Is there a formula to find the number of distinct regular n-gons possible, given n? And by distinct, I mean disregarding anything like reflections or rotations. Working it out, I find the following for the first new n's:

  • n = 3, 1 unique (triangle)
  • n = 4, 1 unique (square)
  • n = 5, 2 unique (regular pentagon, 5-point star)
  • n = 6, 1 unique (regular hexagon)
  • n = 7, 3 unique (regular heptagon, 2 kinds of regular stars, see pic)

regular 7-gons

  • n = 8, 2 unique (regular octogon, 8 point star)
  • ect...

But I cannot seem to find the formula that dictates this number, or even if that formula exists in the first place. This lack of a formula also makes it kind of hard to properly tag this question (under number theory or combinatorics), so my apologies for that.

  • 1
    $\begingroup$ Look-up the concept of "relatively prime" numbers. $\endgroup$
    – Blue
    Oct 12, 2014 at 23:10
  • $\begingroup$ I am aware of the concept of relatively prime numbers, where gcd(a, b) = 1. It was one of the first things I tried to employ to get the pattern, but to no avail. $\endgroup$
    – sm81095
    Oct 12, 2014 at 23:12
  • $\begingroup$ Do you regard a twisted square as identical with a square? By twisted square, I mean a figure like the union of segments in the plane from (0,0) to (1,0) to (0,1) to (1,1) to (0,0). I mention this in view of the figures you have provided. $\endgroup$
    – MPW
    Oct 12, 2014 at 23:12
  • $\begingroup$ The sole identifier is that the polygons are regular (same side length, same angles), and that they are not separated by only a reflection, translation, rotation, ect... $\endgroup$
    – sm81095
    Oct 12, 2014 at 23:15
  • $\begingroup$ A but I see what you mean by a twisted square. Shapes such as those would not count, as the side lengths could not be the same, thus disqualifying it as a regular n-gon. $\endgroup$
    – sm81095
    Oct 12, 2014 at 23:17

1 Answer 1


Suppose you have $n$ points distributed uniformly on a circle. There is a shape corresponding to each value $r$ which is relatively prime to $n$: given a starting point and $r$, just move forward by $r$ points, draw a line between the starting point and this point, and continue on.

The number of values relatively prime to $n$ is given by Euler's totient function $\phi(n)$. However, both $r$ and $n - r$ produce the same figure, so we must divide by two to avoid double-counting. This means that the number of figures with $n$ sides is $\phi(n)/2$.

  • $\begingroup$ You posted this just as I started to look at applying Euler's totient. Many thanks for the straightforward answer. $\endgroup$
    – sm81095
    Oct 12, 2014 at 23:39

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