TITLE: Shortest path connecting two opposite points on a cube
QUESTION [38 upvotes]: Is it true, that a path connecting two opposite points (i.e. such that the segment joining them passes through the centre of mass of the cube) on the surface of the $d$-dimensional unit cube (with $d>1$)  is not shorter than $2$?

REPLY [4 votes]: The impressive and beautiful  @AntonPetrunin solution is external. Let me present an internal argument. The goal is to prove that the Euclidean length $\ |p|\ $ of any (rectifiable) path $\ p:[-\frac 12;\frac 12]\rightarrow F^{d-1}\ $ is at least $\ 2,\ $ where
  $\ p(-\frac 12) = -p(\frac 12),\ $ and $\ F^{d-1}\ $ is the boundary of
$\ [-\frac 12;\frac 12]^{\,d},\ $ for every integer $\ d\ge2.$ (My solution will provide a little more).
Let $\ p\ :=\    \triangle_{k=1}^d\, p_k, $ where $\ p_1\ \ldots\ p_d\ $ are the coordinate functions. Define the set of the essential coordinates:
$$ D\ :=\ \{\, k : |\max p_k| = 1\, \} $$
and let $\ \delta:=|D|\ $ be the essential dimension; we see that $\ \delta\ge 2.\ $ Finally, define the essential projection
$$ q\ :=\ \triangle_{k\in D}\, p_k $$
The range of the essential path $\ q\ $ is contained in $\ F^{\delta-1},\ $ where $\ F^{\delta-1}\ $ is defined as the boundary of cube
  $\ [-\frac 12;\frac 12]^D.$ Of course $\ q(-\frac 12) = -q(\frac 12). $ Since
$$ |p|\ge |q| $$
it is enough to prove that $\ |q|\ge 2$. Observe that
$\forall_{k\in D}\, var(p_k)\ge 1 $ hence
$$ \sum_{k\in D}\, var(p_k)\ \ge D $$
Let us also consider the $\ell^1$-norm in Euclidean spaces, namely
$\ ||x|| = \sum |x_k|.\ $ Let $\ ||q||\ $ be the
$\ell^1$-length of $\ q.\ $ We see that
$$ ||q||\ =\ \sum_{k\in D}\, var(p_k)\ \ge D $$
Let's apply the comparison between the two norms in any $n$-dimensional euclidean space: $\ |x|\ge \frac{||x||}{\sqrt{n}}\ $ for arbitrary
$\ n=1\ 2\ \ldots\ $. It follows that
$$\forall_{k\in D}\,\forall_{x\ y\in F^{\delta-1}: p_k(x)=p_k(y)}\quad
|x-y|\ \ge \frac{||x-y||}{\sqrt{\delta-1}} $$ 
and $\ |q|\ \ge\ \frac{||q||}{\sqrt{\delta-1}} $ hence 
 $\ |p|\ge|q| \ge \frac D{\sqrt{\delta-1}}.\ $ Thus,
THEOREM   
    $\ \forall_{d\ge 2}\ |p|\ \ge\ \frac D{\sqrt{\delta-1}} $
 
COROLLARY    
    $\ \forall_{d\ge 2}\ |p|\ \ge\ 2 $

REMARK   When considering above implication Thm => Cor,
  only $\ D=3\ $ makes you stop for a moment (unless you are a relative of John von Neumann).