A polyhedron whose faces are congruent regular polygons, and whose polyhedral angles are equal, is called a regular polyhedron.
Polyhedrons classified as to Faces
A polyhedron of four faces is called a tetrahedron; one of six faces, a hexahe-dron; one of eight faces, an octahedron; one of twelve faces, a dodecahedron; one of twenty faces, an icosahedron.
Definitions
- Tetrahedron
A tetrahedron is a solid figure contained by four equilateral triangles.
- Cube (Hexahedron)
A cube is a solid figure contained by six equal squares.
- Octahedron
An octahedron is a solid figure contained by eight equal and equilateral triangles.
- Icosahedron
An icosahedron is a solid figure contained by twenty equal and equilateral triangles.
- Dodecahedron
A dodecahedron is a solid figure contained by twelve equal, equilateral and equiangular pentagons.PROPOSITION
To determine the number of regular convex polyhedrons possible.
A convex polyhedral angle must have at least three faces, and the sum of its face angles must be less than 360.
- Since each angle of an equilateral triangle is 60°, convex polyhedral angles may be formed by combining three, four, or five equilateral triangles. The sum of six such angles is 360°, and therefore is greater than the sum of the face angles of a convex polyhedral angle. Hence three regular convex polyhedrons are possible with equilateral triangles for faces.
- Since each angle of a square is 90°, a convex polyhedral angle may be formed by combining three squares. The sum of four such angles is 360°, and therefore is greater than the sum of the face angles of a convex polyhedral angle. Hence one regular convex polyhedron is possible with squares.
- Since each angle of a regular pentagon is 108°, a convex polyhedral angle may be formed by combining three regular pentagons. The sum of four such angles is 432°, and therefore is greater than the sum of the face angles of a convex polyhedral angle. Hence one regular convex polyhedron is possible with regular pentagons.
- The sum of three angles of a regular hexagon is 360°, of a regular heptagon is greater than 360°, and so on. Hence only five regular convex polyhedrons are possible.
The regular polyhedrons are the regular tetrahedron, the regular hexahedron, or cube, the regular octahedron, the regular dodecahedron, and the regular icosahedron. Since these solids were extensively studied by the pupils of Plato, the great Greek philosopher, they are often called the Platonic Bodies or Platonic Solids.The above argument shows that regular polyhedra, if they exist, fall into five types, but it does not prove that regular polyhedra of each of the five types exist. In order to establish their existence it suffices to point out a construction of each of the five Platonic solids. In the case of the cube, which was defined as a rectangular parallelepiped all of whose three dimensions are congruent, such a construction is familiar to us. We will show here how each of the remaining four Platonic solids can be constructed from a cube. The following are not proper “geometric constructions” strictly speaking, but general notions about how it can be done.
- A regular tetrahedron can be constructed by taking four of the eight vertices of a cube for the vertices of the tetrahedron. Namely, pick any vertex A of the cube, and in the three square faces adjacent to this vertex, take the vertices B, C, and D opposite to A.The six edges connecting the vertices A, B, C, D pairwise are diagonals of the cube’s faces (one diagonal in each face), and are therefore congruent. This shows that all faces of the tetrahedron are congruent regular triangles.
- A regular octahedron can be constructed by taking the six centers of cube’s faces for the vertices. Each edge of the resulting polyhedron connects centers of two adjacent faces of the cube and, as easily computed, has the length a where a denotes the cube’s dimension. In particular, all edges have the same length, and hence all faces of the octahedron are congruent regular triangles.
- A dodecahedron can be constructed by drawing 12 planes, one through each of the 12 edges of a cube, and choosing the slopes of these planes in such a way that the resulting polyhedron is regular. The fact that it is possible to achieve this is not obvious at all, and proving it will require some preparation.
- Once the existence of the dodecahedron is established, a regular icosahedron can be constructed by taking the centers of 12 faces of the dodecahedron for the vertices.
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The following construction is from Euclid, and guides through the reasoning and steps involved in the geometric construction of the tetrahedron (referred to by Euclid as pyramid).
PROPOSITION
To construct a pyramid, to comprehend it in a given sphere; and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.
