What is the minimum length of the edge of a cube that can fit 2 unit-radius spheres, which do not have any common interior points?
Solution. Let's consider the minimal-edged cube sought in the problem. Its edge length is greater than 2 units, since the edge length of a cube circumscribing a sphere with a radius of 1 unit is exactly this much. Therefore, there exists a smaller cube inside the minimal-edged cube, which is bounded by planes that are 1 unit away from the faces of the larger cube, and these planes separate the center of the cube from the face planes.

![](https://cdn.mathpix.com/cropped/2024_05_02_7fc5086b22abd123f1dcg-1.jpg?height=805&width=808&top_left_y=346&top_left_x=648)

A sphere with a radius of 1 unit will not protrude from the larger cube if and only if its center is inside or on the boundary of the smaller cube. Furthermore, the two spheres do not have any common interior points, so the distance between their centers is at least 2 units. Therefore, the smaller cube has two points whose distance is at least 2 units. Among the points of a cube, the farthest apart are the endpoints of a space diagonal, so the space diagonal of the smaller cube is at least 2 units, meaning its edge is at least $2 / \sqrt{3}$ units. The edge of the larger cube is exactly 2 units longer, so it is at least $2 + 2 / \sqrt{3}$ units.

In a cube with an edge of $2 + 2 / \sqrt{3}$, two spheres can already be placed such that their centers lie on a space diagonal, 1 unit away from the center of the cube.

Based on the work of Gergely Faragó (Bp., Fazekas M. Gyak. Gimn., I. o. t.)