# Mock AIME 4 2006-2007 Problems/Problem 11

## Problem

Let $\triangle ABC$ be an equilateral triangle. Two points $D$ and $E$ are chosen on $\overline{AB}$ and $\overline{AC}$, respectively, such that $AD = CE$. Let $F$ be the intersection of $\overline{BE}$ and $\overline{CD}$. The area of $\triangle ABC$ is 13 and the area of $\triangle ACF$ is 3. If $\frac{CE}{EA}=\frac{p+\sqrt{q}}{r}$, where $p$, $q$, and $r$ are relatively prime positive integers, compute $p+q+r$.

## Solution

Let $AD = CE = a$, and $EA = DB = b$. Note that we want to compute the ratio $\frac{a}{b}$.

Assign a mass of $a$ to point $A$. This gives point $C$ a mass of $b$ and point $D$ a mass of $a + \frac{a^2}{b}$. Thus, $\frac{CF}{FD} = \frac{a+\frac{a^2}{b}}{b}$.

Since the ratio of areas of triangles that share an altitude is simply the ratio of their bases, we have that:

$13 \cdot \frac{a}{a+b} \cdot \frac{a+\frac{a^2}{b}}{a+\frac{a^2}{b} + b} = 3 \newline \newline \newline 13 \cdot \frac{a}{a+b} \cdot \frac{ab + a^2}{ab + a^2 + b^2} = 3 \newline \newline \newline 13 \cdot \frac{a^2}{ab+a^2+b^2} = 3 \newline \newline \newline \frac{a^2}{ab+a^2+b^2} = \frac{3}{13} \newline \newline \newline \frac{b^2 + a^2 + ab}{a^2} = \frac{13}{3} \newline \newline \newline (\frac{b}{a})^2 + \frac{b}{a} = \frac{10}{3}$.

By the quadratic formula, we find that $\frac{b}{a} = \frac{\sqrt{129}-3}{6}$, so $\frac{a}{b} = \frac{6}{\sqrt{129}-3} = \frac{3 + \sqrt{129}}{20}$.

Thus, our final answer is $3 + 129 + 20 = \boxed{152}$.