2000 IMO Problems/Problem 1
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
.kusu
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.