# 2020 CIME II Problems/Problem 1

## Problem

Let $ABC$ be a triangle. The bisector of $\angle ABC$ intersects $\overline{AC}$ at $E$, and the bisector of $\angle ACB$ intersects $\overline{AB}$ at $F$. If $BF=1$, $CE=2$, and $BC=3$, then the perimeter of $\triangle ABC$ can be expressed in the form $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution

For simplicity, let $AE=x$ and $AF=y$. By the angle bisector theorem, we have that $$\frac{AB}{AE}=\frac{BC}{CE}\Longrightarrow\frac{y+1}{x}=\frac{3}{2}$$ using $\angle B$ as the bisected angle. Using $\angle C$ as the bisected angle, we have that $$\frac{AC}{AF}=\frac{BC}{BF}\Longrightarrow\frac{x+2}{y}=3$$These two equations form a system of equations: $$\left\{\begin{matrix} 2y+2=3x \\ x+2=3y \end{matrix}\right.\Longrightarrow x=\frac{10}{7}, y=\frac{8}{7}$$ Therefore, the perimeter is $1+2+3+\frac{10}{7}+\frac{8}{7}=\frac{60}{7}\Longrightarrow\boxed{067}$\\ ~bhargavakanakapura

The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 