Question

In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $\overline{AB}$ 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 $IC = \frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

Answer

Solution:

The incenter $I$ lies on the bisector $\overline{CL},$ between $C$ and $L.$ In triangle $ACI,$ the exterior angle at $I$ gives $\angle DIA = \angle IAC + \angle ICA.$ On the other hand, $\angle DAB = \angle DCB$ (both subtend arc $DB$ ) and $\angle DCB = \angle ICA$ (the bisector), so $\angle DAI = \angle DAB + \angle BAI = \angle ICA + \angle IAC = \angle DIA.$ Hence triangle $DAI$ is isosceles with $DA = DI = DL + LI = 5.$

Triangles $DAL$ and $DCA$ have a common angle at $D,$ and $\angle DAL = \angle DAB = \angle DCB = \angle DCA,$ so they are similar. Therefore $\frac{DA}{DC} = \frac{DL}{DA},$ giving $DC = \frac{DA^2}{DL} = \frac{25}{3}.$

Finally $IC = DC - DI = \frac{25}{3} - 5 = \frac{10}{3},$ so $p + q = 10 + 3 = 13.$

2016 AIME I Problem 6

Solution:

Problem 6 in Other Years