Difference between revisions of "2009 AIME I Problems/Problem 15"
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− | From Law of Cosines on <math>\triangle{ABC}</math>, <cmath>\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.</cmath>Now, <cmath>\angle{CI_CD}+\angle{BI_BD}=180^\circ+\ | + | From Law of Cosines on <math>\triangle{ABC}</math>, <cmath>\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.</cmath>Now, <cmath>\angle{CI_CD}+\angle{BI_BD}=180^\circ+\frac{\angle{A}}{2}=210^\circ.</cmath>Since <math>CI_CDP</math> and <math>BI_BDP</math> are cyclic quadrilaterals, it follows that <cmath>\angle{BPC}=\angle{CPD}+\angle{DPB}=(180^\circ-\angle{CI_CD})+(180^\circ-\angle{BI_BD})=360^\circ-210^\circ=150^\circ.</cmath>Next, applying Law of Cosines on <math>\triangle{CPB}</math>, |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} |
Revision as of 19:44, 16 July 2020
Contents
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 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 | |
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.