MHB Can Two Points in a Cube Be Less Than √(3)/2 Apart?

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In a cube with a side length of 1, there are nine points within its interior. It can be demonstrated that at least two of these points are less than \( \frac{1}{2}\sqrt{3} \) apart. The discussion raises the question of whether this distance can be reduced further. A suggested solution hints at exploring smaller distances and their feasibility. Ultimately, the inquiry centers on the spatial relationships of points within a defined geometric space.
lfdahl
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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?
 
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Hint:

Divide the unit cube into $8$ cubes with side length $\frac{1}{2}$
 
Suggested solution:
(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).
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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