# Constructing a Möbius strip using a square paper? Is it possible?

I understand that, from a topological perspective, it is irrelevant whether we choose the quotient of the square $[0,1]\times [0,1]$ (by identifying points $(0,t)$ and $(1,1-t)$) or the quotient of the rectangle $[0,1]\times[0,a]$, $a>1$ (where the obvious points are identified), as a model of the Möbius strip. However, it seems pretty hard, if not impossible, to actually physically construct a Möbius stirp using a square like paper.

I wonder if it is possible to determine the least $a\geq 1$, such that he Möbius strip is physically constructible using a paper congruent to $[0,1]\times[0,a]$. I guess that this answer may depend on factors such as the size and the thickness of the paper that we use. This may be related to the mathematics of paper folding. Is there a general answer to this question, taking relevant qualities of a paper into account? Does anyone know of a source answering this question?

• A perfectly rigid sheet of paper can not be bent, but a perfectly stretchable sheet of paper means we don't need to worry about dimensions at all, so you need to be more precise with what you mean by 'physically construct'. – Dan Rust Mar 31 '14 at 14:34
• @Daniel There is a mathematical notion of a “developable surface” which is what is exactly what is wanted here. Intuitively, the idea is that the developable surface is one that can be obtained from a flat sheet by bending, but not by stretching. For example, a cylinder is developable, but a sphere is not. – MJD Mar 31 '14 at 14:41
• And perhaps I should add that such surfaces are exactly those whose gaussian curvature is identically zero, a property possessed by the standard immersion of the Möbius strip into $\Bbb R^3$. – MJD Mar 31 '14 at 14:48
• I believe if we're allowed sharp folds, then it can be done with any rectangle, as we can reduce the problem to a thin rectangle by applying an even accordian fold to the original rectangle. Once the rectangle is suitably thin, we may then glue it together to get something resembling the Google drive logo which is topologically a Mobius strip. – Dan Rust Mar 31 '14 at 15:04
• @MJD the accordian folds should not be 'unfolded' as it were. The angle created is 0 - this of course means it doesn't satisfy any embedding condition you might impose. Nate's answer encounters the same problem. – Dan Rust Mar 31 '14 at 15:23

(apologies for the lack of mathematics in this answer, am posting as wiki, please do improve it)

If you've got any square paper handy, you can fold a (slightly unconventional) Mobius strip in it like this pictured below. It's like a Mobius strip in that:

• Using a twist, the whole of one side connects to the whole of the opposite side inverted ($AB>CD$), the top left of the front connecting to the bottom right of the back, the same as if it was a long strip with one twist.
• You can fold it and unfold it
• It's a ring with a twist in it (not an open-topped "ice cream" cone)
• If you follow a line along the strip with a pencil in an uninterrupted line, when you unfold it, both sides will be marked.

It's different to a standard mobius strip in that it's folded over itself (but all the above still seems to be true). There's more to this answer than just folding a square in quarters then twisting it: as you'll see if you try it, that wouldn't match up the whole of one side against the whole of the other. This does.

I'm not sure if this is strictly a Möbius strip, but it has the main properties and can be folded from a square. Colours are purely as a visual aid. The end shows where on the front side (1) connects to where on the back side (2). If you were to label a conventional 'ribbon' Möbius strip in the same way ($A>D$ down each edge), the connection points would come out the same.

This is really a comment (or at best, a partial answer), but too long and illustrated for a comment.

It suffices if $a > 2$. Here's a proof by picture, for $a = 3$:

Essentially, a "half-turn" of the paper takes one width. Making two half-turns (to reverse the paper) takes two widths, and then returning to the starting point takes one width.

This results in a nice loop (it can be unbent and is flexible.)

You can reduce the length of the last flap, and just bend before you tape. You can do this all the way down to $a = 2$. In this case, in the last step you have

and you just fold the whole thing in half and tape the ends. But with $a = 2$ exactly, you can't unbend it without ripping the paper.

• Out of curiosity - what software did you use to create these images? I had the same idea after playing with a few strips of paper, but didn't know how to easily create such drawings, so I didn't post it. – fgp Apr 1 '14 at 15:10
• @fgp: LaTeX with TikZ! – Nick Matteo Apr 2 '14 at 1:46

Suppose you have a square $ABCD$, fold point $A$ onto point $C$, giving you an isosceles right triangle. Next fold $D$ onto $B$ giving you a smaller isoceles right triangle. If you did this correctly sides $AB$ and $CD$ should be lined up (as the hypotenuse) if you were to tape them together the resulting figure would be a Mobius strip.

• You can't unfold this from a triangle, except to make an ice-cream cone. Effectively all four edges of the square are identified. The resulting surface (of the cone) doesn't have any of the unusual properties we associate with the Möbius strip. It doesn't really address the spirit of the question. – Nick Matteo Mar 31 '14 at 16:50
• Admittedly, when I came up with that I had an 8.5 x 11 piece of paper in front of me. So I really ended up with an isosceles right triangle plus a "tab" which I treated as the "tape" to connect the two appropriate faces, but not the other two. But yes, I agree it is cheating somewhat. – Nate Mar 31 '14 at 16:57