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I was looking at this problem and was wondering if you could use Descartes Theorem on it. The only problem I had was that this problem uses tangent spheres instead of circles and I have never used Descartes Theorem for 3D and don't know if it is possible. I've tried looking at the cross section, but the spheres are not at the same height so you can't really do it. Here is the problem:

"Three spheres of radius 2 sit on a flat surface touching one another. A smaller sphere sits on the same surface, in the middle and touching all three of the bigger spheres. What is its radius?"

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    $\begingroup$ Descartes' theorem is very general, for circles. But you have a very specific problem: use the fact that all three of the large spheres have the same radius. $\endgroup$
    – Chris Lewis
    Commented Jan 17 at 11:20
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    $\begingroup$ Due to symmetry, you don't really need the full power of 3D Descartes en.wikipedia.org/wiki/Soddy%27s_hexlet for this problem. But if you do want to use that approach then you can treat the plane as a sphere of zero curvature. Disclosure: I've only ever used the 2D version. raw.githubusercontent.com/gist/PM2Ring/… gist.github.com/PM2Ring/… $\endgroup$
    – PM 2Ring
    Commented Jan 17 at 11:23

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I don't want to give the game away, but here's a partial solution.

Let $R=2$ be the radius of the big spheres and $r$ the radius of the small sphere. Let the surface be the XY plane.

The centres of the big spheres form an equilateral triangle. We can align one side of the triangle with the Y axis, with the perpendicular bisector of that side aligned with the X axis, passing over the origin.

Eg, let the coords of the centres of the big spheres be $$(-u, -R, R), (-u, R, R), (s, 0, R)$$

Now find $u, s$ so that the spheres are equidistant from each other, and are all $s$ from $(0, 0, R)$. They will also be equidistant from the origin.

By symmetry, the small sphere sits on the origin, so the coords of its centre are $(0, 0, r)$. We now need to find $r$ such that it touches the big sphere at $(s, 0, R)$. (And by symmetry it will also be tangent to the other two big spheres).

The line connecting the centres of these two spheres is the hypotenuse of a right triangle. Its length is $R+r$. The other two sides are $s$ and $R-r$.

Thus $$(R+r)^2 - (R-r)^2 = s^2$$ $$2R\cdot2r = s^2$$ Therefore $$r = \frac{s^2}{4R}$$

Here's a 3D diagram, but please don't peek before you've attempted the calculations.

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