Is there an equation to describe regular polygons?

Any polygon (regular or not) can be described by an equation involving only absolute values and polynomials. Here is a small explanation of how to do that.

Let's say that a curve $C$ is given by the equation $f$ if we have $C = \{(x,y) \in \mathbb{R}^2, \, f(x,y) = 0\}$.

• If $C_1$ and $C_2$ are given by $f_1$ and $f_2$ respectively, then $C_1 \cup C_2$ is given by $f_1 . f_2$ and $C_1 \cap C_2$ is given by $f_1^2 + f_2^2$ (or $|f_1| + |f_2|$). So if $C_1$ and $C_2$ can be described by an equation involving absolute values and polynomials, then so do $C_1 \cup C_2$ and $C_1 \cap C_2$.

• If $C = \{(x,y) \in \mathbb{R}^2, \, f(x,y) \ge 0\}$, then $C$ is given by the equation $|f|-f$.

Now, any segment $S$ can be described as $S = \{(x,y) \in \mathbb{R}^2, \, a x + b y = c, \, x_0 \le x \le x_1, \, y_0 \le y \le y_1\}$, which is given by a single equation by the above principles. And since union of segments also are given by an equation, you get the result.

EDIT : For the specific case of the octagon of radius $r$, if you denote $s = \sin(\pi/8)$, $c = \cos(\pi/8)$, then one segment is given by $|y| \le rs$ and $x = rc$, for which an equation is

$$f(x, y) = \left||rs - |y|| - (rs - |y|)\right| + |x-rc| = 0$$

So I think the octagon is given by

$$f(|x|,|y|) \ f(|y|,|x|) \ f\left(\frac{|x|+|y|}{\sqrt{2}}, \frac{|x|-|y|}{\sqrt{2}}\right) = 0$$

To get a general formula for a regular polygon of radius $r$ with $n$ sides, denote $c_n = \cos(\pi/n)$, $s_n = \sin(\pi/n)$ and

$$f_n(x+iy) = \left||rs_n - |y|| - (rs_n - |y|)\right| + |x-rc_n|$$

then your polygon is given by

$$\prod_{k = 0}^{n-1} f_n\left(e^{-\frac{2 i k \pi}{n}} (x+iy)\right) = 0$$

Depending on $n$, you can use symmetries to lower the degree a bit (as was done with $n = 8$).

Here's a parametric equation I have made for a regular $n$-gon, coded in R:

n=5;
theta=(0:999)/1000;
r=cos(pi/n)/cos(2*pi*(n*theta)%%1/n-pi/n);
plot(r*cos(2*pi*theta),r*sin(2*pi*theta),asp=1,xlab="X",ylab="Y",
main=paste("Regular ",n,"-gon",sep=""));

And picture:

The formula I used is

$$\displaystyle r=\frac{\cos\left(\frac{\pi}{n}\right)}{\cos\left(\left(\theta \mod \frac{2\pi}{n}\right) -\frac{\pi}{n}\right)} \; .$$

This equation is actually just the polar equation for the line through the point $(1,0)$ and $(\cos(2\pi/n),\sin(2\pi/n))$ which contains one of the edges. By restricting the range of the variable $\theta$ to the interval $[0,2\pi/n[$, you will in fact just get that edge. Now, we want to replicate that edge by rotating it repeatedly through an angle $2\pi/n$ to get the full polygon. But this can also be achieved by using the modulus function and reducing all angles to the interval $[0,2\pi/n[$. This way, you get the polar equation I propose.

So, using polar plots and the modulo function, it's pretty easy to make regular $n$-gons.

Here is another parametric equation for a regular $n$-gon with unit radius:

$$\begin{align*}x&=\cos\left(\frac{\pi}{n}\right)\cos\left(\frac{\pi}{n}(2\lfloor u\rfloor+1)\right)-(2u-2\lfloor u\rfloor-1)\sin\left(\frac{\pi}{n}\right)\sin\left(\frac{\pi}{n}(2\lfloor u\rfloor+1)\right)\\y&=\cos\left(\frac{\pi}{n}\right)\sin\left(\frac{\pi}{n}(2\lfloor u\rfloor+1)\right)+(2u-2\lfloor u\rfloor-1)\sin\left(\frac{\pi}{n}\right)\cos\left(\frac{\pi}{n}(2\lfloor u\rfloor+1)\right)\end{align*}$$

for $0 \leq u \leq n$.

The provenance of this set is a bit more transparent if we switch to matrix-vector notation:

$$\begin{pmatrix}\cos\left(\frac{\pi}{n}(2\lfloor u\rfloor+1)\right)&-\sin\left(\frac{\pi}{n}(2\lfloor u\rfloor+1)\right)\\\sin\left(\frac{\pi}{n}(2\lfloor u\rfloor+1)\right)&\cos\left(\frac{\pi}{n}(2\lfloor u\rfloor+1)\right)\end{pmatrix}\begin{pmatrix}\cos\left(\frac{\pi}{n}\right)\\(2u-2\lfloor u\rfloor-1)\sin\left(\frac{\pi}{n}\right)\end{pmatrix}$$

and we see that the construction involves rotating and copying the line segment joining the points $(\cos\left(\frac{\pi}{n}\right),\pm\sin\left(\frac{\pi}{n}\right))$ $n$ times around a circle.

---

Here's sundry Mathematica code:

GraphicsGrid[Partition[Table[
  ParametricPlot[Through[{Re, Im}[
        (Cos[Pi/n] + I (2u - 2 Floor[u] - 1)Sin[Pi/n])*
         Exp[(I Pi/n) (2 Floor[u] + 1)]],
     {u, 0, n}], {n, 3, 11}], 3]]