Difference between revisions of "2000 IMO Problems/Problem 1"
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==Problem== | ==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>. |
==Solution== | ==Solution== |
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.