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

Revision as of 01:40, 2 October 2022

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 $C_1$ ?

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

Diagram

$[asy] size(8cm,8cm); path circ1, circ2; circ1=circle((0,0),5); circ2=circle((3,4),3); pair O, Z; O=(3,4); Z=(3,-4); pair [] x=intersectionpoints(circ1,circ2); pair [] y=intersectionpoints(x[1]--Z,circ2); pair B; B=midpoint(x[1]--y[0]); draw(B--O); draw(x[0]--Z); draw(O--Z); draw(x[1]--Z); draw(O--x[0]); draw(circ1); draw(circ2); draw(rightanglemark(Z,B,O,15)); draw(x[1]--O--y[0]); label("$O$",O,NE); label("$Y$",x[0],SE); label("$X$",x[1],NW); label("$Z$",Z,S); label("$A$",y[0],SW); label("$B$",B,SW);[/asy]$

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

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

Note

Instead of using the Extended Law of Sines, you can note that $OX = OY \implies \text{arc}\ OX =\text{arc}\ OY \implies \angle{OZY} \cong \angle{OZX}$ , since the angles inscribe arcs of the same length.

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

Notice that $\angle YZO=\angle XZO$ as they subtend arcs of the same length. Let $A$ be the point of intersection of $C_1$ 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 $XZ$ at point $B$ above. By the Pythagorean Theorem, $\begin{align*}BZ^2+BO^2&=OZ^2\\(BA+AZ)^2+OA^2-BA^2&=11^2\\(3+7)^2+r^2-3^2&=121\\r^2&=30\end{align*}$ Thus, $r=\sqrt{30}\implies\boxed{\textbf{E}}$

Solution 6

Use the diagram above. Notice that $\angle YZO=\angle XZO$ as they subtend arcs of the same length. Let $A$ be the point of intersection of $C_1$ and $XZ$ . We now have $AZ=YZ=7$ and $XA=6$ . Consider the power of point $Z$ with respect to Circle $O,$ we have $13\cdot 7 = (11 + r)(11 - r) = 11^2 - r^2,$ which gives $r=\boxed{\sqrt{30}}.$

Solution 7 (Only Law of Cosines)

Note that $OX$ and $OY$ are the same length, which is also the radius $R$ we want. Using the law of cosines on $\triangle OYZ$ , we have $11^2=R^2+7^2-2\cdot 7 \cdot R \cdot \cos\theta$ , where $\theta$ is the angle formed by $\angle{OYZ}$ . Since $\angle{OYZ}$ and $\angle{OXZ}$ are supplementary, $\angle{OXZ}=\pi-\theta$ . Using the law of cosines on $\triangle OXZ$ , $11^2=13^2+R^2-2 \cdot 13 \cdot R \cdot \cos(\pi-\theta)$ . As $\cos(\pi-\theta)=-\cos\theta$ , $11^2=13^2+R^2+\cos\theta$ . Solving for theta on the first equation and substituting gives $\frac{72-R^2}{14R}=\frac{48+R^2}{26R}$ . Solving for R gives $R=\textbf{(E)}\ \boxed{\sqrt{30}}$ .

