2004 USAMO Problems/Problem 1
Problem
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
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 |
- <url>viewtopic.php?p=17439&sid=d212b9d95317a1fad7651771b6efa5bb Discussion on AoPS/MathLinks</url>
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