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Platonic solids inside and outside a unit sphere

Q: What are the volumes of the platonic solids inscribed in/circumscribed about a unit sphere?
[TB, 11 Apr 1996]

A: Consider a given solid as a set of pyramids on the faces. Suppose there are f faces each with k sides of length d, and let a = pi / k. The area of the pyramid's base, ie the face, is k(s/2)² / tan a. For the inscribed solid the pyramid's slant height (measured from a vertex of the base) is 1, so its height measured normal to the base is sqrt(4sin²a - s²) / (2sin a). Hence its volume is fks²sqrt(4sin²a - s²) / (24sin a tan a).

For the circumscribed solid we can just divide the volume of the inscribed solid by the cube of the above normal height.

Now if p = (sqrt(5) - 1) / 2 = 0.618034 is the golden ratio then we have

	`f`	`k`	`s`
Tetrahedron	4	3	sqrt(8/3)
Cube	6	4	2 / sqrt(3)
Octahedron	8	3	sqrt(2)
Dodecahedron	12	5	2 / sqrt(5 + 4p + p²)
Icosahedron	20	3	2p / sqrt(1 + p²)

We can also calculate the volumes for the rhombic dodecahedron (2 and 4sqrt(2)) quite easily. Thus we have the following - percentages indicate volume relative to that of the sphere, ie 4.188790:

	# vertices	# faces	Volume of inscribed solid		Volume of circumscribed solid
Tetrahedron	6	4	0.513200	12.3%	13.856420	330.8%
Cube	8	6	1.539601	36.8%	8.000000	191.0%
Octahedron	6	8	1.333333	31.8%	6.928203	165.4%
Rhombic dodecahedron	14	12	2.000000	47.7%	5.656854	135.0%
Dodecahedron	20	12	2.785164	66.5%	5.550290	132.5%
Icosahedron	12	20	2.536151	60.5%	5.054058	120.7%

Among the regular solids the inscribed solids' volumes are in order of # vertices and the circumscribed solids' are in order of # faces [David Cartwright].

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