Euler's Polyhedral Formula, math that surprises you lesson

A cube has 8 corners, 12 edges and 6 faces. A tetrahedron has 4 corners, 6 edges and 4 faces, and the classic soccer ball pattern has 60, 90 and 32. The lists look unrelated, yet in every case the corners minus the edges plus the faces come to exactly 2. This page explains why that happens for every convex polyhedron, by flattening its network of edges into the plane and pruning it one edge at a time, and shows what changes on a surface with a tunnel through it.

Key terms

Euler’s polyhedral formula

Definition For every convex polyhedron with $V$ vertices, $E$ edges and $F$ faces,

V - E + F = 2 .

Each vertex, edge and face is counted once, including those at the back. More generally the formula holds for a polyhedron whose surface can be bent and stretched into a sphere without tearing, provided every face is a polygon without holes. Both conditions matter. On the surface of a solid with a tunnel through it, with every face a polygon without holes, the count is $0$ instead, and a face with a hole can throw the count off on any surface. The proof has three steps. First, remove one face and flatten the rest of the surface into the plane without breaking or crossing any edges, and count the region outside the drawing as the removed face, so that $V$ , $E$ and $F$ do not change. Second, as long as two or more faces are left, delete an edge that has different faces on its two sides. This lowers $E$ and $F$ by $1$ each, keeps $V - E + F$ the same and keeps the network connected. Third, when one face is left the network is a tree, which has $V - 1$ edges, so

V - (V - 1) + 1 = 2 .

The formula is named after Leonhard Euler, who stated it in 1750.

In plain words Count the corners, subtract the edges and add the faces. For any convex solid with flat faces the result is $2$ , whatever its shape and however many faces it has.

Example The tetrahedron gives $4 - 6 + 4 = 2$ , the cube $8 - 12 + 6 = 2$ and the octahedron, with 6 vertices, 12 edges and 8 triangular faces, $6 - 12 + 8 = 2$ . The soccer ball pattern, called the truncated icosahedron, has 12 pentagons and 20 hexagons, so $F = 32$ . Every edge borders exactly two faces and every vertex lies on exactly three, so counting the corners of all the faces counts each edge twice and each vertex three times,

\begin{aligned} E &= \frac{12 \cdot 5 + 20 \cdot 6}{2} = 90, \\[0.4em] V &= \frac{12 \cdot 5 + 20 \cdot 6}{3} = 60, \end{aligned}

and $60 - 90 + 32 = 2$ . Careless counting breaks the pattern. Counting the four edges of each face of a cube separately gives $6 \cdot 4 = 24$ edges, every edge twice, and $8 - 24 + 6 = -10$ .

Euler characteristic

Definition Draw a network of vertices and edges on a closed surface, such as a sphere or the surface of a doughnut, so that every edge is a curve that ends at vertices and meets other edges only at vertices, and so that every piece the network cuts the surface into, called a face, is a disk, a region that can be flattened into a polygon without holes. The Euler characteristic of the surface is

\chi = V - E + F .

Its value depends only on the surface and not on the network chosen, as long as the network meets these conditions. Every such network on a sphere, and so every convex polyhedron, gives $\chi = 2$ . Every such network on the surface of a doughnut, called a torus, gives $\chi = 0$ . More generally, the surface of a solid with $g$ tunnels through it has $\chi = 2 - 2g$ .

In plain words The number $V - E + F$ is a fingerprint of the surface, not of the way it is cut into pieces. A sphere has fingerprint $2$ and a doughnut has $0$ .

Example Draw a 3 by 3 grid of squares on a flexible square sheet. Glue the left edge of the sheet to the right edge to make a tube, then bend the tube around and glue its two end circles together, which gives the surface of a doughnut. The $16$ grid points become $9$ vertices, since points on glued edges come together, in pairs along the sides of the sheet and all four corners into one, and the $24$ grid segments become $18$ edges, since the $3$ segments on the left edge join those on the right and the $3$ on the top join those on the bottom. The $9$ squares stay $9$ faces, each a disk, so

\chi = 9 - 18 + 9 = 0 .

Changing the network does not change this. Cutting one square in two along a diagonal adds one edge and one face, and putting a new vertex in the middle of an edge adds one vertex and one edge, so in both cases $\chi$ stays $0$ .

