Platonic Solids

The Platonic Solids are five very special polyhedra. Consider a plane. It is flat and two dimensional. It is easy enough to construct polygons, i.e. Triangles, Quadrilaterals, Pentagons, and so forth. Furthermore, we may require that all their sides and angles are equal. We call such a figure a regular polygon. And it should be pretty obvious that one can construct a regular polygon with any number of sides that we want. So we say there are an infinite number of different regular polygons.

Now say we wanted to construct an analogous figure in three dimensions - a regular polyhedron. For a polyhedron to be regular, it should have the same regular polygon as each of its sides, and each vertex (corner) should have the same number of polygons around it. One may think at first that one can construct an infinite number of regular polyhedra in three-dimensions, just as we could construct an infinite number of regular polygons in two-dimensions. However, this does not turn out to be the case. In fact, there are only five regular polyhedra! They are called the platonic solids.

With a bit of thought, one can see that no more than five regular polyhedrons. A corner of a polyhedron must have at least three faces, the simplest of which is an equilateral triangle. We can form a vertex by putting together three, four, or five equilateral triangles. If we use more than five, the angles total 360 degrees or more and therefore cannot form a corner. So there are three possible ways to construct a regular polyhedra with triangles. Three and only three squares will form a corner, indicating the possibility of one regular solid with square faces. The same reasoning yields one possible polyhedron with three pentagons at each corner. We cannot go beyond the pentagon, because when we try to put three hexagons together at a corner, they equal 360 degrees. So we have a total of five regular polyhedra. A slightly more mathematically rigorous proof is at the bottom of the page.

Are here are the the five platonic solids:
4 triangular sides
Hexahedron (cube)
6 square sides
8 triangular sides
12 pentoganal sides
20 triangular sides

The slightly-more-rigorous proof:

Euler's formula (for Algebraic Topolgy) states that for a given polyhedron (not neccessarily regular) the number of Faces minus the number of Edges plus the number of Vertices equals two, i.e.
F - E + V = 2
Now we look at a regular polyhedron. It has n faces, each one a k-gon; and at each vertex, p faces are touching. Euler's formula becomes
n - (nk/2) + (nk/p) = 2
We factor this into
n( 1 - (k/2) + (k/p) ) = 2
Since n is positive, we know that 1 - (k/2) + (k/p) must be greater than zero (since its product with n is a positive number). Thus we have
1 - (k/2) + (k/p) > 0
We simplify this to
(2/k)+(2/p) > 1
Now for some critical thinking. A k-gon is meaningless when k = 1 or 2. Additionally, we must have more than two polygons meeting at a corner, so p > 2 as well.
The only remaining values of k and p that satisfy the equation are
k = 3, p = 3tetrahedron
k = 3, p = 4octahedron
k = 3, p = 5dodecahedron
k = 4, p = 3cube
k = 5, p = 3isocahedron


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