triangle is the number of 2faces of the (
n

1)triangle, plus the number of
edges. That is, letting
F
2
(
n
) denote the number of 2faces of the
n
triangle,
F
2
(
n
) =
F
2
(
n

1) +
E
(
n

1)
and working backwards,
F
2
(
n
) = 1 + 3 + 6 +
. . .
+ (
n

1)
n/
2
.
This is in fact
n
(
n
+ 1)(
n

1)
/
6, but this is hard to guess.
Similarly, 3faces of the
n
triangle come from 3faces of the (
n

1)triangle and 2faces
of the (
n

1)triangle. Letting
F
3
(
n
) denote their number,
F
3
(
n
) =
F
3
(
n

1) +
F
2
(
n

1)
.
In fact, all of these numbers are binomial coefficients, the numbers from Pascal’s triangle!
This occurs since there is a face of an
n
triangle corresponding to
any
set of its vertices we
might pick.
2.
Consider the five regular solids with their vertices cut off. These objects are called
truncated solids
. For each truncated solid, count the number of vertices, edges, and faces,
and verify that the Euler characteristic is correct. (See p. 370 of the text for pictures. Also,
note that the things you should be “counting” here include those vertices, edges, and faces
which are hidden in the drawings.) (B+S 5.3.22)
Solution.
For the truncated tetrahedron, there are three vertices where each vertex of
the original tetrahedron used to be. Thus there are 4
·
3 = 12 vertices. The six edges of the
original tetrahedron remain, and in addition there are three “new” edges surrounding each
vertex of the original tetrahedron, giving 6 + 4
·
3 = 18 edges. Finally, the number of faces
is the number of faces of the original tetrahedron, plus one new face for each vertex, giving
4 + 4 = 8 faces. The Euler characteristic is therefore 12

18 + 8 = 2.
This pattern holds in general. The number of vertices of a truncated solid is the number of
vertices of the original solid, times the number of faces meeting at each vertex of the original
solid. The number of edges of a truncated solid is the number of edges of the original solid,
plus the number of vertices of the
truncated
solid which we have already computed. Finally,
the number of faces of the truncated solid is the number of faces of the original solid, plus
the number of vertices of the original solid. This gives the values
solid
vertices
edges
faces
trunc. tetrahedron
4
·
3 = 12
6 + 4
·
3 = 18
4 + 4 = 8
trunc. cube
8
·
3 = 24
12 + 8
·
3 = 36
6 + 8 = 14
trunc. octahedron
6
·
4 = 24
12 + 6
·
4 = 36
8 + 6 = 14
trunc. dodecahedron
20
·
3 = 60
30 + 20
·
3 = 90
12 + 20 = 32
trunc. icosahedron
12
·
5 = 60
30 + 12
·
5 = 90
20 + 12 = 32
We can easily check that
V

E
+
F
= 2 for each solid. Note that the numbers which
are duplicated between the truncated cube and truncated octahedron, and between the
2
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truncated dodecahedron and truncated icosahedron, are not duplicated by coincidence; this
is yet another consequence of duality.
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 Summer '09
 Lugo
 Math, Angles, triangle, vertices, Cantor, Truncation, Archimedean solid

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