2012 AMC 12A Problems/Problem 16

Problem

Circle $C_1$ has its center $O$ lying on circle $C_2$ . The two circles meet at $X$ and $Y$ . Point $Z$ in the exterior of $C_1$ lies on circle $C_2$ and $XZ=13$ , $OZ=11$ , and $YZ=7$ . What is the radius of circle $C_1$ ?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ \sqrt{26}\qquad\textbf{(C)}\ 3\sqrt{3}\qquad\textbf{(D)}\ 2\sqrt{7}\qquad\textbf{(E)}\ \sqrt{30}$

Solution

Let $r$ denote the radius of circle $C_1$ . Note that quadrilateral $ZYOX$ is cyclic. By Ptolomys Theorem we have that $11XY=13r+7r$ so that $XY=20r/11$ . Let t be the measure of angle $YOR$ . Since $YO=OX=r$ by the law of cosines on triangle $YOX$ we obtain $\cos t =-79/121$ . Again since $ZYOX$ is cyclic, the measure of angle $YZX=180-t$ . We apply the law of cosines to triangle $ZYX$ so that $XY^2=7^2+13^2-2(7)(13)\cos(180-t)$ . Since $\cos(180-t)=-\cos t=79/121$ we obtain $XY^2=12000/121$ . But $XY^2=400r^2/121$ so that $r=\sqrt{30}$ .

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2012 AMC 12A Problems/Problem 16

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