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2002 AIME I Problems/Problem 13: Difference between revisions

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In [[triangle]] <math>ABC</math> the [[median]]s <math>\overline{AD}</math> and <math>\overline{CE}</math> have lengths <math>18</math> and <math>27</math>, 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 <math>18</math> and <math>27</math>, 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 1==
<center><asy>
<center><asy>
size(150); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C));
size(150); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C));
Line 23: Line 23:
\end{align*}</cmath>
\end{align*}</cmath>
</center>
</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>.  
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>.
 
== Solution 2 ==
 
Let <math>AD</math> and <math>CE</math> intersect at <math>P</math> and construct median <math>BF</math>. Since medians split one another in a 2:1 ratio, we have
<center>
<cmath>\begin{align*}
AP = 12, PE = 9
\end{align*}</cmath>
</center>
This gives isosceles <math>APE</math> and thus an easy area calculation. After extending the altitude to <math>PE</math> and using the fact that it is also a median, we find
<center>
<cmath>\begin{align*}
[APE] = \frac{27\sqrt{55}}{4}
\end{align*}</cmath>
</center>
Since <math>P</math> is the centroid, we have that
<center>
<cmath>\begin{align*}
[CBE] = 3 [APE] = \frac{81\sqrt{55}}{4}
\end{align*}</cmath>
</center>
Using Power of a Point, we have
<center>
<cmath>\begin{align*}
EF=\frac{16}{3}
\end{align*}</cmath>
</center>
By Same Height Different Base,
<center>
<cmath>\begin{align*}
\frac{[EFB]}{[CBE]}=\frac{[EFB]}{(\frac{81\sqrt{55}}{4})}=\frac{(\frac{16}{3})}{27}=\frac{16}{81}
\end{align*}</cmath>
</center>
Solving gives
<center>
<cmath>\begin{align*}
[EFB] = 4\sqrt{55}
\end{align*}</cmath>
</center>
and
<center>
<cmath>\begin{align*}
[AFB]=2[EFB]=8\sqrt{55}
\end{align*}</cmath>
</center>
Thus, our answer is <math>8+55=\boxed{063}</math>.
 
'''-Solution by thecmd999'''


== See also ==
== See also ==

Revision as of 21:02, 28 August 2012

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 1

[asy] size(150); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C)); D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,NW))--cycle); D(circumcircle(A,B,C)); D(MP("F",F)); D(A--D); D(C--F); D(A--F--B); D(MP("E",E,NE)); D(MP("D",D,NE)); MP("12",(A+E)/2,SE,f);MP("12",(B+E)/2,f); MP("27",(C+E)/2,SW,f); MP("18",(A+D)/2,SE,f); [/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 $8 + 55 = \boxed{063}$.

Solution 2

Let $AD$ and $CE$ intersect at $P$ and construct median $BF$. Since medians split one another in a 2:1 ratio, we have

\begin{align*} AP = 12, PE = 9 \end{align*}

This gives isosceles $APE$ and thus an easy area calculation. After extending the altitude to $PE$ and using the fact that it is also a median, we find

\begin{align*} [APE] = \frac{27\sqrt{55}}{4} \end{align*}

Since $P$ is the centroid, we have that

\begin{align*} [CBE] = 3 [APE] = \frac{81\sqrt{55}}{4} \end{align*}

Using Power of a Point, we have

\begin{align*} EF=\frac{16}{3} \end{align*}

By Same Height Different Base,

\begin{align*} \frac{[EFB]}{[CBE]}=\frac{[EFB]}{(\frac{81\sqrt{55}}{4})}=\frac{(\frac{16}{3})}{27}=\frac{16}{81} \end{align*}

Solving gives

\begin{align*} [EFB] = 4\sqrt{55} \end{align*}

and

\begin{align*} [AFB]=2[EFB]=8\sqrt{55} \end{align*}

Thus, our answer is $8+55=\boxed{063}$.

-Solution by thecmd999

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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