The triangle a strip becomes

Everything up to this point has been about what a glued band cannot do. This one is about the band that folds, and about the number that comes out of it.

A strip of paper joined into a Möbius band, creased three times at sixty degrees to its edges, presses flat into an equilateral triangle. Three layers everywhere, three creases forming the three sides, and the whole thing turns over into itself because the object has one side. It is a pleasing thing to hold, it takes about a minute to make, and the shortest strip it can be made from is $\sqrt3$ times as long as it is wide.

None of those numbers is chosen. Each falls out of the closure condition in turn, and the interest is in the order they fall out in.

Where the angles come from #

The composition of the reflections has to equal the map that glues the band’s ends together, and that equation has two halves that behave differently.

The linear half depends only on the crease angles. Reflecting in a line at angle $\phi$ has a linear part determined by $\phi$ alone, and composing three of them gives a reflection in a line at the alternating sum $\phi_1 - \phi_2 + \phi_3$. The Möbius band’s gluing map is a slide along the band together with a flip across its own axis, and the flip is a reflection in a line at nought degrees.

So the angles have to satisfy

$$\phi_1 - \phi_2 + \phi_3 \equiv 0 \pmod{180^\circ}.$$

That is one equation among three angles, so it leaves a two-parameter family — and it is a curve rather than a region, which is why almost every triple of angles folds nothing at all.

Sixty, a hundred and twenty and sixty is on the curve: $60 - 120 + 60 = 0$. So is forty-five, a hundred and thirty-five and ninety. So is seventy, a hundred and forty and seventy. Nothing so far singles out the equilateral one.

Where the positions come from #

With the angles fixed, the linear half of the equation is satisfied identically and what is left is the translation — two numbers, which have to match the slide the gluing demands.

The translation of the composition depends on where the creases sit, and it depends on them affinely. Reflecting in a line through a point $p$ contributes a translation that is a linear function of $p$; composing three such reflections gives a translation that is a linear function of the three positions. So the condition is two linear equations in three unknowns.

Two equations and three unknowns leaves a line of solutions, and the line is easy to identify: sliding all three creases along the band by the same amount moves the drawing and not the motion, because the composition is conjugated by a translation it commutes with. So the free parameter is where along the band the whole arrangement sits, which is not a degree of freedom in any interesting sense, and the arrangement itself is determined.

This is worth dwelling on, because it is unusual for this subject. Almost every question here is a search — which letters can be assigned, which layer goes on top, which packing fits. This one is a linear solve, and it is a linear solve because the sheet has no interior vertices and therefore exactly one loop to close.

Where the length comes from #

The closure condition holds at every length. Given a strip of any length at all, the two equations have a solution, and the solution puts the three creases somewhere.

What fails at short lengths is not the algebra. It is that the creases the algebra asks for do not fit.

A crease at sixty degrees crossing a strip of width one spans $1/\tan 60^\circ \approx 0.577$ along the band. Three of them, at the positions the solve produces, occupy a certain span, and the strip has to be at least that long — and it has to hold them in order, with each crease entirely to the left of the next, since a pair of creases that cross inside the band is not a band with panels between its creases.

Bisecting on the length, feasibility is monotone — the strip grows faster than the solution does — and the crossing sits at

$$\frac{L}{W} = 1.7320508\ldots = \sqrt3.$$

At exactly that ratio the creases meet the ends of the strip and each other; a hair above it they fit with room to spare.

Why √3 is the right answer to expect #

The number is not a surprise to anyone who has the folded object in front of them, and the check takes two lines.

Call the triangle’s side $s$. The band has three panels, and at the shortest length each of them is an equilateral triangle of side $s$ — the solve puts the creases so that consecutive ones meet on the strip’s edges, and the pieces between them are exactly the triangle. The three stack onto each other, three layers everywhere, which is why the flattened object is opaque and why turning it over shows the same triangle.

The strip’s width is therefore the triangle’s height, $s\sqrt3/2$. Its area is three triangles, $3 \cdot s^2\sqrt3/4$. Divide the second by the first and the strip’s length is $\tfrac32 s$ — one and a half sides. So

$$\frac{L}{W} = \frac{\tfrac32 s}{s\sqrt3/2} = \frac{3}{\sqrt3} = \sqrt3.$$

The solved drawing agrees to seven figures: at a width of one it puts the creases at $0$ and $1.1547$, and $1.1547$ is $2/\sqrt3$, which is the side of an equilateral triangle of height one.

