2000 AMC 12 Problems/Problem 24
Problem
If circular arcs and
have centers at
and
, respectively, then there exists a circle tangent to both
and
, and to
. If the length of
is
, then the circumference of the circle is
Solution
Since are all radii, it follows that
is an equilateral triangle.
Draw the circle with center and radius
. Then let
be the point of tangency of the two circles, and
be the intersection of the smaller circle and
. Let
be the intersection of the smaller circle and
. Also define the radii
(note that
is a diameter of the smaller circle, as
is the point of tangency of both circles, the radii of a circle is perpendicular to the tangent, hence the two centers of the circle are collinear with each other and
).
By the Power of a Point Theorem,
Since , then
. Since
is equilateral,
, and so
. Thus
and the circumference of the circle is
.
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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All AMC 12 Problems and Solutions |