Difference between revisions of "2011 USAMO Problems/Problem 5"
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Applying the law of sines to the triangles with vertices at P yields <math>S=\frac{|PC|}{|PB|}\frac{|PA|}{|PD|}\frac{|PD|}{|PC|}\frac{|PB|}{|PA|}=1</math>. | Applying the law of sines to the triangles with vertices at P yields <math>S=\frac{|PC|}{|PB|}\frac{|PA|}{|PD|}\frac{|PD|}{|PC|}\frac{|PB|}{|PA|}=1</math>. | ||
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+ | ==Solution 2== | ||
+ | Lemma. If <math>AB</math> and <math>CD</math> are not parallel, then <math>AB, CD, Q_1 Q_2</math> are concurrent. | ||
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+ | Proof. Let <math>AB</math> and <math>CD</math> meet at <math>R</math>. Notice that with respect to triangle <math>ADR</math>, <math>P</math> and <math>Q_2</math> are isogonal conjugates. With respect to triangle <math>BCR</math>, <math>P</math> and <math>Q_1</math> are isogonal conjugates. Therefore, <math>Q_1</math> and <math>Q_2</math> lie on the reflection of <math>RP</math> in the angle bisector of <math>\angle{DRA}</math>, so <math>R, Q_1, Q_2</math> are collinear. Hence, <math>AB, CD, Q_1 Q_2</math> are concurrent at <math>R</math>. | ||
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+ | Now suppose <math>Q_1 Q_2 // AB</math> but <math>Q_1 Q_2</math> is not parallel to <math>CD</math>. Then <math>AB</math> and <math>CD</math> are not parallel and thus intersect at a point <math>R</math>. But then <math>Q_1 Q_2</math> also passes through <math>R</math>, contradicting <math>Q_1 Q_2 // AB</math>. A similar contradiction occurs if <math>Q_1 Q_2 // CD</math> but <math>Q_1 Q_2</math> is not parallel to <math>CD</math>, so we can conclude that <math>Q_1 Q_2 // AB</math> if and only if <math>Q_1 Q_2 // CD</math>. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:03, 20 May 2015
Contents
Problem
Let be a given point inside quadrilateral . Points and are located within such that , , , . Prove that if and only if .
Solution
First note that if and only if the altitudes from and to are the same, or . Similarly iff .
If we define , then we are done if we can show that S=1.
By the law of sines, and .
So,
By the terms of the problem, . (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.)
Rearranging yields .
Applying the law of sines to the triangles with vertices at P yields .
Solution 2
Lemma. If and are not parallel, then are concurrent.
Proof. Let and meet at . Notice that with respect to triangle , and are isogonal conjugates. With respect to triangle , and are isogonal conjugates. Therefore, and lie on the reflection of in the angle bisector of , so are collinear. Hence, are concurrent at .
Now suppose but is not parallel to . Then and are not parallel and thus intersect at a point . But then also passes through , contradicting . A similar contradiction occurs if but is not parallel to , so we can conclude that if and only if .
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2011 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |