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### 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

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Polyhedra

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