2012 AMC 12A Problems/Problem 16
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
.
.
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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