Difference between revisions of "2008 AIME II Problems/Problem 5"
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\frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\sin{37}}{\cos{37}}\end{align*}</cmath> | \frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\sin{37}}{\cos{37}}\end{align*}</cmath> | ||
− | However, we know that <math>\cos{90-x} = sin{x}</math> and <math>\sin{90-x} = cos{x}</math> are co-functions. Applying this, | + | However, we know that <math>\cos{90-x} = \sin{x}</math> and <math>\sin{90-x} = \cos{x}</math> are co-functions. Applying this, |
<cmath>\begin{align*}\frac{(1008-x)}{x} &= \frac{\sin^2{53}}{\cos^2{53}} \\ | <cmath>\begin{align*}\frac{(1008-x)}{x} &= \frac{\sin^2{53}}{\cos^2{53}} \\ |
Revision as of 11:55, 27 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 and to , and call the points if you drop perpendiculars from and to and call the points where they meet , and respectively and call and , then you can solve an equation in tangents. Since and , you can solve the equation [by cross-multiplication]:
However, we know that and are co-functions. Applying this,
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 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 |