The limitation to five such three-dimensional solids is easily demonstrated using elementary geometry:
[*]1 Each vertex of the solid must coincide with one vertex each of at least three faces.
[*]2 At each vertex of the solid, the total, among the adjacent faces, of the angles between their respective adjacent sides must be less than 360°.
[*]3 The angles at all vertices of all faces of a Platonic solid are identical, so each vertex of each face must contribute less than 360°/3=120°.
[*]4 Regular polygons of six or more sides have only angles of 120° or more, so the common face must be the triangle, square, or pentagon. And for:
Triangular faces: each vertex of a regular triangle is 60°, so a shape may have 3, 4, or 5 triangles meeting at a vertex; these are the tetrahedron, octahedron, and icosahedron respectively.
Square faces: each vertex of a square is 90°, so there is only one arrangement possible with three faces at a vertex, the cube.
Pentagonal faces: each vertex is 108°; again, only one arrangement, of three faces at a vertex is possible, the dodecahedron.
In fact, the problem of classifying the Platonic polyhedra is not truly geometric, but merely topological. That is, the same list of five polyhedra can be deduced using only combinatorial information about a regular polyhedron: the number v of vertices, the number e of edges, the number f of faces, the number n of edges bounding each face, and the number d of edges meeting each vertex. These numbers are related as follows:
[*]all are positive, and n and d are at least 3, as above;
[*]since each edge is adjacent to two faces and meets two vertices, n f = 2 e = v d;
[*]v - e + f = 2. (This is an elementary fact of algebraic topology, that the Euler characteristic of the sphere is 2.)
Using these facts, it is not difficult to show algebraically that only the five classical regular polyhedra are possible.
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