 ## The Platonic Solids

A platonic solid is a polyhedron all of whose faces are congruent regular polygons, and where the same number of faces meet at every vertex. The best know example is a cube (or hexahedron ) whose faces are six congruent squares.

### Manipulating the shapes on this page.

If you have a Java compatible browse you can click on any of the figures of this page and bring up a new window in which you can rotate the polyhedron and so get a better feeling for its shape. The applets are special cases of a simple 3D geometry viewer that comes with instructions.

### There are only five!

The Greeks recognized that there are only five platonic solids. But why is this so? The key observation is that the interior angles of the polygons meeting at a vertex of a polyhedron add to less than 360 degrees. To see this note that if such polygons met in a plane, the interior angles of all the polygons meeting at a vertex would add to exactly 360 degrees. Now cut an angle out of paper, and fold another piece of paper to that angle along a line. The first piece will fit into the second piece when it is perpendicular to the fold. Think of the fold as a line coming out of our polyhedron. The faces of the polyhedron meet at the fold at angles less than 90 degrees. How can this be possible? Try wiggling your first piece of paper within the second. To be able to incline it with respect to the fold you have to decrease the angle of the first piece, or increase the angle of the second.

Next we'll consider all possibilities for the number of faces meeting at a vertex of a regular polyhedron. For each possibility we actually construct such a polyhedron, a picture of which you can see close by on this page. Here are the possibilities:

• Triangles. The interior angle of an equilateral triangle is 60 degrees. Thus on a regular polyhedron, only 3, 4, or 5 triangles can meet a vertex. If there were more than 6 their angles would add up to at least 360 degrees which they can't. Consider the possibilities:
• 3 triangles meet at each vertex. This gives rise to a Tetrahedron.
• 4 triangles meet at each vertex. This gives rise to an Octahedron.
• 5 triangles meet at each vertex. This gives rise to an Icosahedron
• Squares. Since the interior angle of a square is 90 degrees, at most three squares can meet at a vertex. This is indeed possible and it gives rise to a hexahedron or cube.
• Pentagons. As in the case of cubes, the only possibility is that three pentagons meet at a vertex. This gives rise to a Dodecahedron.
• Hexagons or regular polygons with more than six sides cannot form the faces of a regular polyhedron since their interior angles are at least 120 degrees.

But now things get a little more subtle. We have looked at all possibilities of congruent regular polygons meeting at a vertex of a polyhedron, but how do we know that there isn't another regular polyhedron for some of these cases? For example, why is the cube the only polyhedron for which three squares meet at each vertex? The rest of this page gives the answer to this question, but the going will be much harder!

Before continuing, let us collect some data. Let
• m be the number of polygons meeting at a vertex,
• n the number of vertices of each polygon,
• f the number of faces of the polyhedron,
• e the number of edges of the polyhedron, and
• v the number of vertices of the polyhedron.

The values of these numbers for each of the polyhedra are listed in this table:

 n m f e v Tetrahedron 3 3 4 6 4 Octahedron 3 4 8 12 6 Icosahedron 3 5 20 30 12 Hexahedron 4 3 6 12 8 Dodecahedron 5 3 12 30 20 Table: Combinatorics of Regular Polyhedra

Our aim now is to show that for any pair of number n and m the values of the other parameters, f, e, and v are determined uniquely.

First we note that since two faces meet in one edge, we must have

e = nf/2

Next, since every vertex is shared by m faces, we must have

v = nf/m

It is apparent from the Table that for all five regular polyhedra

f=2+e-v             (E)

We'll see below that this equations actually holds for all convex polyhedra. Given m and n the above three equations determine f, e, and v uniquely, and so there are only five possible regular polyhedra.

The result (E) is known as

### Euler's Polyhedron Theorem

To see why it is true we proceed in several steps. First we remove one face from the polyhedron. Let
F=f-1
be the new number of faces. We need to show
F=1+e-v             (*)

Now think of the remaining faces of the polyhedron as made of rubber and stretched out on a table. This will certainly change the shape of the polygons and the angles involved, but it will not alter the number of vertices, edges, and faces. Now we draw diagonals in the stretched faces out of the Polygons. Every diagonal increase the number e of edges by one, and also the number F of faces, so that our equation (*) remains valid. We continue this process until all polygons have been changed into triangles.

In the final stage we remove triangles until we are left with only one triangle for which (*) is obviously true. How do we do that? If the removed triangle has exactly one edge on the boundary then F and e are both decreased by 1 and (*) remains true. If it has two edges on the boundary then F is reduced by 1, e is reduced by 2, and v is reduced by 1, so that (*) remains true.

There is one final subtlety. Can we really dismantle the triangles as described? The answer is yes. But as an exercise you may wish to modify the dismantling procedure to remove all doubts in your mind. A similar dismantling procedure could be designed for a tessellation of a polyhedron by polyhedra, but in that case it is not always possible. For an illustration you may want to visit my page that describes Rudin's example of an unshellable triangulation.

If you'd like to play with polyhedron with many more faces, here is a crude rendering of a sphere, which is of course not a platonic solid!

[22-Jan-1997]