Mock AIME II 2012 Problems/Problem 4

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 $7+4=\boxed{011}$ .

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Mock AIME II 2012 Problems/Problem 4

Problem

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