## Enumerating the Platonic Solids

The past few weeks, I outlined my approach to a series of lectures on polyhedra.  One of my constraints is that students will not have seen a lot of trigonometry yet, and will not have been exposed to three-dimensional Cartesian coordinates.  But there is Euler’s Formula!  I just finished a pair of lectures on the algebraic enumeration of the Platonic solids using Euler’s Formula, and I thought others might be interested as well.

As a reminder, Euler’s Formula states that if $V,$ $E,$ and $F$ are the number of vertices, edges, and faces, respectively, on a convex polyhedron, then

$V-E+F=2.$

How might we use this formula to enumerate the Platonic Solids?  We need to make sure we agree on what a Platonic Solid is:  a convex polyhedron with all the same regular polygon for faces, and with the same number meeting at each vertex.

To use this definition, we will define a few more variables:  let $p$ denote the number of sides on the regular polygons, and let $q$ denote the number of polygons meeting at each vertex of the Platonic solid.  (Those familiar with polyhedra will recognize these as the usual variables.)

The trick is to count the number of sides and vertices on all the polygons in two different ways.  For example, since there are $F$ polygons on the Platonic solid, each having $p$ sides, there are a total of $pF$ sides on all of the polygons.

But notice that when we build a cube from six squares, two sides of the squares meet at each edge of the cube.  This implies that $2E$ also counts all of the sides on the polygons.  Since we are counting the same thing in two different ways, we have

$pF=2E.$

We may similarly count all the vertices on the polygons as well.  Of course since a regular polygon with $p$ sides also has $p$ vertices, there are $pF$ vertices on all of the polygons.

But notice that when we put the squares together, three vertices from the squares meet at a vertex of the cube.  Thus, if there are $V$ vertices on a Platonic Solid, and if $q$ vertices of the polygons come together at each one, then it must be that $qV$ is the total number of vertices on all of the polygons.  Again, having counted the same thing in two different ways, we have

$pF=qV.$

Thus, so far we have

$V-E+F=2,\quad pF=2E,\quad pF=qV.$

Note that we have three equations in five variables here; in general, such a system has infinitely many solutions.  But we have additional constraints here — note that all variables are counting some feature of a Platonic Solid, and so all must be positive integers.

Also, since a regular polygon has at least three sides, we must have $p\ge3,$ and since at least three polygons must come together at the vertex of a convex polyhedron, we must also have $q\ge3.$

These additional constraints will guarantee a finite (as we know!) number of solutions.  So let’s go about solving this system.  The simplest approach is to solve the last two equations above for $E$ and $V$ and substitute into Euler’s Formula, yielding

$\dfrac{pF}q-\dfrac{pF}2+F=2.$

Now divide through by $F$ and observe that $F>0,$ so that

$\dfrac pq-\dfrac p2+1>0.$

Multiply through by $2q$ and rearrange terms, giving

$pq-2p-2q<0.$

How should we go about solving this inequality?  There’s a nice trick here:  add $4$ to both sides so that the left-hand side factors nicely:

$(p-2)(q-2)<4.$

Now we are almost done!  Since $p,q\ge3,$ then $p-2$ and $q-2$ must both be integers at least $1;$ but since their product must be less than $4,$ they can be at most $3.$

This directly implies that $p$ and $q$ must be $3, 4,$ or $5.$

This leaves only nine possibilities — but of course, not all options need be considered.  For example, if $p=q=5,$ then

$(p-2)(q-2)=9>4,$

and so does not represent a valid solution.  But when $p=3$ and $q=4,$ we have the octahedron, since $p=3$ means that the polygons on the Platonic Solid are equilateral triangles, and $q=4$ means that four triangles meet at each vertex.

So out of these nine possibilities to consider, there are just five options for $p$ and $q$ which satisfy the inequality $(p-2)(q-2)<4.$  And since each pair corresponds to a Platonic Solid, this implies that there are just five of them, as enumerated in the following table:

Actually, this implies that there are at most five Platonic Solids.  How do we know that twelve pentagons actually fit together exactly to form a regular dodecahedron?  A further argument is necessary here to be complete.  But for the purposes of my lectures, I just show images of these Platonic Solids, with the presumption that they do, in fact, exist.

Now keep in mind that in an earlier lecture, I enumerated the Platonic Solids using a geometrical approach; that is, by looking at those with triangular faces, square faces, etc.  I like the problem of enumerating the Platonic Solids since the geometric and algebraic methods are so different, and emphasize different aspects of the problem.  Further, both methods are fairly accessible to good algebra students.  The question of when to take an algebraic approach rather than a geometric approach to a geometry problem is frequently difficult for students to answer; hopefully, looking at this problem from both perspectives will give students more insight into this question.