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From: gjk@talcott.UUCP (Greg J Kuperberg)
Newsgroups: net.puzzle,net.math
Subject: Re: 3D Geometric Puzzle (*spoiler*)
Message-ID: <117@talcott.UUCP>
Date: Mon, 19-Nov-84 14:50:58 EST
Article-I.D.: talcott.117
Posted: Mon Nov 19 14:50:58 1984
Date-Received: Tue, 20-Nov-84 07:21:45 EST
References: <107@talcott.UUCP>
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> If I'm lucky, this one will be a stumper:
> 
> What is the angle between two faces of a regular octahedron?
> 
> And an even trickier one:
> 
> What is the angle between two faces of a regular icosahedron?

I only got one reply to my puzzle, but here's the solution anyway:

The Shadow Principle:  If you have a planar figure of area a which is
elevated at an angle of theta, and the sun's rays are perpendicular to the
ground, then the shadow of the planar figure has area cos(theta)*a.

Now, consider a regular octahedron.  Divide the octahedron into eight
tetrahedra, each of whose vertex is the center of the octahedron and whose
base is one side of the octahedron.  Note that three sides of one such
"octant" are isoceles right triangles, while the fourth is equilateral.
Let a be the area of the equilateral side and b be the area of one of the
smaller sides.  Let phi be the angle between the larger side and one of the
other sides.  If we set one of these tetrahedra on the ground with the
equilateral side on the bottom, then by the shadow principle:

3*cos(phi)*b=a

If on the other hand we let a right-isoceles side be on the bottom, then by
the shadow principle:

b=cos(phi)*a.

By algebra,

cos(phi)^2=1/3

Now I originally asked for 2*phi.  So we have:

cos(2*phi)=2*cos(phi)^2-1=-1/3         .: 2*phi=arccos(-1/3)

Thus we are in agreement here.

For the icosahedron, things are not that much trickier.  Divide the
icosahedron into twenty meta-isoceles tetrahedra, each of which has a
vertex in the middle of the icosahedron and whose base is one side.  Call
one of these tetrahedra I.  Note that three edges of I have an angle of 72
degrees, while the other three have an angle which I will call alpha.  Let
c be the area of the equilateral side, and d be the area of one of the
isoceles sides.  By the shadow principle, we have:

c=3*cos(alpha)*b
d=2*cos(72)*d+cos(alpha)*a

Some algebra:

cos(alpha)^2=(1-2*cos(72))/3

Consulting our paleo-Galois theory, we find that cos(72)=(sqr(5)-1)/4.  So,

cos(alpha)^2=1/2-sqr(5)/6

Again, I originally asked for 2*alpha.

cos(2*alpha)=2*cos(alpha)^2-1=-sqr(5)/3    .: 2*alpha=arccos(-sqr(5)/3)

Note:  The only platonic solids which can pack space are the cube by
itself, and the regular tetrahedron combined with the regular octahedron.

			Greg