2011 AIME II Problems/Problem 4
In triangle , and . The angle bisector of intersects at point , and point is the midpoint of . Let be the point of the intersection of and . The ratio of to can be expressed in the form , where and are relatively prime positive integers. Find .
Let be on such that . It follows that , so by the Angle Bisector Theorem. Similarly, we see by the Midline Theorem that . Thus, and .
Solution 2 (mass points)
Assign mass points as follows: by Angle-Bisector Theorem, , so we assign . Since , then , and , so .
We will use barycentric coordinates. Let , , . By the Angle Bisector Theorem, . Since is the midpoint of , . Therefore, the equation for line BM is . Let . Using the equation for , we get Therefore, so the answer is .
Let . Then by the Angle Bisector Theorem, . By the Ratio Lemma, we have that Notice that since their bases have the same length and they share a height. By the sin area formula, we have that Simplifying, we get that Plugging this into what we got from the Ratio Lemma, we have that
Solution 6 (quick Menelaus)
First, we will find . By Menelaus on and the line , we have This implies that . Then, by Menelaus on and line , we have Therefore, The answer is . -brainiacmaniac31
Solution 7 (Visual)
Solution 8 (Cheese)
Assume is a right triangle at . Line and . These two lines intersect at which have coordinates and thus has coordinates . Thus, the line . When , has coordinate equal to = which equals giving an answer of
Video Solution by OmegaLearn
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