2008 AIME II Problems/Problem 5
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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 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:
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 
.
See also
| 2008 AIME II (Problems • Resources) | ||
| Preceded by Problem 4 | Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
