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 $4+3+11 = \boxed{018}$ .

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2012 AIME I Problems/Problem 12

Problem 12

Solution

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