Класифікація Платонових тіл за Формулою Ейлера

We know that, in a Platonic solid, all of the faces must be regular polygons, all of the faces must be congruent, and the same number of faces must meet at each vertex. Using these properties, as well as Euler's formula and the notion of angular defect, we can classify the Platonic solids.

Suppose that several faces meet at a vertex. The angular defect of that vertex equals $360^{\circ}$ minus the sum of the angles at that vertex. A theorem states that, in any polyhedron, the angular defect must be strictly greater than $0$.

Define the following variables for convenience:

$F$ is the total number of faces;

$E$ is the total number of edges;

$V$ is the total number of vertices;

$S$ is the number of edges (and thus also vertices) surrounding each face (for example, if the faces were triangles, then $S=3$);

$N$ is the number of faces (and thus also edges) meeting at each vertex. It is obvious that $N\ge 3$.

Each vertex will have $N$ edges emanating out of it. However, each edge will be counted twice, since it has $2$ endpoints. Therefore, $E=NV/2$.

Each vertex will have $N$ faces surrounding it. However, each face will be counted $S$ times. Therefore, $F=NV/S$.

Euler's formula tells us that: $$V-E+F=2$$ $$V-\frac{N}{2}V+\frac{N}{S}V=2$$ $$V=\frac{4S}{2S+2N-NS}$$ Then, we can solve for $E$ and $F$: $$E=\frac{2NS}{2S+2N-NS}$$ $$V=\frac{4N}{2S+2N-NS}$$

First suppose that the faces of our Platonic solids are triangles, i.e. $S=3$. Since each triangle has an interior angle measure of $60^{\circ}$, we can have $N=3, 4, 5$. There are $3$ cases: $$S=3, N=3\rightarrow V=4, E=6, F=4$$ $$S=3, N=4\rightarrow V=6, E=12, F=8$$ $$S=3, N=5\rightarrow V=12, E=30, F=20$$

Now, suppose that the faces of our Platonic solids are squares, i.e. $S=4$. Since each square has an interior angle measure of $90^{\circ}$, we can only have $N=3$. There is only $1$ case: $$S=4, N=3\rightarrow V=8, E=12, F=6$$

Then, suppose that the faces of our Platonic solids are pentagons, i.e. $S=5$. Since each pentagon has an interior angle measure of $108^{\circ}$, we can only have $N=3$. There is only $1$ case: $$S=5, N=3\rightarrow V=20, E=30, F=12$$

Finally, suppose that the faces of our Platonic solids have at least $6$ sides, i.e. $S\ge 6$. Since each $S$-gon has an interior angle measure of at least $120^{\circ}$, we cannot fit at least $3$ of them into less than $360^{\circ}$. Therefore, this case is not possible.

We thus have only $5$ possible cases: $$S=3, N=3\rightarrow V=4, E=6, F=4$$ $$S=4, N=3\rightarrow V=8, E=12, F=6$$ $$S=3, N=4\rightarrow V=6, E=12, F=8$$ $$S=5, N=3\rightarrow V=20, E=30, F=12$$ $$S=3, N=5\rightarrow V=12, E=30, F=20$$ These Platonic solids are called the regular tetrahedron, hexahedron (cube), octahedron, dodecahedron, and icosahedron, respectively. Since "-hedron" refers to faces, these names come directly from the number of faces in each Platonic solid. The existence and uniqueness of each of these cases are not proved here.