Difference between revisions of "2002 AIME I Problems/Problem 13"
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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>. | ||
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+ | == Solution 2 == | ||
== Solution 2 == | == Solution 2 == | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
[APE] = \frac{27\sqrt{55}}{4} | [APE] = \frac{27\sqrt{55}}{4} | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
</center> | </center> | ||
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<center> | <center> | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \frac{[ | + | \frac{[AFE]}{[APE]}=\frac{[AFE]}{(\frac{27\sqrt{55}}{4})}=\frac{PE}{EF}=\frac{(\frac{16}{3})}{9}=\frac{16}{27} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
</center> | </center> | ||
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<center> | <center> | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | [ | + | [AFE] = 4\sqrt{55} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
</center> | </center> | ||
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<center> | <center> | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | [AFB]=2[ | + | [AFB]=2[AFE]=8\sqrt{55} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
</center> | </center> |
Revision as of 23:23, 7 March 2013
Problem
In triangle the medians
and
have lengths
and
, respectively, and
. Extend
to intersect the circumcircle of
at
. The area of triangle
is
, where
and
are positive integers and
is not divisible by the square of any prime. Find
.
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]](http://latex.artofproblemsolving.com/2/3/6/236d2c7ecccc097482080a6f0ffbf91d00c72995.png)
Applying Stewart's Theorem to medians , we have:
Substituting the first equation into the second and simplification yields
.
By the Power of a Point Theorem on , we get
. The Law of Cosines on
gives
Hence . Because
have the same height and equal bases, they have the same area, and
, and the answer is
.
Solution 2
Solution 2
Let and
intersect at
. Since medians split one another in a 2:1 ratio, we have
This gives isosceles and thus an easy area calculation. After extending the altitude to
and using the fact that it is also a median, we find
Using Power of a Point, we have
By Same Height Different Base,
Solving gives
and
Thus, our answer is .
-Solution by thecmd999
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 |