Conic-Constructible Regular Polygons
John Conway
conway at math.Princeton.EDU
Wed Dec 15 21:57:08 CET 1999
On Wed, 15 Dec 1999, Antreas P. Hatzipolakis wrote:
> John Conway wrote:
>
> >On Tue, 14 Dec 1999, Antreas P. Hatzipolakis quoted:
> >
> > I think it's obvious that the answer is those for which n has the form
> >2^a.3^b.p.q.r.... , where p,q,r,... are distinct primes of the
> >form 2^x.3^y + 1. Yes, that IS obvious.
and wrote:
> And I think that what is obvious for JHC is not obvious for everyone else! :-)
Well, let me explain this one. The intersections of two conics
(finding which I take as the additional "move") are found by solving a
quartic equation, and indeed you get the general quartic this way.
So the difference is that we can solve quartics as well as quadratics.
But solving a quartic can be done by solving a cubic and some
quadratics, so we can instead regard the new move as solving a cubic.
Now it's "obvious" (if you know Galois theory) that the degree of any
equation that can be solved by quadratics must have form 2^a.3^b
and applying this to the equation
1 + z + z^2 + ... + z^(p-1) = 0
whose roots are the primitive pth roots of 1 (p prime), we see that
there's no hope of constructing the regular p-gon (p prime) unless
p-1 has this form.
But if it DOES, then Gauss' method shows that this particular equation
CAN so be solved. This settles everything except for the question of
repeated primes. The number of primitive p^2'th roots of one is
phi(p^2) = p(p-1), and since this degree has a factor of p, we can only
hope to construct the regular (p^2 or higher)gon if p = 2 or 3. But we
can actually construct any 2^a.3^b gon by repeated angle bisections and
trisections. Summing up, this gives the answer I gave above.
> There is a theorem (by Vieta) which says:
>
> (conic-constructibility) <=> (marked ruler constructibility: trisect an angle)
I don't think there is any credible sense in which this theorem can
be ascribed to Vieta, although I'm prepared to believe that he did
something towards it.
John Conway
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