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In-Depth Information
Proof.
Let
A
,
B
, and
C
be three noncollinear points and let M be a motion with
M(
A
,
B
,
C
) = (
A
,
B
,
C
). Let
P
be any other point in the plane. We would like to show that
M(
P
) =
P
. By Theorem 2.2.5.3, M is the identity on the three lines determined by the
points
A
,
B
, and
C
. If
P
lies on these lines we are done; otherwise, Lemma 2.2.3 implies
that
P
lies on a line through two distinct points that lie on two of these lines. Using
Theorem 2.2.5.3 we can again conclude that M fixes
P
.
2.2.5.5. Corollary.
Two motions of the plane that agree on three noncollinear points
must be identical.
Proof.
Let M and M¢ be motions and assume that M(
A
,
B
,
C
) = M¢(
A
,
B
,
C
) for three
noncollinear points
A
,
B
, and
C
. Consider the motion T = M
-1
M¢. Since T(
A
,
B
,
C
) =
(
A
,
B
,
C
), Theorem 2.2.5.4 implies that T is the identity, that is, M = M¢.
2.2.5.6. Corollary.
Every motion of the plane is a composite of a translation, a rota-
tion, and/or possibly a reflection.
Proof.
This follows from the construction in Case 3 above and Corollary 2.2.5.5.
Theorem 2.2.5.3 raises the question whether a motion of the plane that fixes two
distinct points is actually the identity map. That is not the case. Reflections, such
as the map T(x,y) = (x,-y), can leave all the points of a line fixed but still not be the
identity.
2.2.5.7. Theorem.
A motion M of the plane that fixes two distinct points
A
and
B
is either the identity map or the reflection about the line
L
determined by
A
and
B
.
Proof
. By Theorem 2.2.5.3, M fixes all the points on the line
L
. Let
C
be any point
not on
L
. Lemma 2.2.5.1 shows that
C
gets mapped by M either to itself or to its reflec-
tion
C
¢ about the line
L
. The theorem now follows from the Corollary 2.2.5.5 since we
know what M does on three points.
2.2.6
Rigid Motions in the Plane
2.2.6.1. Lemma.
Every rotation R of the plane can be expressed in the form R =
R
0
T
1
= T
2
R
0
, where R
0
is a rotation about the origin and T
1
and T
2
are translations.
Conversely, if R
0
is any rotation about the origin through a
nonzero
angle and if T is
a translation of the plane, then both R
0
T and TR
0
are rotations.
Proof.
Suppose that R = TR
0
T
-1
, where R
0
is a rotation about the origin and T is a
translation. By Theorem 2.2.4.2 we can move the translations to either side of R
0
,
which proves the first part of the lemma. The other part can be proved by showing
that certain equations have unique solutions. For example, to show that TR
0
is a rota-
tion, one assumes that it is a rotation about some point (a,b) and tries to solve the
equations
(
xa
-
)
cos
q
--
(
yb
)
sin
q
+ a = x cos
q
-
y
sin
q
+
c
