Difference between revisions of "2008 AIME II Problems/Problem 5"
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Since <math>\overline{BC} \parallel \overline{AD}</math>, then <math>BC</math> and <math>AD</math> are [[homothecy|homothetic]] with respect to point <math>E</math> by a ratio of <math>\frac{BC}{AD} = \frac{125}{251}</math>. Since the homothety carries the midpoint of <math>\overline{BC}</math>, <math>M</math>, to the midpoint of <math>\overline{AD}</math>, which is <math>N</math>, then <math>E,M,N</math> are [[collinear]]. | Since <math>\overline{BC} \parallel \overline{AD}</math>, then <math>BC</math> and <math>AD</math> are [[homothecy|homothetic]] with respect to point <math>E</math> by a ratio of <math>\frac{BC}{AD} = \frac{125}{251}</math>. Since the homothety carries the midpoint of <math>\overline{BC}</math>, <math>M</math>, to the midpoint of <math>\overline{AD}</math>, which is <math>N</math>, then <math>E,M,N</math> are [[collinear]]. | ||
− | As <math>\angle AED = 90^{\circ}</math>, note that the midpoint of <math>\overline{AD}</math>, <math>N</math>, is the center of the [[circumcircle]] of <math>\triangle AED</math>. | + | As <math>\angle AED = 90^{\circ}</math>, note that the midpoint of <math>\overline{AD}</math>, <math>N</math>, is the center of the [[circumcircle]] of <math>\triangle AED</math>. We can do the same with the circumcircle about <math>\triangle BEC</math> and <math>M</math> (or we could apply the homothety to find <math>ME</math> in terms of <math>NE</math>). It follows that |
+ | <cmath> | ||
+ | NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500 | ||
+ | </cmath> | ||
+ | Thus <math>MN = NE - ME = \boxed{504}</math>. | ||
=== Solution 2 === | === Solution 2 === |
Revision as of 19:17, 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 .
Since and , it follows that . By the Pythagorean Theorem on , and the answer is .
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 |