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

Suppose we label the angles as shown below. $[asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label("$A$",A,SW,fontsize(8)); label("$B$",B,SE,fontsize(8)); label("$C$",C,N,fontsize(8)); label("$D$",D,S,fontsize(8)); label("$I$",I,NE,fontsize(8)); label("$L$",L,SW,fontsize(8)); label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy]$ 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 $\angle LAI$ , so by the Angle Bisector Theorem $\frac{CI}{IL} =\frac{AC}{AL}= \frac{5}{3}$ . Thus $CI = \frac{10}{3}$ , and the answer is $\boxed{013}$ .

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 $\boxed{013}$

Solution 3

WLOG assume $\triangle ABC$ is isosceles (with vertex $C$ ). Let $O$ be the center of the circumcircle, $R$ the circumradius, and $r$ the inradius. A simple sketch will reveal that $\triangle ABC$ must be obtuse (as an acute triangle will result in $LI$ being greater than $DL$ ) and that $O$ and $I$ are collinear. Next, if $OI=d$ , $DO+OI=R+d$ and $R+d=DL+LI=5$ . Euler gives us that $d^{2}=R(R-2r)$ , and in this case, $r=LI=2$ . Thus, $d=\sqrt{R^{2}-4R}$ . Solving for $d$ , we have $R+\sqrt{R^{2}-4R}=5$ , then $R^{2}-4R=25-10R+R^{2}$ , yielding $R=\frac{25}{6}$ . Next, $R+d=5$ so $d=\frac{5}{6}$ . Finally, $OC=OI+IC$ gives us $R=d+IC$ , and $IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}$ . Our answer is then $\boxed{013}$ .

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

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Solution

Solution 1

Solution 2

Solution 3

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