The diameter of a bounded set is the supremum of the distances between any two points of the set, and the circumradius is the infimum of the radii of balls containing the set. Obviously, the diameter is never greater than twice the circumradius.

A hyperplane passing through the center of the unit cube $[0,1]^n$ cuts the cube into congruent halves. We want the halves to have the minimum diameter, or the minimum circumradius, respectively.


(a) Which hyperplane(s) produce the smallest diameter of the halves?

(b) Which hyperplane(s) produce the smallest circumradius of the halves?


I believe that the hyperplane perpendicular to some edge of the cube minimizes both the diameter and the circumradius of the two pieces. However, the hyperplane minimizing the diameter need not be perpendicular to any edge. Already in dimension 3, the plane perpendicularly bisecting the main diagonal produces the same minimum diameter of the pieces as the plane perpendicularly bisecting an edge, but does not produce the minimum circumradius.

  • $\begingroup$ If the hyperplane cuts an edge with ratio $x:1-x$ it will cut the opposite edge (i.e. the edge symmetric to the center of the cube) with ratio $1-x:x$. It follows that the diameter of a half is at least $\sqrt{(n-1)+x^2}$. Hence we see that the smallest possible diameter is $\sqrt{n-3/4}$ and that if a hyperplane with the smallest possible diameter cuts an edge it must cut it in half. Hence we are left with only finitely many cases. $\endgroup$ – Markus Sprecher Dec 28 '16 at 9:12

Answer to (a): We consider first $n$ odd and the hypercube $[-1,1]^n$. Let $(a_1,\dots,a_n)\in \mathbb{R}^{n}$ be a normal to the hyperplane. W.L.O.G we can assume that $0\leq a_1 \dots \leq a_n=1$. Consider the edge $(1,-1,\dots,1,-1,x), x\in [-1,1]$. The hyperplane will cut this edge at $x=\frac{a_1-a_2+\dots+a_{n-2}-a_{n-1}}{a_n}$. Note that $x\in [-1,0]$. By the comment above that the hyperplane can cut the edges only at the midpoints we must have $x=0$ and therefore $a_1=a_2,\dots ,a_{n-2}=a_{n-1}$. For $n$ even we get analogously $a_1=0, a_2=a_3,\dots,a_{n-2}=a_{n-1}$. Now assume that $a_{n-1}>0$ and let $i$ be the smallest index such that $a_i\neq 0$. Consider the edge $(\underbrace{1,\dots,1}_{i+1},1,-1,\dots,1,-1,x)$. By the argument that the edges can only be cut at the midpoints we must have $a_n<2a_i$. Now consider the edge $(\underbrace{1,\dots,1}_{i},x,1,-1,\dots,1,-1,-1)$. This edge will be cut at $x=\frac{a_n-a_i}{a_i}\in [0,1]$. Hence we must have $a_i=a_n$ and therefore $a_i=a_{i+1}=\dots=a_n=1$. It follows that the normal to a hyperplane is of the form $(0,\dots,0,\underbrace{1,\dots,1}_{u})$ with $u$ odd. Vice versa one can check that the corresponding halfes have the optimal diameter.

  • $\begingroup$ ...And it can be easily seen that only for $u=1$ the circumradius is half the diameter, which settles (b). $\endgroup$ – Ilya Bogdanov Dec 28 '16 at 16:13

This seems to be just a rephrasing of Markus's answer, with some more details on the circumradius.

The minimal diameter is $d=\sqrt{n-3/4}$, the minimal circumradius is half the minimal diameter. An example is indeed provided by the hyperplane parallel to some $(n-1)$-dimensional face.

Consider any edge $ab$ which is crossed by the hyperplane at point $c$. The vertices $a'$ and $b'$ opposire to $a$ and $b$ are also in different parts. So, if $c$ is not a midpoint of $ab$, the diameter is not minimal.

On the other hand, of all crossing points are the midpoints, we reach the minimal diameter of $d=\sqrt{n-3/4}$. Indeed, the diameter is the maximal distance between the vertices of the part, which are the vertices of the cube together with the crossing points, and it is easily seen that these distances do not exceed $d$. THus this is equivalent condition for minimizing the diameter.

This condition indeed shows that for any two edges, the hyperplane is either parallel to either one of them or to their bisector line, thus the normal vector indeed contains only zeroes and ones. Since the hyperplane does not pass through vertices, there should be an odd number of ones.

Finally, if the circumradius is $d/2$< then the diameter of the part (connecting a vertex and a midpoint o an opposite edge)is also a diameter of the circumball. Thus, all crossing points should lie on the boundary of this ball, which yields that the normal vector has exactly one one.


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