# 2000 IMO Problems/Problem 1

## Problem

Two circles and intersect at two points and . Let be the line tangent to these circles at and , respectively, so that lies closer to than . Let be the line parallel to and passing through the point , with on and on . Lines and meet at ; lines and meet at ; lines and meet at . Show that .

## Solution

Lemma: Given a triangle ABC and a point P in its interior, assume that the circumcircles of ACP and ABP are tangent to BC. Prove that ray AP bisects BC. Proof: Call the intersection of AP and BC D. By power of a point, and , so BD=CD. Proof of problem: Let ray NM intersect AB at X. By our lemma (the two circles are tangent to AB), X bisects AB. Since triangles NAX and NPM are similar, and NBX and NQM are similar, M bisects PQ. Now, since CD is parallel to AB, we know that triangle BMD is isoceles, so and by simple parallel line rules, . Similarily, , so by ASA triangle ABM and triangle EAB are congruent, so EBMA is a kite, so EM is perpendicular to AB. That means EM is perpendicular to PQ, so EPQ is isoceles, and EP=EQ.