Polygons, including the Art Gallery Problem, should have been treated before

Here ares some examples. The empty red triangles in the third example still count as faces.

The flat polygons used as sides of the polyhedron are called theFor the moment, let us define polyhedra aas follows:
A **polyhedron** is a three-dimensional object created by putting
flat polygons together such that two polygons are joined by a common edge, and
every edge bounds two polygons. Obviously every polyhedron could also
be viewed as a graph, since we have vertices---the vertices of the polygons---
that are connected by edges. The **degree** of a vertex in the polygon is
the number of edges it is incident with. It is the same as the degree in the
graph-theoretical sense.

For some famous polyhedra, all faces have the same shape---they are regular n-gons for a fixed value of n. This is true for the five Platonic Solids, where all faces are regular polygons of only one shape, and every vertex has the same degree.

Name (click for animation) | Tetrahedron | Cube | Octahedron | Dodecahedron | Icosahedron |

Look | |||||

degree (number of faces or edges) at each vertex | 3 | 3 | 4 | 3 | 5 |

V=number of vertices | 4 | 8 | 6 | 20 | 12 |

E=number of edges | 6 | 12 | 12 | 30 | 30 |

F=number of faces | 4 | 6 | 8 | 12 | 20 |

degree (number of edges) of each face | 3 | 4 | 3 | 5 | 3 |

Since polyhedra are also graph, we know from the graph page:

Therefore the number of edges doesn't have to be counted but can be computed from the other parameters. In particular, since the graphs of Platonic solids are r-regular, (all vertices have the same degree r) we know that 2E=rV for them, i.e. E=rV/2. For instance, for the cube we have 3·8/2 edges, for the icosahedron we have 5·12/2 edges.

We want to be able to compute the number of edges also for non-Platonic polyhedra,
i.e. for nonregular graphs. For this purpose we use the abbreviation of
**V _{n}** for the number of vertices of degree
n in every polyhedron, and in the same way

Name | doghouse | double pyramid | Napoleon's hat | UFO | ||

Look | The base is considered to be one 6-gon, instead of six triangles. |
|||||

V_{3},V_{4},V_{5....} |
6,3 | 2,3 | 2,2,2 | 2,5 | 0,6,0,2 | 6,0,0,1 |

V=number of vertices | 9 | 5 | 6 | 7 | 8 | 7 |

E=number of edges | 15 = (3·6+4·3)/2 = (3·5+4·0+5·3)/2 | 9 = (3·2+4·3)/2 = 3·6/2 | 12 = (3·2+4·2+5·2)/2 = 3·8/2 | 13 = (3·2+4·5)/2 = (3·6+4·2)/2 | 18 = (3·0+4·6+5·0+6·2)/2 = 3·12/2 | 12 = (3·6+4·0+5·0+6·1)/2 =(3·6+4·0+5·0+6·1)/2 |

F=number of faces | 8 | 6 | 8 | 8 | 12 | 7 |

F_{3},F_{4},F_{5,...} |
5,0,3 | 6 (all triangles) | 8 (all triangles) | 6,2 | 12 (all triangles) | or: 6,0,0,1 |

nonconvex | self-dual |

Now the sum of all degrees of all vertices equals 3·V_{3}
+4·V_{4}+5·V_{5}+ .... .
Therefore the above remark can be restated as:

Start with any polyhedron. The middle of each face will become a new vertex. Now connect those new vertices by straight lines---the new edges---provided the corresponding faces share an edge. We get another polyhedron, having as many edges as the old one, as many vertices as the old one had faces, and as many faces as the old one had vertices. The dual of the cube is the octahedron, and the dual of the dodecahedron is the icosahedron. The tetrahedron is its own dual. The dual of the dual is the old poyhedron.

Recall that the degree of a face is the number of edges bounding it. It equals the degree of the corresponding vertex in the dual.

**Truncating** a polyhedron means: All vertices are sliced off (with a sharp knife).
The cuts form new faces, with new vertices and new edges.
Here you can see the truncated
dodecahedron, and here the truncated icosahedron---the soccer ball,
also displayed above in the logo.

How do V, E, F of the original polyhedron relate to
vertex number V_{T},
edge number E_{T},
and face number F_{T} of the truncated polyhedron?

- All former faces remain, although in modified form. Additionaly, at each former vertex
a face from the cut pops up. Therefore F
_{T}=F+V. - All former edges also remian, although shortened.
Additionaly, at each cut at former vertex x, d(x) new edges pop up,
where d(x) is the degree of that vertex.
Therefore we get as many new edges as the sum of all former degrees.
According to our previous formula, this number equals 2E.
Therefore E
_{T}= E+2E=3E. - All old vertices are cut off, but at every former vertex x,
d(x) new vertices pop up at the cut face. therefore we get as many new vertices
as the sum of all former degrees is, which again equals 2E.
Therefore V
_{T}=2E.

In the same way as above, let us investigate how the
vertex number V_{G}, edge number E_{G}, and face number F_{G}
of the glued polyhedron
relate to these parameters V_{1}, E_{1}, F_{1} of the first polyhedron and
V_{2}, E_{2}, F_{2} of the second polyhedron.
Assume both polyhedra are glued together along an n-gon.

- Two faces are lost in order to make the glueing possible, one from each former
polyhedron. Therefore
F
_{G}= F_{1}-1+F_{2}-1 = F_{1}+F_{2}-2 - All edges remain, but the n edges in each of the glued n-gon are identified.
Thus overall n edges get lost by the process. We get E
_{G}= E_{1}+E_{2}-n. - In the same way than with the edges, n vertices get lost too, since the vertices on
the two n-gons are identified. Thus we get V
_{G}= V_{1}+V_{2}-n.

Our data proved that you cannot make a polyhedron with less edges than faces.

- Wikipedia's page on Polyhedra, containing unfortunately more than you want to know about them.
- Paper Models of Polyhedra: Contains models of many of the most famous polyhedra for creating them, and also nice pictures.
- The Platonic Solids, A rather elementary explanation of much what is done also on the present pages.
- The Uniform Polyhedra Describes the 75 plus some so-called uniform polyhedra.
- Another small page about uniform polyhedra, displaying all of them in movable models. I find it rather nice.
- Abigail Kirk, Euler's polyhedron formula, a very well-written article on the subject.
- Konrad Polthier, Imaging maths - Unfolding polyhedra,about paper models of polyhedra.
- ................

- You glue two octahedra together along a face. How many vertices, edges, and faces does the resulting polyhedron have. What are the degree of the vertices? Why is the resulting polyhedron not Platonic?
- How many vertices, edges, and faces does the truncated icosahedron have? Wjy is it not a Platonic solid?
- How many vertices, edges, and faces does the truncated dodecahedron have? Wjy is it not a Platonic solid?
- Draw the graph of the truncated cube. Draw also the graph of the dual of the truncated cube. How many vertices, edges, and faces does the dual of the truncated cube have?