What fraction of a regular dodecahedron is closer to center than any surface
he same method was used as was used in "What fraction of a regular tetrahedron is closer to center than any surface".
This problem's question is its title. Since the volumes of both the regular dodecahedron and the interior closer space scale with the cube of the edge the fraction remains the same regardless of the chosen edge length. The answer is in parts per billion rounded to an integer, i.e., Round[1000000000 value]. I used an edge length of 1.
In the solution the integral and a closed form are given.
By the way, this is the most difficult of the Platonic solids to solve because of the pentagonal faces.
The answer is 107751438.
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1 solution
∫ 8 + 8 5 1 1 + 5 5 − 6 ( 6 5 + 2 9 5 ) 0 ∫ 2 0 4 7 + 2 1 5 ( 2 5 + 1 1 5 ) 3 / 2 − 1 0 − 4 ( 4 7 + 2 1 5 ) x 2 + 3 7 7 5 + 8 4 3 1 + 5 2 x ∫ 4 2 5 0 + 1 1 0 5 4 0 x 2 + 4 0 y 2 − 1 1 5 − 2 5 2 1 1 + 5 2 1 0 + 5 2 2 y 1 d z d y d x
4 8 1 1 0 1 ( 2 1 5 + 4 7 ) 1 5 3 6 0 4 2 5 1 5 − 4 5 6 ( 9 3 4 9 5 + 2 0 9 0 5 ) + 4 π 1 8 6 9 8 5 + 4 1 8 1 0 + 9 5 0 5 → 3 2 1 0 ( 2 1 5 + 4 7 ) 4 2 5 1 5 − 4 5 6 ( 9 3 4 9 5 + 2 0 9 0 5 ) + 4 π 1 8 6 9 8 5 + 4 1 8 1 0 + 9 5 0 5 ≈ 1 0 7 7 5 1 4 3 8