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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.

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    $\begingroup$ Dear unknown (yahoo), someone (probably you) has already asked this question a few days ago (I cannot find that question now, perhaps it has been deleted). In a comment there, I proposed a counter-example: $A=diag[1,1,100]$. All central sections of $A(Q)$ passing through vertices have diameters at least 100, so their minimal cubes are rather large. On the other hand, the horizontal section of $A(Q)$ is a unit square and it fits in a unit cube. So no plane through a vertex minimize the size of the cube. Is there anything wrong with this example? $\endgroup$
    – Sergei Ivanov
    Commented Nov 28, 2010 at 0:12
  • $\begingroup$ I didn't see the previous version myself, but Sergei's example is convincing. I was going to propose an analogous 2D counterexample. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 28, 2010 at 0:21
  • $\begingroup$ Dear Sergei, your example is a couter-example for the previous question. But the previous question did not describe the situation I really wanted to ask. Now I have corrected the question. Is there always one of the planes $P_0$ such that the plane does not intersect with the interior of any two neighbor edges? Thanks! $\endgroup$
    – user7738
    Commented Nov 28, 2010 at 0:29
  • $\begingroup$ Dear Joe, the previous question did not describe the situation I really wanted to ask. Now I have corrected the question. $\endgroup$
    – user7738
    Commented Nov 28, 2010 at 0:30
  • $\begingroup$ @unkn: Perhaps clearer to say: ...does not intersect the interior of two adjacent edges, where adjacent means they are incident to a common vertex. Or: only intersects the interiors of nonadjacent edges. In any case, I do not see an immediate counterexample... $\endgroup$
    – Joseph O'Rourke
    Commented Nov 28, 2010 at 0:54

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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.

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  • $\begingroup$ If $Q$ itself is "straight", then yes, the answer covers them. $\endgroup$
    – Sergei Ivanov
    Commented Nov 28, 2010 at 3:53
  • $\begingroup$ Dear Sergei, your example is very interesting. If you plot the image of $Q$ under $A$, $A(Q)$ is not a parallelepiped, as it has more than 6 faces. The question should be: If $A(Q)$ is a parallelepiped, is there always one of the planes $P_0$ such that the plane does not intersect with the interior of any two adjacent edges? $\endgroup$
    – user7738
    Commented Nov 28, 2010 at 4:12
  • $\begingroup$ I don't understand your comment. The set $A(Q)$ has 6 faces no matter whether you plot it or not. They are just the images of the faces of $Q$ under the linear map $A$. Also, I'm having trouble finding a definition of the word "parallelepiped" such that an affine image of a cube would not be one. $\endgroup$
    – Sergei Ivanov
    Commented Nov 28, 2010 at 5:19
  • $\begingroup$ For "parallelepiped", you can go to en.wikipedia.org/wiki/Parallelepiped $\endgroup$
    – user7738
    Commented Nov 28, 2010 at 5:43
  • $\begingroup$ I plotted $A(Q)$ for A={{34, 33, 33}, {33, 34, 33}, {33, 33, 34}} by using Mathematica and got an object more than 6 faces, which is strange to me. Did you plot $A(Q)$ of your example? $\endgroup$
    – user7738
    Commented Nov 28, 2010 at 5:50
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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.
Parallelopiped

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