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==Short Solution: Smart Similarity== | ==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 | + | 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>. |
+ | |||
+ | == 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 </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 | ||
+ | |||
+ | == Solution 5 (Barycentric Coordinates) == | ||
+ | Apply barycentric coordinates on <math>\triangle ABC</math>. We know that <math>D=\left(0, \frac{1}{2}, \frac{1}{2}\right), E=\left(\frac{1}{2}, \frac{1}{2}, 0\right)</math>. We can now get the displacement vectors <math>\overrightarrow{AD} = \left(1, -\frac{1}{2}, -\frac{1}{2}\right)</math> and <math>\overrightarrow{CE}=\left(-\frac{1}{2}, -\frac{1}{2}, 1\right)</math>. Now, applying the distance formula and simplifying gives us the two equations | ||
+ | <cmath>\begin{align*} | ||
+ | 2b^2+2c^2-a^2&=1296 \\ | ||
+ | 2a^2+2b^2-c^2&=2916. \\ | ||
+ | \end{align*}</cmath> | ||
+ | Substituting <math>c=24</math> and solving with algebra now gives <math>a=6\sqrt{31}, b=3\sqrt{70}</math>. Now we can find <math>F</math>. Note that <math>CE</math> can be parameterized as <math>(1:1:t)</math>, so plugging into the circumcircle equation and solving for <math>t</math> gives <math>t=\frac{-c^2}{a^2+b^2}</math> so <math>F=(a^2+b^2:a^2+b^2:-c^2)</math>. Plugging in for <math>a,b</math> gives us <math>F=(1746:1746:-576)</math>. Thus, by the area formula, we have<cmath>\frac{[AFB]}{[ABC]}= | ||
+ | \left|\begin{matrix} | ||
+ | 1 & 0 & 0 \\ | ||
+ | 0 & 1 & 0 \\ | ||
+ | \frac{97}{162} & \frac{97}{162} & -\frac{16}{81} | ||
+ | \end{matrix}\right|=\frac{16}{81}.</cmath>By Heron's Formula, we have <math>[ABC]=\frac{81\sqrt{55}}{2}</math> which immediately gives <math>[AFB]=8\sqrt{55}</math> from our ratio, extracting <math>\boxed{063}</math>. | ||
+ | |||
+ | -Taco12 | ||
+ | |||
+ | ==Solution 6 (Law of Cosines + Stewarts)== | ||
+ | Since <math>AD</math> is the median, let <math>BD=BC=x</math>. Since <math>CE</math> is a median, <math>AE=BE=12</math>. Applying Power of a Point with respect to point <math>E</math>, we see that <math>EF=\frac{16}{3}</math>. Applying Stewart's Theorem on triangles <math>\triangle ADC</math> and <math>\triangle ABC</math>, we get that <math>x=3\sqrt{31}</math> and <math>y=3\sqrt{70}</math>. The area of <math>\triangle AFB</math> is simply <math>\frac{1}{2} \cdot \sin{\angle FBA}\cdot FB\cdot AB</math>. We know <math>AB=24</math>. Also, we know that <math>\angle FBA = \angle FCA</math>. Then, applying Law of Cosines on triangle <math>EAC</math>, we get that <math>\cos{\angle FCA}=\frac{3\sqrt{70}}{28}</math> which means that <math>\sin{\angle FCA=\angle FBA}=\frac{\sqrt{154}}{28}</math>. Then, applying Stewart's Theorem on triangle <math>FBC</math> with cevian <math>BE</math> allows us to receive that <math>FB=\frac{4\sqrt{70}}{3}</math>. Now, plugging into our earlier area formula, we receive <math>\frac{1}{2} \cdot \frac{\sqrt{154}}{28} \cdot \frac{4\sqrt{70}}{3} \cdot 24 = 8\sqrt{55}.</math> Therefore, the desired answer is <math>8+55=\boxed{063}</math>. | ||
+ | |||
+ | ~SirAppel | ||
== See also == | == See also == |
Latest revision as of 23:15, 11 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
Solution 6 (Law of Cosines + Stewarts)
Since is the median, let
. Since
is a median,
. Applying Power of a Point with respect to point
, we see that
. Applying Stewart's Theorem on triangles
and
, we get that
and
. The area of
is simply
. We know
. Also, we know that
. Then, applying Law of Cosines on triangle
, we get that
which means that
. Then, applying Stewart's Theorem on triangle
with cevian
allows us to receive that
. Now, plugging into our earlier area formula, we receive
Therefore, the desired answer is
.
~SirAppel
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.