Average of an odd number of equally spaced points on a circle I want to show that the average of an odd number of equally spaced points on the unit circle is equal to 0. More precisely, let $n$ be an odd number, $\theta_1,\ldots, \theta_n\in[0,2\pi)$ and $$
re^{i\psi}:=\frac{1}{n}\sum_{i=1}^ne^{i\theta_i}.$$ We want to show that if the $\theta_i$s are equally spaced then $r=0$. I remark that I do not want to use Vieta's formulas but rather prove it "by hand".
I was able to prove this formula for $r$:
$$r=\frac{1}{n}\left(n+2\sum_{i=1}^{n-1}\sum_{j=i+1}^n\cos(\theta_i-\theta_j)\right)^{1/2}
$$
(I wouldn't know if this is a well known formula or not...). If the angles are equally spaced, upon relabeling if necessary we have $\theta_i=\frac{2(i-1)\pi}{n},\:i=1,\ldots,n$ so that
\begin{align*}
r^2 & = \frac{1}{n^2}\left(n+2\sum_{i=1}^{n-1}\sum_{j=i+1}^n\cos(\theta_i-\theta_j)\right) \\
 & = \frac{1}{n^2}\left(n+2\sum_{i=1}^{n-1}\sum_{j=i+1}^n\cos\left(\frac{2(i-j)\pi}{n}\right)\right)
\end{align*}
Then we would like to show that
$$\sum_{i=1}^{n-1}\sum_{j=i+1}^n\cos\left(\frac{2(i-j)\pi}{n}\right)=-\frac{n}{2}.$$
If we set $n=2k+1\:k\in\mathbb{N}$ we can rewrite the previous formula in terms of $k$:
$$\sum_{i=1}^{2k}\sum_{j=i+1}^{2k+1}\cos\left(\frac{2(i-j)\pi}{2k+1}\right)=-k-\frac{1}{2}.$$
Expanding this double sum gives
\begin{align*}
\sum_{i=1}^{2k}\sum_{j=i+1}^{2k+1}\cos\left(\frac{2(i-j)\pi}{2k+1}\right) & = \cos\left(\frac{-2\pi}{2k+1}\right)+\cos\left(\frac{-4\pi}{2k+1}\right)+\ldots+\cos\left(\frac{-4k\pi}{2k+1}\right) \\
                    & + \cos\left(\frac{-2\pi}{2k+1}\right)+\cos\left(\frac{-4\pi}{2k+1}\right)+\ldots+ \cos\left(\frac{-(4k-2)\pi}{2k+1}\right) \\
                    &\vdots \\
                    & + \cos\left(\frac{-2\pi}{2k+1}\right)+\cos\left(\frac{-4\pi}{2k+1}\right) \\                           & +\cos\left(\frac{-2\pi}{2k+1}\right)
\end{align*}
which can be rearranged as:
$$\sum_{i=1}^{2k}\sum_{j=i+1}^{2k+1}\cos\left(\frac{2(i-j)\pi}{2k+1}\right) =2k\cos\left(\frac{2\pi}{2k+1}\right)+(2k-1)\cos\left(\frac{4\pi}{2k+1}\right)+\ldots+2\cos\left(\frac{(4k-2)\pi}{2k+1}\right)+\cos\left(\frac{4k\pi}{2k+1}\right).$$
Then, my question is the following: is it true that
$$ 2k\cos\left(\frac{2\pi}{2k+1}\right)+(2k-1)\cos\left(\frac{4\pi}{2k+1}\right)+\ldots+2\cos\left(\frac{(4k-2)\pi}{2k+1}\right)+\cos\left(\frac{4k\pi}{2k+1}\right)=-k-\frac{1}{2}?$$
I checked it for some values of $k$ and try induction for the general case, but I couldn't get very far. Also, as I mentioned, one can prove the result using Vieta's formula but it seems that one should be able to prove it this way as well.

Symmetry permits no other average than the center of the circle.  Here's one way to make that somewhat more formal:

Choose an orthonormal coordinate system with its origin at the center of a circle with radius $r$, and the x axis directed such that $\theta_1 = 0$.
Observe that because of the even spacing, the remaining angles group in pairs of the form $\pm \frac{m2\pi}{n}$.  The members of each pair will have the opposite sines (and $\sin \theta_1 = 0$), therefore in the average, all the sine terms, which correspond proportionally to $y$ coordinates, cancel.  It follows that the average of all the points lies on the chosen x axis, a line containing the origin and $(r, \theta_1)_{polar}$.
Now observe that we can apply the same argument to determine that the average lies on each line containing the origin and any of the $(r, \theta_i)$.  These cannot all be the same line, because a line intersects a circle at at most two points, and there are at least three distinct $(r, \theta_i)$.  Choose any two distinct lines from the set. We know that the average lies on both, so it must be at their intersection.  We know that their intersection is unique, because the lines are distinct.  And we know that the origin, which is at the center of the circle, is on both lines, so it must be the intersection.

Note that that does not particularly depend on the number of points being odd.  With only minor changes, it can be generalized to any number of points greater than two.
