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

(→‎Problem)
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<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>
 
<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>
 +
 +
==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==
 
==Solution 1==
Line 33: Line 62:
 
==Solution 5==
 
==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(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>
 
 
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>BZ2+BO2=OZ2(BA+AZ)2+OA2−BA2=112(3+7)2+r2−32=121r2=30</cmath>Thus, <math>r=\sqrt{30}\implies\boxed{\textbf{E}}</math>
 
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>BZ2+BO2=OZ2(BA+AZ)2+OA2−BA2=112(3+7)2+r2−32=121r2=30</cmath>Thus, <math>r=\sqrt{30}\implies\boxed{\textbf{E}}</math>
  

Revision as of 22:04, 11 May 2021

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

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

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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