Difference between revisions of "2008 AIME II Problems/Problem 5"
m (→Solution 2) |
|||
Line 72: | Line 72: | ||
By the [[Pythagorean Theorem]] on <math>\triangle MHN</math>, | By the [[Pythagorean Theorem]] on <math>\triangle MHN</math>, | ||
− | <cmath>MN^{2} = x^2 + h^2 = 504^2,</cmath> so <math>MN = \boxed{504} | + | <cmath>MN^{2} = x^2 + h^2 = 504^2,</cmath> so <math>MN = \boxed{504}</math>. |
== See also == | == See also == |
Revision as of 23:30, 3 April 2008
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |