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 $\frac{AB}{AE}=\frac{BC}{BE}\Longrightarrow\frac{y+1}{x}=\frac{3}{2}$ 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 $1+2+3+\frac{10}{7}+\frac{8}{7}=\frac{60}{7}\Longrightarrow\boxed{067}$

2020 CIME II (Problems • Answer Key • Resources)
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2020 CIME II Problems/Problem 1

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