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

Solution 1

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 $\boxed{013}$

Solution 2

This is a cheap solution.

WLOG assume $\triangle ABC$ is isosceles. Then, $L$ is the midpoint of $AB$ , and $\angle CLB=\angle CLA=90^\circ$ . Draw the perpendicular from $I$ to $CB$ , and let it meet $CB$ at $E$ . Since $IL=2$ , $IE$ is also $2$ (they are both inradii). Set $BD$ as $x$ . Then, triangles $BLD$ and $CEI$ are similar, and $\tfrac{2}{3}=\tfrac{CI}{x}$ . Thus, $CI=\tfrac{2x}{3}$ . $\triangle CBD~\triangle CEI$ , so $\tfrac{IE}{DB}=\tfrac{CI}{CD}$ . Thus $\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}$ . Solving for $x$ , we have: $x^2-2x-15=0$ , or $x=5, -3$ . $x$ is positive, so $x=5$ . As a result, $CI=\tfrac{2x}{3}=\tfrac{10}{3}$ and the answer is $\boxed{013}$

2016 AIME I (Problems • Answer Key • Resources)
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2016 AIME I Problems/Problem 6

Contents

Problem

Solution

Solution 1

Solution 2

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