Difference between revisions of "2009 AIME I Problems/Problem 15"
m (→Solution 3) |
(→Solution 3) |
||
Line 26: | Line 26: | ||
== Solution 3 == | == Solution 3 == | ||
− | First, we notice that triangle ABC is a scaled version of a 5-7-8 triangle (which has a 60 degree angle opposite the side with length 7). So <math>\angle{BAC} = 60^\circ{}</math>. Therefore, let <math>\angle{I_B AB} = \angle{I_B AD} = \alpha,</math> and <math>\angle{I_C AD} = \angle{I_C AC} = 30 - \alpha.</math> Therefore, in triangle <math>ABD</math>, we know that | + | First, we notice that triangle ABC is a scaled version of a 5-7-8 triangle (which has a 60 degree angle opposite the side with length 7). So <math>\angle{BAC} = 60^\circ{}</math>. Therefore, let <math>\angle{I_B AB} = \angle{I_B AD} = \alpha,</math> and <math>\angle{I_C AD} = \angle{I_C AC} = 30 - \alpha.</math> Therefore, in triangle <math>ABD</math>, we know that <math>\angle{BI_BD} = 90^\circ{} + \frac{\angle{BAD}}{2} = 90 + \alpha</math> and <math>\angle{CI_CD} = 90^\circ{}+ \frac{\angle{CAD}}{2} = 90^\circ{} + (30^\circ{} - \alpha) = 120^\circ{} - \alpha</math>. |
− | |||
(I will work on finishing solution maybe sometime later today...) | (I will work on finishing solution maybe sometime later today...) | ||
Revision as of 11:01, 9 August 2019
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 1
First, by 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
.
Solution 3
First, we notice that triangle ABC is a scaled version of a 5-7-8 triangle (which has a 60 degree angle opposite the side with length 7). So . Therefore, let
and
Therefore, in triangle
, we know that
and
.
(I will work on finishing solution maybe sometime later today...)
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.