Is a quadrilateral whose vertices' centroid is the center of the circumscribed circle a rectangle? Let $\mathbb{S}^1$ be the unit circle in $\mathbb{R}^2$.
Let $x_1,x_2,x_3,x_4 \in \mathbb{S}^1$ and suppose that $\sum_i x_i=0$, where we sum the vectors $x_i$ in $\mathbb{R}^2$.
Question: Do the $x_i$ form a rectangle? Equivalently, do the $x_i$ form two pairs of antipodal points?

Here is an attempt at a proof:
$$
 \sum_i x_i=0 \iff 0=\langle \sum_i x_i,\sum_j x_j \rangle=4+2\sum_{i<j}\langle x_i, x_j \rangle,
$$
or
$$
 \sum_i x_i=0 \iff \sum_{i<j}\langle x_i, x_j \rangle=\sum_{i<j}\cos \theta_{ij}=-2,
$$
where $\theta_{ij}$ is the angle between the vectors $x_i,x_j$.
The sum $\sum_{i<j}\cos \theta_{ij}$ contain $6$ summands. In the case of a rectangle, these angles are $\alpha, \pi-\alpha, \alpha, \pi-\alpha,\pi,\pi$, so the sum of cosines is indeed $-2$.

Now, in general:
$\theta_{12}+\theta_{23}+\theta_{34}+\theta_{41}=2\pi$,
$\theta_{13}=\theta_{12}+\theta_{23}$, $\theta_{24}=\theta_{34}+\theta_{23}$.
(Not exactly, since $\theta_{12}+\theta_{23}$ may be greater than $\pi$, but that doesn't matter, since then $\theta_{13}=2\pi-(\theta_{12}+\theta_{23})$, and this doesn't change the cosine value).
I am not sure how to proceed.
 A: Since $\sum x_i=0$ the diagram of the vector sum forms a closed quadrilateral. Assume the quadrilateral be not degenerate (no two adjacent vectors sum to 0). Since the opposite sides of the quadrilateral are equal it is a parallelogram  (more precisely it is a rhombus). This means $x_i$ form two pairs of opposite (and equal) vectors. This holds obviously  in the degenerate case as well.
Can you finish from here?
A: Let's do it in complex notation. Without loss of generality, let's rotate the axes such that $x_1 = e^{i\theta}$ and $x_2 = e^{-i\theta}$. It follows that
$$
x_1 + x_2 = 2\,\cos\theta,
$$
hence
$$
x_3 + x_4 = -2\,\cos\theta.
$$
Since their sum is real-valued, we may write once again $x_3= e^{i\phi}$ and $x_4= e^{-i\phi}$. Thus,
$$
2\cos\phi = -2\cos\theta,
$$
from which we deduce that $\phi = \pi + \theta$ or $\phi = \pi - \theta$. In either case, we obtain that the four points form a rectangle.
A: No, choose any four points in a circle. Then choose their centroid as origin of coordinates. Then their sum is $0$.
