Problem:
In △ABC let I be the center of the inscribed circle, and let the bisector of ∠ACB intersect AB at L. The line through C and L intersects the circumscribed circle of △ABC at the two points C and D. If LI=2 and LD=3, then IC=qp​, where p and q are relatively prime positive integers. Find p+q.
Solution:
Because I is the incenter of △ABC,AI is the angle bisector of ∠BAC and ∠LAI=∠IAC. It is also clear that ∠BAD=21​BD=∠BCD. Thus ∠AID=∠DCA+∠IAC=∠DCB+∠LAI=∠BAD+∠LAI=∠DAI and DA=DI=5. Because ∠LAD=∠BAD=∠BCD=∠ACD and ∠ADL=∠ADC, △DAL is similar to △DCA. Hence DLDA​=DADC​, or 35​=55+IC​ and IC=310​. The requested sum is 10+3=13​.
