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Difference between revisions of "2009 AIME I Problems/Problem 15"

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& \implies PC\cdot PB = 196\left(\frac{1}{\frac{PC}{PB}+\frac{PB}{PC}+\sqrt{3}}\right).
 
& \implies PC\cdot PB = 196\left(\frac{1}{\frac{PC}{PB}+\frac{PB}{PC}+\sqrt{3}}\right).
 
\end{align*} </cmath> By AM-GM, <math>\frac{PC}{PB}+\frac{PB}{PC}\geq{2}</math>, so <cmath>PB\cdot PC\leq 196\left(\frac{1}{2+\sqrt{3}}\right)=196(2-\sqrt{3}).</cmath>Finally, <cmath>[\triangle{BPC}]=\frac12 \cdot PB\cdot PC\cdot\sin{150^\circ}=\frac14 \cdot PB\cdot PC,</cmath>and the maximum area would be <math>49(2-\sqrt{3})=98-49\sqrt{3},</math> so the answer is <math>\boxed{150}</math>.
 
\end{align*} </cmath> By AM-GM, <math>\frac{PC}{PB}+\frac{PB}{PC}\geq{2}</math>, so <cmath>PB\cdot PC\leq 196\left(\frac{1}{2+\sqrt{3}}\right)=196(2-\sqrt{3}).</cmath>Finally, <cmath>[\triangle{BPC}]=\frac12 \cdot PB\cdot PC\cdot\sin{150^\circ}=\frac14 \cdot PB\cdot PC,</cmath>and the maximum area would be <math>49(2-\sqrt{3})=98-49\sqrt{3},</math> so the answer is <math>\boxed{150}</math>.
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== Solution 3 ==
 
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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>. Now note that quadrilaterals <math>BI_BDP</math> and <math>CI_CDP</math> are both cyclic. This means that <math>\angle{BPD} = 180^\circ{} - \angle{BI_BD} = 90^\circ{} + \alpha</math> and <math>\angle{DPC} = 180^\circ{} -  \angle{DI_CC} = 60^\circ + \alpha</math>. Therefore, <math>\angle{BPC} = 150^\circ{}</math>.
 
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Now note that in order to maximize the area of <math>BPC</math>, we have to maximize the distance from <math>P</math> to line <math>BC</math>. Note that since the second intersection of the Circles must lie below line <math>BC</math>, we try to find the locus of all points P under BC such that <math>\angle{BPC} = 150^\circ{}</math>. Let the circumcenter of triangle <math>BPC</math> be <math>O</math>. Then major <math>\arc{BC} = 2 * \angle{BCP} = 300^\circ{}</math>, minor <math>\arc{BC} = 60^\circ{}</math> which means <math>OBC</math> is an equilateral triangle.
 
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(I will work on finishing solution maybe sometime later today...)
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=14|after=Last Question}}
 
{{AIME box|year=2009|n=I|num-b=14|after=Last Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:33, 9 August 2019

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 $a + b + c$.

Solution 1

First, by 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 $\overline{LC} = \overline{LB} = \overline {BC} = 14$; $P$ 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 $7\sqrt {3}$.

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, \[\angle{CI_CD}+\angle{BI_BD}=180^\circ+\angle{\frac{A}{2}}=210^\circ.\]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 $\boxed{150}$.

See also

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