2016 AIME I Problems/Problem 6

Problem

In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Solution

It is well known that $DA = DI = DB$ and so we have $DA = DB = 5$ . Then $\triangle DLB \sim \triangle ALC$ and so $\frac{AL}{AC} = \frac{3}{5}$ and from the angle bisector theorem $\frac{CI}{IL} = \frac{5}{3}$ so $CI = \frac{10}{3}$ and our answer is $\boxed{013}$

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2016 AIME I Problems/Problem 6

Problem

Solution