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$\newcommand{\S}{\mathbb{S}^1}$ $\newcommand{\la}{\lambda}$

While solving an optimization problem, I reached the following question:

Let $x_1,x_2,x_3,x_4 \in \S$ be four distinct points on the unit circle.

Suppose that there exist strictly positive real numbers $\la_{ij}=\la_{ji}, 1\le i\le j\le 4$ such that $$ \sum_{j \neq i} \la_{ij}x_j \in \text{span}\{x_i\} $$ for every $i \in \{1,2,3,4\}$.

Question: Are the $x_i$ the vertices of a rectangle?

It suffices to prove that $\sum_i x_i=0$.

A rectangle does satisfy the requirement, with $\la_{ij}=1$; Note that $x_2=-x_4, x_1=-x_3$ are antipodal.

Edit:

If we omit the symmetric condition $\la_{ij}=\la_{ji}$ and the positivity condition $\lambda_{ij}>0$, then any "non-degenerate" configuration satisfies this; if any three of the vertices are linearly independent, then we can choose the (not necessarily symmetric coefficients) that satisfy the requirement. But with the symmetric condition this is not so clear.

Ted Shifrin
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Asaf Shachar
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  • Isn't it "obvious" that there are other such quadrilaterals? A small perturbation of the vertices of a rectangle should only perturb the $\lambda_{ij}$ slightly, leaving them positive. My intuition is that this property holds for any such quadrilateral such that no half-circle contains all four vertices. – Greg Martin May 26 '22 at 18:14
  • Are you requiring that $\lambda_{ij}=\lambda_{ji}$? It seems like you only define $\lambda_{ij}$ for $i<j$, but use it for all pairs $i\neq j$. – Carl Schildkraut May 26 '22 at 18:21
  • @CarlSchildkraut Yes, sorry. I added this to the question. – Asaf Shachar May 26 '22 at 18:25
  • @GregMartin I am not sure about this. The fact that the points are antipodal is crucial for the rectangle solution. Any perturbation of this won't have this symmetry. But I am not sure. – Asaf Shachar May 26 '22 at 18:26
  • @GregMartin On a second thought, I think you are right if we didn't require the $\lambda_{ij}=\lambda_{ji}$. If we omit the symmetric condition, then indeed any three vertices of a rectangle are linearly independent. Thus there exist (not necessarily symmetric coefficients) that satisfy the requirement. But with the symmetric condition this is not so clear. – Asaf Shachar May 26 '22 at 19:26
  • You're right that I didn't see the symmetric condition earlier. – Greg Martin May 26 '22 at 21:14

1 Answers1

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The answer is no. One counterexample is $$ x_1=(-1,0),\; x_2=\bigl(\tfrac12,\tfrac{\sqrt3}2\bigr),\; x_3=(1,0),\; x_4=\bigl(\tfrac12,-\tfrac{\sqrt3}2\bigr), $$ for which $$ \lambda_{12}=\lambda_{14}=2,\; \lambda_{23}=\lambda_{24}=\lambda_{34}=1 $$ works (for any positive $\lambda_{13}$—this coefficient never matters if the respective points are antipodal).

Greg Martin
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  • Thanks, that is an interesting answer. I liked the symmetry $x_3 \iff x_4$. BTW, you have small typos: It should be $ \lambda_{13}=\lambda_{14}=2$, and $\lambda_{12}$ is the one who could be arbitrary. (This makes the construction invariant under replacing the indices $3,4$ and the points $x_3,x_4$). Finally, you wrote $x_3$ twice. Again, thank you very much for this nice solution! I really liked that you tried the "next symmetric thing" after a rectangle: A rectangle has two pairs of antipodal points, and your construction has one such pair, and an additional swapping symmetry $3 \iff 4$. – Asaf Shachar May 27 '22 at 05:06
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    Thanks for catching the typos. I still feel that quadrilaterals with even less symmetry might still be in this class.... – Greg Martin May 27 '22 at 07:00