All of us agree on the fairness of using a cube, or indeed any of the regular polyhedra, to play dice, because any face has the same chance to appear. From a mathematical viewpoint, we say that the faces are indistinguishable. But what does "indistinguishable" mean exactly?

This adjective indicates that if we have a regular polyhedron and we choose two of its faces, we can always rotate it so that the first face goes to occupy the place of the second one and, as a whole, the polyhedron goes back onto itself.

For example, to move the front face of the cube in the figure to the position of the top face, it is enough to rotate the cube through 90 degrees about the straight line labelled r in the figure.

The same thing happens for two vertices or two edges:
to send edge a onto edge b we can rotate the cube through 120 degrees
about the straight line s; the same rotation can be used to send vertex P onto vertex Q.

The table below describes the five regular polyhedra.

In the table, the first three columns of numbers give the number of faces,
edges, and vertices for each polyhedron; the two following columns can be interpreted as the "instructions" to build the skeleton of the polyhedron if sticks of the same length together with something to connect their ends are available (for example some drinking straws with pipe-cleaners for making joints). So the numbers that appear in the last row tell us that in an icosahedron there are 20 faces, 30 edges, 12 vertices, 5 edges meeting at each vertex, and 3 edges per face: if we join straws five at each vertex and form triangular faces, the object will "close itself" and give rise to an icosahedron.

The last column in the table also gives instructions for building the corresponding polyhedron: in this case, starting from the drawing of the net on cardboard we can obtain the surface of the polyhedron, rather than just its skeleton.

There is other information that we can extract from the table.
For example, the fact that the cube and the octahedron are "relatives":
they have the same number of edges, and each of them has as many faces as the other has vertices.
Also, the number of edges meeting at any vertex of one is the same as the number of edges around a face of the other.
This relationship is made clearer in the figures on the left: if we consider the centres of the faces of a cube and we join two of them when the corresponding faces have an edge in common, we will obtain an octahedron.

And vice versa, if we perform the same procedure starting from an octahedron, we will obtain a cube.

Also the dodecahedron and the icosahedron are relatives, in the sense above, as we can notice both from the numbers in the table and the figures.

Now what is "relative" of the tetrahedron?
There are no more regular polyhedra, but the fact that the regular tetrahedron has as many faces as vertices may help you to find an answer.

When we throw a die we expect, since its shape is evidently symmetric, that any of the six numbers marked on its faces has the same probability to appear.

In some toy shops, besides the usual six-faced dice, you can also buy dice with four, eight, twelve e twenty faces.

What is special about the five polyhedra represented by these dice? To see what properties they have, we first consider some other examples of polyhedra.

We could build a die from a geoid, for example. Such a polyhedron could even seem more "regular" than the five dice above since its shape is "closer" to that of a sphere, but, if we study it carefully, we notice that its vertices are not all of the same type: at some vertices (like A) five edges radiate and so five triangles come together; at others (like B) six edges start and so six triangles meet.

We call a polyhedron regular if all its faces are equal regular polygons and the same number of faces meet at every vertex. In such a polyhedron all the faces, edges and vertices are indistinguishable.

None of the polyhedra in the second figure above is regular, while the five dice are all examples of regular polyhedra. Are there any more? (After all, in the plane we can build regular polygons with as many edges as we like.) The answer is: no. In space, there are only the five regular polyhedra of the diceshown in the first figure.

Even the ancient Greeks knew this fact; it so impressed them that the Pythagoreans, and later Plato, built their cosmogonical theories on it, associating the fundamental constituents of nature to the five regular polyhedra. Because of this, the regular polyhedra are also called Platonic solids.

In his `Timaeus', Plato describes the tetrahedron as "element and germ" of fire, the octahedron of air, the icosahedron of water, the cube of earth, and the dodecahedron represents the image of the entire universe: "There still remained a fifth construction, which the god used for embroidering the constellations on the whole heaven".
The last book of the Euclid's "Elements" is dedicated to the five Platonic solids, to some of their properties and relationships, and to the proof that there are no other regular polyhedra.

Models of regular polyhedra have been found from pre-Greek civilisations. The harmony of their proportions and their mathematical properties give them a charm that has been captivating artists and scientists ever since.
Dodecahedra had a religious meaning in the Etruscan civilisation, and were also used as dice by the Romans.
In the Renaissance, regular polyhedra were used as a good subject for studies of perspective: we can find them in works by Paolo Uccello, Piero della Francesca, Albrecht Dürer, Leonardo da Vinci, Luca Pacioli and Leonardo Pisano (also known as Fibonacci).

In 1595 Kepler believed that he had "penetrated the secrets of the Creator" because he had elaborated a model of a planetary system (that later became discredited) by using the Platonic solids to describe the distances between the elliptic orbits of the six planets then known.

Shapes inspired by regular polyhedra are also common in modern art and design, from works of Escher to modular architecture, to objects by Munari.
One last curiosity: in France they built dodecahedral bottle banks.

