2011 USAMO Problems/Problem 5
Contents
Problem
Let be a given point inside quadrilateral . Points and are located within such that , , , . Prove that if and only if .
Solution 1
Lemma. If and are not parallel, then are concurrent.
Proof. Let and meet at . Notice that with respect to triangle , and are isogonal conjugates (this can be proven by dropping altitudes from to , , and or depending on where is). 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 .
Solution 2
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 .
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 |
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