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

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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 Ptolemy's Theorem, we have $11XY=13r+7r$ and $XY=20r/11$ . Let t be the measure of angle $YOX$ . Since $YO=OX=r$ , the law of cosines on triangle $YOX$ gives us $\cos t =-79/121$ . 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 $r = \sqrt{30}$ . $\boxed{E}$ .

Solution 3

Let $r$ denote the radius of circle $C_1$ . Note that quadrilateral $ZYOX$ is cyclic. By Ptolemy's Theorem, we have $11XY=13r+7r$ and $XY=20r/11$ . Consider isosceles triangle $XOY$ . Pulling an altitude to $XY$ from $O$ , we obtain $\cos(\angle{OXY}) = \frac{10}{11}$ . Since quadrilateral $ZYOX$ is cyclic, we have $\angle{OXY}=\angle{OZY}$ , so $\cos(\angle{OXY}) = \cos(\angle{OZY})$ . Applying the Law of Cosines to triangle $OZY$ , we obtain $\frac{10}{11} = \frac{7^2+11^2-r^2}{2(7)(11)}$ . Solving gives $r=\sqrt{30}$ . $\boxed{E}$ .

-Solution by thecmd999

Solution 4

Let $P = XY \cap OZ$ . Consider an inversion about $C_1 \implies C_2 \to XY, Z \to P$ . So, $OP \cdot OZ = r^2 \implies OP = r^2/11 \implies PZ = \dfrac{121 - r^2}{11}$ . Using $\triangle YPZ \sim OXZ \implies r = \sqrt{30} \implies \boxed{E}$ .

-Solution by IDMasterz

Solution 5

$[asy] size(8cm,8cm); path circ1, circ2; circ1=circle((0,0),5); circ2=circle((3,4),3); pair O, Z; O=(3,4); Z=(3,-4); pair [] x=intersectionpoints(circ1,circ2); pair [] y=intersectionpoints(x[1]--Z,circ2); pair B; B=midpoint(x[1]--y[0]); draw(B--O); draw(x[0]--Z); draw(O--Z); draw(x[1]--Z); draw(circ1); draw(circ2); draw(rightanglemark(Z,B,O,15)); draw(x[1]--O--y[0]); label("$O$",O,NE); label("$Y$",x[0],NW); label("$X$",x[1],NW); label("$Z$",Z,S); label("$A$",y[0],SW); label("$B$",B,SW);[/asy]$ Notice that $\angle YZO=\angle XZO$ as they subtend arcs of the same length. Let $A$ be the point of intersection of $C_2$ and $XZ$ . We now have $AZ=YZ=7$ and $XA=6$ . Furthermore, notice that $\triangle XAO$ is isosceles, thus the altitude from $O$ to $XA$ bisects $XW$ at point $B$ above. By the Pythagorean Theorem, $\begin{align*}BZ^2+BO^2&=OZ^2\\(BA+AZ)^2+OA^2-BA^2&=11^1\\(3+7)^2+r^2-3^2&=121\\r^2&=30\end{align*}$ Thus, $r=\boxed{\sqrt{30}}$

2012 AMC 12A (Problems • Answer Key • Resources)
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