geometric-constructiongeometry
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!
It's because the area of a circle is really the sum of infinitely many circles' circumferences, which is the integral: $$ \int^r_0 \! 2\pi{}t \, dt $$
See http://betterexplained.com/articles/a-gentle-introduction-to-learning-calculus/ for a nice explanation
Your solution fails because you give up a move creating the center point of your bounding circle. To complete the challenge in eight moves, you need to use one of the preexisting intersections as your center point, and then rely on the fact that extending a line is not considered a move to create the fourth vertex of the square.
The following diagram illustrates these steps, where step 4b is the non-move extension of the line:
Best Answer
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.