# Formula for vertices of a Polygon with only 1 vertex at the top and y-axis symmetric

I'm trying to find the formula for the vertices of a polygon with n-sides such that there is always only 1 vertex at the top and the polygon is symmetric with respect to the y-axis...

so generally the formula for the vertices of a polygon is

$$x_i = r \cos \Big ( \psi + \frac{ 2 \pi i }{n}\Big )$$

and

$$y_i = r \sin \Big ( \psi + \frac{ 2 \pi i }{n}\Big )$$

where $(x_i,y_i)$ is the $i$th vertex of a polygon of $n$-sides and $r$ is the radius, and $\psi$ is the angle by which the polygon is rotated.

So the question is to find the function $\psi$. I would assume $\psi$ is a function of $i$ and $n$

I've tried a few examples but can't seem to find a general formula for $\psi(i,n)$

so for the example of the triangle we want this:

## migrated from mathoverflow.netApr 2 '16 at 22:15

This question came from our site for professional mathematicians.

For my convenience, let $p_i = (x_i, y_i)$ be the point labeled $i$.
If $\psi = 0$, then $p_0 = (x_0, y_0) = (1, 0)$ starts us on the $x$-axis. The constant $\psi = \frac{\pi}{2}$ should always work by rotating that point $(1, 0)$ a quarter turn counterclockwise. Your collection of vertices will then be $\{p_0, p_1, \ldots, p_{n-1}\}$, where $p_0 = (0, 1)$ is on the $y$-axis.
If you insist on using $\{1, 2, \ldots, n\}$ as the index set, you'll need to unshift $\psi$ by the amount $\frac{2\pi}{n}$. That is, use $\psi = \frac{\pi}{2} - \frac{2\pi}{n} = \frac{\pi(n - 4)}{2n}$.