# Difference between revisions of "2000 IMO Problems/Problem 1"

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+ | ==Problem== | ||

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

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+ | ==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, <math>BD^2=AD(PD)</math> and <math>CD^2=AD(PD)</math>, 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. |

## Revision as of 23:08, 12 April 2018

## 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. Since EM is perpendicular to AB (proof needed) EQ=EP.