That is the shape of the whole result. The closure condition did not know it was making a triangle; it composed three reflections, matched them against a glide, and solved two linear equations. The triangle is what the answer turns out to be.

The other triples, and how much worse they are #

Every point on the admissible curve gives a band that folds. They do not all need the same amount of paper, and the equilateral one is not merely the tidiest.

The gap is not marginal. Sixty, a hundred and twenty and sixty needs 1.732 widths; the next-best triples need 1.92, and most of the curve needs two and a half or more. That is a genuine minimum with a margin, and it is the reason the equilateral fold is the one people arrive at by experiment rather than one convention among several.

It is also a reminder that the admissible curve is not a curve of equivalent answers. Every point on it satisfies the same linear condition, and what separates them is the geometry of fitting rather than the algebra of closing.

What the parity was not doing #

The counting condition refuses every even crease count and it approves every odd one, and every odd square-creased band is wrong.

So the sequence of conditions runs: the parity refuses half the counts, the alternating sum refuses all but a curve of the angle triples, and the fitting refuses all but a half-line of the lengths. Each is a genuine constraint, each is cheaper than the one after it, and the first two are the ones a counting argument can reach.

The proportion, beside the others #

A strip that has to be $\sqrt3$ long to fold is a paper proportion, and this subject has a small collection of those.

The comparison is instructive because the two numbers are different kinds of number. The A series’ √2 is a shape: a rectangle has that proportion or it does not, and the property it buys is that halving it returns the same proportion. The band’s $\sqrt3$ is a minimum: any strip at least that long works, and longer strips work with room to spare.

That distinction is taken up separately, because it changes what the number is for — a shape is something to cut paper to, and a bound is something to check paper against.

Making one #

The instructions are short and the object is worth having.

Cut a strip whose length is a little more than $\sqrt3$ times its width — twenty-one centimetres by twelve is comfortable, and anything above about 1.74 to 1 will do. Mark the long edges lightly at the two points that are a third and two thirds of the way along.

Fold the strip at sixty degrees to the long edges, running from a corner to the first mark; unfold. Fold again from that mark to the second, slanting the other way; unfold. Fold a third time from the second mark to the far corner, slanting back. The three creases zigzag across the strip and meet the edges at the marks.

Now join the ends with the half twist, so that the front of one end meets the back of the other, and let the three creases collapse. The band settles into the triangle without persuasion. Tape the join if it is to survive being handled.

The last step is the one worth watching. As the paper closes, the two ends arrive at each other already in the right relative position, which is what the two linear equations were about: the creases were placed so that the composed folding motion carries one end exactly onto the other, the other way up.

Cut the strip too short — 1.6 times its width, say — and the same three creases can be marked but the join no longer meets, because the solution the algebra wants has its creases running off the ends of the paper. The failure is visible before anything is folded.

What a longer strip does #

Above $\sqrt3$ the band still folds, and it does not fold into a triangle any more.

The two linear equations still have a solution at every length, and the solution slides the three creases apart as the strip grows. The panels between them stop being equilateral triangles and become trapezoids, and the flattened object becomes a triangle with its corners cut off, or a hexagon, depending on how much extra paper there is.

That is a family rather than an exception, and it is the ordinary situation. The equilateral triangle is the boundary case where the trapezoids have degenerated into triangles and the creases have run into each other at the strip’s edges, which is precisely why it is the shortest: any less paper and the degeneration would have to continue past the point where the creases cross, and crossing creases are not a band with panels between them.

So $\sqrt3$ is a boundary of feasibility rather than an optimum of anything, and the object sitting on it is prettier than its neighbours for the same reason that a tangent circle is prettier than a secant one.

Two numbers that are easy to swap #

There are two ratios in play and confusing them produces an answer that is wrong by a factor of two, so it is worth pinning both down.

The strip’s proportion is $\sqrt3 \approx 1.732$: length over width. That is the number the solve reports and the number to cut paper to.

The triangle’s proportion is not a proportion at all — an equilateral triangle has only one shape — but its side is $2/\sqrt3 \approx 1.155$ times the strip’s width, because the width is the triangle’s height and a triangle’s height is $\sqrt3/2$ of its side.

