and the homothety ratio is at most $2$ . Hence $PA_1/AA_1 \le 1$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.