2002 AIME I Problems/Problem 13
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
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 angle 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
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 | ||
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