Question

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 $m + n + p.$

Answer

Answer: 18

Concepts:angle bisector theorem law of cosines trigonometry

Difficulty rating: 2840

Small Hint:

In triangle $DCB,$ ray $CE$ bisects angle $DCB,$ which measures $60^\circ,$ so $\frac{CD}{CB} = \frac{DE}{EB} = \frac{8}{15}$

Big Hint:

Take $CD = 8$ and $CB = 15;$ the Law of Cosines gives $BD,$ and a second application gives $\cos B$

Solution:

The trisectors make $\angle ACD = \angle DCE = \angle ECB$ $= 30^\circ.$ In triangle $DCB,$ ray $CE$ bisects angle $DCB,$ which measures $60^\circ,$ so the angle bisector theorem gives $\frac{CD}{CB} = \frac{DE}{EB} = \frac{8}{15}.$ Scale the triangle so that $CD = 8$ and $CB = 15.$

By the Law of Cosines in triangle $DCB,$ $\begin{aligned} BD^2 &= 8^2 + 15^2 \\ &\quad {}- 2 \cdot 8 \cdot 15 \cos 60^\circ \\ &= 169, \end{aligned}$ so $BD = 13.$ Applying the Law of Cosines again in the same triangle, $8^2 = 13^2 + 15^2 - 2 \cdot 13 \cdot 15 \cos B,$ which gives $\cos B = \frac{11}{13}.$

Then $\sin B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13},$ so $\tan B = \frac{4\sqrt{3}}{11}$ and $m + n + p = 4 + 11 + 3 = 18.$

2012 AIME I Problem 12

Answer: 18

Difficulty rating: 2840

Small Hint:

Big Hint:

Solution:

Problem 12 in Other Years