# Optimal present wrapping with a rectangle

This is a (slightly silly) version of Penguino's Ernie gift-wrapping puzzle.

You need to wrap a cubical present with side length 1 meter. As in the linked question, when you wrap a present, you are not allowed to cut the wrapping paper. You can only fold it to cover the cube.

Unfortunately, unlike in the linked question, here you have no triangle. You can only use one rectangle of wrapping paper. The rectangle can have whatever dimensions you like. You want to use the least paper possible. How small can you make the area of the rectangle?

• Keyword; Rectangle. Squares are allowed? Dec 29, 2014 at 2:14
• @warspyking Yes, squares are allowed. Dec 29, 2014 at 3:16
• I believe you can in fact do it with six square meters plus epsilon, but am having trouble specifying the construction.
– xnor
Dec 29, 2014 at 3:55
• I dont have the maths to back this up, but my gut tells me you can make this work with nearly no overhead using a long long and very fine strip (with width approaching 0) of paper then winding it around from corner to opposite corner in a sort of spiral Dec 29, 2014 at 8:36
• My guess would by that the overhead would then be 6W where W was the width of the strip, thus arbitrarily low, but I would be interested to see if anyone can prove if this is really possible. Dec 29, 2014 at 8:42

I can do it in $6 + \epsilon$ square meters, where $\epsilon$ is any number greater than zero. Let the rectangle be of dimensions $w \times (6/w + 3/2)$, where $w$ is a very, very small width. Lay it out as follows:

Have the rectangle border the edge of the above figure, turn at the edge, trace the border back agian, and so on. In each repetition, which covers a width of w, there are 3 90 degree turns. Each 90 degree turn causes an overhead of $w^2/2$, the area of the folded over triangle created this way. Thus, there is an overhead of $3w^2/2$ in each repetition, and $3w/2$ in constructing the entire figure, which has $1/w$ repetitions. The above figure, once covered, can easily be assembled into a full cube, so covering the full cube also has an overhead of $3w/2$. Since w can be chosen arbitrarily, I will choose it to be $2\epsilon/3$. Thus, there is an overhead of $\epsilon$, for a full area of $6+\epsilon$.

I can do it with 7 square meters of paper.

Make the paper 1m by 7m. Mark it every 1m, making seven squares:

_____________________________
| 1 | 2/| 3 | 4 | 5 | 6/| 7 |
|___|/__|___|___|___|/__|___|


Fold squares 2 and 6 in half diagonally, making each into a pair of equal triangles. It now looks like this:

                    _____
| 7 |
____________|___|
/| 3 | 4 | 5 |6 /
/_2|___|___|___|/
| 1 |
|___|


Now fold the four inner markings (the sides of 3, 4, and 5) by 90 degrees, toward each other. The diagonal edges of 2 and 6 now meet each other, and you have a square tube comprising 2-3-4-5-6 (half of 2 and half of 6 are "wasted"), plus two "flanges" 1 and 7. Bend 1 and 7 by 90 degrees as well, and you have a cube.

• Interesting. Isn't that giving you exactly the same waste as the initially proposed 45 rotated- centered square which was proposed on the Ernie-question in a first place? (Has the answer been deleted there? Can't see it anymore.) Dec 29, 2014 at 7:44
• @BmyGuest: as far as I can tell you're talking about this: mathpuzzle.com/wrap.htm which uses 8 square meters of paper (same as Len's answer to this question). My solution uses 7 square meters, which is better. I manage this by using a rectangle rather than a square piece of paper. Dec 29, 2014 at 8:03
• Absolutely right, my bad. Nice answer! Dec 29, 2014 at 8:05

OK, I'm not very clever, but I think I'll take a shot at this. Center the box on a square piece of paper. The distance from the center of the paper/box to the edge of the paper needs to be 2 meters (.5 for the bottom, 1 for the side, and .5 for the top) so the pieces meet on the opposite side of the box. Any less and it'll fall short. The way to minimize paper use is to rotate the box 45 degrees so it's "diamond shaped" on the square piece of paper. That way the required 2 meter length will be on the diagonal of the wrapping paper, the longest distance from centre to edge of the paper.

So the entire diagonal will be 4 meters, each side is sqrt(8)....

So, 8 square meters of paper? One square of paper the square root of 8 on each side? (Square square square.)

But geometry is not my long suit. Anyone out there know a better way to do this with some creative folding?

(Edit: Alternately, if it was a 1x8 sheet, you could wrap 4 sides of it with the middle 4m and then do 45 degree folds to cover the last two sides with the ends, or wrap 4 sides starting at one end, then do a 45 degree fold and wrap the last two sides with the last 3m.)

Also, I like the way this d6 I wrapped up in a square of toilet paper looks.

• +1 for wrapping a d6 in toilet paper, and also for calling it a d6. Dec 29, 2014 at 4:18
• What can I say, I'm a diehard boardgamer. My Christmas gift to myself this year was a set of Sicherman dice from Eric Harshbarger's site. Dec 29, 2014 at 4:32