2012 AMC 12A Problems/Problem 16
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 ?
Diagram
Solution 1
Let denote the radius of circle . Note that quadrilateral is cyclic. By Ptolemy's Theorem, we have and . Let 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 . .
Note
Instead of using the Extended Law of Sines, you can note that , since the angles inscribe arcs of the same length.
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 .
Solution 8
We first note that is the circumcircle of both and . Thus the circumradius of both the triangles are equal. We set the radius of as , and noting that the circumradius of a triangle is and that the area of a triangle by Heron's formula is with as the semi-perimeter we have the following, Now substituting , This gives us 2 values for namely and .
Now notice that we can apply Ptolemy's theorem on to find in terms of . We get Here we substitute our values of receiving . Notice that the latter of the cases does not satisfy the triangle inequality for as . But the former does thus our answer is .
~Aaryabhatta1
Solution 9 (Similar Triangles)
We first apply Ptolemy's Theorem on cyclic quadrilateral to get . Since and . From this, we can see and . That means . So, if you let , you will get . Continuing in this fashion, we can get and . Since , we have which gives us . Plugging it into gives
Solving for yields .
~sml1809
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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All AMC 12 Problems and Solutions |
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