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Is the following true?

CONJECTURE: $\,$ Let $\ B\ C\subseteq\mathbb R^n\ $ be convex bodies in $\mathbb R^n$ such that $\ C\ $ is centrally symmetric, $\ B\subseteq C,\ $ and $\ t\!\cdot\! B\ $ cannot be isometrically embedded in $\ C,\ $ for no $\ t>1.\ $ Then the center $c(C)$ of $C$ must belong to $B$, $\ c(C)\in B$.

Thank you Douglas Z. for pointing out the mess in my earlier formulation.

Sorry for a series of additional omissions. (Now the text is complete, I hope).

EDIT (after solutions of the original conjecture, by katz and Wlodek K)--As Wlodek Kuperberg has observed, an additional assumption about the n-dim volume (or area in 2-dim) of $B$ and $C$:

$$ |C|\ \le\ 2\cdot|B| $$

makes the conjecture true even without the earlier assumption about enlargement of $B$ (nor about symmetry).

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    $\begingroup$ From the title I would guess that you have an inclusion reversed, and you want to allow translations of $tB$. $\endgroup$
    – Douglas Zare
    Commented Jan 10, 2016 at 23:46
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    $\begingroup$ What does " $\ B\ C\subseteq\mathbb R^n\ $" mean? That both $B$ and $C$ are subsets of $\mathbb{R}^n$, or that some product of $B$ and $C$ is? $\endgroup$
    – Joseph O'Rourke
    Commented Jan 10, 2016 at 23:51
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    $\begingroup$ For any $B$ not containing the center of $C$, we can expand it to almost half of $C$, on one side of a hyperplane through the center. The question is whether any such almost-half of a centrally symmetric figure can be repositioned inside $C$ to cover the center. This should not be the case if you take a skew slice through the center of a cube, but I haven't yet proved that no rotation works. $\endgroup$
    – Douglas Zare
    Commented Jan 11, 2016 at 0:10
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    $\begingroup$ It is much more conventional and understandable to write $B, C \subseteq \mathbb{R}^n$. $\endgroup$
    – Todd Trimble
    Commented Jan 11, 2016 at 0:11
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    $\begingroup$ It is unclear to me where the assumptions end and the statement (that you want to prove) begins. A full stop (.) and/or a connecting word (e.g. 'then') would be useful. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 11, 2016 at 0:44

1 Answer 1

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Let $C$ be a $3\times 1$ rectangle centered at the origin, and $B$ a $1\times 1$ square "filling" the left side of $C$. Then $tB$ cannot be embedded in $C$ if $t>1$ but $B$ does not contain the origin, disproving the conjecture.

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    $\begingroup$ It turns out that Wlodek Kuperberg has posted a similar counter-example to my conjecture on FB about 3h before the katz's solution on MO. He had a unit disk inside a $5x2$ rectangle. Each example has its own subtle advantages (while my conjecture was rather easy--after the fact). $\endgroup$
    – Włodzimierz Holsztyński
    Commented Jan 11, 2016 at 17:51

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