```From:           Robin Chapman <rjc@noether.ex.ac.uk>
Date:           Fri, 17 Oct 1997 08:01:44 GMT
Newsgroups:     sci.math.research
Subject:        Re: Volumes of pieces of dodecahedron.
```

```David Epstein wrote:
>
> Take a regular dodecahedron and set it down on a horizontal surface.
> Then there are four levels of vertices, one at the bottom face,
> one the top face, and two levels in between. Now saw through the
> dodecahedron horizontal at the two interior levels. The dodecahedron
> is sawn into three pieces, of which the top and bottom piece are
> congruent. How can one show that the three pieces each have the same
> volume without doing a long and horrible integration?
>
> Is this a special case of some more general theorem?

Let the dodecahedron have height h and face area A. One can
decompose it into 12 pentagonal pyramids of base area A and
height h/2. Then the volume of the dodecahedron is
12((h/2)A/3) = 2Ah. It suffices to show that the bottom piece
has volume 2Ah/3. This piece is a frustum of a pentgonal pyramid.
Its top face has area tau^2 A where tau =(1+sqrt(5))/2.
(This comes from tau being the ratio of diagonal to side of a
regular pentagon.) Let h_1 and h_2 be the heights of the bottom
and central pieces. Then 2h_1 + h_2 = h. We need to determine h_1.
But the ratio of (h_1 + h_2) to h_1 is the same as the ratio of the
distance of the vertices A_1 and A_2 from the side A_3 A_4 in the
regular pentagon A_1 A_2 A_3 A_4 A_5, and is seen to be tau.
Then we find h_1 = tau^{-2}h and h_2 = tau^{-3}h. (Naturally
everything is in golden ratio). We can express the frustum in
question as the "difference" of two pentagonal pyramids. Let d be
the height of the smaller of these, so that d + h_1 is the height
of the larger. Then the ratio of d + h_1 to d is the same as the
ratio of the sidelengths of the bases of the pyramids, i.e., it is
tau. Thus d = tau^{-1}h and d + h_1 = h. Thus the large pyramid
has volume (tau^2 A)h/3 and the small one has volume A tau^{-1}h/3.
Thus the frustum has volume (tau^2-tau^{-1})Ah/3 = 2Ah/3.

Not an integration in sight! I'm sure Euclid could have come up
with such an argument.

--
Robin Chapman			 	"256 256 256.
Department of Mathematics		 O hel, ol rite; 256; whot's
University of Exeter, EX4 4QE, UK	 12 tyms 256? Bugird if I no.
rjc@maths.exeter.ac.uk             	 2 dificult 2 work out."
http://www.maths.ex.ac.uk/~rjc/rjc.html	 Iain M. Banks - Feersum Endjinn
```