[Math] Is the “Napkin conjecture” open? (origami)

discrete geometrymg.metric-geometryorigami-folding

The falsity of the following conjecture would be a nice counter-intuitive fact.

Given a square sheet of perimeter $P$, when folding it along origami moves, you end up with some polygonal flat figure with perimeter $P'$.
Napkin conjecture: You always have $P' \leq P$.

In other words, you cannot increase the perimeter using any finite sequence of origami folds.

Question 1: Intuition tells us it is true (how in hell can it increase?). Yet, I think I read somewhere that there was some weird folding (perhaps called "mountain urchin"?) which strictly increases the perimeter. Is this true?
Note 1: I am not even sure that the initial sheet's squareness is required.

I cannot find any reference on the internet. Maybe the name has changed; I heard about this 20 years ago.

The second question is about generalizing the conjecture.

Question 2: With the idea of generalizing the conjecture to continuous folds or bends (using some average shadow as a perimeter), I stumble on how you can mathematically define bending a sheet. Alternatively, how do you say "a sheet is untearable" in mathematical terms?
Note 2: It might also be a matter of physics about how much we idealize bending mathematically.

Best Answer

There is a general version of this question which is known as "the rumpled dollar problem". It was posed by V.I. Arnold at his seminar in 1956. It appears as the very first problem in "Arnold's Problems":

Is it possible to increase the perimeter of a rectangle by a sequence of foldings and unfoldings?

According to the same source (p. 182),

Alexei Tarasov has shown that a rectangle admits a realizable folding with arbitrarily large perimeter. A realizable folding means that it could be realized in such a way as if the rectangle were made of infinitely thin but absolutely nontensile paper. Thus, a folding is a map $f:B\to\mathbb R^2$ which is isometric on every polygon of some subdivision of the rectangle $B$. Moreover, the folding $f$ is realizable as a piecewise isometric homotopy which, in turn, can be approximated by some isotopy of space (which corresponds to the impossibility of self-intersection of a paper sheet during the folding process).

Have a look at

  • A. Tarasov, Solution of Arnold’s “folded rouble” problem. (in Russian) Chebyshevskii Sb. 5 (2004), 174–187.

  • I. Yashenko, Make your dollar bigger now!!! Math. Intelligencer 20 (1998), no. 2, 38–40.

A history of the problem is also briefly discussed in Tabachnikov's review of "Arnold's Problems":

It is interesting that the problem was solved by origami practitioners way before it was posed (at least, in 1797, in the Japanese origami book “Senbazuru Orikata”).

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