11. Solid geometry¶

Flat geometry has infinitely many regular polygons: a regular triangle, square, pentagon, and so on for every number of sides. Step up one dimension and ask for solids whose faces are all the same regular polygon, the same number meeting at every corner, and only five exist. Why five? And why does every such solid, indeed every convex solid with flat faces, satisfy the same strange equation, corners minus edges plus faces equals two? This chapter answers both, then measures solids: the volumes of prisms, pyramids, cones and spheres, found by slicing as Cavalieri and Archimedes did, and the surface of a sphere. It ends with Hilbert's third problem, where three dimensions turn out to be genuinely harder than two: a cube and a regular tetrahedron of the same volume can never be cut into each other.

In this chapter

You will learn:

the words for polyhedra (faces, edges, vertices, convex, prisms, pyramids, nets)
why there are exactly five Platonic solids, with two proofs, and how to build each one
Euler's formula \(V - E + F = 2\), with a proof, and Descartes' total angle defect of \(720°\)
Cavalieri's principle, and from it the volumes of the prism, pyramid, cone and sphere
Archimedes' discovery that a sphere's surface is \(4\pi r^2\), and his "hat-box" theorem about zones
Archimedean solids and duality, and why a football has exactly 12 pentagons
Hilbert's third problem and the Dehn invariant: why a tetrahedron cannot be cut up into a cube

You need: Chapter 3 (angle sums of polygons), Chapter 4 (area, and the Bolyai-Gerwien theorem in 4.7), Chapter 6 (the area of a circle) and Chapter 9 (coordinates in space). Gentler versions: Faces, edges and corners, Nets of a cube, Volume of pyramids and cones and Volume of a sphere.

Levels: 11.1 is beginner to intermediate; 11.2 to 11.5 are intermediate (with complete proofs); 11.6 is intermediate to advanced; 11.7 is advanced to expert.

11.1 Polyhedra¶

A polyhedron (Greek: "many seats", plural polyhedra) is a solid bounded by flat polygons. The polygons are its faces; two faces meet along an edge; edges meet at corners called vertices (singular vertex). A cube has 6 faces, 12 edges and 8 vertices. We write \(F\), \(E\), \(V\) for these counts.

A polyhedron is convex if the segment between any two of its points stays inside it: no dents, no tunnels. Every polyhedron in this chapter is convex unless we say otherwise. Convexity gives a useful test: a polyhedron is convex exactly when it lies entirely on one side of the plane of each face.

Families¶

A prism has two congruent parallel polygons (the bases) joined by parallelograms. If the side faces are rectangles it is a right prism. An \(n\)-gonal prism has \(V = 2n\), \(E = 3n\), \(F = n + 2\). A cube is a square prism; a box is a rectangular prism.
A pyramid has one polygon base and triangles joining its edges to one point, the apex. An \(n\)-gonal pyramid has \(V = n + 1\), \(E = 2n\), \(F = n + 1\). A tetrahedron is a triangular pyramid.
An antiprism has two parallel copies of an \(n\)-gon, one turned by half a step, joined by a zigzag band of \(2n\) triangles: \(V = 2n\), \(E = 4n\), \(F = 2n + 2\).

The limiting cases with curved surfaces are the cylinder (a prism with a circular base) and the cone (a pyramid with a circular base). They are not polyhedra, but the same volume formulas will hold.

Nets¶

Cut a polyhedron along enough edges and you can unfold it flat into one piece, a net. Going the other way, a net folds up into the solid. A cube has 11 different nets (the Topics lesson Nets of a cube lists them). Nets show something important: the faces around a vertex, laid flat, do not fill \(360°\). There is always a gap, and the gap is what lets the faces fold up into a corner. That gap will be the key to the next two sections.

Definition. The angle defect at a vertex is \(360°\) minus the sum of the face angles there. For a convex polyhedron each defect is positive. At a corner of a cube three right angles meet, so the defect is \(360° - 270° = 90°\).

11.2 The five Platonic solids¶

A regular polyhedron, or Platonic solid, is a convex polyhedron whose faces are all congruent regular polygons, with the same number of faces meeting at every vertex. Five are known since antiquity:

solid	face	faces at a vertex	\(V\)	\(E\)	\(F\)
tetrahedron	triangle	3	4	6	4
cube	square	3	8	12	6
octahedron	triangle	4	6	12	8
dodecahedron	pentagon	3	20	30	12
icosahedron	triangle	5	12	30	20

The Schläfli symbol \(\{p, q\}\) records the type: regular \(p\)-gons, \(q\) at each vertex. The cube is \(\{4, 3\}\), the icosahedron \(\{3, 5\}\).

Why there are only five¶

Theorem. There are at most five Platonic solids: \(\{3,3\}\), \(\{4,3\}\), \(\{3,4\}\), \(\{5,3\}\), \(\{3,5\}\).

Proof. At a vertex, at least 3 faces must meet (two polygons meeting along an edge make no corner). The face angles there must add up to less than \(360°\), because a convex corner has positive defect: if they made exactly \(360°\) the faces would lie flat, and more than \(360°\) cannot even be fitted round a point of a convex solid.

