Prove that a rectangular box can be inscribed with all white vertices in the sphere

Question

Twelve percent of a sphere is painted black, the remainder is white. Prove that one can insert a rectangular box with all white vertices into the sphere.

This is supposed to be a box principle problem, but I am unable to cast this into that. I need some help.

Answer 1

Put the centre of the sphere at the origin of a 3D Cartesian coordinate system, where three orthogonal planes intersect. Take a black point on the spherical surface, and consider the other $3$ points of the surface that are symmetric to the initial one with respect to each of the orthogonal planes. The $4$ points are vertices of a box. Complete the box by taking the remaining symmetric points so that you have $8$ points on the surface. Now paint these points as black. Repeating this process for all black points, at most $8*12=96\%\;$ of the surface will become black. Now choose any point in the region $(\geq 4\%)$ that has remained white. Constructing a box from this point in the same way as above necessarily yields a box with $8$ white edges.

Answer 2

A hint: Use $8\cdot 0.12<1$.

Next hint: You can find an axes-aligned box with white vertices without even rotating the painted sphere.