Proof of the Classification Theorem for Platonic Solids
Theorem
In three-dimensional Euclidean space $\mathbb{R}^{3}$, a polyhedron whose every face is the same regular polygon and at which the same number of faces meet at each vertex is called a regular polyhedron.
There are only five convex regular polyhedra, and these are called the Platonic solids. The Platonic solids are the regular tetrahedron, the regular cube, the regular octahedron, the regular dodecahedron, and the regular icosahedron.
Proof
Existence
If each face of a regular polyhedron is a regular $p$-gon, then the measure of one interior angle $\theta$ is given by: $$ \theta = {\frac{(p-2) \pi}{p}} $$
Let the number of faces meeting at each vertex be $q$. For the regular polyhedron to be convex, at each vertex $q$ faces whose interior angles are $\theta$ meet, and the sum of these interior angles must be less than $2\pi$; hence $(p, q)$ must satisfy the following inequality for a regular polyhedron to exist. $$ \begin{align*} q \theta <& 2 \pi \\ \implies q {\frac{(p-2) \pi}{p}} <& 2 \pi \\ \implies q {\frac{(p-2)}{p}} <& 2 \end{align*} $$ Starting from the equilateral-triangle case $p = 3$, substitute each possibility one by one to determine all cases. $$ \begin{align*} q {\frac{ 1 }{ 3 }} <& 2 \\ \implies q <& 6 \end{align*} $$ Since $q$ is the number of faces meeting at a vertex, it must satisfy $q \ge 3$. Therefore, in the case $p = 3$, $q = 3, 4, 5$ is possible. In the square case $p = 4$, since $$ \begin{align*} q {\frac{ 1 }{ 2 }} <& 2 \\ \implies q <& 4 \end{align*} $$ only $q = 3$ is possible, and in the regular pentagon case $p = 5$, since $$ \begin{align*} q {\frac{ 3 }{ 5 }} <& 2 \\ \implies q <& {\frac{10}{3}} \\ \implies q <& 3.\dot{3} \end{align*} $$ again only $q = 3$ is possible. For n-gons with n ≥ 6 ($p \ge 6$), since $$ \begin{align*} q {\frac{(p-2)}{p}} <& 2 \\ \implies 3 \le q <& {\frac{2p}{p-2}} \le 3 \end{align*} $$ there is no $q$ corresponding to $p$. Thus the only possible cases that can form a regular polyhedron are the following five:
- $(p, q) = (3, 3)$
- $(p, q) = (3, 4)$
- $(p, q) = (3, 5)$
- $(p, q) = (4, 3)$
- $(p, q) = (5, 3)$
Classification
Let the number of vertices of the polyhedron be $n$, the number of edges be $m$, and the number of faces be $f$. By the first condition of a regular polyhedron, if all faces are regular $p$-gons then each face has $p$ edges, so $$ pf = 2m $$ must hold. By the second condition of a regular polyhedron, if $q$ faces meet at each vertex, then the following holds. $$ qn = 2m $$
Euler’s polyhedron theorem: For a connected planar graph $G$, if $n:=|V(G)|$, $m:=|E(G)|$, $f$ are the numbers of vertices, edges, and faces respectively, then $$ n-m+f=2 $$
Substituting $f$ and $n$ into Euler’s polyhedron theorem yields: $$ \begin{align*} {\frac{ 2 }{ q }} m - m + {\frac{ 2 }{ p }} m =& 2 \\ \implies \left( {\frac{ 2 }{ q }} - 1 + {\frac{ 2 }{ p }} \right) m =& 2 \\ \implies \left( {\frac{ 2p + 2q - pq }{ pq }} \right) m =& 2 \\ \implies m =& {\frac{ 2pq }{ 2p + 2q - pq }} \end{align*} $$ Now substitute all instances of $(p , q)$ to compute the number of edges $m$ forming each regular polyhedron, and thereby identify each regular $f$-hedron. $$ \begin{align*} (3,3) \implies & m = 6 \implies 3 f = 12 & \left( 정사면체 \right) \\ (3,4) \implies & m = 12 \implies 3 f = 24 & \left( 정팔면체 \right) \\ (3,5) \implies & m = 30 \implies 3 f = 60 & \left( 정이십면체 \right) \\ (4,3) \implies & m = 12 \implies 4 f = 24 & \left( 정육면체 \right) \\ (5,3) \implies & m = 30 \implies 5 f = 60 & \left( 정십이면체 \right) \end{align*} $$
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