Difference between revisions of "2000 IMO Problems/Problem 1"
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<math>\textbf{Proof of problem:}</math> Let ray <math>NM</math> intersect <math>AB</math> at <math>X</math>. By our lemma, <math>\textit{(the two circles are tangent to AB)}</math>, <math>X</math> bisects <math>AB</math>. Since <math>\triangle{NAX}</math> and <math>\triangle{NPM}</math> are similar, and <math>\triangle{NBX}</math> and <math>\triangle{NQM}</math> are similar implies <math>M</math> bisects <math>PQ</math>. | <math>\textbf{Proof of problem:}</math> Let ray <math>NM</math> intersect <math>AB</math> at <math>X</math>. By our lemma, <math>\textit{(the two circles are tangent to AB)}</math>, <math>X</math> bisects <math>AB</math>. Since <math>\triangle{NAX}</math> and <math>\triangle{NPM}</math> are similar, and <math>\triangle{NBX}</math> and <math>\triangle{NQM}</math> are similar implies <math>M</math> bisects <math>PQ</math>. | ||
− | + | Now, <math>\angle{ABM} = \angle{BMD}</math> since <math>CD</math> is parallel to <math>AB</math>. But <math>AB</math> is tangent to the circumcircle of <math>\triangle{BMD}</math> hence <math>\angle{ABM} = \angle{BDM}</math> and that implies <math>\angle{BMD} = \angle{BDM} . </math>So<math>\triangle{BMD}</math> is isosceles and <math>BM=BD</math>. | |
− | + | By simple parallel line rules, <math>\angle{EBA}=\angle{BDM}=\angle{ABM}</math>. Similarly, <math>\angle{BAM}=\angle{EAB}</math>, so by <math>\textit{ASA}</math> criterion, <math>\triangle{ABM}</math> and <math>\triangle{ABE}</math> are congruent. | |
We know that <math>BE=BM=BD</math> so a circle with diameter <math>ED</math> can be circumscribed around <math>\triangle{EMD}</math>. Join points <math>E</math> and <math>M</math>, <math>EM</math> is perpendicular on <math>PQ</math>, previously we proved <math>MP = MQ</math>, hence <math>\triangle{EPQ}</math> is isoscles and <math>EP = EQ</math> . | We know that <math>BE=BM=BD</math> so a circle with diameter <math>ED</math> can be circumscribed around <math>\triangle{EMD}</math>. Join points <math>E</math> and <math>M</math>, <math>EM</math> is perpendicular on <math>PQ</math>, previously we proved <math>MP = MQ</math>, hence <math>\triangle{EPQ}</math> is isoscles and <math>EP = EQ</math> . |
Latest revision as of 14:12, 21 June 2024
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
Given a triangle, and a point in its interior, assume that the circumcircles of and are tangent to . Prove that ray bisects . Let the intersection of and be . By power of a point, and , so .
Let ray intersect at . By our lemma, , bisects . Since and are similar, and and are similar implies bisects .
Now, since is parallel to . But is tangent to the circumcircle of hence and that implies So is isosceles and .
By simple parallel line rules, . Similarly, , so by criterion, and are congruent.
We know that so a circle with diameter can be circumscribed around . Join points and , is perpendicular on , previously we proved , hence is isoscles and .
See Also
2000 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |