Remember This Problem From 2/19/2008?
On a Internet message board, "Mark" posted the following: "I hoping one of you smart people can help me. Here's the problem: You have a sheet of paper 25 units long by 6 wide. With the long side facing you, fold the upper right corner to touch the bottom, so that the crease is minimized. Write a function defining the crease, and find the minimum."
Now, the new stuff....
Note: This diagram was offered as my visual interpretation of Mark's problem:
It makes my day when users of this website send in solutions, which I then try to add to the solution section for the relevant problem. But on this specific problem, T.M. (from Anacortes, WA and a former student of mine) went beyond expectations and offered this solution (after "many pieces of paper had fallen victim" to his efforts):
H = Height of paper
Thanks T.M.! Your efforts, insight, and perseverance are appreciated.
L = Length of paper
x = distance from bottom left corner to corner touching the bottom of the paper (top left)
Crease = Sqrt[x^2 + H^2]Sin[ArcTan[x/H]] / 2Sin[90-ArcTan[x/H]] +
L*Sin[90-2ArcTan[x/H]] / Sin[90 - ArcTan[x/H]]Sin[90 + ArcTan[x/H]]
Here's what I did:
 Take a piece of paper and fold it so that there are 16 equally-sized rectangles (by folding in half vertically twice and horizontally
 Repeat step one with another piece of paper. Then cut out the middle rectangle, made of four similar rectangles.
 Place the cut out triangle on the center of the originial piece of paper.
 Label the four corners of the cut out paper A,B,C, and D from bottom left clockwise to bottom right.
 Fold the top left corner of the cut out piece of paper to the bottom edge of the cut out piece of paper.
 Label the ends of the crease (left and right) E and F and the new position of the top corners G and H 7. Extend the following segments. BC, EF, GH to the right. These
intersect at some point Z. Triangle BZE is congruent to GZE.
 Now make the line segment BG. Because we know the lenght and width of the paper we can find the lenght of BG. And because B and G
are on congruent triangles, BZE and GZE respectively, where segment BG intersects the line (call it I) that goes through EZ they form right angles (this is from a problem in your non-euclidean geometry class).
 Now lots of trig is used to find the distance from E to F which is the crease.
Solution Commentary: Not relebvant...just go back and try to understand T.M.'s solution...and compare it to the solution comments made originally.