The angle of a regular \(p\)-gon is \(\frac{(p - 2)\,180°}{p}\) (Chapter 3): \(60°\) for triangles, \(90°\) for squares, \(108°\) for pentagons, \(120°\) for hexagons. So we need

\[ q \cdot \frac{(p - 2)\,180°}{p} < 360°, \qquad\text{that is}\qquad q(p - 2) < 2p. \]

Rearranging, \(pq - 2p - 2q < 0\), and adding 4 to both sides,

\[ (p - 2)(q - 2) < 4. \]

With \(p, q \ge 3\) both factors are at least 1, so the possibilities are \(1 \cdot 1\), \(1 \cdot 2\), \(2 \cdot 1\), \(1 \cdot 3\), \(3 \cdot 1\): \((p, q) = (3,3), (3,4), (4,3), (3,5), (5,3)\). Hexagons are already too big (\(3 \times 120° = 360°\), flat), and so is everything with more sides. \(\square\)

This proves there are at most five types. To finish, each must actually exist, which needs a construction. (One can also prove that each type is unique once the edge length is fixed, but we will not need that.)

Building them¶

Coordinates in space (Chapter 9) make each one concrete.

Cube: the eight points \((\pm1, \pm1, \pm1)\).
Tetrahedron: four alternate corners of that cube, \((1,1,1)\), \((1,-1,-1)\), \((-1,1,-1)\), \((-1,-1,1)\). Each pair is joined by a face diagonal of the cube, of length \(2\sqrt2\), so all six edges are equal and all four faces are equilateral.
Octahedron: the six points \((\pm1, 0, 0)\), \((0, \pm1, 0)\), \((0, 0, \pm1)\), the centres of the faces of a cube of side 2. Each point is \(\sqrt2\) from the four that are not opposite to it.
Icosahedron: the twelve points \((0, \pm1, \pm\varphi)\), \((\pm1, \pm\varphi, 0)\), \((\pm\varphi, 0, \pm1)\), where \(\varphi = \frac{1 + \sqrt5}{2}\) is the golden ratio (Chapter 5). The shortest distance between two of these points is 2: for example from \((0, 1, \varphi)\) to \((0, -1, \varphi)\). And from \((0, 1, \varphi)\) to \((1, \varphi, 0)\) the squared distance is \(1 + (\varphi - 1)^2 + \varphi^2 = 1 + \varphi^2 - 2\varphi + 1 + \varphi^2\), which equals 4 because \(\varphi^2 = \varphi + 1\). Each point has five neighbours at distance 2, and they form the 20 equilateral triangles.
Dodecahedron: the cube's eight points \((\pm1, \pm1, \pm1)\) together with the twelve points \((0, \pm\tfrac1\varphi, \pm\varphi)\), \((\pm\tfrac1\varphi, \pm\varphi, 0)\), \((\pm\varphi, 0, \pm\tfrac1\varphi)\). The faces are 12 regular pentagons, and a cube sits inside it, each cube edge a diagonal of one pentagon.

Try it in Python. A computer confirms the icosahedron's 12 vertices and 30 edges by listing the pairs of points at distance 2:

from itertools import combinations, product
phi = (1 + 5 ** 0.5) / 2
ico = [p for a, b in product((1, -1), repeat=2)
       for p in [(0, a, b * phi), (a, b * phi, 0), (b * phi, 0, a)]]
d2 = lambda p, q: sum((x - y) ** 2 for x, y in zip(p, q))
edges = [e for e in combinations(ico, 2) if abs(d2(*e) - 4) < 1e-9]
print(len(ico), len(edges))     # 12 30

History¶

The five solids were known to the Greeks. Euclid's Elements ends, in Book XIII, by constructing all five inside a sphere and proving that there are no others, by the angle argument above. Later Greek commentary attributes the octahedron and icosahedron, or the whole theory, to Theaetetus, a contemporary of Plato; how much is his is uncertain. Plato's dialogue Timaeus matched four of them with the elements (fire, air, water, earth) and the dodecahedron with the cosmos, which is how they got his name.

In 1596 the young Johannes Kepler, in Mysterium Cosmographicum, proposed that the six planets then known had orbits on spheres nested around the five solids: Saturn's sphere, cube, Jupiter's, tetrahedron, Mars's, dodecahedron, Earth's, icosahedron, Venus's, octahedron, Mercury's. The fit was only approximate, and when Kepler later worked through Tycho Brahe's careful observations of Mars, the data defeated circles altogether and led him to elliptical orbits (Chapter 8). The beautiful theory was wrong; the honest analysis of Mars was right.

11.3 Euler's formula¶

Count for each Platonic solid: \(4 - 6 + 4 = 2\), \(8 - 12 + 6 = 2\), \(6 - 12 + 8 = 2\), \(20 - 30 + 12 = 2\), \(12 - 30 + 20 = 2\). For a pentagonal prism, \(10 - 15 + 7 = 2\). For a hexagonal pyramid, \(7 - 12 + 7 = 2\). The number \(V - E + F\) never changes.

Theorem (Euler's polyhedron formula). For every convex polyhedron, \(V - E + F = 2\).

Flatten it¶

Imagine the polyhedron made of rubber, and remove one face. Stretch the hole wide and flatten the rest onto a table. The vertices and edges become a drawing in the plane, a planar graph, with no edges crossing. The faces become the regions inside the drawing, and the removed face becomes the outside region. For a convex polyhedron this can be done precisely: put your eye just outside the centre of one face and project everything onto a plane beyond the opposite side (the result is called a Schlegel diagram).

