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

Latest revision as of 18:35, 2 January 2023

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 $a + b + c$ .

Diagram

$[asy] defaultpen(fontsize(11)+0.8); size(300); pair A,B,C,D,Ic,Ib,P; A=MP("A",origin,down+left); B=MP("B",8*right,down+right); C=MP("C",IP(CR(A,5), CR(B,7)),2*up); real t=0.505; D=MP("",B+t*(C-B),SW); draw(A--B--C--A--D); path c1=incircle(A,D,C); path c2=incircle(A,D,B); draw(c1, gray+0.25); draw(c2, gray+0.25); Ic=MP("I_C",incenter(A,D,C),down+left); Ib=MP("I_B",incenter(A,D,B),left); path c3=circumcircle(Ic,D,C); path c4=circumcircle(Ib,D,B); draw(c3, fuchsia+0.2); draw(c4, fuchsia+0.2); P=MP("P",OP(c3,c4),up); draw(arc(circumcenter(B,C,P),B,C), royalblue+0.5+dashed); draw(C--Ic--D--P--C^^P--Ic, black+0.3); draw(B--Ib--D--P--B^^P--Ib, black+0.3); label("10",A--B,down); label("16",A--C,left); [/asy]$

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 (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 $\triangle{ABC}$ , $\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.$ 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 $98+49+3=\boxed{150}$ .

@@ Line 1: / Line 1: @@
 == 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>. Let points <math>I_B</math> and <math>I_C</math> denote the incenters of triangles <math>ABD</math> and <math>ACD</math>, respectively. The circumcircles of triangles <math>BI_BD</math> and <math>CI_CD</math> meet at distinct points <math>P</math> and <math>D</math>. The maximum possible area of <math>\triangle BPC</math> can be expressed in the form <math>a - b\sqrt {c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers and <math>c</math> is not divisible by the square of any prime. Find <math>a + b + c</math>.
+==Diagram==
+<center><asy>
+defaultpen(fontsize(11)+0.8); size(300);
+pair A,B,C,D,Ic,Ib,P;
+A=MP("A",origin,down+left); B=MP("B",8*right,down+right); C=MP("C",IP(CR(A,5), CR(B,7)),2*up); real t=0.505; D=MP("",B+t*(C-B),SW); draw(A--B--C--A--D); path c1=incircle(A,D,C); path c2=incircle(A,D,B); draw(c1, gray+0.25); draw(c2, gray+0.25); Ic=MP("I_C",incenter(A,D,C),down+left); Ib=MP("I_B",incenter(A,D,B),left); path c3=circumcircle(Ic,D,C); path c4=circumcircle(Ib,D,B); draw(c3, fuchsia+0.2); draw(c4, fuchsia+0.2); P=MP("P",OP(c3,c4),up); draw(arc(circumcenter(B,C,P),B,C), royalblue+0.5+dashed); draw(C--Ic--D--P--C^^P--Ic, black+0.3); draw(B--Ib--D--P--B^^P--Ib, black+0.3); label("10",A--B,down); label("16",A--C,left);
+</asy></center>
 == Solution 1 ==
 First, by the [[Law of Cosines]], we have <cmath>\cos BAC = \frac {16^2 + 10^2 - 14^2}{2\cdot 10 \cdot 16} = \frac {256+100-196}{320} = \frac {1}{2},</cmath> so <math>\angle BAC = 60^\circ</math>.
-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.  We first compute <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD.</cmath> 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 can be simplified to <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> Similarly, <math>\angle CO_2D = \angle ACD + \angle CDA</math>.  As a result <cmath>\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*}</cmath>
+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.  We first compute <cmath>\angle BO_1D = \angle BO_1I_B + \angle I_BO_1D = 2\angle BDI_B + 2\angle I_BBD.</cmath> 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 can be simplified to <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> Similarly, <math>\angle CO_2D = \angle ACD + \angle CDA</math>.  As a result
+<cmath>\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*}</cmath>
-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>LC = LB = 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>.
+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>LC = LB = 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 (and only) distance from <math>L</math> to <math>\overline{BC}</math> is <math>7\sqrt {3}</math>.
 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
@@ Line 23: / Line 33: @@
 & \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>.
+== 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 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 ==

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Last Question
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