Difference between revisions of "2000 IMO Problems/Problem 1"
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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. | 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: 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. | + | 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 12:23, 13 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. 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.