The tempting third number is the perimeter, $3 \times 1.155 = 3.46$ widths, which is twice the strip’s length. That factor of two is the source of the usual error: the strip does not run round the triangle, it lies across it, and its length is one and a half sides rather than three. Anyone who reaches $2\sqrt3$ has computed the wrapping picture rather than the folding one.

The check that settles it is the area. Strip area is $L \times W$; folded area is three triangles; and the two agree only for $L = \tfrac32 s$.

The three layers, and what they are not #

A folded object with three layers everywhere invites a question this collection usually has to work for: which layer is on top.

Here the question is nearly empty, and the reason is the sheet rather than the fold. The three panels stack, and on top is a direction relative to a reader; the sheet has one side, so a reader walking round the band arrives back with the stack reversed. What is well defined is the cyclic order — which panel is between which two — and that is definite and easy to read off the paper.

The usual machinery expects better than that. Enumerating the stackings of a pattern starts from the panel with nothing below it and works upward, and a sheet with no edge has no such panel. This band has an edge — the strip’s own long sides survive the gluing and become the triangle’s boundary — so a bottom panel exists locally. What does not exist is a global agreement about which one it is.

That is a smaller loss than it sounds, because nothing about whether the object folds depends on it. The layer order matters when panels would have to pass through each other, and three panels covering the same triangle in some order never do.

Why this could not have been asked before #

The construction needed one thing the collection did not have, and it is worth naming because it is the whole of the new machinery.

Everything here is built out of pieces that already existed: reflections in lines, composition of plane isometries, the comparison of two motions, and a bisection. What was missing was a way to say which points of a sheet’s boundary are the same point, and to hand that statement to the closure condition as the thing the composition has to equal.

Before that, the closure condition compared the composed motion against the identity, because on a disc the identification is trivial and the identity is what one puts on the right-hand side when there is nothing to put there. That comparison is correct for every sheet the collection could build and it refuses every band, including the ones that fold, because on a band the walk does not come back to where it started.

One argument to one function, in other words. The whole of the Möbius band’s arrival in this collection is that a constant became a parameter.

Where this sits in the literature #

The flat-folded Möbius band is not new and the collection did not find it, which is worth saying plainly.

A strip of paper joined with a half twist and creased into a triangle is a standard construction, and the question of how short a strip can be has a substantial history in differential geometry — where the object of study is the smooth Möbius band, made without creases at all, and the answer is a genuinely hard theorem rather than a linear solve. The two problems have the same bound and they are not the same problem: a smooth band bends continuously and a folded one is flat everywhere except on three lines.

The relationship between them is a limit rather than a coincidence, and it is the kind of relationship that is easy to assert and hard to establish. Nothing here establishes it. What is computed above is the folded case, by composing three reflections and solving for three positions, and the agreement with the number the smooth case is known for is reported rather than explained.

That the same ratio turns up in both is the sort of fact that deserves a sentence saying it has not been proved here.

The conditions in order #

For a reader assembling the construction, the sequence of things that had to be true.

The parity: an odd number of creases, which is the condition the seam’s sign produces and which refuses half the counts.

The angles: an alternating sum that is a multiple of a straight angle, which refuses all but a curve.

The positions: two linear equations, always solvable.

The length: at least √3 widths, which is the fitting.

And none of the subject’s four vertex conditions enters anywhere, because the band has no vertices for them to hold at.

What is checked, and what is taken on trust #

The angles condition, the linear solve and the fitting are each computed rather than reasoned about, and the assertions on them are what stops a wrong drawing reaching a page.

The composition is compared against the gluing map at every band drawn, and the comparison is a distance in widths of the strip rather than a test. The ratio is found by bisection, and the bisection is only valid because feasibility is monotone in the length, which is asserted rather than assumed: the band is checked to fail below one and a half widths and to fit comfortably above two and a half.

And a triple that ought not to close is checked not to. Seventy, a hundred and ten and seventy has an alternating sum of thirty degrees, and if the machinery ever reported it as folding, something would be wrong with the machinery rather than with the mathematics.

What is taken on trust is the identification of the folded object as an equilateral triangle. The closure condition says the paper closes up; it does not say what shape the result is, and the triangle above is read off the geometry of the solution rather than computed as an outline. A reader who wants that identification checked has the strip, the tape and ninety seconds, which is the sort of corroboration this subject is unusually well supplied with.