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Difference between revisions of "2002 AIME I Problems/Problem 13"

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== Solution ==
 
== Solution ==
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Let CD=BD=x, AC=y. By Stewart's with cevian AD, we have <math>24^2x+n^2x=18^2\cdot2x+2x^2</math>, so <math>n^2=2x^2+72</math>. Also, Stewart's with cevian CE simplifies to <math>4x^2+n^2=1746</math>. Subtracting the two and solving gives x=<math>3\sqrt{31}</math>. By power of a point on E, EF=16/3. We now use the law of cosines on <math>\triangleEBC</math> to find cos BEC=3/8, so sin BEC=<math>\sqrt{55}/8</math>=sinAEF. But the area of AFB is twice that of AEF, since E is the midpoint of AB, so by the formula A=1/2absinC, the area is <math>2((1/2)(12)(16/3)(\sqrt{55}/8))=8\sqrt{55}</math>, and the answer is 63.
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Let CD=BD=x, AC=y. By Stewart's with cevian AD, we have <math>24^2x+n^2x=18^2\cdot2x+2x^2</math>, so <math>n^2=2x^2+72</math>. Also, Stewart's with cevian CE simplifies to <math>4x^2+n^2=1746</math>. Subtracting the two and solving gives x=<math>3\sqrt{31}</math>. By power of a point on E, EF=16/3. We now use the law of cosines on <math>\triangle EBC</math> to find cos BEC=3/8, so sin BEC=<math>\sqrt{55}/8</math>=sinAEF. But the area of AFB is twice that of AEF, since E is the midpoint of AB, so by the formula A=1/2absinC, the area is <math>2((1/2)(12)(16/3)(\sqrt{55}/8))=8\sqrt{55}</math>, and the answer is 63.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2002|n=I|num-b=12|num-a=14}}

Revision as of 17:37, 7 May 2008

Problem

In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths 18 and 27, respectively, and $AB=24$. Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$. The area of triangle $AFB$ is $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.

Solution

Let CD=BD=x, AC=y. By Stewart's with cevian AD, we have $24^2x+n^2x=18^2\cdot2x+2x^2$, so $n^2=2x^2+72$. Also, Stewart's with cevian CE simplifies to $4x^2+n^2=1746$. Subtracting the two and solving gives x=$3\sqrt{31}$. By power of a point on E, EF=16/3. We now use the law of cosines on $\triangle EBC$ to find cos BEC=3/8, so sin BEC=$\sqrt{55}/8$=sinAEF. But the area of AFB is twice that of AEF, since E is the midpoint of AB, so by the formula A=1/2absinC, the area is $2((1/2)(12)(16/3)(\sqrt{55}/8))=8\sqrt{55}$, and the answer is 63.

See also

2002 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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