Difference between revisions of "2002 AIME I Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | {{ | + | Let CD=BD=x, AC=y. By Stewart's with cevian AD, we have <math>24^2x+n^2x=18^2\cdot2x+2x^2</math>, so <math>n^2=2x^2+72</math>. Also, Stewart's with cevian CE simplifies to <math>4x^2+n^2=1746</math>. Subtracting the two and solving gives x=<math>3\sqrt{31}</math>. By power of a point on E, EF=16/3. We now use the law of cosines to find cos BEC=3/8, so sin BEC=<math>\sqrt{55}/8</math>=sinAEF. But the area of AFB is twice that of AEF, since E is the midpoint of AB, so by the formula A=1/2absinC, the area is <math>2((1/2)(12)(16/3)(\sqrt{55}/8))=8\sqrt{55}</math>, and the answer is 63. |
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=12|num-a=14}} | {{AIME box|year=2002|n=I|num-b=12|num-a=14}} | ||
Revision as of 17:36, 7 May 2008
Problem
In triangle 
the medians 
and 
have lengths 18 and 27, 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
Let CD=BD=x, AC=y. By Stewart's with cevian AD, we have 
, so 
. Also, Stewart's with cevian CE simplifies to 
. Subtracting the two and solving gives x=
. By power of a point on E, EF=16/3. We now use the law of cosines to find cos BEC=3/8, so sin BEC=
=sinAEF. But the area of AFB is twice that of AEF, since E is the midpoint of AB, so by the formula A=1/2absinC, the area is 
, and the answer is 63.
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 | ||
