# 1995 AIME Problems/Problem 9

## Problem

Triangle $\displaystyle ABC$ is isosceles, with $\displaystyle AB=AC$ and altitude $\displaystyle AM=11.$ Suppose that there is a point $\displaystyle D$ on $\displaystyle \overline{AM}$ with $\displaystyle AD=10$ and $\displaystyle \angle BDC=3\angle BAC.$ Then the perimeter of $\displaystyle \triangle ABC$ may be written in the form $\displaystyle a+\sqrt{b},$ where $\displaystyle a$ and $\displaystyle b$ are integers. Find $\displaystyle a+b.$

## Solution

Let $\angle CAM=x$, $\angle CDM=3x$. Then, $(tan 3x)/(tan x)=(CM/1)/(CM/11)=11$. Expanding tan 3x using the angle sum formula gives $tan 3x=(3tan x-tan^3x)/(1-3tan^2x)$. Thus, $(3-tan^2x)/(1-3tan^2x)=11$. Solving, we get $tan x=1/2$. Hence, $CM=11/2$ and AC=$11\sqrt{5}/2$ by Pythag. The total perimeter is double the sum of these, which is $\sqrt{605}+11$. The answer is then $616$.