Why Do Solids Always Obey One Rule?

Here is what you should have found, with a few solids added. In Grade 7 the corners are called vertices (one vertex); we use both words.

Solid	Corners	Edges	Faces	Corners − edges + faces
Tetrahedron (4 triangles)	4	6	4	2
Cube	8	12	6	2
Triangular prism	6	9	5	2
Square pyramid	5	8	5	2
Pentagonal prism	10	15	7	2
Hexagonal pyramid	7	12	7	2
Octahedron (8 triangles)	6	12	8	2
Dodecahedron (12 pentagons)	20	30	12	2
Icosahedron (20 triangles)	12	30	20	2

Look along the rows: the corners and the faces together are always 2 more than the edges. The cube: 8 + 6 = 14 = 12 + 2.

Prisms and pyramids

A prism has two copies of the same polygon, one above the other, joined by rectangles. If the polygon has n sides, there are n corners at the bottom and n at the top; n edges round the bottom, n round the top and n going up; and n rectangles plus the top and bottom faces.

A pyramid has a polygon of n sides at its base and one more corner, the apex, joined to every corner of the base.

	Corners	Edges	Faces
Prism, n-sided	2n	3n	n + 2
Pyramid, n-sided	n + 1	2n	n + 1

For the prism, 2n − 3n + (n + 2): the n’s cancel (2 − 3 + 1 = 0) and 2 is left. For the pyramid, (n + 1) − 2n + (n + 1) = 2 the same way. However many sides the base has, the answer is 2.

Counting without counting

The dodecahedron has 12 faces, each a pentagon. Counting its 30 edges one by one on a turning solid is slow. There is a quicker way.

Count the sides of all the faces: 12 × 5 = 60. Every edge is the side of exactly two faces, so each edge was counted twice: there are 60 ÷ 2 = 30 edges. In the same way, at every corner of a dodecahedron three faces meet, so the 60 corners of the faces, counted face by face, are each corner of the solid three times over: 60 ÷ 3 = 20 corners.

The rule is called Euler’s formula for solids (or Euler’s polyhedron formula). A polyhedron is a solid whose faces are all flat polygons, meeting along straight edges. For any polyhedron without holes, with V vertices (corners), E edges and F faces,

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

Nine solids, or even a hundred, aren’t a proof: a pattern that works many times can still fail (as you can see in Is a Pattern That Works Ten Times Always True?). Here is an argument that works for all of them at once. It has two steps.

Step 1: make the solid flat

Imagine the solid is made of thin rubber. Cut away one face, say the top of the cube. What is left is an open box. Now stretch the open box out flat on a table, pulling the rim wide.

Stretching doesn’t change how many corners, edges or faces there are: each corner becomes a dot, each edge a line (bent or stretched, never cut), each face a region of the paper. The face we cut away is not lost either: we count the paper outside the drawing as one region, standing for it. So the flat drawing in Figure 23.1 has 8 dots, 12 lines and 6 regions, the same numbers as the cube.

Why do no two lines cross? For a solid with no dents, like every solid in the table, picture it made of wire, the wires of one face taken away, and a small lamp just outside that open face, close to its middle. Each ray from the lamp goes in through the open face and comes out through the rest of the surface at exactly one point, because a solid with no dents can’t be gone in and out of twice. So the wires’ shadow on the floor is the flat drawing, with no crossings: where two shadows met, one ray would come out through two points. (A solid with dents but no holes can be stretched flat too, but the proof is harder.)

So now we only need to show that a flat drawing like this always has dots − lines + regions = 2.

Step 2: draw it one line at a time

Start with a single dot on a sheet of paper. There is 1 dot, no lines, and 1 region: the whole sheet. 1 − 0 + 1 = 2.

Now draw the rest of the drawing one line at a time, each new line starting from a dot already drawn, so the drawing stays in one piece. A new line can do only one of two things:

• It reaches a new dot. Then the dots go up by 1 and the lines go up by 1. No region is cut. The +1 and the −1 cancel: dots − lines + regions doesn’t change.
• It joins two dots already drawn. Then the lines go up by 1. The drawing is in one piece, so the two dots were already joined by a path of lines; the new line closes that path into a loop, and a loop drawn on paper always cuts the region it runs through in two, an inside and an outside: the regions go up by 1. The −1 and the +1 cancel again.

