It's well known that angle trisection cannot be done with straightedge and compass alone, as

Theorem 1. If $z \in \mathbb C$ is constructible with straightedge and compass from $\mathbb Q$, then $$\mathbb Q (z) : \mathbb Q = 2^n.$$

But the minimal polynomial of $\cos 20 ^{\circ}$ is $8 x ^ { 3 } - 6 x - 1$, so $$\mathbb Q (\cos 20 ^{\circ}) : \mathbb Q = 3,$$

That proves we cannot trisect $ 60 ^{\circ}$.

However, it's doable with origami, as Huzita Axiom 6 - Computing the Origami Trisection of an Angle shows. My question is:

Exactly what field extensions can be obtained by considering origami constructible number? Is this as well-studied as straightedge and compass, i.e. do we have a similar theorem as Theorem 1?

YuiTo Cheng
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    Just as standard Geometric Construction leads to the Fermat primes, adding a trisector leads to the [Pierpont primes](http://mathworld.wolfram.com/PierpontPrime.html), primes of the form $2^a3^b+1$. It appears that there are a lot more of these. The link contains references, – lulu Feb 18 '19 at 11:04
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    @lulu: Are you saying that origami amounts to compass+straightedge+trisector? – tomasz Feb 18 '19 at 11:05
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    @tomasz I believe that's the case, though my memory of the details is sketchy at best (and certainly may be faulty). [This article](https://en.wikipedia.org/wiki/Pierpont_prime) contains further references, and specifically mentions paper folding. – lulu Feb 18 '19 at 11:07
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    @tomasz It appears to be somewhat murky. [this article](https://langorigami.com/article/huzita-justin-axioms/) appears to suggest that compass+straightedge+trisector is the same as the first six axioms of paper folding, but that there is at least a seventh which does not follow. However, that seventh does not enable the solutions of any new algebraic system. But then, the article suggests that it is unclear whether or not there might be still more structure which is as yet undiscovered. Murky. Hard to believe the question is actually open, but I can't find a definitive statement. – lulu Feb 18 '19 at 11:15
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    Is trisection the same as adding arbitrary cubic extensions? It appears that if K is a subfield of origami-constructible numbers, then extensions of K of degree 2 and 3 are also subfields. But it's not clear if closure under these extensions is all that can be done. – Aravind Feb 18 '19 at 11:25
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    @Aravind Sorry, i didn't see your comment until just now (you need to use the ampersand notation, as I started this comment with, to send a comment to a specified user). In any case: Trisections let you extract cube roots, and [Cardano's formula](https://proofwiki.org/wiki/Cardano%27s_Formula) tells us that that's all you need to solve cubics (in addition to standard algebraic operations up to square roots). – lulu Feb 18 '19 at 16:02
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    @lulu, that's not an ampersand. This is an ampersand: & – Gerry Myerson Mar 03 '19 at 09:09
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    @GerryMyerson Thanks! – lulu Mar 03 '19 at 11:06

1 Answers1


There is a paper posted on ArXiv by Antonio M. Oller Marcén entitled "Origami Constructions" that claims to show that:

If $a \in \mathbb{R}$ is origami-constructible, then $$[\mathbb{Q}(a): \mathbb{Q}] = 2^r3^s$$ for some $0\leq r, s \in \mathbb{Z}$.

Unfortunately, I have not been able to find this particular paper published in any peer-reviewed venue nor am I able to personally vouch for the proof, so I guess caveat lector.


The same result is also found in a Master's thesis by Hwa Young Lee entitled "Origami-Constructible Numbers" (Corollary 4.3.10, pg. 50).

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