2000 IMO Problems/Problem 1

Problem

Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$ . Let $AB$ be the line tangent to these circles at $A$ and $B$ , respectively, so that $M$ lies closer to $AB$ than $N$ . Let $CD$ be the line parallel to $AB$ and passing through the point $M$ , with $C$ on $G_1$ and $D$ on $G_2$ . Lines $AC$ and $BD$ meet at $E$ ; lines $AN$ and $CD$ meet at $P$ ; lines $BN$ and $CD$ meet at $Q$ . Show that $EP=EQ$ .

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, $BD^2=AD(PD)$ and $CD^2=AD(PD)$ , 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. Since EM is perpendicular to AB (proof needed) EQ=EP.

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2000 IMO Problems/Problem 1

Problem

Solution