So it is enough to prove: for a connected drawing in the plane with \(V\) points, \(E\) edges (curves that do not cross) and \(F\) regions, counting the outside region, \(V - E + F = 2\).

Proof by building¶

Proof. Build the drawing one edge at a time, keeping it connected, and watch \(V - E + F\).

Start with a single point: \(V = 1\), \(E = 0\), \(F = 1\) (the whole plane is one region). So \(V - E + F = 2\).
Add an edge to a new point. One new vertex, one new edge, no new region (\(V\) and \(E\) go up by 1). The sum is unchanged.
Add an edge between two points already in the drawing. One new edge; and since the two points were already connected, the new edge closes a loop, which cuts the region it runs through into two. One new edge, one new region. The sum is unchanged.

Every connected drawing can be built like this (grow outward from any starting point: as long as some edge is missing, one of its ends is already drawn). So at the end \(V - E + F = 2\). \(\square\)

One step deserves a flag of honesty: "a closed loop cuts a region into two" is the Jordan curve theorem, which is obvious for polygons and surprisingly hard to prove for wild curves. For the straight edges of a Schlegel diagram the polygon case is enough, and its proof is elementary.

What the formula gives¶

Five again. Let a Platonic solid \(\{p, q\}\) have \(V\), \(E\), \(F\). Each face has \(p\) edges and each edge borders 2 faces, so \(pF = 2E\). Each vertex has \(q\) edges and each edge has 2 ends, so \(qV = 2E\). Put \(F = 2E/p\), \(V = 2E/q\) into Euler's formula:

\[ \frac{2E}{q} - E + \frac{2E}{p} = 2 \qquad\Longrightarrow\qquad \frac1p + \frac1q = \frac12 + \frac1E. \]

The right side is more than \(\tfrac12\), so \(\tfrac1p + \tfrac1q > \tfrac12\), the same five pairs as before; and now the formula even computes \(E\). For \(\{3, 5\}\): \(\tfrac13 + \tfrac15 - \tfrac12 = \tfrac{1}{30}\), so \(E = 30\), \(F = 20\), \(V = 12\). Two different proofs, one by angles and one by counting, give the same list.

Descartes' theorem. The angle defects of a convex polyhedron add up to exactly \(720°\).

Proof. The sum of the defects is \(360°\,V\) minus the sum of all the face angles of all the faces. A face with \(n\) sides has angle sum \((n - 2)180°\), so the total of all face angles is \(180°\sum(n - 2) = 180°(2E - 2F)\), since the numbers of sides of all faces add up to \(2E\) (each edge counted twice). So the total defect is \(360°V - 360°E + 360°F = 360°(V - E + F) = 720°\). \(\square\)

Check: a cube has 8 corners of defect \(90°\), total \(720°\). A dodecahedron has 20 corners of defect \(360° - 3 \times 108° = 36°\), total \(720°\). An icosahedron: 12 corners of \(360° - 300° = 60°\), total \(720°\). The "curvature" of a closed convex surface is concentrated at its corners and always adds up to the same amount; in Chapter 19 this becomes the Gauss-Bonnet theorem, with \(720°\) appearing as \(4\pi\).

A history with counterexamples¶

René Descartes wrote down the angle-defect theorem in notes around 1630 that were lost and only found again, in a copy made by Leibniz, in the nineteenth century. Leonhard Euler stated \(V - E + F = 2\) in a letter to Christian Goldbach in 1750 and published proofs in the 1750s; his proof had gaps. Adrien-Marie Legendre gave the first sound proof in 1794, using spherical geometry (Chapter 16), and Augustin-Louis Cauchy another in 1813.

The formula is false for polyhedra that are not convex in a strong way. A "picture frame" polyhedron, a square ring with a hole through it, can be built with \(V = 16\), \(E = 32\), \(F = 16\), giving \(V - E + F = 0\). The number \(V - E + F\) is the Euler characteristic: 2 for anything shaped like a sphere, 0 for a torus, \(2 - 2g\) for a surface with \(g\) holes. Imre Lakatos's book Proofs and Refutations (1976) is a classic dialogue on how the counterexamples forced mathematicians to sharpen what "polyhedron" means. The Euler characteristic is the first invariant of topology, and returns in Chapters 19 and 21.

11.4 Cavalieri's principle and volume¶

Level: intermediate.

Volume is measured in unit cubes, as area is measured in unit squares (Chapter 4). A box \(a \times b \times c\) holds \(abc\) of them (exactly, for whole numbers; by a limiting argument for all lengths). The next step, slanted and curved solids, needs a new idea.

The principle¶

Take a neat stack of coins, then push it so it leans. It holds the same coins, so it has the same volume. That is the whole idea:

Cavalieri's principle. If two solids lie between the same two parallel planes, and every plane parallel to them cuts the two solids in cross-sections of equal area, then the two solids have equal volume.

Bonaventura Cavalieri published it in 1635 in Geometria indivisibilibus, treating a solid as made of infinitely many thin slices ("indivisibles"). Archimedes had used the same idea earlier, as we will see. Today it is a theorem of calculus: volume is the integral of the cross-section area, \(V = \int A(z)\,dz\), so equal \(A(z)\) gives equal \(V\) (in the language of the U of T guide The Riemann integral in R^n, this is Fubini's theorem). In this chapter we use it as a principle and see how far it reaches.

