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The Euler formula for a sphere, applied to the surface of a convex polyhedron, is $V-E+F=2$. In a regular polyhedron let $q\ge 3$ be the common number of edges incident at a vertex. Counting vertex-edge and face-edge incidences, pentagonal faces give $qV=2E$ and $5F=2E$. Hencewhose right-hand side can be positive only if $q<10/3$. Thus $q=3$. Substitution then yields $2=E/15$, provingThese are the counts of the dodecahedron. Here regular polyhedron is understood in the convex, ordinary-face sense used by the stated Euler formula; self-intersecting star constructions do not share these hypotheses.

For a finite triangulation of a surface of the sphere, Euler formula for a sphere iswhere $V$, $E$ and $F$ count vertices, edges and triangular faces. Since each face has three edges and each edge borders two faces, also $3F=2E$.

Let $p$ and $h$ count the pentagonal and hexagonal faces. Counting incidences of edges with vertices and faces givesThe Euler formula for a sphere also applies to this polygonal decomposition: triangulating a polygon by adding one diagonal increases both $E$ and $F$ by one, preserving $V-E+F$. Substitution givesThus there are exactly 12 pentagons, independently of the number of hexagons.