# classification of Platonic solids

###### Proof.

Each vertex of a Platonic solid is incident   with at least three faces. The interior angles  incident with that vertex must sum to less than $2\pi$, for otherwise the solid would be flat at that vertex. Since all faces of the solid have the same number of sides, this implies bounds on the number of faces which could meet at a vertex.

The interior angle of an equilateral triangle  has measure $\frac{\pi}{3}$, so a Platonic solid could only have three, four, or five triangles meeting at each vertex. By similar  reasoning, a Platonic solid could only have three squares or three pentagons  meeting at each vertex. But the interior angle of a regular hexagon has measure $\frac{2\pi}{3}$. To avoid flatness a solid with hexagons  as faces would thus have to have only two faces meeting at each vertex, which is impossible. For polygons   with more sides it only gets worse.

Since a Platonic solid is uniquely determined by the number and kind of faces meeting at each vertex, there are at most five Platonic solids, with the numbers and kinds of faces listed above. But these correspond to the five known Platonic solids. Hence there are exactly five Platonic solids. ∎

Title classification of Platonic solids ClassificationOfPlatonicSolids 2013-03-22 15:53:07 2013-03-22 15:53:07 mps (409) mps (409) 5 mps (409) Result msc 51-00