# Mock AIME II 2012 Problems/Problem 4

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## Problem

Let $\triangle ABC$ be a triangle, and let $I_A$, $I_B$, and $I_C$ be the points where the angle bisectors of $A$, $B$, and $C$, respectfully, intersect the sides opposite them. Given that $AI_B=5$, $CI_B=4$, and $CI_A=3$, then the ratio $AI_C:BI_C$ can be written in the form $m/n$ where $m$ and $n$ are positive relatively prime integers. Find $m+n$.

## Solution

Let $m=BI_A$ and $n=AB$. We use the Angle Bisector Theorem twice to get two different equations relating $m$ and $n$. From the angle bisector $BI_B$, we have the proportion $\dfrac{AI_B}{CI_B}=\dfrac{AB}{CB}$, or $\dfrac{5}{4}=\dfrac{n}{m+3}$. From the angle bisector $AI_A$, we have the proportion $\dfrac{BI_A}{CI_A}=\dfrac{AB}{AA}$, or $\dfrac{m}{3}=\dfrac{n}{9}$. Multiply the second equation by $9$ to get $n=3m$, then plug this expression for $n$ into the first equation to get $\dfrac{5}{4}=\dfrac{3m}{m+3}$. Solving this equation gives us $m=\dfrac{15}{7}$. Finally, the ratio $AI_C:BI_C$ is, by the Angle Bisector Theorem, equivalent to the ratio $\dfrac{AC}{BC}$, which is equal to $\dfrac{9}{3+15/7}=\dfrac{9}{36/7}=\dfrac{7}{4}$ so the answer is $7+4=\boxed{011}$.