Question

I am writing a program in which I need to draw polygons of an arbitrary number of sides, each one being translated by a given formula which changes dynamically. There is some rather interesting mathematics involved but I am stuck on this probelm.

How can I calculate the coordinates of the vertices of a regular polygon (one in which all angles are equal), given only the number of sides, and ideally (but not neccessarily) having the origin at the centre?

For example: a hexagon might have the following points (all are floats):

( 1.5  ,  0.5 *Math.Sqrt(3) )
( 0    ,  1   *Math.Sqrt(3) )
(-1.5  ,  0.5 *Math.Sqrt(3) )
(-1.5  , -0.5 *Math.Sqrt(3) )
( 0    , -1   *Math.Sqrt(3) )
( 1.5  , -0.5 *Math.Sqrt(3) )

My method looks like this:

void InitPolygonVertexCoords(RegularPolygon poly)

and the coordinates need to be added to this (or something similar, like a list):

Point[] _polygonVertexPoints;

I’m interested mainly in the algorithm here but examples in C# would be useful. I don’t even know where to start. How should I implement it? Is it even possible?!

Thank you.

Answer 1

for (i = 0; i < n; i++) {
  printf("%f %f\n",r * Math.cos(2 * Math.PI * i / n), r * Math.sin(2 * Math.PI * i / n));
}

where r is the radius of the circumsribing circle. Sorry for the wrong language No Habla C#.

Basically the angle between any two vertices is 2 pi / n and all the vertices are at distance r from the origin.

EDIT:
If you want to have the center somewher other than the origin, say at (x,y)

for (i = 0; i < n; i++) {
  printf("%f %f\n",x + r * Math.cos(2 * Math.PI * i / n), y + r * Math.sin(2 * Math.PI * i / n));
}