2009 AIME I Problems/Problem 15

Problem

In triangle $ABC$ , $AB = 10$ , $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area of $\triangle BPC$ can be expressed in the form $a - b\sqrt {c}$ , where $a$ , $b$ , and $c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a + b + c$ .

First, by Law of Cosines, we have

Let $O_1$ and $O_2$ be the circumcenters of triangles $BI_BD$ and $CI_CD$ , respectively.

Because $\angle BDI_B$ and $\angle I_BBD$ are half of $\angle BDA$ and $\angle ABD$ , respectively, the above expression would be,

Therefore $\angle CPB$ is constant ( $150^\circ$ ). Also, $P$ is $B$ or $C$ when $D$ is $B$ or $C$ . Let point $L$ be on the same side of $\overline{BC}$ as $A$ with $\overline{LC} = \overline{LB} = \overline {BC} = 14$ ; $P$ is on the circle with $L$ as the center and $\overline{LC}$ as the radius, which is $14$ . The shortest distance from $L$ to $\overline{BC}$ is $7\sqrt {3}$ .