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For excellent papers on the significance and history of Euler's theorem see [Gott96],
[GrüS94], [HilP94], and [HilP96]. As an example, we shall now give a simple proof of
the following fact that was already known to Euclid.
6.1.2. Theorem.
There are no regular polyhedra other than the ones shown in
Figure 6.1.
Proof. Assume that we have a regular polyhedron for which every face has h edges
and every vertex belongs to k edges. Since every edge has two vertices and belongs to
exactly two faces, it follows that
nh
==
2
n
n k
.
f
e
v
Substituting these identities into Euler's formula, we get
n
k
n
h
e
e
2
-+
n
2
=,
2
e
or equivalently,
1111
2
=+-.
nhk
(6.2)
e
Now h and k are always assumed to be larger than 2 for a polyhedron. On the other
hand, if both h and k were larger than 3, then equation (6.2) would imply that
1111
2
1
4
1
4
1
2
0
<
= +-£+-=
nhk
0
,
e
which is impossible. Therefore, either h or k must equal 3. If h = 3, then
11
3
11
2
0
<=+-
n
k
e
implies that 3 £ k £ 5. By symmetry, if k = 3, then 3 £ h £ 5. It follows that
(
) = (
)
(
)
(
)
(
)
(
)
hkn
,,
336
,, ,
4312
,, , ,, , ,,
3412
5330
,
or
3530
,,
.
e
These values are in fact realized by the tetrahedron, the cube, the octahedron, the
dodecahedron, and the icosahedron, respectively. See Figure 6.1 again.
Other early results in topology dealt with the following:
Map-coloring problems: The problem is to find the smallest number of colors
required to color an arbitrary map in such a way that no two adjacent countries have
the same color. For example, the countries A, B, C, and D in Figure 6.5 show that it
takes at least four colors. Countries such as E and G that meet in a single point are
not considered adjacent. The conjecture that four colors suffice to color any map was
finally proved in 1976 with the help of a computer ([AppH77]). The map-coloring
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