Difference between revisions of "2012 AMC 12A Problems/Problem 16"
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==Solution 7 (Only Law of Cosines)== | ==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\cdot7\cdotR\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-\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)}\ \sqrt{30} </math>. | + | 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\cdot7\cdotR\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-\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 19:37, 20 January 2020
Contents
[hide]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 $11^2=R^2+7^2-2\cdot7\cdotR\cdot\cos\theta$ (Error compiling LaTeX. Unknown error_msg), 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 |
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