2002 AIME I Problems/Problem 13
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
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 .
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.