Dear Francisco,

To make sure that I answer to the right question, let me first summarize your problem as being the construction the regular 60-gon using only ruler and compass, and divide this task into 2 parts : the theoretical possibility, and the practical construction.

Theory : Gauss’ celebrated theorem (nowadays usually presented as a mere corollary of Galois theory) states that the regular n-gon is constructible by ruler and compass iff n is of the form 2p.Fq, where Fq = 22^q+1 (a so called Fermat prime). Here 60 = 22.3.5. So the regular 60-gon is indeed constructible by ruler and compass. 

Practice : in your approach, you try to 5-sect (allow me this expression) the angle 2pi/12 = 30°. It’s much easier to 12-sect the angle 2pi/5 or, even better, to 3-sect first this angle. You rightly recall that the 3-section of a general angle by ruler and compass is impossible, but here we deal with the particular angle 2pi/5, which we can already get by the well known construction of the regular 5-gon. Let us give details. Putting a = exp(2i.pi/5), the last problem amounts to solving the complex equation z3 = a, whose 3 roots are a, j.a, j2.a, where j = exp(2i.pi/3) verifies j2+j+1 = 0, a quadratic polynomial whose discriminant is i.(sqr3)/2. So our problem amounts to constructing (sqr3)/2, and we are done.

Further explanation : things get crystal clear when going back to Galois theory. The construction of the regular n-gon is obviously equivalent to that of the complex roots of the equation zn – 1 = 0. From Galois point of view, this consists in getting a hand on the field L obtained by adding these roots to Q , via its Galois group Gal(L/Q), which is cyclic, isomorphic to (Z/nZ)*, the group of invertible elements of the ring Z/nZ. In our particular case, this is C2 x C2 x C4 . In your approach, in Galois theoretical terms, you first introduce the field K = Q(exp(pi/6)) (which is of degree 4 over Q), and then L, which is of degree 4 over K. Whereas the « practical » step above introduces first the simpler  M = Q(exp(2i.pi/5)), of degree 4 over Q , which is simpler than L. Actually, the « black box » that we used in the construction of the 5-gon is « contained » in the structure of this extension M/Q. ¤

Dear Thong,

Thank you very much for showing me that this is possible. Indeed, a pentagon superimposed on a dodecagon produces the regular 60-gon.

It was really very simple.

Regards

Dear Francisco,

I made a misprint when writing  "... the complex equation z3 = a, whose 3 roots are a, j.a, j2.a... ". Of course the 3 roots are alpha, j.alpha, j2.alpha, where alpha denotes a given 3rd root of a.This renders the end of my answer hardly understandable. Here are the correct details : we aim now to construct alpha, which is a primitive 15-th root of 1. The field N = Q(alpha) = M(alpha) is of degree 2 x 4 over Q , and it  contains j. So we can immediately check that N = M(j), and we are done.   ¤

Added remark : Your drawing (superimposition of a pentagon and a dodecagon ) actually reflects a consequence of the Chinese remainder theorem: if Wn denotes the group of n-th roots of 1, then Wmn is isomorphic to Wm x Wn  .

... isomorphic to Wm x Wn  if m and n are coprime. I'm definitely stuck with "misprints" !

I don't understand why one should be able to trisect the angle 2\pi/5. To me this seems a genuine cubic equation whose roots are not constructible. (E.g. the  20 degrees angle is not constructible.) Am I wrong?

Best regards

Jost

My question is solvable if the whole circle is divided into 12 and 5. So 60 divisions of 6 degrees is obtained (like drawing).

The question is how to divide 6 degrees. By bisecting 3 degrees be obtained, which cannot trisect.

Then, with a ruler and compass it can get the 60-gon. However, how the 360-gon it is obtained? How an angle that is not a multiple of 3 is constructed? (e.g. 20 degrees).

Best regards

Dear Jost,

It seems we have a contradiction here:

1) Your feeling is that the trisection of 2pi/5 is not possible because it amounts to solve a "genuine cubic" equation by ruler and compass

2) The Gauss (and Wantzen) (*) theorem asserts that the 60-gon is constructible, and Francisco actually did it with his "artistic" version of the Chinese remainder theorem. Having the angle 2pi/60, of course we also have 2pi/15 = 4.(2pi/60)

The paradox (= apparent contradiction) lies in the word "genuine cubic". You cite the classical example of the impossibility of the trisection of 2pi/6 = 60°, i;e. of the construction of 20°. The underlying cubic equation is 8x3- 6x - 1= 0, where x = cos (pi/9). This is indeed a genuine (= irreducible) cubic, but with coefficients in Q (because of the particular value of cos(pi/3)). Whereas in my answer to Francisco - granted the preliminary construction of the regular pentagon - we saw that the key base field is M, and the polynomial involved in the adjunction of j to M is the cubic      t3 - 1, yes, but it is not irreducible, since j is a root of t3 - 1/t - 1 = t2+t+1. In other words, we only added to M the roots of a quadratic equation, not of a "genuine cubic". Using further  properties of the Galois group  C2 x C2 x C4, we could even show that in our situation, the adjunctions of j (or of sqr(3)) to M or to Q are equivalent.

