In 1750, in a letter to mathematician Christian Goldbach, Euler noted a relationship between the number of vertices ($V$), edges ($A$) and faces ($F$) of a convex polyhedron:

$$V - A + F = 2.$$

Rotate the polyhedra with your finger or mouse and check the calculation:

V, A and F are counted by the program from the drawing itself. In all cases it gives 2.

Check out the five regular polyhedra:

Polyhedron $V$ $A$ $F$ $V - A + F$
Tetrahedron 4 6 4 2
Cube 8 12 6 2
Octahedron 6 12 8 2
Dodecahedron 20 30 12 2
Icosahedron 12 30 20 2

The formula allows you to deduce surprising things. Think of a ball stitched together with $p$ pentagons and $h$ hexagons, in which each vertex is the meeting point of exactly three faces, as in a soccer ball classic. How many pentagons does it need to have?

  • Faces: $F = p + h$.
  • Edges: each pentagon has 5 sides, each hexagon has 6, and each edge is shared by two faces. Then $A = \frac{5p + 6h}{2}$.
  • Vertices: Each vertex is shared by three faces. Then $V = \frac{5p + 6h}{3}$.

Replacing in $V - A + F = 2$ and multiplying everything by 6:

$$2(5p + 6h) - 3(5p + 6h) + 6(p + h) = 12$$ $$-(5p + 6h) + 6p + 6h = 12$$ $$p = 12.$$

The number of hexagons disappeared from the count! Every ball of this type has exactly 12 pentagons, no It doesn’t matter how many hexagons it has. The classic soccer ball has 12 pentagons and 20 hexagons ($V = 60$, $A = 90$, $F = 32$, and $60 - 90 + 32 = 2$). The same account applies to molecules of fullerene, such as $C_{60}$.

The classic football (truncated icosahedron): 12 pentagons, 20 hexagons and V − A + F = 2.

All topics by Leonhard Euler · ← The Königsberg bridges and the birth of graphs · Infinite prime numbers from a sum →

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