Sphere stacking question?

Question

There are 10 identical spheres of radius R stacked tangentially in a pyramidal formation such that they can be easily circumscribed by a large tetrahedron. What is the volume of the negative space within the tetrahedron?

If you can do that, what is the volume of the negative space for n spheres stacked in the way described given that n is a tetrahedral number, i.e. 1, 4, 10, 20, 35, etc.

Answer 1

The difficulty with this problem is visualizing it.
Let's start with the simplest case of 1 sphere and maybe we can build on that.

A tetrahedron is circumscribed around a sphere. It is easy to see that the points of contact are along the altitudes of each face of the tetrahedron.

Let's label the corner of the tetrahedron A,B,C,D. each edge, AB = a.
Let E = mid point of BD, such that AE is the altitude of face ABD.
AE = a* sqrt(3) / 2
Let F = 1/3 of the way from E to C. We expect that AF is vertical and is equal to h, height.
EF = a*sqrt(3) / 6
AF = a*sqrt(6) / 3

We expect the center of the sphere to lie on AF, call the center O.
Let G = point on AE such that OG is perpendicular to AE. So OG = OF = R = radius of sphere.
After all the analyses, you can determine that
R = a * sqrt(6) / 12
or a = R * 2*sqrt(6)

Volume of tetra hedron = a^3 * sqrt(2) / 12
= R^3 * 8 * sqrt(3)

Volume of sphere = 4*π/3 * R^3
Volume of void = [8*sqrt(3) - 4*π/3] * R^3

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For 4 spheres, I followed a similar principle.
The triangle AEC, in this case has the center of 2 spheres in its plane. Both of these spheres touch the lines AE and EC respectively, and one of them is centered on AF.
Again using trig and geom (I really can't show the working here), but I'm getting
or a = 2R * (sqrt(6) + 1)
Volume of tetrahedron =
R^3 / 3 * (38*sqrt(2) + 36*sqrt(3))
For four spheres, you can find the void volume from here.
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10 spheres:
After all the working,
a = R *[4 + 2*sqrt(6) ]
Volume of tetrahedron = [88*sqrt(2) + 72*sqrt(3)] / 3 * R^3
Volume of spheres = 40π/3 * R^3

Answer 2

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