Solution 8

We first note that $C_2$ is the circumcircle of both $\triangle XOZ$ and $\triangle OYZ$ . Thus the circumradius of both the triangles are equal. We set the radius of $C_1$ as $r$ , and noting that the circumradius of a triangle is $\frac{abc}{4A}$ and that the area of a triangle by Heron's formula is $\sqrt{(S)(S-a)(S-b)(S-c)}$ with $S$ as the semi-perimeter we have the following, $\begin{align*}\dfrac{r \cdot 13 \cdot 11}{4\sqrt{(12 + \frac{r}{2})(12 - \frac{r}{2})(1 + \frac{r}{2})(\frac{r}{2} - 1)}} &= \dfrac{r \cdot 7 \cdot 11}{4\sqrt{(9 + \frac{r}{2})(9 - \frac{r}{2})(2 + \frac{r}{2})(\frac{r}{2} - 2)}} \\ \dfrac{13}{\sqrt{(144- \frac{r^2}{4})(\frac{r^2}{4} - 1)}} &= \dfrac{7}{\sqrt{(81- \frac{r^2}{4})(\frac{r^2}{4} - 4)}} \\ 169 \cdot (81 - \frac{r^2}{4})(\frac{r^2}{4} - 4) &= 49 \cdot (144 - \frac{r^2}{4})(\frac{r^2}{4} - 1) .\end{align*}$ Now substituting $a = \frac{r^2}{4}$ , $\begin{align*}169a^2 - (85) \cdot 169 a + 4 \cdot 81 \cdot 169 &= 49a^2 - (145) \cdot 49 a + 144 \cdot 49 \\ 120a^2 - 7260a + 47700 &= 0 \\ 2a^2 - 121a + 795 &= 0 \\ (2a-15)(a-53) &= 0 \\ a = \frac{15}{2}, 53.\end{align*}$ This gives us 2 values for $r$ namely $r = \sqrt{4 \cdot \frac{15}{2}} = \sqrt{30}$ and $r = \sqrt{4 \cdot 53} = 2\sqrt{53}$ .

Now notice that we can apply Ptolemy's theorem on $XOYZ$ to find $XY$ in terms of $r$ . We get $\begin{align*}11 \cdot XY &= 13r + 7r \\ XY &= \frac{20r}{11}.\end{align*}$ Here we substitute our $2$ values of $r$ receiving $XY = \frac{20\sqrt{30}}{11}, \frac{40\sqrt{53}}{11}$ . Notice that the latter of the $2$ cases does not satisfy the triangle inequality for $\triangle XYZ$ as $\frac{40\sqrt{53}}{11} \approx 26.5 > 7 + 13 = 20$ . But the former does thus our answer is $\textbf{(E)}\ \boxed{\sqrt{30}}$ .

~Aaryabhatta1

Solution 9 (Similar Triangles)

We first apply Ptolemy's Theorem on cyclic quadrilateral $XZYO$ to get $13r+7r = 11 \cdot XY \Longrightarrow XY=\frac{20r}{11}$ . Since $\angle ZXY = \angle ZOY$ and $\angle XZO = \angle XYO$ . From this, we can see $\triangle ZPY \sim \triangle XPO$ and $\triangle ZPX \sim \triangle YPO$ . That means $ZP:PY = 13:r, \: ZP:PX = 7:r, \: XP:PO = 13:r$ . So, if you let $PY=x$ , you will get $ZP = \frac{13x}{r}$ . Continuing in this fashion, we can get $XP = \frac{13x}{r} \cdot \frac{r}{7} = \frac{13x}{7}$ and $PO = \frac{13x}{7} \cdot \frac{r}{13} = \frac{xr}{7}$ . Since $XY = \frac{20r}{11} = XP+PY$ , we have $x+\frac{13x}{7} = \frac{20r}{11}$ which gives us $x=\frac{7r}{11}$ . Plugging it into $ZO = 11 = ZP+PO$ gives

$\frac{13x}{r} + \frac{xr}{7} = \frac{13 \cdot \frac{7r}{11}}{r} + \frac{r \cdot \frac{7r}{11}}{7} = \frac{91}{11} + \frac{r^2}{11} = 11.$

Solving for $r$ yields $r=\boxed{\sqrt{30}}$ .

