2012 AMC 12A Problems/Problem 16
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 ?
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 .
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 . .
Instead of using the Extended Law of Sines, you can note that , since the angles inscribe arcs of the same length.
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
Let . Consider an inversion about . So, . Using .
-Solution by IDMasterz
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,
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 .
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 .
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 .
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