# Polygons inscribed in a circle and stars

If on a circumference they are marked $n$ equally spaced points, those points can be joined by line segments contiguous (without lifting the pencil). If you join the consecutive points, you get a polygon regular of n sides (that's not funny). But if you join non-contiguous points (skipping from one, or two or three, etc.), are obtained polygons crashed sometimes and others Sometimes they are not crashed. Which are the Where are star-studded polygons? The 5-pointed star (so famous) is a example of them.

I came to the conclusion that if $n$ is odd we can build a star without lifting the pencil by jumping from a vertex, what more interesting things can be said? What happens if I jump from two vertices? I will form a star with what characteristics about $n$?

• It has to do with the greatest common divisor of the number of points in the polygon and size of the jumps when you connect the vertexes. – Doug M Aug 7 '18 at 20:25
• @DougM How is the greatest common divisor and the points that are deleted related? – Nash Aug 7 '18 at 20:28
• Suppose you have 12 points. If you connect adjacent points you get a do-decagon. If connect every other point (jumps of 2) you get 2 hexagons. If you make jumps of 3, you get 3 squares. If you make jumps of 4 you get 4 triangles. and 5 and 12 are co-prime and you get a star. – Doug M Aug 7 '18 at 20:31

In fact it is not the "number you skip", as @EthanBolker said, it is more the $k$-th vertex you'd visit next in a total sequance of $n$ vertices, which has or has not a common factor. That is, whenever $\gcd(n,k)=1$, you could trace a complete polygon, which finally visits all of the $n$ vertices.
It is this very number $\gcd(n,k)$, which provides the count of separate circuits.
Hint. $n$ being odd is not enough. Do some experiments with $15$ points.