2009 AIME I Problems/Problem 15
Problem
In triangle ,
,
, and
. Let
be a point in the interior of
. Let
and
denote the incenters of triangles
and
, respectively. The circumcircles of triangles
and
meet at distinct points
and
. The maximum possible area of
can be expressed in the form
, where
,
, and
are positive integers and
is not divisible by the square of any prime. Find
.
Solution
First, by Law of Cosines, we have
Therefore, .
Let and
be the circumcenters of triangles
and
, respectively.
Because and
are half of
and
, respectively, the above expression would be,
Similarly,
Therefore is constant (
). Also,
is
or
when
is
or
. Let point
be on the same side of
as
with
;
is on the circle with
as the center and
as the radius, which is
. The shortest distance from
to
is
.
When the area of is the maximum, the distance from
to
has to be the greatest. In this case, it's
. The maximum area of
is
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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