Given the regular polygon's side count $n$, the circumscribed radius $r$ and the center coordinates $(x,y)$ of the circumscribed circle,

How to calculate the coordinates of all polygon's vertices if one of the vertices coordinates are $(x,?)$?

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    $\begingroup$ Do you know complex numbers? $\endgroup$ – davidlowryduda Mar 6 '12 at 18:16
  • $\begingroup$ @mixedmath i? $\endgroup$ – Cobold Mar 6 '12 at 18:19

One vertex is $(x, y+r)$ or $(x, y-r)$. I'll assume the former (the latter case is similar, just swap $r$ and $-r$). The vertices will have coordinates $(x+r\sin\theta,y+r\cos\theta)$, where $\theta$ is an integer multiple of $\frac{2\pi}{n}$. ($\frac{360}{n}$ if you prefer degrees to radians.)

  • $\begingroup$ I don't need the area, I need the coordinates of the polygon's vertices. $\endgroup$ – Cobold Mar 6 '12 at 18:34
  • $\begingroup$ @Cobold: Silly me. I'll work that out for you in a minute. $\endgroup$ – Brett Frankel Mar 6 '12 at 18:36
  • $\begingroup$ Shouldn't it be $ (x + r cos \theta , y + r sin \theta ) $ (switch $sin$ and $cos$)? $\endgroup$ – jasonszhao Jun 2 '16 at 18:39
  • $\begingroup$ @jasonszhao both will work, it's just a question of the order the vertices are created. As written in the answer, the first vertex drawn (when $\theta = 0$) will be the north-most one. The way you have it, the first vertex would be the west-most one. $\endgroup$ – bgschiller Aug 10 '17 at 0:18

Supposing you know complex numbers, we care only about polygons around the origin which are inscribed in the unit circle. If a vertex is at $e^{i\omega}$, then the other n vertices will be at $e^{i(\omega + 2\pi k/n)}$ for $k$ up to $n$.

Then, if they're not on the unit circle, multiply everything by the radius.

If it's not centered at the origin, translate it to that spot instead.


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