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

(New page: == Problem == In triangle <math>ABC</math>, <math>AB = 10</math>, <math>BC = 14</math>, and <math>CA = 16</math>. Let <math>D</math> be a point in the interior of <math>\overline{BC}</math...)
 
(→‎Solution)
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== Solution ==
 
== Solution ==
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First, by [[Law of Cosines]], we have
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<cmath>\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2}</cmath>
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Therefore, <math>\angle BAC = 60^\circ</math>.
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Let <math>O_1</math> and <math>O_2</math> be the circumcenters of triangles <math>BI_BD</math> and <math>CI_CD</math>, respectively.
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<cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD</cmath>
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Because <math>\angle BDI_B</math> and <math>\angle I_BBD</math> are half of <math>\angle BDA</math> and <math>\angle ABD</math>, respectively, the above expression would be,
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<cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD = \angle ABD + \angle BDA</cmath>
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Similarly, <cmath>\angle CO_2D = \angle ACD + \angle CDA</cmath>
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<cmath>\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</cmath>
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Therefore <math>\angle CPB</math> is constant (<math>150^\circ</math>). Also, <math>P</math> is <math>B</math> or <math>C</math> when <math>D</math> is <math>B</math> or <math>C</math>. Let point <math>L</math> be on the same side of <math>\overline{BC}</math> as <math>A</math> with <math>\overline{LC} = \overline{LB} = \overline {BC} = 14</math>; <math>P</math> is on the circle with <math>L</math> as the center and <math>\overline{LC}</math> as the radius, which is <math>14</math>. The shortest distance from <math>L</math> to <math>\overline{BC}</math> is <math>7\sqrt {3}</math>.
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When the area of <math>\triangle BPC</math> is the maximum, the distance from <math>P</math> to <math>\overline{BC}</math> has to be the greatest. In this case, it's <math>14 - 7\sqrt {3}</math>. The maximum area of <math>\triangle BPC</math> is
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<cmath>\frac {1}{2} \cdot 14 \cdot (14 - 7\sqrt {3}) = 98 - 49 \sqrt {3} = a - b\sqrt {c}</cmath>
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<cmath>a + b + c = 98 + 49 + 3 = \boxed{150}</cmath>
  
 
== 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}}

Revision as of 21:42, 26 March 2009

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

First, by Law of Cosines, we have

\[\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2}\]

Therefore, $\angle BAC = 60^\circ$.


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


\[\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD\]

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

\[\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD = \angle ABD + \angle BDA\]

Similarly, \[\angle CO_2D = \angle ACD + \angle CDA\]


\[\angle CPB = \angle CPD + \angle BPD = \frac {1}{2} \cdot \angle CO_2D + \frac {1}{2} \cdot \angle BO_1D = \frac {1}{2}(\angle ABD + \angle BDA + \angle ACD + \angle CDA) = \frac {1}{2} (2 \cdot 180^\circ - \angle BAC) = \frac {1}{2} \cdot 300^\circ = 150^\circ\]

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 \[\frac {1}{2} \cdot 14 \cdot (14 - 7\sqrt {3}) = 98 - 49 \sqrt {3} = a - b\sqrt {c}\] \[a + b + c = 98 + 49 + 3 = \boxed{150}\]

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
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