Calculate coordinates of a regular polygon's vertices?

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 `float`s): ( 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

A good place to start is by examining the properties of a regular polygon:

1. The angle between every vertex will be equal.
2. The angles within a regular polygon will all sum to 2*PI radians (360 degrees). This is important as combined with the first property, will allow us to find the angle for each vertex given the number of vertices.
3. All polygons have a number of vertices equal to the number of sides they have. This will allow us to finally determine the number of vertices, and with it, the angle between each vertex.

With the angle of each vertex, we can use the trigonometric functions Cosine and Sine to determine the X and Y coordinates respectively! Putting these into code, we can break this into three steps:

// 1. Determine the number of sides

int numberOfVertices = numberOfSides;



for (int vertexIndex = 0; vertexIndex < numberOfVertices; vertexIndex = 0)

{

// 2. Calculate the angle of the current vertex

float vertexAngle = 2 * MathF.PI / numberOfVertices * vertexIndex;

// 3. Calculate the coordinates of the current vertex

Vector2 vertexCoordinate = new Vector2(x: MathF.Cos(vertexAngle), y: MathF.Sin(vertexAngle)) * radiusOfPolygon;

// 4. Store the vertex wherever you need! In your case:

_polygonVertexPoints[vertexAngle] = vertexCoordinate;

}

Note that this will always have a vertex directly to the right. To make all the shapes "standing" as it appears in the coordinates you provide, you can add an offset to "vertexAngle." To make it stand upright, add MathF.PI / 2.