AIME 2009 I · 第 15 题

Solution 1

First, by the Law of Cosines, we have

$$\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2},$$ so $\angle BAC = 60^\circ$.

Let $O_1$ and $O_2$ be the circumcenters of triangles $BI_BD$ and $CI_CD$, respectively. The circles make us think of constant angles, so we first compute

$$\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD.$$ Because $\angle BDI_B$ and $\angle I_BBD$ are half of $\angle BDA$ and $\angle ABD$, respectively, the above expression can be simplified to

$$\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD = \angle ABD + \angle BDA.$$ Similarly, $\angle CO_2D = \angle ACD + \angle CDA$. As a result

$$\begin{aligned}\angle CPB &= \angle CPD + \angle BPD \\&= \frac {1}{2} \cdot \angle CO_2D + \frac {1}{2} \cdot \angle BO_1D \\&= \frac {1}{2}(\angle ABD + \angle BDA + \angle ACD + \angle CDA) \\&= \frac {1}{2} (2 \cdot 180^\circ - \angle BAC) \\&= \frac {1}{2} \cdot 300^\circ = 150^\circ.\end{aligned}$$ 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 $LC = LB = BC = 14$; $P$ is on the circle with $L$ as the center and $\overline{LC}$ as the radius, which is $14$. The shortest (and only) distance from $L$ to $\overline{BC}$ is $7\sqrt {3}$.

When the area of $\triangle BPC$ is the maximum, the distance from $P$ to $\overline{BC}$ has to be the greatest. In this case, it's $14 - 7\sqrt {3}$. The maximum area of $\triangle BPC$ is

$$\frac {1}{2} \cdot 14 \cdot (14 - 7\sqrt {3}) = 98 - 49 \sqrt {3}$$ and the requested answer is $98 + 49 + 3 = \boxed{150}$.

Solution 2

From Law of Cosines on $\triangle{ABC}$,

$$\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.$$ Setting $\angle{DAC}=\theta$ and doing some angle chasing, we find that $\angle{CI_CD}=\frac{\theta}{2}+90^\circ$ and $\angle{BI_BD}=\frac{\angle{A}-\theta}{2}+90^\circ,$ hence

$$\angle{CI_CD}+\angle{BI_BD}=180^\circ+\frac{\angle{A}}{2}=210^\circ.$$ Since $CI_CDP$ and $BI_BDP$ are cyclic quadrilaterals, it follows that

$$\angle{BPC}=\angle{CPD}+\angle{DPB}=(180^\circ-\angle{CI_CD})+(180^\circ-\angle{BI_BD})=360^\circ-210^\circ=150^\circ.$$ Next, applying Law of Cosines on $\triangle{CPB}$,

$$\begin{aligned} & BC^2=14^2=PC^2+PB^2+2\cdot PB\cdot PC\cdot\frac{\sqrt{3}}{2} \\ & \implies \frac{PC^2+PB^2-196}{PC\cdot PB}=-\sqrt{3} \\ & \implies \frac{PC}{PB}+\frac{PB}{PC}-\frac{196}{PC\cdot PB}=-\sqrt{3} \\ & \implies PC\cdot PB = 196\left(\frac{1}{\frac{PC}{PB}+\frac{PB}{PC}+\sqrt{3}}\right). \end{aligned}$$ By AM-GM, $\frac{PC}{PB}+\frac{PB}{PC}\geq{2}$, so

$$PB\cdot PC\leq 196\left(\frac{1}{2+\sqrt{3}}\right)=196(2-\sqrt{3}).$$ Finally,

$$[\triangle{BPC}]=\frac12 \cdot PB\cdot PC\cdot\sin{150^\circ}=\frac14 \cdot PB\cdot PC,$$ and the maximum area would be $49(2-\sqrt{3})=98-49\sqrt{3},$ so the answer is $\boxed{150}$.

Solution 3

Proceed as in Solution 2 until you find $\angle CPB = 150$. The locus of points $P$ that give $\angle CPB = 150$ is a fixed arc from $B$ to $C$ ($P$ will move along this arc as $D$ moves along $BC$) and we want to maximise the area of [$\triangle BPC$]. This means we want $P$ to be farthest distance away from $BC$ as possible, so we put $P$ in the middle of the arc (making $\triangle BPC$ isosceles). We know that $BC=14$ and $\angle CPB = 150$, so $\angle PBC = \angle PCB = 15$. Let $O$ be the foot of the perpendicular from $P$ to line $BC$. Then the area of [$\triangle BPC$] is the same as $7OP$ because base $BC$ has length $14$. We can split $\triangle BPC$ into two $15-75-90$ triangles $BOP$ and $COP$, with $BO=CO=7$ and $OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3$. Then, the area of [$\triangle BPC$] is equal to $7 \cdot OP=98-49\sqrt{3}$, and so the answer is $98+49+3=\boxed{150}$.