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2002 AIME I Problems/Problem 13: Difference between revisions

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


== Solution ==
== Solution ==
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.
<asy>
 
</asy>
 
Applying [[Stewart's Theorem]] to medians <math>AD, CE</math>, we have:
<center>
<cmath>\begin{align*}
BC^2 + 4 \cdot 18^2 &= 2\left(24^2 + AC^2\right) \\
24^2 + 4 \cdot 27^2 &= 2\left(AC^2 + BC^2\right)
\end{align*}</cmath>
</center>
Substituting the first equation into the second and simplification yields <math>24^2 = 2\left(3AC^2 + 2 \cdot 24^2 - 4 \cdot 18^2\right)- 4 \cdot 27^2</math> <math> \Longrightarrow AC = \sqrt{2^5 \cdot 3 + 2 \cdot 3^5 + 2^4 \cdot 3^3 - 2^7 \cdot 3} = 3\sqrt{70}</math>.  
 
By the [[Power of a Point Theorem]] on <math>E</math>, we get <math>EF = \frac{12^2}{27} = \frac{16}{3}</math>. The [[Law of Cosines]] on <math>\triangle ACE</math> gives
<center>
<cmath>\begin{align*}
\cos \angle AEC = \left(\frac{12^2 + 27^2 - 9 \cdot 70}{2 \cdot 12 \cdot 27}\right) = \frac{3}{8}
\end{align*}</cmath>
</center>
Hence <math>\sin \angle AEC = \sqrt{1 - \cos^2 \angle AEC} = \frac{\sqrt{55}}{8}</math>. Because <math>\triangle AEF, BEF</math> have the same height and equal bases, they have the same area, and <math>[ABF] = 2[AEF] = 2 \cdot \frac 12 \cdot AE \cdot EF \sin \angle AEF = 12 \cdot \frac{16}{3} \cdot \frac{\sqrt{55}}{8} = 8\sqrt{55}</math>, and the answer is <math>8 + 55 = \boxed{063}</math>.  


== 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}}
[[Category:Intermediate Geometry Problems]]

Revision as of 19:01, 13 March 2009

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

[asy] [/asy]

Applying Stewart's Theorem to medians $AD, CE$, we have:

\begin{align*} BC^2 + 4 \cdot 18^2 &= 2\left(24^2 + AC^2\right) \\ 24^2 + 4 \cdot 27^2 &= 2\left(AC^2 + BC^2\right) \end{align*}

Substituting the first equation into the second and simplification yields $24^2 = 2\left(3AC^2 + 2 \cdot 24^2 - 4 \cdot 18^2\right)- 4 \cdot 27^2$ $\Longrightarrow AC = \sqrt{2^5 \cdot 3 + 2 \cdot 3^5 + 2^4 \cdot 3^3 - 2^7 \cdot 3} = 3\sqrt{70}$.

By the Power of a Point Theorem on $E$, we get $EF = \frac{12^2}{27} = \frac{16}{3}$. The Law of Cosines on $\triangle ACE$ gives

\begin{align*} \cos \angle AEC = \left(\frac{12^2 + 27^2 - 9 \cdot 70}{2 \cdot 12 \cdot 27}\right) = \frac{3}{8} \end{align*}

Hence $\sin \angle AEC = \sqrt{1 - \cos^2 \angle AEC} = \frac{\sqrt{55}}{8}$. Because $\triangle AEF, BEF$ have the same height and equal bases, they have the same area, and $[ABF] = 2[AEF] = 2 \cdot \frac 12 \cdot AE \cdot EF \sin \angle AEF = 12 \cdot \frac{16}{3} \cdot \frac{\sqrt{55}}{8} = 8\sqrt{55}$, and the answer is $8 + 55 = \boxed{063}$.

See also

2002 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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All AIME Problems and Solutions
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