Given a square piece of paper, show how by creasing and folding only, a square of half the area of the original can be obtained.
(In reply to
Variation: by Gamer)
This is going to be hard to explain with out diagrams, but here goes:
Take original square ABCD (AB = length 1 and AC = length 2)
1. Fold and unfold the diagonal AC.
2. Fold AB onto AC (B lies on AC.)
3. Fold AD onto AC (D lies on AC too.)
4. Fold C onto A.
5. Unfold steps 2, 3, and 4.
There is now a crease from B to D. Call the point where it crosses AC 'O'. Call the small creases perpendicular to AB and AD 'X' and 'Y' respectively. AO = AX = AY = .52.
6. Fold DC up so that D is on AY and C is on BC.
Call the new bottom corner across from Y and below B 'Z'.
7. Fold BZ over so that B is on AX and Z is on ZY.
AXZY is the required square.
This takes 6 folds. I'm not sure if it can be improved.
Jer
(I recommend you actually try this. It will probably make more sense if you do.)
ps. The fold in step 3 is not strictly needed, so this is 5 folds.
Edited on April 8, 2004, 9:59 am

Posted by Jer
on 20040408 09:52:16 