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Mock AIME 2 2006-2007/Problem 9

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Problem

In right triangle \displaystyle ABC, \displaystyle \angle C=90^\circ. Cevians \displaystyle AX and \displaystyle BY intersect at \displaystyle P and are drawn to \displaystyle BC and \displaystyle AC respectively such that \displaystyle \frac{BX}{CX}=\frac23 and \displaystyle \frac{AY}{CY}=\sqrt 3. If \displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d}, where \displaystyle a,b, and \displaystyle d are relatively prime and \displaystyle c has no perfect square divisors excluding \displaystyle 1, find \displaystyle a+b+c+d.

Solution

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