Difference between revisions of "2009 AIME I Problems/Problem 15"
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== Problem == | == 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 <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>. | + | 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 == | == 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>. | 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> | + | 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> | ||
− | 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 | 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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== Solution 2 == | == Solution 2 == | ||
− | From Law of Cosines on <math>\triangle{ABC}</math>, <cmath>\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.</cmath>Now, <cmath>\angle{CI_CD}+\angle{BI_BD}=180^\circ+\ | + | From Law of Cosines on <math>\triangle{ABC}</math>, <cmath>\cos{A}=\frac{16^2+10^2-14^2}{2\cdot 10\cdot 16}=\frac{1}{2}\implies\angle{A}=60^\circ.</cmath>Now, <cmath>\angle{CI_CD}+\angle{BI_BD}=180^\circ+\frac{\angle{A}}{2}=210^\circ.</cmath>Since <math>CI_CDP</math> and <math>BI_BDP</math> are cyclic quadrilaterals, it follows that <cmath>\angle{BPC}=\angle{CPD}+\angle{DPB}=(180^\circ-\angle{CI_CD})+(180^\circ-\angle{BI_BD})=360^\circ-210^\circ=150^\circ.</cmath>Next, applying Law of Cosines on <math>\triangle{CPB}</math>, |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
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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>. | ||
+ | |||
+ | == 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 == | == 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}} | ||
+ | [[Category: Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:35, 2 January 2023
Contents
[hide]Problem
In triangle , , , and . Let be a point in the interior of . Let points and denote the incenters of triangles and , respectively. The circumcircles of triangles and meet at distinct points and . The maximum possible area of can be expressed in the form , where , , and are positive integers and is not divisible by the square of any prime. Find .
Diagram
Solution 1
First, by the Law of Cosines, we have so .
Let and be the circumcenters of triangles and , respectively. We first compute Because and are half of and , respectively, the above expression can be simplified to Similarly, . As a result
Therefore is constant (). Also, is or when is or . Let point be on the same side of as with ; is on the circle with as the center and as the radius, which is . The shortest (and only) distance from to is .
When the area of is the maximum, the distance from to has to be the greatest. In this case, it's . The maximum area of is and the requested answer is .
Solution 2
From Law of Cosines on , Now, Since and are cyclic quadrilaterals, it follows that Next, applying Law of Cosines on , By AM-GM, , so Finally, and the maximum area would be so the answer is .
Solution 3
Proceed as in Solution 2 until you find . The locus of points that give is a fixed arc from to ( will move along this arc as moves along ) and we want to maximise the area of []. This means we want to be farthest distance away from as possible, so we put in the middle of the arc (making isosceles). We know that and , so . Let be the foot of the perpendicular from to line . Then the area of [] is the same as because base has length . We can split into two triangles and , with and . Then, the area of [] is equal to , and so the answer is .
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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