2002 AIME II Problems/Problem 13
In triangle point is on with and point is on with and and and intersect at Points and lie on so that is parallel to and is parallel to It is given that the ratio of the area of triangle to the area of triangle is where and are relatively prime positive integers. Find .
Let be the intersection of and .
Since and , and . So , and thus, .
Using mass points:
WLOG, let .
Thus, . Therefore, , and . Note we can just use mass points to get which is .
First draw and extend it so that it meets with at point .
We have that
By Ceva's, That means that
Now we apply mass points. Assume WLOG that . That means that
Notice now that is similar to . Therefore,
Also, is similar to . Therefore,
Because is similar to , .
As a result, .
- Not the author writing here, but a note is that Ceva's Theorem was actually not necessary to solve this problem. The information was just nice to know :)
Use the mass of point. Denoting the mass of , we can see that the mass of Q is , hence we know that , now we can find that which implies , it is obvious that is similar to so we need to find the ration between PQ and AC, which is easy, it is , so our final answer is which is . ~bluesoul
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