Difference between revisions of "2012 AMC 12A Problems/Problem 16"
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− | 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 = 11^2-r^2,</math> which gives <math>r=\boxed{\sqrt{30}}.</math> | + | 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> |
== See Also == | == See Also == |
Revision as of 17:23, 16 July 2017
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
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
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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