Mock AIME 2 2006-2007 Problems/Problem 9

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


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