A cube with 9 interior points: Show that at least two of them are less than √(3)/2 apart.

[lfdahl]

There are $9$ points in the interior of a cube of side $1$.

a) Show that at least two of them are less than $\frac{1}{2}\sqrt{3}$ apart.

b) Can $\frac{1}{2}\sqrt{3}$ be replaced by a smaller number?

 

[lfdahl]

Hint:

Spoiler

Divide the unit cube into $8$ cubes with side length $\frac{1}{2}$

 

[lfdahl]

Suggested solution:

Spoiler

(a). The cube may be partitioned into eight equal cubes, each with side of $\frac{1}{2}$, so that
at least one of these cubes contains more than one point. The greatest distance
between any two points in the interior of a cube is less than the length of its
longest diagonal, which in the case of the smaller cubes is precisely $\frac{1}{2}\sqrt{3}$.

(b). Placing a point at the center of the unit cube, and the remaining ones arbitrarily
close to the cube's vertices yields, among the distances between any two points,
one arbitrarily close to $\frac{1}{2}\sqrt{3}$. Hence, this number cannot be replaced by a smaller
one in the result of (a).