- #1

- 105

- 33

- A tetrahedron with four equilateral triangular faces, and

- A square pyramid with a square base and four equilateral triangular faces, as shown below.

Figure 1. Tetrahedral ball stack.

Figure 2. Square pyramidal ball stack.

Here are the Excel-friendly algorithms (n = number of spheres along bottom row):

Density of Tetrahedral Stack (Dt):

Dt = (4*(2^0.5)*n*(n+1)*(n+2)*Pi())/(3*(2*n+2*3^0.5-2)^3 or

Density of Square Pyramidal Stack (Dp):

Dp = ( n*(1+n)*(1+2*n)*Pi())/(3*((1+(2^0.5*n)))^3) or

I found that, as the number of spheres approaches infinity for both arrangements, that Dt = Dp = ≈ 0.74048... or π/√18.

I resorted to Wolfram Alpha with the following query:

and got the following result:

True!

I then forwarded my calculations to Dr. Thomas C. Hales, who proved the Kepler Conjecture, asking him if Dt = Dp was correct, and he responded, saying,

"The reason for the equal densities in a tetrahedron and square pyramid is that they can both be viewed as part of the face-centered-cubic packing, each with a different set of exposed facets."

**My question to you is this: can you provide a detailed proof that (1) = (2) as n→∞, i.e., that
is true?**

Thanks for any input!

Fizixfan.