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

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== Solution ==
 
== Solution ==
  
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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>.
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===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>.
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===Solution 2===
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Let us call the <math>r</math> the radius of circle <math>C_1</math>, and <math>R</math> the radius of <math>C_2</math>. Consider <math>\triangle OZX</math> and <math>\triangle OZY</math>. Both of these triangles have the same circumcircle (<math>C_2</math>). From the Extended Law of Sines, we see that <math>\frac{r}{\sin{\angle{OZY}}} = \frac{r}{\sin{\angle{OZX}}}= 2R</math>. Therefore, <math>\angle{OZY} \cong \angle{OZX}</math>. We will now apply the Law of Cosines to <math>\triangle OZX</math> and <math>\triangle OZY</math> and get the equations
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 +
<math>r^2 = 13^2 + 11^2 - 2 \cdot 13 \cdot 11 \cdot \cos{\angle{OZX}}</math>,
 +
 
 +
<math>r^2 = 11^2 + 7^2 - 2 \cdot 11 \cdot 7 \cdot \cos{\angle{OZY}}</math>,
 +
 
 +
respectively. Because <math>\angle{OZY} \cong \angle{OZX}</math>, this is a system of two equations and two variables. Solving for <math>r</math> gives <math>r = \sqrt{30}</math>. <math>\boxed{E}</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2012|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2012|ab=A|num-b=15|num-a=17}}

Revision as of 23:02, 13 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 $YOR$. 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 $r=\sqrt{30}$. $\boxed{E}$.

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

See Also

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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