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.
Best Answer
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?