2008 AIME II Problems/Problem 5
Problem 5
In trapezoid with , let and . Let , , and and be the midpoints of and , respectively. Find the length .
Solution
Solution 1
Extend and to meet at a point . Then .
Since , then and are homothetic with respect to point by a ratio of . Since the homothety carries the midpoint of , , to the midpoint of , which is , then are collinear.
As , note that the midpoint of , , is the center of the circumcircle of . We can do the same with the circumcircle about and (or we could apply the homothety to find in terms of ). It follows that Thus .
Solution 2
Let be the feet of the perpendiculars from onto , respectively. Let , so and . Also, let .
By AA~, we have that , and so
By the Pythagorean Theorem on , so .
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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All AIME Problems and Solutions |