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."
Note: This diagram was offered as my visual interpretation of Mark's problem:
Now, the new stuff....
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
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[90ArcTan[x/H]] +
L*Sin[902ArcTan[x/H]] / Sin[90  ArcTan[x/H]]Sin[90 + ArcTan[x/H]]
Here's what I did:
[1] Take a piece of paper and fold it so that there are 16 equallysized rectangles (by folding in half vertically twice and horizontally
twice)
[2] Repeat step one with another piece of paper. Then cut out the middle rectangle, made of four similar rectangles.
[3] Place the cut out triangle on the center of the originial piece of paper.
[4] Label the four corners of the cut out paper A,B,C, and D from bottom left clockwise to bottom right.
[5] Fold the top left corner of the cut out piece of paper to the bottom edge of the cut out piece of paper.
[6] 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.
[8] 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 noneuclidean geometry class).
[9] Now lots of trig is used to find the distance from E to F which is the crease.
Thanks T.M.! Your efforts, insight, and perseverance are appreciated.
Hint:
Solution Commentary: Not relebvant...just go back and try to understand T.M.'s solution...and compare it to the solution comments made originally.
