2009 AIME I Problems/Problem 15
Contents
Problem
In triangle ,
,
, and
. Let
be a point in the interior of
. Let points
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 1
First, by the Law of Cosines, we have so
.
Let and
be the circumcenters of triangles
and
, respectively. We first compute
Because
and
are half of
and
, respectively, the above expression can be simplified to
Similarly,
. As a result
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
and the requested answer is
.
Solution 2
From Law of Cosines on ,
Now,
Since
and
are cyclic quadrilaterals, it follows that
Next, applying Law of Cosines on
,
By AM-GM,
, so
Finally,
and the maximum area would be
so the answer is
.
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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