TITLE: Equipartition of the circle
QUESTION [3 upvotes]: Browsing an old technical studies pupil's school book, I have found the description of a method to place at equal distance $N$ points on the circumference of a circle. I am looking for a proof of this method, but it doesn't look obvious to me... and I haven't been able to find one on the internet.
The algorithm: draw a circle of center $O$ and diameter $AB$, which is partitioned in $N$ parts $A=X_0$ and $B=X_N$ (using a ruler or any other method). Draw the equilateral triangles $ABC$ and $ABD$. Then the book claims that the intersections $X'_i$ of the circle and lines $CX_{2i}$ outside of segment $CX_{2i}$, $i\leq N/2$ and same with $DX_{N-2i}$, define an equipartition of the circle circumference.
I have tried to prove it starting with the dot product of $OX'_i$ and $OX'_{i+1}$ and trying to show it's not dependent on $i$, but have not succeeded.
Is there a classical proof? Ideas?

REPLY [1 votes]: Not quite an answer, but food for thought.
Consider a curve with N (let's say distinct) distinguished points on it.
Call two such C and D PR (or C PR D, for projectively related) if there is
a point O  and N lines such that OX_iY_i are on one of the N lines, where
the X's are distinct from the Y's and O and are the distinguished points of
C respectively D.  I should think C PR D implies the N points of C are
constructible iff the N points of D are constructible. (Some tweaking of
PR may be needed to handle translations and rigid rotations of curves.)
  Further some nice version of the relation PR may be equivalent to constructibility.
Anyway, this is the notion behind my comments saying that the
described construction is an approximation.
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