If you inscribe a regular polygon into a given circle, the larger the number of sides, the larger the area of the polygon. I guess I always thought that the same would apply to Platonic solids inscribed in a sphere..... It doesn't. I noticed this as I was looking though "The Penny cyclopædia of the Society for the Diffusion of Useful Knowledge

By Society for the Diffusion of Useful Knowledge" (1841). As I browsed the book, I came across the table below:

The table gives features of the Platonic solids when inscribed in a one-unit sphere. At first I thought they must have made a mistake, but not so. The Dodecahedron fills almost 10% more of a sphere (about 66%) than the icosahedron (about 60%). So the Dodecahedron is closer to the sphere than the others.

Interestingly, if you look at the radii of the inscribing spheres, it is clear that solids which are duals are tangent to the same internal sphere.

But if you look at the table of volumes when the solids are inscribed with a sphere inside tangent to each face:

When you put the Platonic solids around a sphere, the one smallest, and thus closest to the sphere is the icosahedron.

This leads to the paradox that when platonic solids are inscribed with a sphere, the icosahedron is closest to the circle in volume (thus most spherical?) but when they are circumscribed by a sphere, the dodecahedron is the closest to the volume of a sphere (and thus most spherical?)... hmmm

Here is a table of the same values when the surface area (superficies) is one square unit. Notice that for a given surface area, the icosahedron has the largest volume, so it is the most efficient "packaging" of the solids (thus more spherical?).

I guess that makes it 2-1 for the icosahedron, so I wasn't completely wrong all along.

POSTSCRIPT::: Allen Knutson's comments on the likely cause of this reversal of "closeness" to the sphere:

- I think it's about points of contact. On the inside, the dodecahedron touches the sphere at the most points (20), and on the outside, the icosahedron touches the sphere at the most points (again 20).
- Indeed: my recipe would suggest that inside, the 8-vertex cube is bigger than the 6-vertex octahedron, and outside, the 8-face octahedron is smaller than the 6-face cube. Both are borne out by your tables. Thank you, Allen