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The minimal rectangle containing the euclidean ball in the plane is the standard cube $B_\infty = [-1;1]^2$. I would like to know if the euclidean ball is the worst symmetric convex body to be included in a parallelogram.

Is it true that for every $K$ symmetric convex body of $\mathbb{R}^2$ such that $Vol(K) = \pi$ then there exist a parallelogram $R_K$ such that $K \subset R_K$ and $Vol(R_K) \leq 4 ?$ Where $Vol$ is the area.

(It is equivalent that for every such $K$ there is a volume preserving linear map $T$ such that $TK \subset B_\infty$. There are various ways to reformulate the question, by dualizing for instance but I believe this one is the most intuitive)

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    $\begingroup$ Does this work for the regular hexagon? $\endgroup$
    – Guillaume Aubrun
    Nov 29, 2021 at 15:05
  • $\begingroup$ Indeed thats a counterexample. The smallest rectangle enclosing the regular hexagon is $[-\sqrt{3}/2 ; \sqrt{3}/2] \times [-1;1]$ the area ratio is then $4/3$, while for the ball it is $4/\pi$. $\endgroup$
    – Gericault
    Nov 29, 2021 at 15:32
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    $\begingroup$ It may not be related to your maximization problem, but let me point that the Banach-Mazur distance between the regular hexagon and the square is 3/2, which is the maximal value for a pair of centrally symmetric convex bodies. $\endgroup$
    – Guillaume Aubrun
    Nov 29, 2021 at 16:20

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