Convex polyhedron

Definition A polyhedron is a solid whose surface is made of finitely many flat polygons, its faces. The sides of the faces are the edges, each shared by exactly two faces, and the corners of the faces are the vertices. A polyhedron is convex when, for any two of its points, the straight segment joining them lies entirely inside it. Every face of a convex polyhedron is then a convex polygon, and the whole solid lies on one side of the plane of each face. Seen from any point inside, every direction meets the surface exactly once, so the surface can be pushed out onto a sphere around that point without tearing or overlapping, which is why its Euler characteristic is $2$ . Convexity is enough for Euler’s formula but not needed. What matters is that the surface can be bent into a sphere and that every face is a disk.

In plain words A convex polyhedron has no dents and no tunnels. Set it down on any of its faces and the whole solid stays above the table.

Example The cube, the tetrahedron and the truncated icosahedron are convex. Push a dent into a cube by replacing its top face with four triangles that meet at a point inside the cube, below the center of the top face. The solid is no longer convex, but it can still be bent into a sphere, and it has $9$ vertices, $16$ edges and $9$ faces, with $9 - 16 + 9 = 2$ . A square picture frame, a square block with a square hole cut through its middle, is not sphere-like at all. It has $8$ vertices on top and $8$ on the bottom, $16$ faces made of four trapezoids on top, four on the bottom, four outer walls and four inner walls, and $32$ edges, twelve on top, twelve on the bottom and eight running up and down. So

16 - 32 + 16 = 0 ,

the value of a torus, because the hole through the frame is a tunnel.

Planar graph

Definition A graph is a collection of vertices together with edges, each edge joining two vertices. It is connected when any vertex can be reached from any other along edges. A graph is planar when it can be drawn in the plane with no two edges crossing, so that edges meet only at shared endpoints. Such a drawing cuts the plane into regions called faces, and exactly one of them, the outer face, is unbounded. Euler’s formula for planar graphs says that every crossing-free drawing of a connected planar graph has $V - E + F = 2$ , with the outer face counted. Every convex polyhedron gives one. Look at it from a point just outside the middle of one face, so close that no other face can be seen from there, and project every vertex and edge along the lines of sight onto the plane of that face. The result, called a Schlegel diagram, has no crossings. The chosen face is the outer boundary of the drawing and corresponds to the outer face, while every other face of the polyhedron becomes a bounded region, so the drawing has the same $V$ , $E$ and $F$ as the polyhedron.

In plain words A planar graph is a network of dots and lines that can be drawn on paper without crossings. The region outside the drawing counts as a face too.

Example The Schlegel diagram of a cube is a small square inside a large square, with four spokes joining their corners. It has $8$ vertices and $12$ edges. Its faces are the small square, the four four-sided regions between the squares and the outer face, six in all, and $8 - 12 + 6 = 2$ . Leaving out the outer face would give $8 - 12 + 5 = 1$ . Not every graph is planar. Five points, each joined to the other four, cannot be drawn in the plane without a crossing.

Spanning tree

Definition A cycle in a graph is a closed route along edges that returns to its starting vertex without using any edge or any other vertex twice. A tree is a connected graph with no cycles, and a tree with $V$ vertices has exactly $V - 1$ edges. To see why, start from one vertex and reach the others one at a time, each time using an edge from a vertex already reached to a new one, which exists because the tree is connected. Each step adds one vertex and one edge, so when all $V$ vertices are reached there are $V - 1$ edges. No edge is left out, since an edge between two vertices already reached would close a cycle with the route already built between them. A spanning tree of a connected graph is a tree made of all its vertices and some of its edges. Every connected graph with finitely many edges has one. As long as some edge lies on a cycle, delete it. The graph stays connected, because the rest of the cycle still joins the two ends of the deleted edge, and when no cycle is left, what remains is a spanning tree. In a crossing-free drawing, an edge lies on a cycle exactly when it has different faces on its two sides, so each deletion merges two faces into one. An edge with the same face on both sides must not be deleted, because that would cut the network in two.

In plain words A spanning tree is a skeleton of the network that still reaches every vertex but has no loops. Cutting the loops open one edge at a time merges the faces until only the outside is left.