~sml1809

@@ Line 5: / Line 5: @@
 <math> \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} </math>
-== Solution ==
+==Diagram==
-===Solution 1===
-Let <math>r</math> denote the radius of circle <math>C_1</math>. Note that quadrilateral <math>ZYOX</math> is cyclic. By Ptolemy's Theorem, we have <math>11XY=13r+7r</math> and <math>XY=20r/11</math>. Let t be the measure of angle <math>YOX</math>. Since <math>YO=OX=r</math>, the law of cosines on triangle <math>YOX</math> gives us <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===
-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
-<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>.
-===Solution 3===
-Let <math>r</math> denote the radius of circle <math>C_1</math>. Note that quadrilateral <math>ZYOX</math> is cyclic. By Ptolemy's Theorem, we have <math>11XY=13r+7r</math> and <math>XY=20r/11</math>. Consider isosceles triangle <math>XOY</math>. Pulling an altitude to <math>XY</math> from <math>O</math>, we obtain <math>\cos(\angle{OXY}) = \frac{10}{11}</math>. Since quadrilateral <math>ZYOX</math> is cyclic, we have <math>\angle{OXY}=\angle{OZY}</math>, so <math>\cos(\angle{OXY}) = \cos(\angle{OZY})</math>. Applying the Law of Cosines to triangle <math>OZY</math>, we obtain <math>\frac{10}{11} = \frac{7^2+11^2-r^2}{2(7)(11)}</math>. Solving gives <math>r=\sqrt{30}</math>. <math>\boxed{E}</math>.
--Solution by '''thecmd999'''
-===Solution 4===
-Let <math>P = XY \cap OZ</math>. Consider an inversion about <math>C_1 \implies C_2 \to XY, Z \to P</math>. So,  <math>OP \cdot OZ = r^2 \implies OP = r^2/11 \implies PZ = \dfrac{121 - r^2}{11}</math>. Using <math>\triangle YPZ \sim OXZ \implies  r = \sqrt{30} \implies \boxed{E}</math>.
--Solution by '''IDMasterz'''
-===Solution 5===
 <asy>
 size(8cm,8cm);
@@ Line 52: / Line 22: @@
 draw(O--Z);
 draw(x[1]--Z);
+draw(O--x[0]);
 draw(circ1);
 draw(circ2);
@@ Line 57: / Line 28: @@
 draw(x[1]--O--y[0]);
 label("$O$",O,NE);
-label("$Y$",x[0],NW);
+label("$Y$",x[0],SE);
 label("$X$",x[1],NW);
 label("$Z$",Z,S);
 label("$A$",y[0],SW);
 label("$B$",B,SW);</asy>
-Notice that <math>\angle YZO=\angle XZO</math> as they subtend arcs of the same length. Let <math>A</math> be the point of intersection of <math>C_2</math> and <math>XZ</math>. We now have <math>AZ=YZ=7</math> and <math>XA=6</math>. Furthermore, notice that <math>\triangle XAO</math> is isosceles, thus the altitude from <math>O</math> to <math>XA</math> bisects <math>XW</math> at point <math>B</math> above. By the Pythagorean Theorem, <cmath>\begin{align*}BZ^2+BO^2&=OZ^2\\(BA+AZ)^2+OA^2-BA^2&=11^2\\(3+7)^2+r^2-3^2&=121\\r^2&=30\end{align*}</cmath>Thus, <math>r=\sqrt{30}\implies\boxed{\textbf{E}}</math>
+==Solution 1==
+Let <math>r</math> denote the radius of circle <math>C_1</math>. Note that quadrilateral <math>ZYOX</math> is cyclic. By Ptolemy's Theorem, we have <math>11XY=13r+7r</math> and <math>XY=20r/11</math>. Let <math>t</math> be the measure of angle <math>YOX</math>. Since <math>YO=OX=r</math>, the law of cosines on triangle <math>YOX</math> gives us <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=\boxed{(E)\sqrt{30}}</math>.
+==Solution 2==
+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
+<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>.
+===Note===
