Beyond angle trisection: Constructing regular polygons by dividing angles into 5, 7, 11, (et cetera) equal parts

Question

I've read a paper by Andrew Gleason where he was able to come up with a way to construct heptagons and tridecagons using angle trisection to supplement the usual compass and straightedge. This post questions the ability to use angle quintisection (dividing into five) to construct an undecagon (11 sides).

Gleason also mentioned that a 19-gon (enneadecagon) requires 2 angle trisections to construct using compass/straightedge/trisector, since $18=2*3^2$ and the 3 is raised to the power of two. My questions are as follows:

1. Given a circle of radius $19-1=18$, how would one proceed with the construction? I can't seem to follow along with Gleason in his methods and I'm totally lost beyond constructing $\sqrt{19}$. Could someone help me work out a construction?

2. In addition to compass and straightedge, does this mean:

   a. A regular 41-gon can be constructed with one angle quintisection? $41-1=2^3*5$

   b. A regular 61-gon can be constructed with one angle trisection and one quintisection? $61-1=2^2*3*5$

   c. A 101-gon with two quintisections? $101-1=2^2*5^2$

   d. A 433-gon with three trisections? $433-1=2^4*3^3$

Quintisection, heptasection, et cetera can be done with an Archimedean spiral. So if the above statements are true, one can construct a regular polygon with any number of sides even without "cheating" (using the spiral to construct 360/n). An 89-gon can be constructed using 1 angle undecasection (dividing into 11 equal parts), a 331-gon / 661-gon / 1321-gon can all be constructed with one each of trisection / quintisection / undecasection, et cetera.

My goal is to use Gleason's principles to come up with a way to construct a 433-gon using compass, straightedge, and angle trisector. It would be extremely long, but I would like to see it worked out.

Answer 1

1)) Gleason wrote, that although Theorem 2 tells us that the regular triskaidecagon can be constructed using one angle trisection and there are many ways to proceed, but none seem geometrically perspicuous. But, since enneadecagon requires two angle trisections to construct, its construction should be more complicated and requires two steps.

2)) A general construction of an $n$-gon in the proof of Theorem 2 (and, I expect, its possible generalizations at p. 194) follows the composition series of the Galois group of $\Bbb Q(\eta)$, consecutively constructing segments, whose lengths generate the respective fields extensions. Understanding of this requires a knowledge of Galois theory, which can be studied, for instance, by the references. In particular, the last sentence in the previous to last paragraph of the proof provides a positive answer to Question 2.d.

2.a–2.c)) Consider a generalization of the theorem to any natural $n\ge 3$. The second paragraph of its proof is still valid, see, for instance, [vdW, §60]. If $\varphi(n)=2p_1\dots p_\ell$, where $p_i$ are (not necessarily distinct) prime numbers, then the Galois group of $\Bbb Q(\eta)$ has a composition series of length $\ell$ with the quotients isomorphic to cyclic groups $\Bbb Z_{p_i}$. This series corresponds to a tower of consecutive cyclic extensions of $\Bbb Q$ to $\Bbb Q(\eta)$. But the problem is that in general case we don’t have a counterpart of Lemma (based on Theorem 1, special for the degree three), assuring that we can construct a segment, whose length generates the extension by ruler, compass, and angle $p_i$-sector.

On the other hand, the claim and the discussion at the last page follow that we can construct a regular $n$-gon by ruler, compass, and a collection of angle $p_i$-sectors. But there is not stated clearly, how many times we should use angle $p_i$-sectors. According to Gauss’ statement, we should divide an arc into $n-1$ equal parts twice, and each such subdivision can be done using $p_i$-sectors $\ell$ times in total.

References

[Lan] Serge Lange, Algebra, Addison-Wesley, 1965 (Russian translation, Moskow, Mir, 1968).

[vdW] B. L. van der Waerden, Algebra (Russian translation).