# Equation for calculating the vertices of a regular polygon given a line segment

Given a line segment AB.

For simplicity, let it be (-1,0) and (1,0).

and n = number of sides, how can you calculate the vertices of a regular polygon which contains AB?

I'm trying to find an equation for it with no luck, where if one feeds in a value for n, and a vertex number, the equation returns (x,y) for the vertex.

So far, I've been trying this way: 1) find the center of the polygon by the intersection of angle bisectors at A, B. 2) Trying to find the equation relative to the center.

• Whoops, changed. – Neo Nov 15 '18 at 19:52
• Try the rotation matrix applied (n-1) times. – Phil H Nov 15 '18 at 20:34

Denote the number of sides with $$n$$. The central angle is:

$$\alpha=\frac{2\pi}{n}\tag{1}$$

Obviously:

$$x_G=0\tag{2}$$

$$y_G=1\times\cot\frac\alpha2=\cot\frac\alpha2=\cot\frac\pi{n}\tag{3}$$

$$AG=r=\frac1{\sin\frac\alpha2}=\frac{1}{\sin\frac\pi{n}}\tag{4}$$

Denote points $$A,B,C...$$ with $$P_0(x_0,y_0),P_1(x_1,y_1),P_2(x_2,y_2)...$$

$$x_{i}=x_G+r\sin((i-\frac12)\alpha)\tag{5}$$

$$y_{i}=y_G-r\cos((i-\frac12)\alpha)\tag{6}$$

$$i=0,1,\dots n-1$$

Replace (1-4) into (5) and (6) and you are done:

$$x_i=\frac{\sin\frac{(2i-1)\pi}{n}}{\sin\frac\pi{n}}$$

$$y_i=\cot\frac\pi{n}-\frac{\cos\frac{(2i-1)\pi}{n}}{\sin\frac\pi{n}}$$

...for $$i=0,1,\dots n-1$$

For simplicity don't put the side on the $$x$$ axis; put the center at the origin $$(0,0)$$.

The polygon is enscribed in a circle. For simplicity we can assume the circle has radius $$1$$ and each of the vertices will be $$(\cos \theta, \sin \theta)$$ for some angle $$\theta$$.

Assuming the polygon is an $$n$$-gon, The $$n$$ radii connected to the center create $$n$$ central angles of $$\frac {2\pi}{n}$$ radians or $$\frac {360^{\circ}}{n}$$ degrees.

If we assume vertex $$v_0$$ is at $$(1,0)$$ then the remaining vertices $$v_k$$ will be $$(\cos k\frac {2\pi}{n}, \sin k\frac {2\pi}{n})$$ for $$k = 0,.... , n-1$$.

.....

If we scale the $$n$$-gon by a scale of $$R$$ then the vertices will be $$(R\cos k\frac {2\pi}{n}, R\sin k\frac {2\pi}{n})$$.

If we rotate the $$n$$-gon by an angle of $$A$$ then the vertices will be $$(R\cos (k\frac {2\pi}{n}+A), R\sin (k\frac {2\pi}{n}+A))$$

If we move the center of the $$n$$-gon to point $$(X,Y)$$ then the vertices will be $$(X + R\cos (k\frac {2\pi}{n}+A), Y+R\sin (k\frac {2\pi}{n}+A))$$

....

So you instead you are given that one vertex is $$(x_1,y_1)$$ and the (counter-clockwise) adjacent vertex is $$(x_2,y_2)$$

The you need to solve $$X,Y, R, A$$ from:

$$(x_1,y_1) = (X + R\cos (0\frac {2\pi}{n}+A), Y+R\sin (0\frac {2\pi}{n}+A))= (X + R\cos A, Y + R\sin A)$$

and $$(x_2, y_2) = (X + R\cos (\frac {2\pi}{n}+A), Y+R\sin (\frac {2\pi}{n}+A))$$

====

Some things that may help: The side of a unit $$n$$-gon is $$2\sin \frac {\pi}{n}$$ so $$R = \frac {\sqrt{(x_2-x_1)^2 - (y_2-y_1)^2}}{2\sin \frac {\pi}{n}}$$.

The angle of the side of a unit $$n-$$gon from $$(1,0)$$ to $$(\cos \frac {2\pi}n, \sin \frac {2\pi}n)$$ is $$\pi - \frac {\pi - \frac {2\pi}n}2=\frac \pi 2 + \frac \pi n$$. So $$A=\arctan \frac{y_2-y_1}{x_2-x_1} - \frac \pi 2 -\frac \pi n$$.

Locate the center at the origin and apply the rotation matrix in sequence such that the next point's coordinates are determined from the previous point and where $$\theta = \frac{360}{n}$$.

$$\begin{bmatrix} x_2\\ y_2\\ \end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{bmatrix} \begin{bmatrix} x_1\\ y_1\\ \end{bmatrix}$$