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

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==Problem==
 
==Problem==
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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>.kusu
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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>.
  
 
==Solution==
 
==Solution==

Revision as of 08:39, 26 May 2020


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. Now, since CD is parallel to AB, we know that triangle BMD is isoceles, so $\angle{BMD}=\angle{BDM}$ and by simple parallel line rules, $\angle{ABM}=\angle{EBA}$. Similarily, $\angle{BAM}=\angle{EAB}$, 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.

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