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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