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Difference between revisions of "2009 AIME I Problems/Problem 15"
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== Solution 3 ==
 
== Solution 3 ==
−
Proceed as in Solution 2 until you find <math>\angle CPB = 150</math>. The locus of points <math>P</math> that give <math>\angle BPC = 150</math> is a fixed arc from <math>B</math> to <math>C</math> (<math>P</math> will move along this arc as <math>D</math> moves along <math>BC</math>) and we want to maximise the area of [<math>\triangle BPC</math>]. This means we want <math>P</math> to be farthest distance away from <math>BC</math> as possible, so we put <math>P</math> in the middle of the arc (making <math>\triangle BPC</math> isosceles). We know that <math>BC=14</math> and <math>\angle CPB = 150</math>, so <math>\angle PBC = \angle PCB = 15</math>. Let <math>O</math> be the foot of the perpendicular from <math>P</math> to line <math>BC</math>. Then the area of [<math>\triangle BPC</math>] is the same as <math>7OP</math> because base <math>BC</math> has length <math>14</math>. We can split <math>\triangle BPC</math> into two <math>15-75-90</math> triangles <math>BOP</math> and <math>COP</math>, with <math>BO=CO=7</math> and <math>OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC</math>] is equal to <math>7 \cdot OP=98-49\sqrt{3}</math>, and so the answer is <math>98+49+3=\boxed{150}</math>.   
+
Proceed as in Solution 2 until you find <math>\angle CPB = 150</math>. The locus of points <math>P</math> that give <math>\angle CPB = 150</math> is a fixed arc from <math>B</math> to <math>C</math> (<math>P</math> will move along this arc as <math>D</math> moves along <math>BC</math>) and we want to maximise the area of [<math>\triangle BPC</math>]. This means we want <math>P</math> to be farthest distance away from <math>BC</math> as possible, so we put <math>P</math> in the middle of the arc (making <math>\triangle BPC</math> isosceles). We know that <math>BC=14</math> and <math>\angle CPB = 150</math>, so <math>\angle PBC = \angle PCB = 15</math>. Let <math>O</math> be the foot of the perpendicular from <math>P</math> to line <math>BC</math>. Then the area of [<math>\triangle BPC</math>] is the same as <math>7OP</math> because base <math>BC</math> has length <math>14</math>. We can split <math>\triangle BPC</math> into two <math>15-75-90</math> triangles <math>BOP</math> and <math>COP</math>, with <math>BO=CO=7</math> and <math>OP=7 \tan 15=7(2-\sqrt{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC</math>] is equal to <math>7 \cdot OP=98-49\sqrt{3}</math>, and so the answer is <math>98+49+3=\boxed{150}</math>.   
  
 
== See also ==
 
== See also ==

Revision as of 17:34, 20 August 2020

Problem

In triangle $ABC$, $AB = 10$, $BC = 14$, and $CA = 16$. Let $D$ be a point in the interior of $\overline{BC}$. Let points $I_B$ 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 $a + b + c$.

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. 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{align*}\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{align*}

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 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 $\triangle{ABC}$, \[\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.\]Now, \[\angle{CI_CD}+\angle{BI_BD}=180^\circ+\frac{\angle{A}}{2}=210^\circ.\]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{align*} & 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{align*} 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 $98+49+3=\boxed{150}$.

See also

2009 AIME I (Problems • Answer Key • Resources)
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