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Suppose $P_{1}$ is a $\frac{n}{2}$-dimension polytope in $ R^{n}$ with barycenter $c$, and $P_{2}$ is a $(n-1)$-simplex in $R^{n}$ with the same barycenter as $P_{1}$, i.e $c$ . And also suppose they do not contain each other and both lie in the hyperpalne $x_{1}+...+x_{n} =d$

I know these two polytope intersect each other because they have the same barycenter.

but Since my intuition for high dimensional geometry is not always right is this statement true that their facets intersect each other?

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  • $\begingroup$ No, the facets needn't intersect. It's simpler just to think inside the $n-1$-dimensional hyperplane. There, $P_2$ is full-dimensional, so we could think of it as being quite large, with $P_1$ entirely contained inside it. But you would probably do better with this question on math.stackexchange.com. $\endgroup$
    – Hugh Thomas
    Commented Apr 13, 2018 at 12:15
  • $\begingroup$ But I mensioned in the question that they are not contains each other $\endgroup$
    – shere
    Commented Apr 14, 2018 at 14:27
  • $\begingroup$ Sorry, missed that. Since some of the boundary of $P_1$ is contained in $P_2$ and some of it isn't, there must be some intersection between the facets of $P_1$ and $P_2$. But really, you would do better with this question at math.stackexchange.com. $\endgroup$
    – Hugh Thomas
    Commented Apr 14, 2018 at 18:43

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