Difference between revisions of "2008 AIME II Problems/Problem 5"
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=== Solution 3 === | === 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 | + | If you drop perpendiculars from <math>B</math> and <math>C</math> to <math>AD</math>, and call the points if you drop perpendiculars from <math>B</math> and <math>C</math> to <math>\overline{AD}</math> and call the points where they meet <math>\overline{AD}</math>, <math>E</math> and <math>F</math> respectively and call <math>FD = x</math> and <math>EA = 1008-x</math> , then you can solve an equation in tangents. Since <math>\angle{A} = 37</math> and <math>\angle{D} = 53</math>, you can solve the equation [by cross-multiplication]: |
− | < | + | <cmath>\begin{align*}\tan{37}\times (1008-x) &= \tan{53} \times x\\ |
+ | \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, | |
− | < | + | <cmath>\begin{align*}\frac{(1008-x)}{x} &= \frac{\sin^2{53}}{\cos^2{53}} \\ |
− | + | x\sin^2{53} &= 1008\cos^2{53} - x\cos^2{53}\\ | |
− | + | x(\sin^2{53} + \cos^2{53}) &= 1008\cos^2{53}\\ | |
− | + | x = 1008\cos^2{53} &\Longrightarrow 1008-x = 1008\sin^2{53} | |
− | + | \end{align*}</cmath> | |
− | + | Now, if we can find <math>1004 - (EA + 500)</math>, and the height of the trapezoid, we can create a right triangle and use the [[Pythagorean Theorem]] to find <math>MN</math>. | |
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− | Now, if we can find <math>1004 - (EA + 500)</math>, and the height of the trapezoid, we can create a right triangle and | ||
The leg of the right triangle along the horizontal is: | The leg of the right triangle along the horizontal is: | ||
− | < | + | <cmath>1004 - 1008\sin^2{53} - 500 = 504 - 1008\sin^2{53}.</cmath> |
Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression: | Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression: | ||
− | < | + | <cmath>\begin{align*}\tan{37} \times 1008 \sin^2{53} |
− | = | + | = \tan{37} \times 1008 \cos^2{37} |
− | = | + | = 1008\cos{37}\sin{37} |
− | = | + | = 504\sin74\end{align*}</cmath> |
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− | + | Now we used Pythagorean Theorem and get that <math>MN</math> is equal to: | |
− | <math> | ||
− | + | <cmath> | |
+ | \begin{align*}&\sqrt{(1008\sin^2{53} + 500 -1004)^2 + (504\sin{74})^2} = 504\sqrt{1-2\sin^2{53} + \sin^2{74}} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
− | so now we end up with: | + | However, <math>1-2\sin^2{53} = \cos^2{106}</math> and <math>\sin^2{74} = \sin^2{106}</math> so now we end up with: |
− | < | + | <cmath>504\sqrt{\cos^2{106} + \sin^2{106}} =\fbox{504}.</cmath> |
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− | = | ||
== See also == | == See also == |
Revision as of 12:06, 8 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 |