+Instead of using the Extended Law of Sines, you can note that <math>OX = OY \implies \text{arc}\ OX =\text{arc}\ OY \implies \angle{OZY} \cong \angle{OZX}</math>, since the angles inscribe arcs of the same length.
+==Solution 3==
+Let <math>r</math> denote the radius of circle <math>C_1</math>. Note that quadrilateral <math>ZYOX</math> is cyclic. By Ptolemy's Theorem, we have <math>11XY=13r+7r</math> and <math>XY=20r/11</math>. Consider isosceles triangle <math>XOY</math>. Pulling an altitude to <math>XY</math> from <math>O</math>, we obtain <math>\cos(\angle{OXY}) = \frac{10}{11}</math>. Since quadrilateral <math>ZYOX</math> is cyclic, we have <math>\angle{OXY}=\angle{OZY}</math>, so <math>\cos(\angle{OXY}) = \cos(\angle{OZY})</math>. Applying the Law of Cosines to triangle <math>OZY</math>, we obtain <math>\frac{10}{11} = \frac{7^2+11^2-r^2}{2(7)(11)}</math>. Solving gives <math>r=\sqrt{30}</math>. <math>\boxed{E}</math>.
+-Solution by '''thecmd999'''
+==Solution 4==
+Let <math>P = XY \cap OZ</math>. Consider an inversion about <math>C_1 \implies C_2 \to XY, Z \to P</math>. So,  <math>OP \cdot OZ = r^2 \implies OP = r^2/11 \implies PZ = \dfrac{121 - r^2}{11}</math>. Using <math>\triangle YPZ \sim OXZ \implies  r = \sqrt{30} \implies \boxed{E}</math>.
+-Solution by '''IDMasterz'''
+==Solution 5==
+Notice that <math>\angle YZO=\angle XZO</math> as they subtend arcs of the same length. Let <math>A</math> be the point of intersection of <math>C_1</math> and <math>XZ</math>. We now have <math>AZ=YZ=7</math> and <math>XA=6</math>. Furthermore, notice that <math>\triangle XAO</math> is isosceles, thus the altitude from <math>O</math> to <math>XA</math> bisects <math>XZ</math> at point <math>B</math> above. By the Pythagorean Theorem, <cmath>\begin{align*}BZ^2+BO^2&=OZ^2\\(BA+AZ)^2+OA^2-BA^2&=11^2\\(3+7)^2+r^2-3^2&=121\\r^2&=30\end{align*}</cmath>Thus, <math>r=\sqrt{30}\implies\boxed{\textbf{E}}</math>
+==Solution 6==
+Use the diagram above. Notice that <math>\angle YZO=\angle XZO</math> as they subtend arcs of the same length. Let <math>A</math> be the point of intersection of <math>C_1</math> and <math>XZ</math>. We now have <math>AZ=YZ=7</math> and <math>XA=6</math>. Consider the power of point <math>Z</math> with respect to Circle <math>O,</math> we have <math>13\cdot 7 = (11 + r)(11 - r) = 11^2 - r^2,</math> which gives <math>r=\boxed{\sqrt{30}}.</math>
+==Solution 7 (Only Law of Cosines)==
+Note that <math>OX</math> and <math>OY</math> are the same length, which is also the radius <math>R</math> we want. Using the law of cosines on <math>\triangle OYZ</math>, we have <math>11^2=R^2+7^2-2\cdot 7 \cdot R \cdot \cos\theta</math>, where <math>\theta</math> is the angle formed by <math>\angle{OYZ}</math>. Since <math>\angle{OYZ}</math> and <math>\angle{OXZ}</math> are supplementary, <math>\angle{OXZ}=\pi-\theta</math>. Using the law of cosines on <math>\triangle OXZ</math>, <math>11^2=13^2+R^2-2 \cdot 13 \cdot R \cdot \cos(\pi-\theta)</math>. As <math>\cos(\pi-\theta)=-\cos\theta</math>, <math>11^2=13^2+R^2+\cos\theta</math>. Solving for theta on the first equation and substituting gives <math>\frac{72-R^2}{14R}=\frac{48+R^2}{26R}</math>. Solving for R gives <math>R=\textbf{(E)}\ \boxed{\sqrt{30}} </math>.
+==Solution 8==
