I apoligize in advance if the question might appear trivial, but there is something unclear to me related to convex polytopes in high dimensions. Let $P$ be a convex polytope in $\mathbb{R}^d$ for $d\ge 2$, and let $\mathbf{p}$ an interior point of $P$. Let $L$ be straight line through $\mathbf{p}$.

When $d=2$, $L$ intersects $P$ at one of the its edges (more than one edge only if the intersection point is a vertex of $P$).

When $d=3$, $L$ intersects $P$ at one of of its $2$-dimensional faces. Hence, the set of all edges (which are ($d-2$)-dimensional objects) of $P$ does not contain all possible intersection points over the choice of $L$ given any $\mathbf{p}$ inside $P$.

Question: When $d\ge 4$, it is always the case that a subset of polygons (which are $d'$-dimensional objects for $d'\le d-2$) having as vertices the ones of $P$ and containing no interior point of $P$, contains all possible intersection points of $L$?

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    $\begingroup$ There is a line that passes through any two points, so in particular there is a line through $\mathbf{p}$ that passes through any point of $P$. $\endgroup$ Feb 4, 2021 at 15:43

1 Answer 1


A ray from an interior point $p$ of a $d$-dimensional convex polytope will meet the boundary of $P$ at exactly one point $q$.

With probability $1$ the point will $q$ be in the interior of one of the $d-1$-dimensional facets of $P$. But it might be on the boundary of some facet, in which case it will be on several facets.

Any point on the boundary of $P$ will be the $q$ for some ray since you can always join $p$ to $q$.

Think about the cube in space for your intuition. A ray from the center might leave the cube at the interior of a face, at the interior of an edge (in which case it will be on two of the faces) or at a vertex (in which case it will be on three of the faces and three of the edges).


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