Best,  Thong

(*) Me and my "misprints" ! In the Gauss-Wantzel theorem, you should read  "...iff n is the product of 2p by distinct Fermat primes Fq " (and not a single such prime). This answers to Francisco's last question : since 360 = 23.32.5, the regular 360-gon is not constructible by ruler and compass.

Is possible divided an arc in n parts if you using the theorem of Thales (constructing regular polygons of n sides inscribed in tothe cercle).

There is another graphic system (based on the bisection system) that allows divided an arc in n equal parts (if n is not and odd number)

In this tutorial you can see this https://youtu.be/ncUIjRJgcOQ

Thank you, Thong and Francisco! (Your figure is clear enough, Francisco, I did not see it right away). Here is my elementary resolution of this paradox, following Thongs hints: Constructing 6 degrees is as good as constructing 24 degrees, thus we have to trisect the 72 degree angle, that is, solving the equation  x^3 = w  where  w = e^{2i\pi/5}. We try do divide  x^3-w  through a degree-one polynomial  x-a   which gives  x^2+ ax + a^2 + a remainder  a^3-w. Fortunately we can draw this cubic root in Q(w): it is  x = w^2 (then x^3 = w^6 = w  due to  w^5 = 1). Thus  x^3-w = (x^2+wx+w^2)(x-w^2) (decompostion over Q(w)).

Best regards

Jost

Dear Joseph Maria,

Could you please make more precise your hint about Thales' ttheorem ?

Dear colleague

It is difficult to explain. But I think it's can understand whit this tutorial: https://youtu.be/8yjajy3vOdc.

In this case is to draw a decagon, but it's possible draw any tipe of regular poligon. Only you have is to divide the initial line by the number of sides the polygon has you wants.

Dear colleague,

I have just seen such a video, and the solution presented there was obviously flawed. In all cases, any positive answer (for any n) must be false, since it contradicts the Gauss-Wantzen theorem. But some"amateurs" keep trying (*), because, unfortunately, that theorem is not "elementary", its proof requires a mathematical knowledge at the level of a graduate or master degree.

(*) There is a whole page of Wikipedia which is devoted to the discussion of such "amateur" (all false) solutions solutionshttps://fr.wikipedia.org/wiki/Discussion:Trisection_de_l'angle

Dear Francesco,

To construct 6° note first that it is 60° - 54°, and it is easy to construct 60°. Now 54° is half of 108°, which is the interior angle of a regular pentagon. Euclid showed, over 2000 years ago, that we can construct a regular pentagon with ruler and compasses.

Putting these ideas together, we have the following exact construction:

1. Start with a circle, centre O and radius r, and take a point A on its circumference.

2. Construct a line through A perpendicular to OA.

3. Draw a circle with centre A and radius r, cutting this line at B and C.

4. Construct D, the midpoint of AC.

5. With centre D draw a circle that passes through O and cuts AB at a point E (between A and B).

6. With centre O draw a circle that passes through E and cuts the semicircle COB at F. Then angle AOF is 54°.

7. The semicircle COB cuts the original circle at G, near F, and angle AOG is 60 °.

8. The angle FOG is exactly 6°.

I leave you to research any proof you need.

Michael Fox

Dear colleague

It is true. Gauss theorem limits the real number of inscribed polygons can be drawn with a ruler and compass.

But is possible to do a geometric approach. I have proposed the solution from the architectural point of view. Here, in certain cases, these errors are aceptables.

In the architectural projection, often, this errors disappearing into the natural uncertainties of an architectural construction.

Now, from the strictly mathematical point of view, you're right, the solution is incorrect.

Dear colleague

The conclusion is that by rule and compass it can build the 60-gon (6°), including 120-gon (3°). However, it cannot make the 180-gon (2°), nor 360-gon (1°).

It is right?

Dear Francisco,

Yes, you are right. The smallest angle with a whole number of degrees that can be constructed with ruler and compasses is 3°.

If you try to work through such a construction using a pencil, a ruler and compasses, errors creep in from the start, and can become large enough to be easily visible. In my own work I use geometrical software that simulates the use of ruler and compasses. The errors then are almost undetectable.

Michael Fox