Difference between revisions of "2011 USAMO Problems/Problem 5"
(→Solution) |
m (→Solution) |
||
Line 6: | Line 6: | ||
− | If we define <math>S =\frac{|Q_1B|\sin \angle ABQ_1}{|Q_2A|\sin \angle BAQ_2}\ | + | If we define <math>S =\frac{|Q_1B|\sin \angle ABQ_1}{|Q_2A|\sin \angle BAQ_2} \times \frac{|Q_2D|\sin \angle CDQ_2}{|Q_1C|\sin \angle DCQ_1}</math>, then we are done if we can show that S=1. |
Revision as of 09:37, 11 July 2018
Contents
[hide]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 |