The total starts at 2 and never changes, so when the whole drawing is finished it is still 2. Try it.

That proves it: every polyhedron without holes has V − E + F = 2 (for a solid with dents, take step 1 on trust). Notice what step 2 never used: lengths, angles, flat faces, straight edges. Only how the corners, edges and faces are joined. That is why the rule is so stubborn, and also, as you’ll see, where it can break.

Look at an old-style football: black pentagons and white hexagons, sewn together so that three panels meet at every corner. Count: 12 pentagons and 20 hexagons, 32 panels.

The edges, counting without counting: the panels have 12 × 5 + 20 × 6 = 180 sides between them, and each edge joins two panels, so there are 180 ÷ 2 = 90 edges. Three panels meet at every corner, so there are 180 ÷ 3 = 60 corners. And 60 − 90 + 32 = 2. The ball is a polyhedron blown round.

Now a harder question. Could a ball like this, three panels at every corner, use more hexagons and still close up? Say 30 hexagons. How many pentagons would it need?

Suppose it has p pentagons and h hexagons. The panels have 5p + 6h sides between them. Then:

• edges = (5p + 6h) ÷ 2, because each edge joins two panels;
• corners = (5p + 6h) ÷ 3, because three panels meet at each corner;
• faces = p + h.

Each hexagon brings 6 sides: that is 6 ÷ 3 = 2 corners and 6 ÷ 2 = 3 edges, and 1 face. In corners − edges + faces, one hexagon adds 2 − 3 + 1 = 0. Hexagons make no difference at all to the sum!

Each pentagon brings 5 sides: 5 ÷ 3 corners, 5 ÷ 2 edges, and 1 face, adding 5/3 − 5/2 + 1 = 10/6 − 15/6 + 6/6 = 1/6. So the whole sum is p ÷ 6. Euler’s formula says it must be 2, so p ÷ 6 = 2:

A ball like this always has exactly 12 pentagons, however many hexagons it has. With 0 hexagons it is the dodecahedron. With 20, the football. With 30 hexagons, still 12 pentagons. (The count allows any number of hexagons, but one combination can’t actually be built: 12 pentagons with exactly one hexagon.)

The same reasoning works for balls made of triangles. A geodesic sphere, a ball built of triangles (a geodesic dome, the kind of round building made of steel triangles, is part of one), has 6 triangles meeting at most of its corners and 5 at exactly 12 of them. Six triangles round a corner lie almost flat; the 12 corners with 5 are where the ball bends round. Euler’s formula says there must be exactly 12 of them, however many triangles the sphere has.

Every proof rests on something. Ours rested on being able to cut away one face and stretch the rest flat in one piece. What if a solid won’t do that?

Think of a wooden picture frame lying on a table: a square bar going round a square hole. Its top is a square ring, and a ring isn’t a polygon, so let us cut the top into four trapezoids, and the bottom too, to make every face a proper polygon.

The frame gives 0. Its hole is the trouble: cut away any face and the frame still has a loop of wood going round the hole, so it can’t be stretched flat into a drawing on paper without tearing. A solid with two holes, like a block with two square tunnels through it, gives −2, and one with three holes −4: every hole takes away 2. This number, corners − edges + faces, became one of the first ways mathematicians had to count holes without looking. It is now called the Euler characteristic of a shape.

The cube on a cube gives 3, and here the trouble is subtler. The top of the big cube is a square with a square hole in the middle, where the small cube stands. That face isn’t a polygon: it is a ring, like the frame’s top before we cut it. Draw two lines across the ring, each from a corner of the small cube to the nearest corner of the big one, and the ring splits into two proper polygons: two more edges and one more face, and 16 − 26 + 12 = 2. The rule is safe again. (Lhuilier mended such faces the same way, with lines drawn across them.)

• Can You Cross Every Bridge Just Once? Dots and lines again, and the other famous puzzle Euler solved by caring only about what is joined to what.
• Can Three Houses Reach Three Wells Without Paths Crossing? Euler’s formula for flat drawings proves that one famous drawing is impossible.
• Why Are There Only Five Perfect Dice? (Math, Going Further 9–12) Solids whose faces are all the same regular polygon, and why Euler’s formula leaves room for only five.
• Can Four Colors Paint Any Map of Nepal? (Computing, Going Further 6–8) A map is a flat drawing of regions, like a flattened solid.