Construction:
Set out the diameter AB of the given sphere, cut it at the point C so that AC is double CB, describe the semicircle ADB on AB, draw CD from the point C at right angles to AB, and join DA.
Set out the circle EFG with radius equal to DC, inscribe the equilateral triangle EFG in the circle EFG, take the center H of the circle, and join EH, HF, and HG.
Set HK up from the point H at right angles to the plane of the circle EFG, cut off HK equal to the straight line AC from HK, and join KE, KF, and KG.
Now, since KH is at right angles to the plane of the circle EFG, therefore it makes right angles with all the straight lines which meet it and are in the plane of the circle EFG. But each of the straight lines HE, HF, and HG meets it, therefore HK is at right angles to each of the straight lines HE, HF, and HG.
And, since AC equals HK, and CD equals HE, and they contain right angles, therefore the base DA equals the base KE. For the same reason each of the straight lines KF and KG also equals DA. Therefore the three straight lines KE, KF, and KG equal one another.
And, since AC is double CB, therefore AB is triple BC.
But that AB is to BC as the square on AD is to the square on DC will be proved afterwards.
Therefore the square on AD is triple the square on DC. But the square on FE is also triple the square on EH, and DC equals EH, therefore DA also equals EF.
But DA was proved equal to each of the straight lines KE, KF, and KG, therefore each of the straight lines EF, FG, and GE also equals each of the straight lines KE, KF, and KG. Therefore the four triangles EFG, KEF, KFG, and KEG are equilateral.
Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex.
It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.
Produce the straight line HL in a straight line with KH, and make HL equal to CB.
Now, since AC is to CD as CD is to CB, while AC equals KH, CD equals HE, and CB equals HL, therefore KH is to HE as EH is to HL. Therefore the rectangle KH by HL equals the square on EH.
And each of the angles KHE, EHL is right, therefore the semicircle described on KL passes through E also.
If then, KL remaining fixed, the semicircle is carried round and restored to the same position from which it began to be moved, then it also passes through the points F and G, since, if FL and LG are joined, then the angles at F and G similarly become right angles, and the pyramid is comprehended in the given sphere. For KL, the diameter of the sphere, equals the diameter AB of the given sphere, since KH was made equal to AC, and HL to CB
I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.
Since AC is double CB, therefore AB is triple BC, and, in conversion, BA is one and a half times AC.
But BA is to AC as the square on BA is to the square on AD. Therefore the square on BA is also one and a half times the square on AD. And BA is the diameter of the given sphere, and AD equals the side of the pyramid. Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.
Q.E.F.
Summary of the construction
Standardize the radius of the sphere at 1 unit, so that AB = 2. Then cut AB at C so that AC = 4/3 and BC = 2/3. Let DC be their mean proportional (2/3)√2. Then AD = (2/3)√6. This line AD will end up being the length of the side of the tetrahedron. Note that it has the correct value so that “the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.”
Set out the circle EFG of radius EH = (2/3)√2, and inscribe in that circle an equilateral triangle. Then each side of the triangle will be (2/3)√6, the same as AD. Make HK of length 4/3 and perpendicular to the plane of the triangle, and connect KE, KF, and KG. Then K lies on the surface of the sphere. And since the triangle HKE is a right triangle, therefore its hypotenuse KE = (2/3)√6, the same as AD. Likewise KF and KG have the same length. That constructs the tetrahedron in the sphere. |
We accepted a very demanding definition of regular polyhedra and found that, up to scale, there are only five such polyhedra. One may ask if the same conclusion can be derived from milder requirements of regularity. It turns out that the answer is “yes”: in order to conclude that a polyhedron is regular, it suffices to require that all of the faces are congruent regular polygons and polyhedral angles are congruent (but assume nothing about dihedral angles). In fact, many attempts to relax the definition even further lead to mistakes. First, merely assuming that all faces are congruent regular polygons is not enough (to construct a counterexample, attach two congruent regular tetrahedra to each other by their bases). Next, the class of polyhedra, all of whose polyhedral angles are congruent and faces regular, includes regular prisms with square lateral sides. This class was first systematically explored by a German mathematician and astronomer Johannes Kepler. In 1619, he found that in addition to the prisms, it also includes antiprisms, and 15 Archimedean solids (13, if symmetric polyhedra are not distinguished), which were described in the 4th century A.D. by a Greek mathematician Pappus and attributed by him to Archimedes. The Archimedean, or semi-regular polyhedra, are ‘facially’ regular. Every face is a regular polygon, though the faces are not all of the same kind. Every vertex, however, is to be congruent to every other vertex, i.e. the faces must be arranged in the same order around each vertex.Notation For Describing Polyhedra
The symbol used to describe the regular polyhedra is a combination of integers and exponents. The base indicates the number of sides of the regular polygonal face: 3.4.5 would mean that a triangle, a square and a pentagon were involved. The exponent indicates the number of these polygonal faces meeting at a vertex.