Even nature shows us examples of regular polyhedra: some cristals , viruses which often have icosahedral form, simple living organisms like radiolaria, ...

How many uniform polyhedra are there, and what is their structure?

To begin with, there are five regular polyhedra.

Another standard example (more precisely a family of infinitely many examples) is given by prisms. A right prism with basis a regular polygon is a (4,4,...) uniform polyhedron as long as the height of the prism makes all the lateral faces squares.

There are infinitely many uniform polyhedra among prisms, one for every regular polygon which is fixed like its basis - in particular, if the basis is a square you get a cube). Here on the left you may see a (4,4,6).

There is another family of polyhedra (maybe less known), which contains infinitely many uniform polyhedra: it is the family of antiprisms. Here on the right you may see a uniform antiprism of type (3,3,3,5): each vertex is adjacent to three equilateral triangles and a pentagon.
It is easy to imagine how you can get a uniform polyhedron if you start from any regular polygon and not just from a pentagon. In particular, if the basis is an equilateral triangle, you get the octahedron).

Are there other uniform polyhedra in addition to prisms, antiprisms and regular polyhedra? How many and which ones? In fact there are some of them, but not too many: thirteen all together, which are sometimes called archimedean polyhedra. If we want to list them, as we did for the others, with a numerical symbol, the full list can be given in the following table:

The table has been made following a pattern which can be useful to describe the corresponding polyhedra. The polyhedra in the first column are related to the cube, those in the second are related with the dodecahedron and those in the third column with the tetrahedron.

Almost all the polyhedra in the first column have the same symmetry of the cube. They can be seen in a kaleidoscope, in particular that associated with the cube. Only the polyhedron on the last row has a symmetry which is similar to that of a cube, but differs from it because it contains only rotations - thus it can not be seen in a kaleidoscope.

Analogously, all the polyhedra in the second column - except the last one -, scan be seen in the kaleidoscope of the dodecahedron; those in the last column can be seen in the kaleidoscope of the tetrahedron.

The row layout is not random. For instance, the (3,8,8) polyhedron represented below can be obtained from a cube by "smoothing" the vertices so that one gets equilateral triangles and regular octagons.
The other two polyhedra in the same row of the (3,8,8) are obtained in the same way, but starting from a different polyhedron. To obtain a (3,10,10) one starts from a dodecahedron and to obtain a (3,6,6) one starts from a tetrahedron.
Something similar occurs for the other rows of the table.

People who like numerical patterns may have noticed that the symbols of the second column can be obtained from the corresponding ones of the first column by substituting some "4" in "5" and some "8" in "10". How can we explain this phenomenon? And why is only the central "4" in (3,4,4,4) exchanged with a "5" and not the other two "4"? If you observe the polyhedra that you find below, you may find an answer...

Regular polyhedra are polyhedra where vertices, edges and faces can not be distinguished. This requirement is so strong that there are only five regular polyhedra. At any rate, other polyhedra - although not regular - can strike us for their symmetry.

Here on the left you may see football and, below, a polyhedron with the same structure. It is not a regular polyhedron because the faces are not all equal - there are hexagons and pentagons. Yet, it does have a symmetry. The faces - although not all alike - are regular polygons. Moreover, if we look at a vertex, we realize that at each vertex there are a pentagon and two hexagons. We can imagine to make a plaster cast around a vertex and obtain a sort of three-dimensional "stamp", which can be adapted to any other vertex.

The football is an example of a uniform polyhedron, i.e., a polyhedron which has faces that are regular polygons and all vertices can not be "distinguished". This latter property allows one to identify the uniform polyhedra with numerical symbols, which represent the polygons adjacent to every vertex. For instance, (5,6,6) is the symbol of the football.
We could not use a symbol of this type for the polyhedron here on the right. Indeed, it has vertices with four triangles, others with two squares and two triangles, and others with three squares.

What do we mean by saying that the vertices can’t be distinguished? After all, vertices are points and,
undoubtedly, all points are equal! Already the example of the "three-dimensional stamp" tells us something about all the faces adjacent to a vertex and not only the vertex!

Actually, the problem is even subtler, and to say that the vertices can’t be distinguished, it is not even enough that a three-dimensional "stamp" adapts to any vertex with the same arrangement of faces.

The polyhedron on the right (the Miller polyhedron) has three squares and an equilateral triangle at every vertex; so a convenient three-dimensional "stamp" would adapt to any vertex. Yet, the vertices can be distinguished!
It is possible to see it by looking at the chain of squares that rounds the "equator" of the polyhedron. There is just one chain of this type because, if we start from other pairs of adjacent squares , we can’t make a full round. So, two vertices may be distinguished because they do not have the same position with respect to this chain like, for instance, the two vertices A and B.
It is impossible to move the polyhedron so that the vertex A maps to the vertex B and the polyhedron stays in the same position.

