Algebraic Proof of the Existence of only 5 Platonic Solids

# Algebraic Proof of the Existence of only 5 Platonic Solids

We have already seen an intuitive proof for this, but now let's apply some of our recent results from planar graphs to prove that there are only $5$ platonic solids.

Theorem 1: There are only $5$ platonic solids. |

**Proof:**We break this proof up into cases.

**CASE 1:**Let $v$, $e$, and $f$ denote the number of vertices, edges, and faces in a regular polyhedron containing triangular faces. We know that the sum of the face degrees equals twice the number of edges, that is:

\begin{align} \sum \deg f = 2e \\ 3f = 2e \\ f = \frac{2e}{3} \end{align}

- We also know that exactly $d$ edges meet at each vertex. By the Handshaking lemma, we get that:

\begin{align} \sum_{x \in V(G)} \deg (v) = 2e \\ dv = 2e \\ v = \frac{2e}{d} \end{align}

- Applying this to Euler's formula and we have:

\begin{align} v - e + f = 2 \\ \frac{2e}{d} - e + \frac{2e}{3} = 2 \end{align}

- We now divide each side by $2e$ and we obtain that:

\begin{align} \frac{1}{d} - \frac{1}{2} + \frac{1}{3} = \frac{1}{e} \\ \frac{1}{d} - \frac{1}{6} = \frac{1}{e} \end{align}

- We note that the number of edges $e$ is positive. Hence this is only true when $d < 6$. But we know that a vertex can only be formed when $d \geq 3$. Hence we get that $3 \leq d < 6$, or more precisely, $d = 3, 4, 5$. If $d = 3$, we obtain the tetrahedron. If $d = 4$, we obtain the octahedron, and if $d = 5$, we obtain the icosahedron.

**CASE 2:**Let $v$, $e$, and $f$ denote the number of vertices, edges, and faces in a regular polyhedron containing square faces. We know that the sum of the face degrees equals twice the number of edges, that is:

\begin{align} 4f = 2e \\ 2f = e \\ f = \frac{e}{2} \end{align}

- And once again, we also know that exactly $d$ edges meet at each vertex. By the Handshaking lemma, we get that:

\begin{align} \sum_{x \in V(G)} \deg (v) = 2e \\ dv = 2e \\ v = \frac{2e}{d} \end{align}

- Substituting these into Euler's formula and we have:

\begin{align} v - e + f = 2 \\ \frac{2e}{d} - e + \frac{e}{2} = 2 \end{align}

- And dividing this equation by $2e$ we obtain that:

\begin{align} \frac{1}{d} - \frac{1}{2} + \frac{1}{4} = \frac{1}{e} \\ \frac{1}{d} - \frac{1}{4} = \frac{1}{e} \end{align}

- Once again $e$ is positive, which only happens if $d < 4$. But $d \geq 3$, hence we get that $3 \leq d < 4$, and more precisely, $d = 3$. When $d = 3$, we obtain the cube.

**CASE 3:**Let $v$, $e$, and $f$ denote the number of vertices, edges, and faces in a regular polyhedron containing pentagonal faces. We know that the sum of the face degrees equals twice the number of edges, that is:

\begin{align} 5f = 2e \\ f = \frac{2e}{5} \end{align}

- We also know that exactly d edges meet at each vertex. By the Handshaking lemma, we get that:

\begin{align} \sum_{x \in V(G)} \deg (v) = 2e \\ dv = 2e \\ v = \frac{2e}{d} \end{align}

- Applying these both to Euler's formula we get:

\begin{align} v - e + f = 2 \\ \frac{2e}{d} -e + \frac{2e}{5} = 2 \end{align}

- And yet again, division by $2e$ results in:

\begin{align} \frac{1}{d} - \frac{1}{2} + \frac{1}{5} = \frac{1}{e} \\ \frac{1}{d} - \frac{3}{10} = \frac{1}{e} \end{align}

- Since $e$ is positive, this formula is only true when $d \leq 3$. But we also know that $d \geq 3$. Hence $3 \leq d \leq 3$, so $d = 3$. When $d = 3$ in this case, we get the dodecahedron. [[$ \blacksquare