## Answer to the Platonic Puzzle

#### John Baez

In this puzzle, we were systematically constructing regular polytopes in 4d by taking Platonic solids, pushing the centers of their faces in, getting pyramid-shaped "dents", and sticking suitable Platonic solids into these dents. It all worked nicely until we got to the case of the dodecahedron, where we naturally want to stick *icosahedra* into the dents, since the dents are pentagonal pyramids.

But there's NOT enough room to stick icosahedra into these dents. So there isn't any hope of getting a 4-dimensional regular polytope with isosahedral faces by this trick.

In fact, there is no 4d regular polytope with icosahedral faces. Here's the easiest way to see it: the angle at which adjacent triangular faces of a regular icosahedron meet is greater than 2 pi / 3, or 120 degrees. To be precise, it's arcos(-sqrt(5)/3), or about 138 degrees. So we can't have a polytope where 3 regular icosahedra meet along each edge - and certainly not one where *more* than 3 meet along each edge.

But suppose you just want to show there's not enough room to stick icosahedra into those dents. Here's one way to prove it:

Let's compare two pyramids which have a regular pentagon as their base:

1) the pyramid formed by a vertex of the regular icosahedron and its 5 neighboring vertices

and

2) the pyramid formed by the center of the regular dodecahedron and the 5 corners of one of its faces.

After all, we are trying to stuff pyramid 1) inside pyramid 2). If the pyramid 2) is "shorter and squatter in shape" than pyramid 1), we will succeed - and have wiggle room to boot. If pyramid 2) is "taller and skinnier", we will not.

One way to compare these pyramids is to look at their triangular faces, which are isosceles triangles. In pyramid 1) they are equilateral triangles, by definition. How about in pyramid 2)? Are they shorter and squatter than equilateral triangles - or taller and skinnier?

To figure this out, let's compute the lengths of the sides of the triangles in pyramid 2).

These points are the vertices of a regular dodecahedron:

```                (+-1/G, +-G, 0),
(+-G, 0, +-1/G),
(0, +-1/G, +-G)
(+-1, +-1, +-1)
```
where G is the golden ratio, (sqrt(5) + 1)/2.

The distance from any vertex to the orgin is sqrt(3), while the distance between neighboring vertices is

```sqrt((G-1)^2 + (1/G - 1)^2 + 1^2) = sqrt(8 - 4G) = 2 sqrt(2 - G) = 2/G
```
where I used the nice facts
```G^2   = G + 1,
1/G   = G - 1,
1/G^2 = 2 - G.
```
Now

sqrt(3) ~ 1.7320508075688772935

while

2/G ~ 1.2360679774997896964

so the triangular faces of pyramid 2) are TALL AND SKINNY compared to equilateral triangles. So we cannot succeed in stuffing the corner of an icosahedron into the dent left by pushing in the center of one face of the dodecahedron.

Thanks go to Jim Heckman, David Eppstein and other denizens of the newsgroup sci.math for helping me out with this.