Largest Cube that fits the space between two Spheres?

Question

Two spheres, one of radius 15 and the other of radius 33, have the same center. Find the side length of the largest cube that fits between them.

Answer 1

Tuesday: the wire frame case gives $41 S^4 - 48960 S^2 + 11943936 = 0.$ I got this by squaring a square root, so the larger positive root is fake; the actual figure is the smaller one, just a bit larger than my answer below, that is $S \approx 18.48696.$

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

In order to get everything in a single plane, we must rotate by 45 degrees...

I got $S^2 + 20 S - 576 = 0, $ or $(S + 36)(S - 16) = 0.$

So $S=16.$ Give me a few minutes, I will give coordinates for the vertices of the cube...

The bottom square of the cube has vertices $(8,8,15), (8,-8,15),(-8,8,15), (-8,-8,15).$ The top square has vertices $(8,8,31), (8,-8,31),(-8,8,31), (-8,-8,31).$

We get confirmation that it all works from $$ 8^2 + 8^2 + 31^2 = 33^2, $$ sometimes called a Pythagorean Quadruple.

Answer 2

After $Will$ solved the side length of the solid cube, here is my solution to the wire frame cube. The midpoints of the four sides of a square would lie on the 15 sphere and the four vertices of the opposite square of the cube would lie on the 33 sphere.