Difference between revisions of "2009 AIME I Problems/Problem 15"
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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+\angle{\frac{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>, | 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+\angle{\frac{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> | ||
\begin{align*} | \begin{align*} | ||
& BC^2=14^2=PC^2+PB^2+2\cdot PB\cdot PC\cdot\frac{\sqrt{3}}{2} \ | & BC^2=14^2=PC^2+PB^2+2\cdot PB\cdot PC\cdot\frac{\sqrt{3}}{2} \ | ||
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& \implies \frac{PC}{PB}+\frac{PB}{PC}-\frac{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). | & \implies PC\cdot PB = 196\left(\frac{1}{\frac{PC}{PB}+\frac{PB}{PC}+\sqrt{3}}\right). | ||
− | \end{align*}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>. |
== 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 01:45, 10 September 2018
Contents
[hide]Problem
In triangle , , , and . Let be a point in the interior of . Let 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 .
Solution 1
First, by 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 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 .
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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