Polyhedron inside Sphere

Date: 5/24/96 at 21:54:49
From: Keays
Subject: Polyhedron Inside Sphere

Dear Dr. Math,

A mate of mine is trying to build a 12-faced shape, each face being a 
perfect pentagon. (I have forgotten the correct name for it). He wants 
to know how long each side of the pentagon should be, to fit it in a 
sphere of diameter 2.9 m.

My calculations say about 0.8 m, but I estimate the answer to be 
closer to 1 m. What is the real answer?

Hope this all makes sense to you,
Thanks heaps,
Roger Keays

Date: 5/28/96 at 19:38:39
From: Doctor Pete
Subject: Re: Polyhedron Inside Sphere

The polyhedron you are referring to is called the dodecahedron.  It 
has many interesting properties, one of which is the fact that you can 
inscribe a cube in it.  That is, you can pick eight vertices (corners) 
of the dodecahedron which are also vertices of a cube.  Then 12 edges 
of the cube will touch the 12 faces of the dodecahedron.  Since this 
cube is also inscribed in the same sphere as that of the dodecahedron, 
it gives an easy way to find the length of the dodecahedron's edge in 
terms of the radius of the circumscribed sphere.

Let R be the radius of the given sphere (R = 1.49m in your case), and 
let x be the edge length of the inscribed cube.  Then x*sqrt(3) = 2*R 
(use the Pythagorean Theorem twice), so x = 2*R/sqrt(3).  But since 
the edge is coplanar with the face of a pentagon, we have the 
following picture:


              A  _________________ C
                /\         _. -
              /   \   _. -  
            /    _. -
          / _. -    D
        / - 
      /-
   B

where BC = x, the edge of the cube, AB = BC, two sides of a pentagonal 
face, and BAC = 3*pi/5 = 108 degrees.  AD is the angle bisector of 
BAC, so ADC is a right angle.  Then sin(DAC) = DC/AC = (x/2)/AC, 
so AC = x/(2*sin(3*pi/10)) = R/(sin(3*pi/10)*sqrt(3))
= 4*R/(sqrt(15)+sqrt(3)), or R*0.7136.  So for your case, it should be 
about 1.06333 m, which agrees very closely with your result.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/