2008 AIME II Problems/Problem 5
Extend and to meet at a point . Then .
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 .
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 .
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:
so now we end up with:
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