11) What is the length of the shortest path that passes through all the vertices of a cube with side $1 \mathrm{~m}$? N.B. The path can also pass through the interior of the cube.
(A) $6 \mathrm{~m}$
(B) $7 \mathrm{~m}$
(C) $(6+\sqrt{2}) \mathrm{m}$
(D) $(6+\sqrt{3}) \mathrm{m}$
(E) $8 \mathrm{~m}$
11. The answer is $(\mathbf{B})$.

Let's label the vertices of the cube with the letters from $A$ to $H$, such that $A B C D$ (in this order) is one face; we call $E, F, G, H$ the vertices of the opposite face, so that vertex $A$ is connected by an edge to $E$, vertex $B$ to $F$, vertex $C$ to $G$, and finally $D$ to $H$. The path $A B C D H G F E$ is a path of length 7 that touches all the vertices of the cube.

To show that there is no shorter one, observe that any path that touches all 8 vertices of the cube consists of (at least) 7 segments that start from one vertex of the cube and reach another. But since any two vertices of the cube are at least 1 unit apart, the total length must be at least 7.

[Problem proposed by F. Poloni]