Prisms and cylinders¶

A right prism with base area \(B\) and height \(h\) has volume \(Bh\): for a rectangular base this is the box formula, and a general base can be cut into pieces approximated by rectangles. Every cross-section of a slanted prism is a copy of the base, so by Cavalieri a slanted prism also has volume \(Bh\), with \(h\) the perpendicular height, not the slanted edge. The same holds for a cylinder: \(V = \pi r^2 h\).

Pyramids¶

Theorem. A pyramid (or cone) with base area \(B\) and height \(h\) has volume \(\tfrac13 Bh\).

The \(\tfrac13\) is the surprise, and it needs proof. We do it in three steps.

Step 1: one special pyramid. A cube of side \(a\) can be cut into three congruent square pyramids, each with a corner of the cube as apex and one of the three faces not touching that corner as base. Each has base \(a^2\) and height \(a\), and three of them make the cube, so each has volume \(\tfrac13 a^3 = \tfrac13 Bh\).

Step 2: same base and height, same volume. Two pyramids with bases of equal area and equal heights have equal volume. Cut both at height \(t\) above the base. The cross-section is the base shrunk towards the apex by the factor \(\frac{h - t}{h}\) (similar figures, Chapter 5), so its area is \(B\left(\frac{h - t}{h}\right)^2\), the same for both. By Cavalieri, the volumes are equal.

Step 3: scaling. Stretching a solid in one direction by a factor \(k\) multiplies its volume by \(k\) (each small box is stretched by \(k\)). A square pyramid with base \(a^2\) and height \(h\) is the pyramid of step 1 stretched vertically by \(h/a\), so its volume is \(\tfrac13 a^3 \cdot \tfrac{h}{a} = \tfrac13 a^2 h\). By step 2, any pyramid with base area \(B\) and height \(h\) has the volume of a square pyramid with the same \(B\) and \(h\), namely \(\tfrac13 Bh\). \(\square\)

For a cone with radius \(r\) and height \(h\): \(V = \tfrac13\pi r^2 h\). A cone fills exactly a third of the cylinder around it, which you can check with sand and a cone-shaped cup.

This was known to the Greeks: Euclid's Elements, Book XII, proves that a pyramid is a third of the prism on the same base, and Archimedes credits the discovery to Democritus and the first proof to Eudoxus. Euclid's proof avoids indivisibles by a careful limiting argument, the method of exhaustion.

The sphere¶

Theorem (Archimedes). A ball of radius \(r\) has volume \(\tfrac43\pi r^3\).

Proof. Compare a hemisphere of radius \(r\) with a different solid: a cylinder of radius \(r\) and height \(r\), with a cone removed whose apex is at the centre of the cylinder's base and whose base is the cylinder's top.

Cut both at height \(z\) above the base (\(0 \le z \le r\)).

The hemisphere's slice is a disc. Its radius \(\rho\) satisfies \(\rho^2 + z^2 = r^2\) (Pythagoras, from the centre to the edge of the slice), so its area is \(\pi(r^2 - z^2)\).
The other solid's slice is a ring: the cylinder's disc of radius \(r\) minus the cone's disc. The cone widens at \(45°\) (height \(r\), radius \(r\)), so at height \(z\) its radius is \(z\). The ring's area is \(\pi r^2 - \pi z^2\).

Equal at every height, so by Cavalieri the volumes are equal:

\[ V_{\text{hemisphere}} = \pi r^2 \cdot r - \tfrac13\pi r^2 \cdot r = \tfrac23\pi r^3. \]

The whole ball is twice that, \(\tfrac43\pi r^3\). \(\square\)

Archimedes was so pleased with the result in its form "the sphere is two thirds of the cylinder around it" (a cylinder of radius \(r\) and height \(2r\) has volume \(2\pi r^3\)) that, according to Cicero, a sphere inside a cylinder was carved on his tomb, and Cicero found the tomb in Syracuse by that sign. Archimedes' own route to the formula, a thought experiment balancing slices on a lever, is described in his Method, a text known only from a palimpsest identified by Johan Ludvig Heiberg in 1906.

Example (Mars). Mars has mean radius about 3389.5 km, Earth about 6371 km. Their volumes are about \(\tfrac43\pi(3389.5)^3 \approx 1.63 \times 10^{11}\) km\(^3\) and \(\tfrac43\pi(6371)^3 \approx 1.08 \times 10^{12}\) km\(^3\). The ratio is \((3389.5/6371)^3 \approx 0.151\): Mars has about 15% of Earth's volume, although its radius is just over half. Volume goes with the cube of length, the square-cube law of Chapter 5.

11.5 The surface of a sphere¶

Level: intermediate.

How much paint covers a ball? The surface is curved and cannot be flattened (that is Chapter 16's theme), so we cannot cut it into flat pieces. Archimedes found the answer two ways, and both are worth seeing.

From the volume¶

Cover the sphere with a fine mesh of tiny patches, and join each patch to the centre. The ball is cut into many thin "pyramids" whose bases are the patches and whose heights are all (nearly) \(r\). Their volumes add up:

\[ \tfrac43\pi r^3 = \sum \tfrac13 (\text{patch area})\, r = \tfrac13 r \cdot A, \]

so the surface area is \(A = 4\pi r^2\). In words: the surface of a sphere is four times the area of its great circle. (This is a sketch: making "nearly" precise is a limit argument, done properly with calculus.)

