Difference between revisions of "2011 USAMO Problems/Problem 5"
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So, <math>S=\frac{\sin \angle ABQ_1}{\sin \angle BAQ_2}\frac{\sin \angle CDQ_2}{\sin \angle DCQ_1}\frac{\sin \angle BCQ_1}{\sin \angle CBQ_1}\frac{\sin \angle DAQ_2}{\sin \angle ADQ_2}</math> | So, <math>S=\frac{\sin \angle ABQ_1}{\sin \angle BAQ_2}\frac{\sin \angle CDQ_2}{\sin \angle DCQ_1}\frac{\sin \angle BCQ_1}{\sin \angle CBQ_1}\frac{\sin \angle DAQ_2}{\sin \angle ADQ_2}</math> | ||
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+ | By the terms of the problem, <math>S=\frac{\sin \angle PBC}{\sin \angle PAD}\frac{\sin \angle PDA}{\sin \angle PCB}\frac{\sin \angle PCD}{\sin \angle PBA}\frac{\sin \angle PAB}{\sin \angle PDC} = \frac{\sin \angle PBC}{\sin \angle PCB}\frac{\sin \angle PDA}{\sin \angle PAD}\frac{\sin \angle PCD}{\sin \angle PDC}\frac{\sin \angle PAB}{\sin \angle PBA}</math>. | ||
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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>. | ||
==See also== | ==See also== |
Revision as of 16:29, 8 June 2011
Problem
Let be a given point inside quadrilateral
. Points
and
are located within
such that
,
,
,
. Prove that
if and only if
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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, .
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 |