AIME I #2 2008

[stupidityismygam]

Square [unparseable or potentially dangerous latex formula] has sides of length [unparseable or potentially dangerous latex formula] units. Isosceles triangle [unparseable or potentially dangerous latex formula] has base [unparseable or potentially dangerous latex formula], and the area common to triangle [unparseable or potentially dangerous latex formula] and square [unparseable or potentially dangerous latex formula] is [unparseable or potentially dangerous latex formula] square units. Find the length of the altitude to [unparseable or potentially dangerous latex formula] in [unparseable or potentially dangerous latex formula].

[collegebookworm]

For the following proof, consult my picture (I know, you're probably tired of all my pictures, but I like them).
http://newtonmusic4.googlepages.com/AIME082.png

As can be seen, the shaded area is a parallelogram with area:
[unparseable or potentially dangerous latex formula]
And solving gives us w = 6. Now we can set up a ratio of two similar triangles (since they share similar angles):
[unparseable or potentially dangerous latex formula]

[unparseable or potentially dangerous latex formula]
Solving gives us [unparseable or potentially dangerous latex formula], with altitude [unparseable or potentially dangerous latex formula].
Answer: 25 (bubble in 025)