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We can create stars within polygons by picking a vertex and drawing a segment extending a certain consistent number of vertices from it. Most $2n$-gons seem to have one or two ways to make a star, e.g. the decagonal star, being segments connected every three vertices.

However, I am primarily focusing on odd-numbered polygons or $(2n+1)$-gons. It seems, that most odd polygons apart from the pentagon and triangle have at least two ways to start, one where each inner segment intersects at least three others, and a lesser one, where each segment overlaps at least two others.

Polygons with a prime number of vertices will often have the most variations, e.g. the $11$-gon with at least $4$ stars, spaced $5, 4, 3$, or $2$ vertices away from a given start vertex. Below are more $(2n+1)$-gons with the possible stars (I know of) inscribed: Click here.

If anyone can help give me a formula that will predict or explain the possible variations of inscribed stars for any n-gon, including the triangle and pentagon, please do so.

Benjamin Wang
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Jrt4294967296
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    Consider how many vertices of the polygon do each edge of your star (or polygram) skips. – peterwhy Oct 26 '23 at 18:31
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    See @peterwhy 's comment and have a look at Euler's Totient Function. – Benjamin Wang Oct 26 '23 at 18:41
  • Thanks for the enlightenment! ϕ(n) certainly has given me a great understanding of this idea. edit It seems that the number of points (p) may be p=(n-ϕ(n))+1. Evidence: There are two unique stars (not counting mirrored versions). (7-ϕ(7)+1=2. Is this correct? – Jrt4294967296 Oct 26 '23 at 18:47
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    That can't be right. Let's look at $11=:p$ a prime. We have $\phi(p)=p-1$ for primes. But we need to halve this and disregard stepping by "$1$" which is not a star. So $(p-1)/2-1$. And in general $\phi(k)/2-1$ because $1$ is always coprime to $k$ and so is always overcounted, and because you're considering $k=2n+1$, we can safely divide ${1,\dots,2n}$ into halves without worrying about a "middle" element. – Benjamin Wang Oct 26 '23 at 20:52

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