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 .
Key observation. .
Proof 1. By the triangle inequality, we can immediately see that . However, notice that , so by the law of sines, when , is right and the circle centered at with radius , which we will call , is tangent to . Thus, if were increased, would have to be moved even farther outwards from to maintain the angle of and could not touch it, a contradiction.
Proof 2. Again, use the triangle inequality to obtain . Let and . By the law of cosines on , . Viewing this as a quadratic in , the discriminant must satisfy . Combining these two inequalities yields the desired conclusion.
This observation tells us that , , and are collinear, in that order.
Then, and are triangles. Hence , and
Finally, the answer is .
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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