# 2009 IMO Problems/Problem 2

## Problem

Let $ABC$ be a triangle with circumcentre $O$. The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$ respectively. Let $K,L$ and $M$ be the midpoints of the segments $BP,CQ$ and $PQ$, respectively, and let $\Gamma$ be the circle passing through $K,L$ and $M$. Suppose that the line $PQ$ is tangent to the circle $\Gamma$. Prove that $OP=OQ$.

Author: Sergei Berlov, Russia

## Solution

### Diagram

$[asy] dot("O", (50, 43), W); dot("A", (40, 100), N); dot("B", (0, 0), S); dot("C", (100, 0), S); dot("Q", (24, 60), W); dot("P", (52, 80), E); dot("L", (62, 30), SE); dot("M", (38, 70), N); dot("K", (26, 40), S); draw((100, 0)--(24, 60), dotted); draw((0, 0)--(52, 80), dashed); draw((0, 0)--(100, 0)--(40, 100)--cycle); draw((24, 60)--(52, 80)); draw((26, 40)--(38, 70)--(62, 30)--cycle); draw(circle((47, 48), 23)); [/asy]$ Diagram by qwertysri987

By parallel lines and the tangency condition, $$\angle APM\cong \angle LMP \cong \angle LKM.$$ Similarly, $$\angle AQP\cong \angle KLM,$$ so AA similarity implies $$\triangle APQ\sim \triangle MKL.$$ Let $\omega$ denote the circumcircle of $\triangle ABC,$ and $R$ its circumradius. As both $P$ and $Q$ are inside $\omega,$

\begin{align*} R^2-QO^2&=\text{Pow}_{\omega}(Q)\\ &=QB\cdot AQ \\ &=2AQ\cdot MK\\ &=2AP\cdot ML\\ &=AP\cdot PC\\ &=\text{Pow}_{\omega}(P)\\ &=R^2-PO^2. \end{align*} It follows that $OP=OQ.$ $\blacksquare$