Investigate!
A cube is an example of a convex polyhedron. It contains 6 identical squares for its faces, 8 vertices, and 12 edges. The cube is a
regular polyhedron (also known as a
Platonic solid) because each face is an identical regular polygon and each vertex joins an equal number of faces.
There are exactly four other regular polyhedra: the tetrahedron, octahedron, dodecahedron, and icosahedron with 4, 8, 12 and 20 faces respectively. How many vertices and edges do each of these have?
Another area of mathematics where you might have heard the terms “vertex,” “edge,” and “face” is geometry. A
polyhedron is a geometric solid made up of flat polygonal faces joined at edges and vertices. We are especially interested in
convex polyhedra, which means that any line segment connecting two points on the interior of the polyhedron must be entirely contained inside the polyhedron.
Notice that since
\(8 - 12 + 6 = 2\text{,}\) the vertices, edges and faces of a cube satisfy Euler’s formula for planar graphs. This is not a coincidence. We can represent a cube as a planar graph by projecting the vertices and edges onto the plane. One such projection looks like this:
In fact,
every convex polyhedron can be projected onto the plane without edges crossing. Think of placing the polyhedron inside a sphere, with a light at the center of the sphere. The edges and vertices of the polyhedron cast a shadow onto the interior of the sphere. You can then cut a hole in the sphere in the middle of one of the projected faces and “stretch” the sphere to lie down flat on the plane. The face that was punctured becomes the “outside” face of the planar graph.
The point is, we can apply what we know about graphs (in particular planar graphs) to convex polyhedra. Since every convex polyhedron can be represented as a planar graph, we see that Euler’s formula for planar graphs holds for all convex polyhedra as well. We also can apply the same sort of reasoning we use for graphs in other contexts to convex polyhedra. For example, we know that there is no convex polyhedron with 11 vertices all of degree 3, as this would make 33/2 edges.
Proof.
Recall that all the faces of a regular polyhedron are identical regular polygons, and that each vertex has the same degree. Consider four cases, depending on the type of regular polygon.
Case 1: Each face is a triangle. Let \(f\) be the number of faces. There are then \(3f/2\) edges. Using Euler’s formula we have \(v - 3f/2 + f = 2\) so \(v = 2 + f/2\text{.}\) Now each vertex has the same degree, say \(k\text{.}\) So the number of edges is also \(kv/2\text{.}\) Putting this together gives
\begin{equation*}
e = \frac{3f}{2} = \frac{k(2+f/2)}{2}\text{,}
\end{equation*}
which says
\begin{equation*}
k = \frac{6f}{4+f}\text{.}
\end{equation*}
Both
\(k\) and
\(f\) must be positive integers. Note that
\(\frac{6f}{4+f}\) is an increasing function for positive
\(f\text{,}\) bounded above by a horizontal asymptote at
\(k=6\text{.}\) Thus the only possible values for
\(k\) are 3, 4, and 5. Each of these are possible. To get
\(k = 3\text{,}\) we need
\(f = 4\) (this is the tetrahedron). For
\(k = 4\) we take
\(f = 8\) (the octahedron). For
\(k = 5\) take
\(f = 20\) (the icosahedron). Thus there are exactly three regular polyhedra with triangles for faces.
Case 2: Each face is a square. Now we have \(e = 4f/2 = 2f\text{.}\) Using Euler’s formula we get \(v = 2 + f\text{,}\) and counting edges using the degree \(k\) of each vertex gives us
\begin{equation*}
e = 2f = \frac{k(2+f)}{2}\text{.}
\end{equation*}
Solving for \(k\) gives
\begin{equation*}
k = \frac{4f}{2+f} = \frac{8f}{4+2f}\text{.}
\end{equation*}
This is again an increasing function, but this time the horizontal asymptote is at
\(k = 4\text{,}\) so the only possible value that
\(k\) could take is 3. This produces 6 faces, and we have a cube. There is only one regular polyhedron with square faces.
Case 3: Each face is a pentagon. We perform the same calculation as above, this time getting \(e = 5f/2\) so \(v = 2 + 3f/2\text{.}\) Then
\begin{equation*}
e = \frac{5f}{2} = \frac{k(2+3f/2)}{2}\text{,}
\end{equation*}
so
\begin{equation*}
k = \frac{10f}{4+3f}\text{.}
\end{equation*}
Now the horizontal asymptote is at
\(\frac{10}{3}\text{.}\) This is less than 4, so we can only hope of making
\(k = 3\text{.}\) We can do so by using 12 pentagons, getting the dodecahedron. This is the only regular polyhedron with pentagons as faces.
Case 4: Each face is an \(n\)-gon with \(n \ge 6\text{.}\) Following the same procedure as above, we deduce that
\begin{equation*}
k = \frac{2nf}{4+(n-2)f}\text{,}
\end{equation*}
which will be increasing to a horizontal asymptote of \(\frac{2n}{n-2}\text{.}\) When \(n = 6\text{,}\) this asymptote is at \(k = 3\text{.}\) Any larger value of \(n\) will give an even smaller asymptote. Therefore no regular polyhedra exist with faces larger than pentagons.