Example Name the corners of the large square in the cube’s Schlegel diagram $A$ , $B$ , $C$ and $D$ in order, and the corners of the small square $a$ , $b$ , $c$ and $d$ , so that the spokes are $Aa$ , $Bb$ , $Cc$ and $Dd$ . Delete the edges $ab$ , $bc$ , $cd$ , $AB$ and $DA$ one at a time. Each has different faces on its two sides when it is deleted, and the ledger keeps the value $2$ at every step.

Edges deleted so far	$V$	$E$	$F$	$V - E + F$
none	8	12	6	2
$ab$	8	11	5	2
$ab$ , $bc$	8	10	4	2
$ab$ , $bc$ , $cd$	8	9	3	2
those and $AB$	8	8	2	2
those and $DA$	8	7	1	2

The $7$ edges left, $BC$ , $CD$ , $da$ , $Aa$ , $Bb$ , $Cc$ and $Dd$ , form a spanning tree, with $7 = 8 - 1$ . Other choices of deleted edges give other trees, but always ones with $7$ edges and one face.

Check yourself

Count the vertices, edges and faces of a pyramid with a hexagonal base and of a prism with hexagonal ends, and check Euler’s formula. Then do the same for a pyramid and a prism over a polygon with $n$ sides, where $n \ge 3$ .

Show the answer

The hexagonal pyramid has $6$ base corners and the apex, so $V = 7$ . It has $6$ base edges and $6$ slanted edges, so $E = 12$ , and $6$ triangles plus the base, so $F = 7$ . Then $7 - 12 + 7 = 2$ . The hexagonal prism has $V = 12$ , $E = 6 + 6 + 6 = 18$ and $F = 6 + 2 = 8$ , and $12 - 18 + 8 = 2$ . Over an $n$ -sided polygon the pyramid has $n + 1$ vertices, $2n$ edges and $n + 1$ faces, and the prism has $2n$ vertices, $3n$ edges and $n + 2$ faces, so

\begin{aligned} (n + 1) - 2n + (n + 1) &= 2, \\[0.4em] 2n - 3n + (n + 2) &= 2. \end{aligned}

Glue a small cube onto the middle of the top face of a large cube, so that the small cube’s bottom face lies inside the large cube’s top face without touching its edges. Count the vertices, edges and faces of the surface of this solid and compute $V - E + F$ . Which condition of Euler’s formula fails, and how can you fix it?

Show the answer

Each cube contributes $8$ vertices and $12$ edges, so $V = 16$ and $E = 24$ . The faces are the $5$ untouched faces of the large cube, the $5$ visible faces of the small cube and what is left of the large cube’s top face, so $F = 11$ , and

16 - 24 + 11 = 3 .

The surface can still be bent into a sphere, so the trouble is elsewhere. The leftover top face is a square ring with a hole in it, not a disk, and the network of edges falls into two pieces that do not touch. Draw four new edges from the corners of the small cube’s base to the nearest corners of the large cube’s top face. The ring splits into four four-sided faces, each a disk, the network becomes connected, and the count is $V = 16$ , $E = 28$ , $F = 14$ , with $16 - 28 + 14 = 2$ .

Extend the pattern. A convex polyhedron has only pentagons and hexagons as faces, with exactly three faces meeting at every vertex. Use Euler’s formula to show that it has exactly 12 pentagons, however many hexagons it has.

Show the answer

Let $p$ be the number of pentagons and $h$ the number of hexagons, so $F = p + h$ , and let $S = 5p + 6h$ be the total number of corners of all the faces. Every edge borders two faces, so $S = 2E$ , and every vertex lies on three faces, so $S = 3V$ . Multiplying Euler’s formula by $6$ gives

\begin{aligned} 12 &= 6V - 6E + 6F \\[0.4em] &= 2S - 3S + 6(p + h) \\[0.4em] &= -5p - 6h + 6p + 6h \\[0.4em] &= p . \end{aligned}

The hexagons cancel, so the formula says nothing about $h$ . The regular dodecahedron has $h = 0$ , with $V = 20$ , $E = 30$ and $F = 12$ , and the truncated icosahedron has $h = 20$ , with $V = 60$ , $E = 90$ and $F = 32$ . Both have exactly 12 pentagons.