Archimedes' hat-box theorem¶

There is a more surprising fact. Put the sphere inside the cylinder that just fits it (radius \(r\), height \(2r\)), and cut both with two horizontal planes a distance \(h\) apart.

Theorem (Archimedes). The band of sphere between the planes has the same area as the band of cylinder between them, \(2\pi r h\), wherever the planes are.

Why. Take a very thin band of sphere at height \(z\), of height \(\Delta z\). It is nearly a slanted strip of a cone: its radius is \(\rho = \sqrt{r^2 - z^2}\) and its slanted width is \(\Delta s\). Let \(\alpha\) be the angle between the radius to the band and the horizontal, so \(\cos\alpha = \rho/r\). The sphere's surface there is perpendicular to the radius, so the slanted width makes the same angle \(\alpha\) with the vertical, and its vertical extent is \(\Delta z = \Delta s\cos\alpha = \Delta s\,\rho/r\). So \(\Delta s = \frac{r}{\rho}\Delta z\). The band's area is nearly circumference times width:

\[ 2\pi\rho\,\Delta s = 2\pi\rho \cdot \frac{r}{\rho}\,\Delta z = 2\pi r\,\Delta z, \]

exactly the area of the matching band of the cylinder. Near the equator the sphere's band is wide and steep; near the pole it is small but slanted; the two effects cancel. Adding up the thin bands gives \(2\pi r h\) for any band. \(\square\)

Taking \(h = 2r\) gives the whole sphere: \(4\pi r^2\), agreeing with the first argument. The theorem also says that projecting the sphere horizontally outwards onto the cylinder keeps areas. This is Lambert's cylindrical equal-area map projection, used for maps that must show areas fairly (Chapter 16).

Example (the polar region of Mars). What part of Mars lies north of latitude \(80°\)? The cap above that latitude is a zone of height \(h = R(1 - \sin 80°)\). With \(R = 3389.5\) km, \(h \approx 51.5\) km and the area is \(2\pi R h \approx 1.10\) million km\(^2\), about \(0.76\%\) of the planet's surface, which is \(4\pi R^2 \approx 1.44 \times 10^8\) km\(^2\) (about the same as all the dry land on Earth).

Cones and cylinders¶

Cylinder: unroll the side into a rectangle \(2\pi r\) by \(h\), area \(2\pi r h\); with the two ends, \(2\pi r h + 2\pi r^2\).
Cone: with radius \(r\) and slant height \(s\), unroll the side into a sector of a circle of radius \(s\) with arc length \(2\pi r\). A sector's area is \(\tfrac12 \times\) radius \(\times\) arc (Chapter 6), so the side has area \(\tfrac12 s \cdot 2\pi r = \pi r s\).

Example. A cone of radius 3 and height 4 has slant \(s = 5\), side area \(15\pi\), total area \(15\pi + 9\pi = 24\pi\), and volume \(\tfrac13\pi \cdot 9 \cdot 4 = 12\pi \approx 37.7\).

11.6 Archimedean solids and duality¶

Level: intermediate to advanced.

Relax the rules. Allow several kinds of regular polygon as faces, but insist that every vertex looks the same: the same polygons in the same cyclic order around each one, and, more strongly, that some symmetry of the solid carries any vertex to any other. Besides the endless families of prisms and antiprisms, there are exactly 13 Archimedean solids.

We describe a vertex by its vertex configuration, the polygons around it in order: the cube is \(4.4.4\), the cuboctahedron (squares and triangles alternating, made by cutting each corner of a cube down to the midpoints of its edges) is \(3.4.3.4\).

Some of the thirteen:

solid	vertex	\(V\)	\(E\)	\(F\)
truncated tetrahedron	\(3.6.6\)	12	18	8
cuboctahedron	\(3.4.3.4\)	12	24	14
truncated octahedron	\(4.6.6\)	24	36	14
icosidodecahedron	\(3.5.3.5\)	30	60	32
truncated icosahedron	\(5.6.6\)	60	90	32

Each satisfies Euler's formula (\(12 - 18 + 8 = 2\), and so on). The truncated octahedron has a special property: copies of it fill space with no gaps.

The truncated icosahedron, made by cutting each corner of an icosahedron a third of the way along its edges, is the pattern of the classic football and of the carbon molecule \(C_{60}\) (buckminsterfullerene), discovered in 1985.

Why a football has exactly 12 pentagons. Suppose a convex polyhedron has only pentagons (\(p\) of them) and hexagons (\(h\) of them), with three faces at every vertex. Count edges by faces: \(5p + 6h = 2E\). Count edges by vertices: \(3V = 2E\). And \(V - E + F = 2\) with \(F = p + h\). Multiply Euler by 6: \(6V - 6E + 6F = 12\), that is \(4E - 6E + 6p + 6h = 12\), so \(6p + 6h - 2E = 12\). Substitute \(2E = 5p + 6h\): \(6p + 6h - 5p - 6h = 12\), so \(p = 12\). Exactly 12 pentagons, whatever the number of hexagons. The same count governs the fullerene molecules and the shells of many viruses.

