Difference between revisions of "2008 AIME II Problems/Problem 5"
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=== Solution 6 === | === Solution 6 === | ||
Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially <math>\boxed{504}</math>. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.) | Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially <math>\boxed{504}</math>. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.) | ||
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+ | == Solution 7 == | ||
+ | Let the height be h. Note that if <math>\overline{NH} = x</math> then if we draw perpendiculars like in solution 1, <math>\overline{FN} = 500 - x, \overline{AF} = 504 + x, \overline{HG} = 500, \overline{GD} = 504 - x.</math> Note that we wish to find <math>\overline{MN} = \sqrt{x^2 + h^2}.</math> Let's find <math>\tan(53)</math> in two ways. Finding <math>\tan(53)</math> from <math>\triangle BAF</math> yields <math>\tan(53) = \frac{504+x}{h}.</math> Finding it from <math>\triangle CDG</math> yields <math>\frac{h}{504-x}.</math> Setting these equal yields | ||
+ | <cmath>\frac{504+x}{h}=\frac{h}{504-x} \rightarrow h^2 = 504^2-x^2 \rightarrow \sqrt{x^2+h^2} = \sqrt{504^2} = \boxed{504}</cmath> | ||
== See also == | == See also == |
Latest revision as of 16:43, 17 June 2021
Problem 5
In trapezoid with , let and . Let , , and and be the midpoints of and , respectively. Find the length .
Contents
Solution
Solution 1
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 .
For purposes of rigor we will show that are collinear. 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.
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 where they meet , and respectively, then and , and so 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:
Solution 4
Plot the trapezoid such that , , , and .
The midpoints of the requested sides are and .
To find the distance from to , we simply apply the distance formula and the Pythagorean identity to get .
Solution 5
Similar to solution 1; Notice that it forms a right triangle. Remembering that the median to the hypotenuse is simply half the length of the hypotenuse, we quickly see that the length is .
Solution 6
Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially . (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)
Solution 7
Let the height be h. Note that if then if we draw perpendiculars like in solution 1, Note that we wish to find Let's find in two ways. Finding from yields Finding it from yields Setting these equal yields
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.