2004 USAMO Problems/Problem 1
Contents
Problem
(Titu Andreescu) Let be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
When does equality hold?
Solution 1
It suffices to show that because the other side of the inequality follows by symmetry. By Pitot's Theorem, we have . WLOG, let . Then we have that , so we must show . If , we are done. Therefore, we will assume for the rest of the proof that , .
Then we would like to show that By the Law of Cosines, we have that where (this is because ). Therefore, Then we wish to show that and we are done.
Solution 2
By a well-known property of tangential quadrilaterals, the sum of the two pairs of opposite sides are equal; hence Now we factor the desired expression into . Temporarily discarding the case where and , we can divide through by the to get the simplified expression .
Now, draw diagonal . By the law of cosines, . Since each of the interior and exterior angles of the quadrilateral is at least 60 degrees, we have that . Cosine is monotonically decreasing on this interval, so by setting at the extreme values, we see that . Applying the law of cosines analogously to and , we see that ; we hence have and .
We wrap up first by considering the second inequality. Because , . This latter expression is of course greater than or equal to because the inequality can be rearranged to , which is always true. Multiply the first inequality by and we see that it is simply the second inequality with the variables swapped; hence by symmetry it is true as well.
Equality occurs when and , or when is a kite.
Resources
2004 USAMO (Problems • Resources) | ||
Preceded by First problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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