The other polyhedron you may see below "looks like" the Miller polyhedron quite a lot, but it is "more symmetric". There are three chains of squares and each vertex belongs to two of these three chains. With a little bit of patience, you may verify that the same three-dimensional "stamp" adapts to any vertex. Furthermore, for any two vertices there exists a symmetry of the polyhedron which maps the two vertices one onto the other. This is the requirement which must be fulfilled to talk about a uniform polyhedron (and to denote it by the symbol (3,4,4,4)).

Some virtual leaflets ("The five regular polyhedra" and "Why only five") go over the regular polyhedra; others explore a family of polyhedra which are not regular but they look "rather" symmetric (the so-called "uniform polyhedra").

Moreover, some go over an incursion to the land of "The regular star polyhedra". Finally, here you find a virtual leaflet on "Euler's formula" which connects the number V of vertices, the number E of edges and that of faces F.

The photograph above shows the blue kaleidoscope, in which a
virtual cube can be produced, with a single ball placed in it.
The arrangement of the reflected images of the ball helps us to discover
the angles between pairs of mirrors of the kaleidoscope: there are some rings of four balls, and each of these corresponds to a pair of mirrors that meet at a right angle; there are some rings of six balls, corresponding to mirrors meeting at 60 degrees; finally, there are some rings of eight balls, which correspond to mirrors that meet at 45 degrees.

We can recover these angles if we consider the sphere corresponding to the kaleidoscope.The surface of the sphere is divided into 48 "triangles", all equal to each other, whose angles are 90, 60 and 45 degrees - in fact there are vertices surrounded by four triangles (hence each of the four angles meeting there is 90=360/4 degrees), vertices surrounded by six triangles, and vertices surrounded by eight triangles.

But 90+60+45 is 195, and not 180, so we have a triangle whose angles do not add up to 180 degrees! We should not be too surprised at this fact, because we are not really dealing with a proper triangle: it is a "fat" triangle, drawn on a sphere, whose edges are not straight lines, but the nearest thing that can be drawn on a sphere, that is arcs of great circles.

What kind of geometry applies to triangles on a sphere? What does it have in common with, and how does it differ from, the usual plane geometry that we are familiar with? We have just seen one difference: the sum of the angles of a triangle is no longer 180 degrees.
Moreover, the angle sum is not even the same for all triangles: by examining the spheres corresponding to the other two kaleidoscopes, we can see that one sphere is divided into 24 triangles, all equal, each of which has one angle of 90 degrees and two of 60, while the other sphere is divided into 120 equal triangles, each of which has an angle of 90 degrees, one of 60, and one of 36. The sum of the angles in the first case is 210 degrees, in the second case it is 186 degrees.

The situation is summarised in the table above: the first column contains the sum of the angles of the triangle; the second column, the angular excess (that is by how much this sum exceeds 180 degrees); the third column, the area of the triangle as a fraction of the area of the sphere.

We can see that the larger the triangle, the larger is the sum of its angles.
For the triangles of the sphere corresponding to the red kaleidoscope, which are the smallest ones (their area is 1/120 of the area of the sphere), the angular excess is only 6 degrees.
For the triangles of the sphere corresponding to the blu kaleidoscope, which are larger (their area is 1/48 of the area of the sphere), the angular excess is 15 degrees.
Lastly, we have the triangles on the sphere corresponding to the yellow kaleidoscope: notice that each triangle is formed by two of the triangles on the "blue" sphere so the area is doubled; this time the angular excess is 30 degrees. So we see that doubling the area of a triangle also doubles its angular excess.

This is not a special case. One of the characteristics of the geometry on the sphere which distinguishes it from the usual geometry on the plane is the fact that similitude does not exist: that is we cannot - as on the plane - draw two figures with the same shape but of different sizes.
The size affects the shape, and vice versa: we cannot find a spherical triangle larger than the one shown here in the figure which also has the same angles.
Of course, we could start from a larger sphere, but if the angles are of 90, 60 and 45 degrees, the area of the triangle will still be 1/48 of the area of the sphere. In fact, the area of a spherical triangle is proportional to its angular excess.

Are there also "thin" triangles besides "fat" ones? That is, are there triangles for which the sum of their angles is less than 180 degrees? The answer is yes, and the corresponding geometry is called hyperbolic geometry. The figure to the right (which recalls the structure of some drawings by Escher) is a companion of the spheres corresponding to the kaleidoscopes (and of the grids corresponding to the planar triangles): here too there are some "triangles" (thin ones), all equal to each other, and this time the angles are of 90, 60 and 360/14 degrees.

The interesting thing is that this situation is totally "symmetric" with respect to what we have seen before for the geometry on the sphere: again, in hyperbolic geometry, there is no similitude; the sum of the angles of a triangle is alway less than 180 degrees; the shape of a triangle determines its size; the area of a triangle is proportional to its angular defect, that is to how much the sum of its angles is less than 180 degrees. But this is just the beginning of a long story ...