Math 55  Homework 9 Solutions
Monday, August 8, 2011
1. Suppose the
G
is a simple, undirected, planar graph and every vertex in
G
has degree
at least 5. Prove that
G
has at least 12 vertices of degree exactly 5.
We will prove this for when
G
is connected. If
G
is not connected, then each connected
component must have at least 12 such vertices, so there are at least 12 overall. Since
G
is connected and planar, by Euler’s formula
r
=
e

v
+2
. Since
G
is simple, each region
has at least 3 edges on it. By counting edgeregion touchings, we then know
2
e
≥
3
r
. Let
v
5
be the number of vertices of degree 5, and let
w
be the number of vertices of degree
6 or more. So
v
=
v
5
+
w
. By counting vertexedge touchings, we get
2
e
≥
5
v
5
+ 6
w
.
Putting this all together, we get:
v
5
=
v
5
+ 6 (
v

v
5

w
) = 6
v

(5
v
5
+ 6
w
)
≥
6(
e

r
+ 2)

2
e
= 2(2
e

3
r
) + 12
≥
12
2. Suppose that
T
is a tree and
e
is an edge which goes between two vertices of
T
, but is
not in
T
. Prove that if you add
e
to
T
, then the new graph has exactly one cycle (which
doesn’t repeat a vertex).
Let
(
v,w
)
be an edge not in
T
. Since
T
is connected,
T
contains a path from
(
v,w
)
. By
adding
(
v,w
)
to this path, we get a cycle. So
T
+ (
v,w
)
has at least one cycle.
Suppose
T
+ (
v,w
)
had two cycles
C
1
and
C
2
. Both
C
1
and
C
2
must contain
(
v,w
)
,
since
T
is a tree and has no cycles on its own. If
C
1
6
=
C
2
, let
(
s,t
)
be some edge in one
but not the other. Without loss of generality, let
(
s,t
)
be in
C
1
. In
C
1
, pick the longest