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.


**Questions.**


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


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


**Comment.**


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.



Splitting the $n$-cube into two small congruent convex halves
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.