The Archimedean solids are named after Archimedes because Pappus of Alexandria reported that Archimedes had described 13 such solids, in a work now lost. Kepler rediscovered and listed all 13 in Harmonices Mundi (1619). An honest footnote: there is a fourteenth convex solid with regular faces and the same configuration \(3.4.4.4\) at every vertex, the elongated square gyrobicupola, but no symmetry carries every vertex to every other (it looks the same at each vertex only locally). That is why the definition above insists on symmetry.

Duality¶

Put a point at the centre of each face of a cube and join the points on neighbouring faces. You get an octahedron.

This is duality: from a polyhedron, make a new one with a vertex for each face of the old, an edge crossing each old edge, and a face for each old vertex. So the dual swaps \(V\) and \(F\) and keeps \(E\), and swaps the Schläfli symbol \(\{p, q\} \leftrightarrow \{q, p\}\):

the cube \(\{4,3\}\) and octahedron \(\{3,4\}\) are duals (\(V = 8, F = 6\) and \(V = 6, F = 8\), both \(E = 12\));
the dodecahedron \(\{5,3\}\) and icosahedron \(\{3,5\}\) are duals;
the tetrahedron \(\{3,3\}\) is its own dual.

Dual solids have the same symmetry group. This explains why there are only three symmetry groups among the five solids. Their rotation groups have \(12\), \(24\) and \(60\) elements, which the orbit-stabilizer theorem (Algebra book, Chapter 16) counts quickly: a cube's rotations can bring any of the 6 faces to the top, and then turn it 4 ways, so \(6 \times 4 = 24\). For the dodecahedron, \(12 \times 5 = 60\); for the tetrahedron, \(4 \times 3 = 12\).

The duals of the Archimedean solids are the Catalan solids, with all faces congruent but not regular, such as the rhombic dodecahedron (dual of the cuboctahedron: 14 vertices, 24 edges, 12 rhombus faces), the shape of garnet crystals and, like the truncated octahedron, a solid that fills space.

11.7 Hilbert's third problem¶

Level: advanced to expert.

In the plane, area is the whole story. The Bolyai-Gerwien theorem (Chapter 4, 4.7) says that two polygons of equal area can always be cut into finitely many polygonal pieces that rearrange into each other. The formula for a triangle's area can be proved purely by cutting and rearranging.

In space, Euclid's proof that a pyramid is a third of a prism used a limiting argument, and nobody had managed a proof by cutting into finitely many pieces. Is that a failure of ingenuity, or a real obstacle? In his famous list of 23 problems at the International Congress of Mathematicians in 1900, David Hilbert asked, as his third problem, for two tetrahedra with equal bases and equal heights (hence equal volumes) that cannot be cut into finitely many congruent polyhedral pieces. He expected such a pair to exist.

Theorem (Dehn, 1900). A regular tetrahedron cannot be cut into finitely many polyhedral pieces that reassemble into a cube of the same volume.

Max Dehn, a student of Hilbert, proved this within the year, making the third the first of Hilbert's problems to be solved. His tool is a new quantity attached to every polyhedron that does not change under cutting and regluing, the Dehn invariant, and which is different for the cube and the tetrahedron.

The idea of the invariant¶

Look at what happens to edges and their dihedral angles (the angle between the two faces meeting at an edge) when you cut a polyhedron into pieces.

A cut along a plane through the middle of a face creates new edges where the angles of the two pieces add up to \(180°\), that is \(\pi\).
A new edge inside the solid is surrounded by pieces whose dihedral angles add up to \(360°\), that is \(2\pi\).
An old edge cut lengthwise splits into two edges of the same length whose angles add to the old angle; an old edge cut crosswise splits into two pieces of edge with the same angle whose lengths add up.

So "length times angle" behaves additively, provided we agree to ignore angles that are rational multiples of \(\pi\) (since new edges come with total angles \(\pi\) or \(2\pi\)). Dehn's invariant packages this:

\[ D(P) = \sum_{\text{edges } e} \ell(e) \otimes \theta(e), \]

where \(\ell(e)\) is the edge's length, \(\theta(e)\) its dihedral angle, and the symbol \(\otimes\) means: a formal sum of pairs, additive in each slot (\((\ell_1 + \ell_2) \otimes \theta = \ell_1 \otimes \theta + \ell_2 \otimes \theta\) and likewise for angles), with rational numbers allowed to move across (\(q\ell \otimes \theta = \ell \otimes q\theta\)), and with \(\ell \otimes \theta = 0\) whenever \(\theta\) is a rational multiple of \(\pi\). (Formally, \(D(P)\) lies in the tensor product \(\mathbb{R} \otimes_{\mathbb{Q}} \mathbb{R}/\pi\mathbb{Q}\); the key fact we use is that \(\ell \otimes \theta = 0\) only if \(\ell = 0\) or \(\theta\) is a rational multiple of \(\pi\).)

The three observations above show that if \(P\) is cut into pieces \(P_1, \dots, P_k\), then \(D(P) = D(P_1) + \dots + D(P_k)\). So two polyhedra that can be cut into each other have the same Dehn invariant.

The cube and the tetrahedron¶

Cube: 12 edges, every dihedral angle \(\pi/2\), a rational multiple of \(\pi\). So \(D(\text{cube}) = 0\).
Regular tetrahedron: 6 edges of length \(\ell\), every dihedral angle \(\theta = \arccos\tfrac13 \approx 70.53°\) (the angle between two faces; compute it from the normals of 11.2's coordinates, or see exercise 11). So \(D = 6\ell \otimes \theta\), which is not zero provided \(\theta\) is not a rational multiple of \(\pi\).

