A rectangular sheet of paper is folded so that two diagonally opposite corners come together. If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?
If I understood well, this should be possible.
If you fold a square diagonally, the crease (the line formed upon folding) is 1.41 times (that is square root of 2) the length of the square. And using a rectangle with a very long length as compared to the width makes the crease shorter and shorter.
Except that I don't think that I can solve this problem... since wonderboy mentioned proof by induction in your other thread (which I don't know), I don't think that this problem is within my reach...
"...two diagonally opposite corners..."
Yes, that's what I did. Let the vertices of the rectangle starting from te upper left, going clockwise be A, B, C and D.
The points A and C will becomes toegther, as per my third diagram.
"meaning the longest diagonal making two congruent right triangles."
No, I don't think that this is what the question is looking for, or else, there is no solution as you yourself said.
Well, I'm misunderstand the problem then... What I understand by it is that the rectangle is folded, so that two diagonally opposite corners (A and C in my previous explanation) meet at a point (or corners B and D meet at a point).
I don't see how you see that the fold must pass through two diagonally opposite corners... the corners won't meet that way... =/
With more challenging problems, my belief is to start from what you already know that's most related to the problem to arrive at a solution so my suggestion to matgrl is to first look at the diagonal of the right triangle and see how it relates to the crease to see if you can derive a solution.
Good luck to you.