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(4sin^{2}`a` - `s`^{2}) / (2sin
`a`). Hence its volume is
`fks`^{2}sqrt(4sin^{2}`a` -
`s`^{2}) / (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 + p^{2}) |

Icosahedron | 20 | 3 | 2p / sqrt(1 + p^{2}) |

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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