Lemma. \(\theta = \arccos\tfrac13\) is not a rational multiple of \(\pi\).

Proof. If \(\theta = \frac{m}{n}\pi\), then \(n\theta\) is a multiple of \(\pi\), so \(\cos(n\theta) = \pm1\). We show \(\cos(k\theta)\) is never \(\pm1\) for \(k \ge 1\). The addition formulas (Chapter 7) give \(\cos((k + 1)\theta) = 2\cos\theta\cos(k\theta) - \cos((k - 1)\theta)\). Write \(\cos(k\theta) = a_k/3^k\). Then \(a_0 = 1\), \(a_1 = 1\), and the formula becomes

\[ a_{k+1} = 2a_k - 9a_{k-1}, \]

so all \(a_k\) are integers, and modulo 3, \(a_{k+1} \equiv 2a_k\). Starting from \(a_1 = 1\), no \(a_k\) is divisible by 3. So \(\cos(k\theta) = a_k/3^k\) is a fraction in lowest terms with denominator \(3^k > 1\), never \(\pm 1\). \(\square\)

(For instance \(\cos 2\theta = -\tfrac79\), \(\cos 3\theta = -\tfrac{23}{27}\), \(\cos 4\theta = \tfrac{17}{81}\).)

So \(D(\text{tetrahedron}) \ne 0 = D(\text{cube})\), and the tetrahedron cannot be cut into a cube. The same invariant also separates suitable pairs of tetrahedra with equal bases and equal heights, which is exactly what Hilbert asked for. The deeper lesson: a theory of volume for polyhedra cannot avoid limits, as the theory of area can.

The full answer¶

Is the Dehn invariant the only obstacle? Yes: in 1965 Jean-Pierre Sydler proved that two polyhedra can be cut into each other if and only if they have the same volume and the same Dehn invariant. The proof is long and uses algebra well beyond this book. The question of what invariants control cutting in higher dimensions, and in spherical and hyperbolic space, is still partly open and connects to deep parts of modern algebra (the "scissors congruence" problem).

Check yourself¶

1. Count \(V\), \(E\), \(F\) for a pentagonal prism, a pentagonal pyramid and a square antiprism, and check Euler's formula for each.

Pentagonal prism: \(V = 10\), \(E = 15\), \(F = 7\); \(10 - 15 + 7 = 2\). Pentagonal pyramid: \(V = 6\), \(E = 10\), \(F = 6\); \(6 - 10 + 6 = 2\). Square antiprism: \(V = 8\), \(E = 16\), \(F = 10\) (2 squares, 8 triangles); \(8 - 16 + 10 = 2\).

2. A convex polyhedron has only triangular faces, and 5 meet at every vertex. Find \(V\), \(E\), \(F\) using only Euler's formula.

Each face has 3 edges, each edge 2 faces: \(3F = 2E\). Each vertex has 5 edges: \(5V = 2E\). So \(\tfrac{2E}{5} - E + \tfrac{2E}{3} = 2\), that is \(E\cdot\tfrac{6 - 15 + 10}{15} = 2\), \(E = 30\). Then \(F = 20\), \(V = 12\): the icosahedron.

3. Show that there is no convex polyhedron with exactly 7 edges.

Each face has at least 3 edges, so \(2E \ge 3F\); each vertex has at least 3 edges, so \(2E \ge 3V\). With \(E = 7\): \(F \le 4\) and \(V \le 4\), so \(V + F \le 8\). But Euler needs \(V + F = E + 2 = 9\). Contradiction. (Every other number of edges from 6 upwards does occur.)

4. A square pyramid has base side 6 and height 4. Find its volume and its total surface area.

Volume \(\tfrac13 \cdot 36 \cdot 4 = 48\). Each triangular face has base 6 and slant height (from the apex to the middle of a base edge) \(\sqrt{4^2 + 3^2} = 5\), so area \(\tfrac12 \cdot 6 \cdot 5 = 15\). Total: \(36 + 4 \cdot 15 = 96\).

5. A cone has radius 3 and height 4. Find its volume, its side area and its total area. What angle is the sector you get by unrolling its side?

Slant \(s = 5\). Volume \(\tfrac13\pi\cdot9\cdot4 = 12\pi \approx 37.7\). Side \(\pi r s = 15\pi\); total \(15\pi + 9\pi = 24\pi \approx 75.4\). The sector has radius 5 and arc \(2\pi \cdot 3 = 6\pi\), which is \(\tfrac{6\pi}{10\pi} = \tfrac35\) of a full circle: \(216°\).

6. A sphere fits exactly inside a cylinder. Show that both its volume and its surface area are two thirds of the cylinder's (counting the cylinder's ends).

Radius \(r\), cylinder height \(2r\). Volumes: \(\tfrac43\pi r^3\) and \(\pi r^2 \cdot 2r = 2\pi r^3\), ratio \(\tfrac23\). Areas: \(4\pi r^2\) and \(2\pi r\cdot 2r + 2\pi r^2 = 6\pi r^2\), ratio \(\tfrac23\). This is the result Archimedes wanted on his tomb.

