Is a quadrilateral whose vertices' centroid is the center of the circumscribed circle a rectangle?

Question

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?

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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 $-2$.

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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.

Answer 1

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?

Answer 2

No, choose any four points in a circle. Then choose their centroid as origin of coordinates. Then their sum is $0$.

Answer 3

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.