Difference between revisions of "2002 AIME I Problems/Problem 13"
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Thus, our answer is <math>8+55=\boxed{063}</math>. | Thus, our answer is <math>8+55=\boxed{063}</math>. | ||
+ | |||
+ | ==Short Solution: Smart Similarity== | ||
+ | |||
+ | Use the same diagram as in Solution 1. Call the centroid <math>P</math>. It should be clear that <math>PE=9</math>, and likewise <math>AP=12</math>, <math>AE=12</math>. Then, <math>\sin \angle AEP = \frac{\sqrt{55}}{8}</math>. Power of a Point on <math>E</math> gives <math>FE=\frac{16}{3}</math>, and the area of <math>AFB</math> is <math>AE * EF* \sin \angle \AEP</math>, which is twice the area of <math>AEF</math> or <math>FEB</math> (they have the same area because of equal base and height), giving <math>8\sqrt{55}</math> for an answer of <math>\boxed{063}</math>. | ||
== See also == | == See also == |
Revision as of 21:26, 19 February 2018
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
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 .
Short Solution: Smart Similarity
Use the same diagram as in Solution 1. Call the centroid . It should be clear that
, and likewise
,
. Then,
. Power of a Point on
gives
, and the area of
is $AE * EF* \sin \angle \AEP$ (Error compiling LaTeX. Unknown error_msg), which is twice the area of
or
(they have the same area because of equal base and height), giving
for an answer of
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.