2002 AIME I Problems/Problem 13
Contents
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
, which is twice the area of
or
(they have the same area because of equal base and height), giving
for an answer of
.
Solution 4 (You've Forgotten Power of a Point Exists)
Note that, as above, it is quite easy to get that (equate Heron's and
to find this). Now note that
because they are vertical angles,
, and
(the latter two are derived from the inscribed angle theorem). Therefore
and so
and
so the area of
is
giving us
as our answer. (One may just get the area via triangle similarity too--this is if you are tired by the end of test and just want to bash some stuff out--it may also serve as a useful check).
~Dhillonr25
Solution 5 (Barycentric Coordinates)
Apply barycentric coordinates on . We know that
. We can now get the displacement vectors
and
. Now, applying the distance formula and simplifying gives us the two equations
Substituting
and solving with algebra now gives
. Now we can find
. Note that
can be parameterized as
, so plugging into the circumcircle equation and solving for
gives
so
. Plugging in for
gives us
. Thus, by the area formula, we have
By Heron's Formula, we have
which immediately gives
from our ratio, extracting
.
-Taco12
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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