Is there always a parallelogram cross-section of parallelepiped contained in the smallest box - MathOverflow

Is there always a parallelogram cross-section of parallelepiped contained in the smallest box

2010-11-27T22:46:19Z

Let $M$ be a centered parallelepiped, the intersection of $M$ and any plane $P$ that passes through the origin is a parallelogram or hexagon. Each parallelogram or hexagon has a cubic box that is the smallest box that can contain the parallelogram or hexagon. Denote the cubic box by $B(P)$. There exist planes $P_{0}$ such that $B(P_{0})$ is the smallest among all the boxes $B(P)$.

Is it true that there is always one of the planes $P_{0}$ such that the cross-section of the centered parallelepiped $M$ by $P_{0}$ is a parallelogram?

Thanks.

Answer by Sergei Ivanov for Is there always a parallelogram cross-section of parallelepiped contained in the smallest box

2010-11-28T02:54:03Z

No, here is a counter-example (to revision 9).

Let $A$ be the linear map that sends the vector $(1,1,1)$ to $V:=(100,100,100)$ and is the identity on the orthogonal complement of this vector. Then any optimal cross-section is a hexagon whose plane intersects the six edges of $A(Q)$ separated from the vertices $V$ and $-V$. Indeed, any plane that avoids this configuration must intersect one of the edges adjacent to $V$. And the edges adjacent to $V$ are contained in the half-space $x+y+z\ge 100/3$, so any such section (and hence its minimal cube) has diameter at least $100/3\sqrt3\ge 10$. On the other hand, the intersection of $A(Q)$ with the plane $x+y+z=0$ is the same as the intersection of this plane with $Q$, so it fits in a unit cube (whose diameter is $2\sqrt3<10$). Hence no section of diameter $\ge 10$ can be optimal.

Answer by Joseph O'Rourke for Is there always a parallelogram cross-section of parallelepiped contained in the smallest box

2010-11-28T18:58:42Z

If I may attempt to interpret Sergei's perspicuous answer for unknown: Here is $A(Q)$ when $A$ is the linear map that sends $(1,1,1)$ (the upper right blue vertex) to $V=(10,10,10)$ (the upper right red vertex). Perhaps this helps make his proof transparent.