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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.

Asaf Shachar
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  • What does addition of points mean? – bubba Apr 17 '22 at 11:34
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    Then the sum depends on the coordinate system used to express the points. – bubba Apr 17 '22 at 11:53
  • What precisely do you coin as centroid? It seems that you ask about an arbitrary cyclic quadrilateral, but the title is about a trapezoid... – user Apr 17 '22 at 20:25
  • The centroid of an inscribed quadrilateral cannot be identified merely by considering the sum of the four radial vectors to the vertices. – David K Apr 17 '22 at 20:32

3 Answers3

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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?

user
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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.

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

jjagmath
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    Thanks. I apologize for not being clear: I was implicitly assuming that the center of the circle is the origin of the coordinates. (Or that the the origin of the coordinates is defined to be at the center of the circle). Otherwise, the question is trivial and uninteresting as you noted. – Asaf Shachar Apr 17 '22 at 12:14
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    I know that this is a bad habit to change the question after an answer was given, and I apologize for that. But really, without this assumption, the question is almost meaningless... – Asaf Shachar Apr 17 '22 at 12:17