# 2012 AIME I Problems/Problem 12

## Problem 12

Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

## Solution

Without loss of generality, set $CB = 1$. Then, by the Angle Bisector Theorem on triangle $DCB$, we have $CD = \frac{8}{15}$. We apply the Law of Cosines to triangle $DCB$ to get $1 + \frac{64}{225} - \frac{8}{15} = BD^{2}$, which we can simplify to get $BD = \frac{13}{15}$.

Now, we have $cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}$ by another application of the Law of Cosines to triangle $DCB$, so $cos \angle B = \frac{11}{13}$. In addition, $sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}$, so $tan \angle B = \frac{4\sqrt{3}}{11}$.

Our final answer is $4+3+11 = \boxed{018}$.