7. Mars has radius about 3389.5 km and Earth about 6371 km. What fraction of Earth's surface area, and of its volume, does Mars have?

Areas scale as the square of the radius: \((3389.5/6371)^2 \approx 0.283\), about 28%. Volumes scale as the cube: \((3389.5/6371)^3 \approx 0.151\), about 15%. (Its area, about \(1.44 \times 10^8\) km\(^2\), is close to the land area of Earth.)

8. Check Descartes' theorem for an \(n\)-gonal right prism.

At each of the \(2n\) vertices meet two rectangles (\(90°\) each) and one \(n\)-gon, with angle \(\frac{(n - 2)180°}{n}\). The defect is \(360° - 180° - \frac{(n - 2)180°}{n} = 180° - 180° + \frac{360°}{n} = \frac{360°}{n}\). The total is \(2n \cdot \frac{360°}{n} = 720°\).

9. Find the area of the part of a sphere of radius \(R\) that lies between latitudes \(30°\) N and \(30°\) S, as a fraction of the whole.

The zone lies between heights \(-R\sin30°\) and \(R\sin30°\), so \(h = R\). By the hat-box theorem its area is \(2\pi R \cdot R = 2\pi R^2\), exactly half of \(4\pi R^2\). Half of every planet's surface lies within \(30°\) of its equator.

10. The rhombic dodecahedron is the dual of the cuboctahedron. Find its \(V\), \(E\), \(F\) and the shapes of its faces and vertices.

The cuboctahedron has \(V = 12\), \(E = 24\), \(F = 14\) (8 triangles, 6 squares). The dual swaps \(V\) and \(F\): \(V = 14\), \(E = 24\), \(F = 12\). Each vertex of the cuboctahedron has 4 faces round it, so each face of the dual has 4 sides (a rhombus). The dual has 8 vertices where 3 faces meet (from the triangles) and 6 where 4 meet (from the squares). Check: \(14 - 24 + 12 = 2\).

11. Show that the dihedral angle of the regular tetrahedron is \(\arccos\frac13\), and that the regular octahedron's is \(\pi - \arccos\frac13\). What is the octahedron's Dehn invariant?

Use the tetrahedron \((1,1,1)\), \((1,-1,-1)\), \((-1,1,-1)\), \((-1,-1,1)\) centred at the origin. The outward normal of the face opposite a vertex points away from that vertex, so the normals are the negatives of the vertex vectors, e.g. \(-(1,1,1)\) and \(-(1,-1,-1)\), at angle \(\arccos\frac{1 - 1 - 1}{3} = \arccos(-\tfrac13)\). The dihedral angle is \(\pi\) minus the angle between outward normals: \(\pi - \arccos(-\tfrac13) = \arccos\tfrac13 \approx 70.53°\). For the octahedron the face normals are \((\pm1, \pm1, \pm1)\); two faces sharing an edge have normals like \((1,1,1)\) and \((1,1,-1)\), at angle \(\arccos\tfrac13\), so the dihedral angle is \(\pi - \arccos\tfrac13 \approx 109.47°\). Its Dehn invariant is \(12\ell \otimes (\pi - \theta) = -12\ell \otimes \theta \ne 0\) (the \(\pi\) part counts as zero). So an octahedron cannot be cut into a cube either.

12. Prove that the inequality \((p - 2)(q - 2) < 4\) with whole numbers \(p, q \ge 3\) has exactly the five solutions listed, and explain what the equality \((p - 2)(q - 2) = 4\) describes.

Both factors are whole numbers at least 1 with product at most 3: \((1,1), (1,2), (2,1), (1,3), (3,1)\), giving \((p,q) = (3,3), (3,4), (4,3), (3,5), (5,3)\); any factor 4 or more, or both at least 2, makes the product at least 4. Equality \((p - 2)(q - 2) = 4\) means the angles at a vertex add to exactly \(360°\): \((4,4)\), \((3,6)\), \((6,3)\). These are the three regular tilings of the plane by squares, triangles and hexagons (Chapter 3), the "flat" limits of the Platonic solids.

What you learned¶

Polyhedra have faces, edges and vertices; at each vertex of a convex polyhedron the face angles leave a positive defect.
Exactly five Platonic solids exist, \(\{3,3\}, \{4,3\}, \{3,4\}, \{5,3\}, \{3,5\}\), by the angle inequality \((p - 2)(q - 2) < 4\) or by Euler's formula, and coordinates build each one.
Euler's formula \(V - E + F = 2\) holds for convex polyhedra (proof by building the planar graph); the defects add up to \(720°\) (Descartes).
Cavalieri's principle gives \(V = Bh\) for prisms, \(\tfrac13 Bh\) for pyramids and cones, and Archimedes' \(\tfrac43\pi r^3\) for the ball.
The sphere's surface is \(4\pi r^2\), and a band of it has the same area as the matching band of the circumscribed cylinder.
Archimedean solids have regular faces of several kinds; duality swaps vertices and faces; a pentagon-hexagon polyhedron with three faces per vertex has exactly 12 pentagons.
The Dehn invariant shows that a regular tetrahedron cannot be cut into a cube: volume in space needs limits.

Next: Chapter 12, the classical depth of the triangle: its centres, the Euler line, the nine-point circle, and the theorems of Ceva, Menelaus, Ptolemy and Simson.