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. Now, since CD is parallel to AB, we know that triangle BMD is isoceles, so <math>\angle{BMD}=\angle{BDM}</math> and by simple parallel line rules, <math>\angle{ABM}=\angle{EBA}</math>. Similarily, <math>\angle{BAM}=\angle{EAB}</math>, 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. | ||
Revision as of 09:39, 26 May 2020
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.
