3.
Three hollowed, triangular prisms are used to make a torus. Carefully count the num
ber of vertices, faces, and edges for this prismatic torus. Compute the Euler characteristic
for this torus. (B+S 5.3.29)
Solution.
This torus has nine vertices, six on the “outside” of the hole and three on the
“inside”. It has nine faces, three arising from each of the three original triangular prisms.
It has 18 edges – these can be decomposed into three triangles going around the torus the
“long” way and three triangles going around the short way. So the Euler characteristic is
V

E
+
F
= 9

18 + 9 = 0.
4.
Carefully count the number of vertices, faces, and edges for a twoholed torus. One
way to view this twoholed torus is as two copies of the torus in problem 3, with one side
removed from each and then the open edges glued together.
This operation is called the
connected sum
. Compute the Euler characteristic for this twoholed torus. (B+S 5.3.30)
Solution.
Glue two copies of the torus together along a foursided face. The number of
vertices of the resulting torus is twice the number of vertices of the original torus, minus
the number which disappear in the gluing operation. In the gluing operation four vertices
disappear – four pairs of corresponding vertices on the two copies become single vertices. So
there are 2
·
9

4 = 14 vertices. Similarly, four edges disappear in the gluing operation, so
there are 2
·
18

4 = 32 edges. Finally, two faces are removed in the course of the operation,
so there are 2
·
9

2 = 16 faces. The Euler characteristic is thus 14

32 + 16 =

2, as it is
for any twoholed torus.
5.
Recall that in class we constructed the Koch snowflake, a shape which had finite area
but infinite perimeter, by starting with an equilateral triangle, replacing each edge with four
edges of onethird the length in a certain way, and repeating this process infinitely many
times. (See p. 435 of the text for a picture of what happens to
one
edge when we do this.)
We can construct a similar “square Koch snowflake” by beginning with a square and
replacing each edge with
five
edges of onethird the length, such that the three middle edges
form a square. (Note that the boundary of this snowflake actually touches itself at various
points.) The first few iterations of this appear as follows.
Assume that the original square has side length 1.
3
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(a) What is the length of the
n
th curve in this series? (Note that the
n
th curve in this
series is made up of a large number of short segments of the same length; you should find
the length of each short segment and the number of them.)
Solution.
The first curve consists of four segments; in each iteration after that there are
five times as many curves, so there are a total of 4
·
5
n

1
segments in the
n
th curve in the
series. The segments in each curve are onethird the length of those in the previous one, so
the length of each segment in the
n
th curve is (1
/
3)
n

1
. Thus the total length is 4
·
(5
/
3)
n

1
.
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 Summer '09
 Lugo
 Math, Angles, triangle, vertices, Cantor, Truncation, Archimedean solid

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