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 $IC=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Suppose we label the angles as shown below.

As $\angle BCD$ and $\angle BAD$ intercept the same arc, we know that $\angle BAD=\gamma$. Similarly, $\angle ABD=\gamma$. Also, using $\triangle ICA$, we find $\angle CIA=180-\alpha-\gamma$. Therefore, $\angle AID=\alpha+\gamma$. Therefore, $\angle DAI=\angle AID=\alpha+\gamma$, so $\triangle AID$ must be isosceles with $AD=ID=5$. Similarly, $BD=ID=5$. Then $\triangle DLB \sim \triangle ALC$, hence $\frac{AL}{AC} = \frac{3}{5}$. Also, $AI$ bisects $\angle LAC$, so by the Angle Bisector Theorem $\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}$. Thus $CI = \frac{10}{3}$, and the answer is $\boxed{013}$.