Difference between revisions of "2012 AMC 12A Problems/Problem 16"

Revision as of 21:28, 17 February 2012

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

Solution 1

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 $YOX$ . 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 $r=\sqrt{30}$ . $\boxed{E}$ .

Solution 2

Let us call the $r$ the radius of circle $C_1$ , and $R$ the radius of $C_2$ . Consider $\triangle OZX$ and $\triangle OZY$ . Both of these triangles have the same circumcircle ( $C_2$ ). From the Extended Law of Sines, we see that $\frac{r}{\sin{\angle{OZY}}} = \frac{r}{\sin{\angle{OZX}}}= 2R$ . Therefore, $\angle{OZY} \cong \angle{OZX}$ . We will now apply the Law of Cosines to $\triangle OZX$ and $\triangle OZY$ and get the equations

$r^2 = 13^2 + 11^2 - 2 \cdot 13 \cdot 11 \cdot \cos{\angle{OZX}}$ ,

$r^2 = 11^2 + 7^2 - 2 \cdot 11 \cdot 7 \cdot \cos{\angle{OZY}}$ ,

respectively. Because $\angle{OZY} \cong \angle{OZX}$ , this is a system of two equations and two variables. Solving for $r$ gives $r = \sqrt{30}$ . $\boxed{E}$ .

@@ Line 8: / Line 8: @@
 ===Solution 1===
-Let <math>r</math> denote the radius of circle <math>C_1</math>. Note that quadrilateral <math>ZYOX</math> is cyclic. By Ptolomys Theorem we have that <math>11XY=13r+7r</math> so that <math>XY=20r/11</math>. Let t be the measure of angle <math>YOR</math>. Since <math>YO=OX=r</math> by the law of cosines on triangle <math>YOX</math> we obtain <math>\cos t =-79/121</math>. Again since <math>ZYOX</math> is cyclic, the measure of angle <math>YZX=180-t</math>. We apply the law of cosines to triangle <math>ZYX</math> so that <math>XY^2=7^2+13^2-2(7)(13)\cos(180-t)</math>. Since <math>\cos(180-t)=-\cos t=79/121</math> we obtain <math>XY^2=12000/121</math>. But<math> XY^2=400r^2/121</math> so that <math>r=\sqrt{30}</math>. <math>\boxed{E}</math>.
+Let <math>r</math> denote the radius of circle <math>C_1</math>. Note that quadrilateral <math>ZYOX</math> is cyclic. By Ptolomys Theorem we have that <math>11XY=13r+7r</math> so that <math>XY=20r/11</math>. Let t be the measure of angle <math>YOX</math>. Since <math>YO=OX=r</math> by the law of cosines on triangle <math>YOX</math> we obtain <math>\cos t =-79/121</math>. Again since <math>ZYOX</math> is cyclic, the measure of angle <math>YZX=180-t</math>. We apply the law of cosines to triangle <math>ZYX</math> so that <math>XY^2=7^2+13^2-2(7)(13)\cos(180-t)</math>. Since <math>\cos(180-t)=-\cos t=79/121</math> we obtain <math>XY^2=12000/121</math>. But<math> XY^2=400r^2/121</math> so that <math>r=\sqrt{30}</math>. <math>\boxed{E}</math>.
 ===Solution 2===

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