Difference between revisions of "2012 AMC 12A Problems/Problem 16"
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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> | ||
− | == Solution == | + | ==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=\boxed{(E)\sqrt{30}}</math>. | ||
− | + | ==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 | 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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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>. | 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>. | 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>. | ||
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-Solution by '''thecmd999''' | -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>. | 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>. | ||
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-Solution by '''IDMasterz''' | -Solution by '''IDMasterz''' | ||
− | + | ==Solution 5== | |
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draw(O--Z); | draw(O--Z); | ||
draw(x[1]--Z); | draw(x[1]--Z); | ||
+ | draw(O--x[0]); | ||
draw(circ1); | draw(circ1); | ||
draw(circ2); | draw(circ2); | ||
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draw(x[1]--O--y[0]); | draw(x[1]--O--y[0]); | ||
label("$O$",O,NE); | label("$O$",O,NE); | ||
− | label("$Y$",x[0], | + | label("$Y$",x[0],SE); |
label("$X$",x[1],NW); | label("$X$",x[1],NW); | ||
label("$Z$",Z,S); | label("$Z$",Z,S); | ||
label("$A$",y[0],SW); | label("$A$",y[0],SW); | ||
label("$B$",B,SW);</asy> | 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> | + | 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>. | ||
== See Also == | == See Also == |
Revision as of 22:21, 6 October 2020
Contents
Problem
Circle has its center lying on circle . The two circles meet at and . Point in the exterior of lies on circle and , , and . What is the radius of circle ?
Solution 1
Let denote the radius of circle . Note that quadrilateral is cyclic. By Ptolemy's Theorem, we have and . Let t be the measure of angle . Since , the law of cosines on triangle gives us . Again since is cyclic, the measure of angle . We apply the law of cosines to triangle so that . Since we obtain . But so that .
Solution 2
Let us call the the radius of circle , and the radius of . Consider and . Both of these triangles have the same circumcircle (). From the Extended Law of Sines, we see that . Therefore, . We will now apply the Law of Cosines to and and get the equations
,
,
respectively. Because , this is a system of two equations and two variables. Solving for gives . .
Solution 3
Let denote the radius of circle . Note that quadrilateral is cyclic. By Ptolemy's Theorem, we have and . Consider isosceles triangle . Pulling an altitude to from , we obtain . Since quadrilateral is cyclic, we have , so . Applying the Law of Cosines to triangle , we obtain . Solving gives . .
-Solution by thecmd999
Solution 4
Let . Consider an inversion about . So, . Using .
-Solution by IDMasterz
Solution 5
Notice that as they subtend arcs of the same length. Let be the point of intersection of and . We now have and . Furthermore, notice that is isosceles, thus the altitude from to bisects at point above. By the Pythagorean Theorem, Thus,
Solution 6
Use the diagram above. Notice that as they subtend arcs of the same length. Let be the point of intersection of and . We now have and . Consider the power of point with respect to Circle we have which gives
Solution 7 (Only Law of Cosines)
Note that and are the same length, which is also the radius we want. Using the law of cosines on , we have , where is the angle formed by . Since and are supplementary, . Using the law of cosines on , . As , . Solving for theta on the first equation and substituting gives . Solving for R gives .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.