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 maximal diameter.

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.

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 $x=\frac{a_1-a_2+\dots}{a_n}$. 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 maximal diameter.

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 maximal diameter.