2008 AIME I Problems/Problem 10
Let be an isosceles trapezoid with whose angle at the longer base is . The diagonals have length , and point is at distances and from vertices and , respectively. Let be the foot of the altitude from to . The distance can be expressed in the form , where and are positive integers and is not divisible by the square of any prime. Find .
Assuming that is a triangle and applying the triangle inequality, we see that . However, if is strictly greater than , then the circle with radius and center does not touch , which implies that , a contradiction. As a result, A, D, and E are collinear. Therefore, .
Thus, and are triangles. Hence , and
Finally, the answer is .
No restrictions are set on the lengths of the bases, so for calculational simplicity let . Since is a triangle, .
The answer is . Note that while this is not rigorous, the above solution shows that is indeed the only possibility.
Extend through , to meet (extended through ) at . is an equilateral triangle because of the angle conditions on the base.
If then , because and therefore .
By simple angle chasing, is a 30-60-90 triangle and thus , and
Similarly is a 30-60-90 triangle and thus .
Equating and solving for , and thus .
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