2018 AMC 12B Problems/Problem 25
Problem
Circles ,
, and
each have radius
and are placed in the plane so that each circle is externally tangent to the other two. Points
,
, and
lie on
,
, and
respectively such that
and line
is tangent to
for each
, where
. See the figure below. The area of
can be written in the form
for positive integers
and
. What is
?
Solution 1
Let be the center of circle
for
, and let
be the intersection of lines
and
. Because
, it follows that
is a
triangle. Let
; then
and
. The Law of Cosines in
gives
which simplifies to
. The positive solution is
. Then
, and the required area is
The requested sum is
.
And we are done.
Solution 2
Let and
be the centers of
and
respectively and draw
,
, and
. Note than
and
are both right. Furthermore, since
is equilateral,
and
. Mark
as the base of the altitude from
to
. By special right triangles,
and
. since
and
, we can find find
. Thus,
. This makes
. This makes the answer
.
Solution 3
Let be the center of circle
for
. Let
be the centroid of
, which also happens to be the centroid of
. Because
and
,
.
is
the height of
, thus
is
.
Applying cosine law on , one finds that
. Multiplying by
to solve for the height of
, one gets
. Simply multiplying by
and then calculating the equilateral triangle's area, one would get the final result of
.
This makes the answer .
~AlbeePach~
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
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