2002 AIME I Problems/Problem 13

Problem

In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths 18 and 27, respectively, and $AB=24$ . Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$ . The area of triangle $AFB$ is $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$ .

Solution

Let CD=BD=x, AC=y. By Stewart's with cevian AD, we have $24^2x+n^2x=18^2\cdot2x+2x^2$ , so $n^2=2x^2+72$ . Also, Stewart's with cevian CE simplifies to $4x^2+n^2=1746$ . Subtracting the two and solving gives x= $3\sqrt{31}$ . By power of a point on E, EF=16/3. We now use the law of cosines on $\triangle EBC$ to find cos BEC=3/8, so sin BEC= $\sqrt{55}/8$ =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 $2((1/2)(12)(16/3)(\sqrt{55}/8))=8\sqrt{55}$ , and the answer is 63.

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
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2002 AIME I Problems/Problem 13

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