Since it has squares meeting at each vertex the cube is written 43.
A snub cube is written 34.4 because 4 triangles (3) meet one square (4) at each vertex.13 Archimedean Solids
It can be proven that there are only 13 Archimedean solids, two of which occur in two forms. These two are the two ‘snubs’, and the two forms of each are related to one another like a left-hand and a right hand glove: they are enantiomorphic. The set of thirteen is illustrated below.
One of these solids, the truncated tetrahedron, can be inscribed in a regular tetrahedron. The next six can be inscribed in either a cube or an octahedron, and the last six in either a dodecahedron or an icosahedron. The ‘truncated’ solids are so called because each can be constructed by cutting off the corners of some other solid, but the truncated cuboctahedron and icosidodecahedron require a distortion in addition to converting rectangles into squares. So the better names for these two solids are ‘Great Rhombicuboctahedron’ and ‘Great Rhombicosidodecahedron’. The solids 34.4 and 3.4.5.4 can then bear the prefix ‘small’. The syllable ‘rhomb-‘ shows that one set of faces lies in the planes of the rhombic dodecahedron and rhombic triacontahedron respectively. All Archimedean solids are inscribable in a sphere.Faces, Edges and Vertices
The type of face is indicated by a subscript to F, e.g. F3 represents a triangle; F4a square.
The number of these faces in a given polyhedron is the coefficient: e.g. 6 F4 for a cube.
The number of Vertices and Edges are also given as coefficients e.g. a cube would have 8V and 12E
The Five Platonic Regular Solids
Name | faces
sides |
#/kind
of faces |
Vertices | Edges |
Tetrahedron | 33 | 4 F3 | 4V | 6E |
Hexahedron (cube) | 43 | 6 F4 | 8V | 12E |
Octahedron | 34 | 8 F3 | 6V | 12E |
Dodecahedron | 53 | 12 F5 | 20V | 30E |
Icosahedron | 35 | 20 F3 | 12V | 30E |
The 13 Archimedean semi regular solids
Name | faces
sides |
#/kind
of faces |
Vertices | Edges |
1 Truncated tetrahedron | 3.62 | 4 F3, 4 F6 | 12V | 18E |
2 Cuboctahedron | (3.4)2 | 8 F3, 6 F4 | 12V | 24E |
3 Truncated cube | 3.82 | 8 F3, 6 F8 | 24V | 36E |
4 Truncated octahedron | 4.62 | 6 F4, 8 F6 | 24V | 36E |
5 Small rhombicu-boctahedron | 3.43 | 8 F3, 18 F4 | 24V | 48E |
6 Great rhombicuboctahedron
or Truncated cuboctahedron |
4.6.8 | 12 F4, 8 F6, 6 F8 | 48V | 72E |
7 Snub cube | 34.4 | 32 F3, 6 F4 | 24V | 60E |
8 Icosidodecahedron | (3.5)2 | 20 F3, 12 F5 | 30V | 60E |
9 Truncated dodecahedron | 3.102 | 20 F3, 12 F10 | 60V | 90E |
10 Truncated Icosahedron | 5.62 | 12 F5, 20 F6 | 60V | 90E |
11 Small rhombicosidodecahedron | 3.4.5.4 | 20 F3, 30 F4, 12 F5 | 60V | 120E |
12 Truncated Icosidodecahedron | 4.6.10 | 30 F4, 20 F6, 12 F10 | 120V | 180E |
13 Snub dodecahedron | 34.5 | 80 F3, 12 F5 | 60V | 150E |