I'm a secondary school maths teacher, currently on my holidays working through some maths problems for fun. Here is one I have done, but it felt too easy, so if you could check if there's any mistakes, I'd be grateful.
Suppose there are 8 spheres, radius r. They are positioned touching each other such that the centre points of each sphere would join to make a cube.
The question is "What is the largest possible radius of a ninth sphere in the centre of this arrangement?" I think I have answered it, but I would be grateful for any input.
I don't know how to draw this digitally, so I hope my description is sufficient.
So far I have imagine the problem in 2d... four circles touching to form a square. The radius is r. The square created from the centre points will have side length 2r, therefore using pythag the diagonal is $2r\sqrt2$. Half of this will give you the distance from the centre of a circle to the centre of the arrangement: $r\sqrt2$. Therefore, subtract the initial radius, r and you have the maximum possible radius of the smaller circle: $r\sqrt2-r=r(\sqrt2 - 1)$.
Taking to 3D, the cube created by the centres 8 spheres will have a diagonal of $2r\sqrt2$ along the base and height $2r$, so again using pythag, the longest diagonal of the cube will be:
$\sqrt((2r\sqrt2)^2+(2r)^2)=\sqrt(8r^2 + 4r^2)=2r\sqrt3$
Divide this by 2 to find the distance from centre-of-sphere to centre-of-cube: $r\sqrt3$
Subtract the initial radius to find radius of smaller sphere: $r(\sqrt3 - 1)$
That's correct. You could do it a bit more directly, without factors of $2$ and without iterated application of the Pythagorean theorem: The centres of the spheres are at $(\pm r,\pm r,\pm r)$. The length of each of those vectors is $\sqrt{3r^2}=\sqrt3r$, so that leaves $\sqrt3r-r=r\left(\sqrt3-1\right)$ for the radius of the inner sphere.