Difference between revisions of "2008 AIME II Problems/Problem 5"
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− | Let <math>F,G,H</math> be the feet of the [[perpendicular]]s from <math>B,C,M</math> onto <math>\overline{AD}</math>, respectively. Let <math>x = NH</math>, so <math>DG = 1004 - 500 - x = 504 - x</math> and <math>AF = 1004 - ( | + | Let <math>F,G,H</math> be the feet of the [[perpendicular]]s from <math>B,C,M</math> onto <math>\overline{AD}</math>, respectively. Let <math>x = NH</math>, so <math>DG = 1004 - 500 - x = 504 - x</math> and <math>AF = 1004 - (500 - x) = 504 + x</math>. Also, let <math>h = BF = CG = HM</math>. |
By AA~, we have that <math>\triangle AFB \sim \triangle CGD</math>, and so | By AA~, we have that <math>\triangle AFB \sim \triangle CGD</math>, and so |
Revision as of 15:28, 3 June 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 .
Solution 3
If you drop perpendiculars from B and C to AD, and call the points if you drop perpendiculars from B and C to AD and call the points where they meet AD E and F respectively and call FD = x and EA = , then you can solve an equation in tangents. Since and , you can solve the equation:
.
Now if you cross multiply, you get the equation:
However, we know that and . So if we apply that, we end up with the equation:
.
so if we cross multiply again, we get:
, .
Now, if we can find , and the height of the trapezoid, we can create a right triangle and use the Pythagorean Theorem to find .
The leg of the right triangle along the horizontal is:
.
Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression:
= = =
Now we used Pythagorean Theorem and get that MN is equal to:
=
However, =
and
so now we end up with:
= =
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 |