19
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ Just as standard Geometric Construction leads to the Fermat primes, adding a trisector leads to the Pierpont primes, primes of the form $2^a3^b+1$. It appears that there are a lot more of these. The link contains references, $\endgroup$ – lulu Feb 18 at 11:04
  • 2
    $\begingroup$ @lulu: Are you saying that origami amounts to compass+straightedge+trisector? $\endgroup$ – tomasz Feb 18 at 11:05
  • 2
    $\begingroup$ @tomasz It appears to be somewhat murky. this article 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. $\endgroup$ – lulu Feb 18 at 11:15
  • 2
    $\begingroup$ @lulu, that's not an ampersand. This is an ampersand: & $\endgroup$ – Gerry Myerson Mar 3 at 9:09
  • 2
    $\begingroup$ @GerryMyerson Thanks! $\endgroup$ – lulu Mar 3 at 11:06
3
+50
$\begingroup$

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.

EDITED TO ADD:

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.