Difference between revisions of "2002 AIME I Problems/Problem 13"
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== Solution 4 (You've Forgotten Power of a Point Exists) == | == Solution 4 (You've Forgotten Power of a Point Exists) == | ||
− | Note that, as above, it is quite easy to get that <math>\sin \angle AEP = \frac{\sqrt{55}}{8}</math> (equate Heron's and <math>\frac{1}{2}ab\sin C</math> to find this). Now note that <math>\angle FEA = \angle BEC</math> because they are vertical angles, <math>\angle FAE = \angle ECB</math>, and <math>\angle EFA = \angle ABC</math> (the latter two are derived from the inscribed angle theorem). Therefore <math>\Delta AEF ~ \Delta CEB</math> and so <math>FE = \frac{144}{27}</math> and <math>\sin \angle FEA = \frac{\sqrt{55}}{8}</math> so the area of <math>\Delta BFA</math> is <math>8\sqrt{55}</math> giving us <math>\boxed{063}</math> 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). | + | Note that, as above, it is quite easy to get that <math>\sin \angle AEP = \frac{\sqrt{55}}{8}</math> (equate Heron's and <math>\frac{1}{2}ab\sin C</math> to find this). Now note that <math>\angle FEA = \angle BEC</math> because they are vertical angles, <math>\angle FAE = \angle ECB</math>, and <math>\angle EFA = \angle ABC</math> (the latter two are derived from the inscribed angle theorem). Therefore <math>\Delta AEF </math>~<math> \Delta CEB</math> and so <math>FE = \frac{144}{27}</math> and <math>\sin \angle FEA = \frac{\sqrt{55}}{8}</math> so the area of <math>\Delta BFA</math> is <math>8\sqrt{55}</math> giving us <math>\boxed{063}</math> 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 | ~Dhillonr25 |
Revision as of 15:59, 1 July 2024
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.