+We first note that <math>C_2</math> is the circumcircle of both <math>\triangle XOZ</math> and <math>\triangle OYZ</math>. Thus the circumradius of both the triangles are equal. We set the radius of <math>C_1</math> as <math>r</math>, and noting that the circumradius of a triangle is <math>\frac{abc}{4A}</math> and that the area of a triangle by Heron's formula is <math>\sqrt{(S)(S-a)(S-b)(S-c)}</math> with <math>S</math> as the semi-perimeter we have the following, <cmath>\begin{align*}\dfrac{r \cdot 13 \cdot 11}{4\sqrt{(12 + \frac{r}{2})(12 - \frac{r}{2})(1 + \frac{r}{2})(\frac{r}{2} - 1)}} &= \dfrac{r \cdot 7 \cdot 11}{4\sqrt{(9 + \frac{r}{2})(9 - \frac{r}{2})(2 + \frac{r}{2})(\frac{r}{2} - 2)}} \\ \dfrac{13}{\sqrt{(144- \frac{r^2}{4})(\frac{r^2}{4} - 1)}} &= \dfrac{7}{\sqrt{(81- \frac{r^2}{4})(\frac{r^2}{4} - 4)}} \\ 169 \cdot (81 - \frac{r^2}{4})(\frac{r^2}{4} - 4) &= 49 \cdot (144 - \frac{r^2}{4})(\frac{r^2}{4} - 1) .\end{align*}</cmath>
+Now substituting <math>a = \frac{r^2}{4}</math>, <cmath>\begin{align*}169a^2 - (85) \cdot 169 a + 4 \cdot 81 \cdot 169 &= 49a^2 - (145) \cdot 49 a + 144 \cdot 49 \\ 120a^2 - 7260a + 47700 &= 0 \\ 2a^2 - 121a + 795 &= 0 \\ (2a-15)(a-53) &= 0 \\ a = \frac{15}{2}, 53.\end{align*}</cmath>
+This gives us 2 values for <math>r</math> namely <math>r = \sqrt{4 \cdot \frac{15}{2}} = \sqrt{30}</math> and <math>r = \sqrt{4 \cdot 53} = 2\sqrt{53}</math>.
+Now notice that we can apply Ptolemy's theorem on <math>XOYZ</math> to find <math>XY</math> in terms of <math>r</math>. We get <cmath>\begin{align*}11 \cdot XY &= 13r + 7r \\ XY &= \frac{20r}{11}.\end{align*}</cmath>
+Here we substitute our <math>2</math> values of <math>r</math> receiving <math>XY = \frac{20\sqrt{30}}{11}, \frac{40\sqrt{53}}{11}</math>. Notice that the latter of the <math>2</math> cases does not satisfy the triangle inequality for <math>\triangle XYZ</math> as <math>\frac{40\sqrt{53}}{11} \approx 26.5 > 7 + 13 = 20</math>. But the former does thus our answer is <math>\textbf{(E)}\ \boxed{\sqrt{30}}</math>.
+~Aaryabhatta1
+==Solution 9 (Similar Triangles)==
+We first apply [[Ptolemy's Theorem]] on [[cyclic quadrilateral]] <math>XZYO</math> to get <math>13r+7r = 11 \cdot XY \Longrightarrow XY=\frac{20r}{11}</math>.  Since <math>\angle ZXY = \angle ZOY</math> and <math>\angle XZO = \angle XYO</math>. From this, we can see <math>\triangle ZPY \sim \triangle XPO</math> and <math>\triangle ZPX \sim \triangle YPO</math>. That means <math>ZP:PY = 13:r, \: ZP:PX = 7:r, \: XP:PO = 13:r</math>. So, if you let <math>PY=x</math>, you will get <math>ZP = \frac{13x}{r}</math>. Continuing in this fashion, we can get <math>XP = \frac{13x}{r} \cdot \frac{r}{7} = \frac{13x}{7}</math> and <math>PO = \frac{13x}{7} \cdot \frac{r}{13} = \frac{xr}{7}</math>. Since <math>XY = \frac{20r}{11} = XP+PY</math>, we have <math>x+\frac{13x}{7} = \frac{20r}{11}</math> which gives us <math>x=\frac{7r}{11}</math>. Plugging it into <math>ZO = 11 = ZP+PO</math> gives
+<cmath>\frac{13x}{r} + \frac{xr}{7} = \frac{13 \cdot \frac{7r}{11}}{r} + \frac{r \cdot \frac{7r}{11}}{7} = \frac{91}{11} + \frac{r^2}{11} = 11.</cmath>
+Solving for <math>r</math> yields <math>r=\boxed{\sqrt{30}}</math>.
+~sml1809
 == 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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