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Can a circle's circumference be divided into arbitrary number of equal parts using straight edge and compass only? In other words, are all the $\frac{2\pi}{k} , k \in \mathbb N^+$ angles constructible?

Edit : Added the "Equal" into the title, to be more specific but certainly I do not mean to restrict the answers by parts being "Equal", hell if there is anything intresting not requiring the parts to be equal I still want to know!

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...on that note, I'd be interested to see a video of somebody drawing a heptadecagon or a dihectapentacontakaiheptagon with straightedge and compass... –  J. M. Aug 15 '11 at 8:18
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"I do not mean to restrict the answers by parts being Equal" - then I can just randomly cut up your circle... –  J. M. Aug 15 '11 at 8:30
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@Geoff: When I think about it, I don't think there's even a display on Earth with enough pixels to show a 65537-gon... :) –  anon Aug 15 '11 at 8:36
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@Geoff: Oh, that's Johann Gustav Hermes's thesis. No legend, I've seen it with my own eyes. Impressive. You may want to check this link later (the GDZ-server is down yet again) –  t.b. Aug 15 '11 at 8:36
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No, they are not. You can only construct the angles with $k=2^{\alpha}p_1...p_s$, where the $p_i$'s are (distinct) Fermat primes.

The proof is not hard but a bit long, make comment if you want more details. I denote by $\mathcal{P}~$the set of $k$ such that $\frac{2\pi}{k}$ is constructible :

1) Show that if $k\in \mathcal{P}$, then $2k\in\mathcal{P}$.

2) Show that if $n\in \mathcal{P}$, then any divisor $d$ of $n$ (different from $1$) lies also in $\mathcal{P}$.

3) Show that if $n$ and $m$ are coprime and both belong to $\mathcal{P}$, then $mn$ also belnogs to $\mathcal{P}$.

All these elementary and easy questions show that it remains to answer the following question :

Let $p$ be an odd prime. When does $p^\alpha$ belong to $\mathcal{P}$ ?

The answer is that $p^{\alpha}$ lies in $\mathcal{P}~$ if and only if $p$ is a Fermat prime and $\alpha=1$.

Answering this question is less elementary and (as far as i know) needs to use some Galois theory.

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"needs to use some Galois theory" - not really, all you need to show is that you can't express inconstructible angles with repeated square-root extractions... Wantzel proved this before Galois machinery became available. –  J. M. Aug 15 '11 at 8:22
    
@J.M. that is why i wrote "as far as i know". –  Louis La Brocante Aug 15 '11 at 8:41
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Any text on Galois theory will tell you that the angle of $\frac{\pi}{9}$ can't be constructed by straight edge and compass ( in other words, the angle $\frac{\pi}{3}$ can't be trisected). Classical Galois theory tells us that a positive real number $\alpha$ can't be constructed unless $[\mathbb{Q}[\alpha]:\mathbb{Q}]$ is finite and a power of $2$. When $\alpha = \cos(\frac{\pi}{9}),$ the formula $\cos(3\theta) = 4\cos^3(\theta) - 3 \cos(\theta)$ tells us that $\alpha = \cos(\frac{\pi}{9})$ is a root of the cubic $8x^{3}-6x -1 =0$, and this is irreducible of degree 3, so we have $[\mathbb{Q}[\alpha]:\mathbb{Q}] = 3.$

If you go a little further into Galois theory, you will see that ( as Gauss discovered), the angles $\frac{2 \pi}{k}$ are constructible precisely when $k$ is either a power of 2, or the product of a power of 2 with a product of distinct Fermat primes ( recall that an odd prime $p$ is a Fermat prime if and only if $p-1$ is a power of $2$). Hence $k = 18$ is not a permissible value.

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As has been described in earlier answers, a circle can be divided by compass and straightedge into $n$ equal parts if and only if $$n=2^a p_1p_2\cdots p_k,$$ where $a$ is a non-negative integer and the $p_i$ are distinct Fermat primes. (We allow the possibility $k=0$.)

The Fermat primes are the primes of the form $2^{2^m}+1$. There are only five Fermat primes known, $3$, $5$, $17$, $257$, and $65537$. A large amount of work has been expended to find additional Fermat primes, so far without success. It has even been conjectured, with some heuristic justifications, that there are no Fermat primes beyond these five.

Thus we have an essentially complete answer for the "equal parts" part of your question.

What about unequal parts? One can give an answer, but it is not nearly as satisfactory. Let $A$ and $B$ be points on a circle, for definiteness the unit circle. Let $d(A,B)$ be the (shorter) distance between $A$ and $B$ along the circle. The ratio $d(A,B)/\pi$ tells us what "fraction" of the circumference of the circle is taken up by the shorter arc joining $A$ and $B$. Then we have the following result.

Given a real number $r$, with $0\le r \le 1/2$, there exist constructible points $A$ and $B$ such that $d(A,B)=r$ if and only if the number $r$ is Euclidean.

A real number $x$ is Euclidean if it can be obtained, starting with the number $1$, using a finite number of additions, subtractions, multiplications, divisions, and applications of the square root function.

There is a very detailed analysis of all these matters, with proofs, in these University of Utah notes. All the equal parts stuff is there, and much more. Inevitably, the details require a certain amount of algebra. However, the presentation does not use Galois Theory.

Commensurable parts: Two arcs on the unit circle are called commensurable if the ratio of their lengths is a rational number. We can ask for a characterization of the possible straightedge and compass divisions into parts that are all commensurable with the circumference of the circle. It turns out that we can divide a circle into commensurable parts if and only if the ratio of each part to the full circumference is of the shape $$\frac{m}{2^a p_1p_2\cdots p_k}$$ where $m$ is a positive integer, and, as before, the $p_i$ are distinct Fermat primes. The proof is straightforward, if we take for granted the characterization of those $n$ for which the regular $n$-gon is constructible.

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Very nice answer! +1 –  Jon Aug 16 '11 at 0:56
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