Graphics Reference
In-Depth Information
P
3
P
i
P
3
¢
P
1
¢
P
2
P
1
P
i
¢
P
2
¢
(a)
(b)
Figure 2.10.
Case 4 of the existence theorem.
Similarly, one can show that
CD
•
BD
= 0. It follows easily from this that the
vectors
AD
and
BD
are parallel and that
D
lies on
L
. This proves the claim and the
lemma.
Case 4.
k > 3.
We claim that if the first three points
P
1
,
P
2
, and
P
3
are linearly independent, then
the map defined in Case 3 that sends them to
P
1
¢,
P
2
¢, and
P
3
¢, respectively, will already
send all the other points
P
i
, i > 3, to
P
i
¢. Figure 2.10 shows how the argument pro-
ceeds. In Figure 2.10(a) we show three circles with centers
P
1
,
P
2
, and
P
3
and radius
P
1
P
i
,
P
2
P
i
, and
P
3
P
i
, respectively. The point
P
i
lies on the intersection of these circles.
Figure 2.10(b) shows the corresponding circles around the image points. One has to
show that
P
i
will get sent to the intersection of those circles and that this is the same
as the point
P
i
¢.
We have just given a constructive proof of the following theorem.
2.2.5.2. Theorem.
(The Existence Theorem for Motions) Given points
P
1
,
P
2
,...,
P
k
and
P
1
¢,
P
2
¢,...,
P
k
¢ with the property that |
P
i
P
j
| = |
P
i
¢
P
j
¢| for all i and j, then there
is motion M, so that M(
P
i
) =
P
i
¢.
Proof.
See [Gans69] for missing details in the discussion above.
Now that we have answered the question of the existence of certain motions, let
us look at the issue of uniqueness more closely?
2.2.5.3. Theorem.
A motion that has two distinct fixed points fixes every point on
the line determined by those points.
Proof.
Let M be a motion and assume that M(
A
,
B
) = (
A
,
B
) for two distinct points
A
and
B
. Let
L
be the line determined by
A
and
B
and let
C
be any point of
L
. If
C
=
A
+ t
AB
, then Lemma 2.2.2 implies that M(
C
) = M(
A
) + tM(
A
)M(
B
). In other words,
M(
C
) =
C
and the theorem is proved.
2.2.5.4. Theorem.
Any motion of the plane that leaves fixed three noncollinear
points must be the identity.
