What's a minimal origami construction realizing a cube root?

The constructible numbers are those that can be achieved as lengths of line segments via compass and straightedge, starting with a segment of length $$1$$. The origami (constructible) numbers are those that can be achieved as lengths of line segments by folding paper, starting with a segment of length $$1$$.

We say a length has been achieved when it lies between two points occurring as intersections, of drawn lines and circles for the constructible numbers, and of intersecting folds for the origami numbers. I believe that we also allow points to be identified with origami by marking the image of an existing one when it's folded onto a new location (anyone more familiar with the axioms, please correct me if necessary).

It turns out $$r\in\mathbb{R}^+$$ is constructible if and only if it's contained in some chain of real quadratic field extensions of $$\mathbb{Q}$$ (Wentzel, 1837), and $$r\in\mathbb{R}^+$$ is origami-constructible it it's contained in some chain of real degree $$2$$ or $$3$$ extensions of $$\mathbb{Q}$$ (Haga, 1999). So the difference is that origami-constructible numbers are closed under taking cube roots.

I'm interested in seeing the simplest possible origami construction that realizes a cube root. By simplest, I mean minimal number of folds.

• You jump back and forth between "figure" and "number" (or "length"). Which do you want? They may not be the same thing. – Gerry Myerson May 10 at 4:51
• @GerryMyerson You're quite right. Upon looking more closely, I see there are various definitions of these. To discuss the algebra, it would make more sense to define them as points you can get to in $\mathbb{C}$, but I'd rather appeal to the classical notion, as lengths of line segments. I've made the question more precise now, I hope. – j0equ1nn May 13 at 22:59

Copying from http://mathandmultimedia.com/2011/08/02/paper-folding-cube-root/ (but the diagram seems to be missing).

"Get a rectangular piece of paper and fold it in the middle, horizontally and vertically, and let the creases represent the coordinate axes. Let $$M$$ denote $$(0,1)$$ and let $$R$$ denote $$(-r,0)$$. Make a single fold that places $$M$$ on $$y = -1$$ and $$R$$ on $$x=r$$. The $$x$$-intercept of the fold is $$\sqrt[3]{r}$$."

You can't do better than one fold, nor better than degree 3, so this would seem to be the winner.

EDIT: Since the mathandmultimedia page leaves something to be desired, I looked for other duscussions of folding cube roots. Maybe https://www.researchgate.net/figure/Belochs-origami-construction-of-the-cube-root-of-two_fig2_233592288 will be helpful, as well as the link there to https://www.researchgate.net/publication/233592288_Solving_Cubics_With_Creases_The_Work_of_Beloch_and_Lill

There's a simple step-by-step $$\root3\of2$$ at http://www.cutoutfoldup.com/409-double-a-cube.php

An older question here on math.stackexchange is Solving Cubic Equations (With Origami)

The Hull article in the Monthly is also available at http://sigmaa.maa.org/mcst/documents/ORIGAMI.pdf

One more source for $$\root3\of2$$: http://www.math.ubc.ca/~cass/courses/m308-05b/projects/ayuen/doublecube.html

• "You can't do better than one fold" I see three. – Arthur May 10 at 6:08
• Yes, I forgot to count the two that make the axes. Three it is. Still, probably hard to beat. – Gerry Myerson May 10 at 6:20
• Thanks for that, but I find that description kind of vague. (Admittedly, my question was pretty vague too -- I've just edited it to try to clear that up.) For one, you have to make additional folds in order to identify the lines $y=-1$ and $x=r$. Then you have to get pretty violent with the paper to do that last step. Have you tried it? It's hard. But I believe it's the essence of the answer: to get a cube root, you must be able to send 2 distinct points to 2 distinct lines. – j0equ1nn May 13 at 23:08
• I'd also like to a nice way to see why that intersection point is $\sqrt[3]r$. I think I can do this for myself, but I'm interested in a figure I can do on a board where students will follow it. – j0equ1nn May 13 at 23:15
• Any thoughts on my edits, j0e? – Gerry Myerson May 15 at 6:00