Transformations and Isometries
Definition: A transformation in absolute geometry is a function f that
associates with each point P in the plane some other point PN in the
plane such that (1) f is one-to-one (that is, if
for any
two points P and Q, then P = Q). and (2) f is onto (that is, if Q is any
point in the plane, then there is a point P such that
). We
will often denote f(P) by PN
Definition: A transformation of the plane is said to have a point A as a
fixed point iff f(A) = A.
Definition: A transformation of the plane is said to be the identity
mapping if every point of the plane is a fixed point. The identity
mapping is denoted by e. Thus e(P) = P for all points P of the plane.
Definition: An isometry is a transformation of the plane that preserves
distances; that is, if P and Q are two points, then
.
Theorem: An isometry preserves collinearity, betweenness, and
angles. That is, if A, B, and C are three points in the plane and their
images under an isometry are AN, BN, and CN , then:
1.
2.
3.
If A, B, and C are collinear, then AN, BN, and CN are also
collinear.
If A*B*C, then AN*BN*CN.
If A, B, and C are non-collinear, mpABC = mpANBNCN
~ For (1), suppose A, B, and C are collinear. One of the points must
be between the others; WLOG, suppose A*B*C. Then AB + BC =
AC, so because we have an isometry ANBN + BNCN = ANCN. By the
triangle inequality, the only way AN, BN and CN could fail to be
collinear is if ANBN + BNCN > ANCN, which we have shown is not true.
In addition, we also conclude AN*BN*CN, which proves (2). For (3),
note that AB = ANBN and AC = ANCN and BC = BNC N so by SSS ªABC
is congruent to ªANBNCN. By CPCF, the corresponding angles have
equal measure. €
Theorem: The inverse of an isometry is an isometry.
~ Let f be an isometry and
plane. Let
Then:
thus
. Let R and S be points of the
, so
.
preserves distances.
Theorem: The composition of two isometries is an isometry.
~ Choose points A and B, and isometries f and g. Then
and
. €
Theorem: In addition to preserving lines, angles, betweenness, angles,
and distances, an isometry f also preserves:
a.
b.
c.
d.
Segments: for points A and B,
(and
)
Betweenness of rays: If
, then
.
Triangles: for noncollinear A, B, and C,
is a
triangle and
.
Circles: If " is a circle with radius r and center O, then
is a circle with radius r and center
.
~ For (a), if X is a point such that A*X*B, then AN*XN*BN so XN is on
segment
. Now f–1 is an isometry as well, so if AN*Y*BN,
, so
, so for every Y on
segment
there is an X (namely
) on
such that
. Statement (b) follows from preservation of angle
measures. Statement (c) follows from SSS. Statement (d) is almost
immediate from the definitions of isometry and circle. €
Definition: Let l be a line. Define the reflection over line l as a
function sl that assigns to each point P a point PN defined as follows:
1.
2.
If P is on l, then sl(P) = P.
Otherwise, drop a perpendicular from P to l, with foot
F. Find a point PN on the other side of l from P that lies
on the perpendicular and such that PF = FPN. Let sl(P)
= PN.
Note that sl has the property that l is the perpendicular bisector of
every segment formed by a point and its image under the function.
Theorem: A reflection is an isometry.
~ We show that for any two points P and Q, PQ = PNQN. We consider
cases:
Case 1: If P and Q are both on l they are fixed so P = PN and Q = QN
and the result is trivial.
Case 2: P is on l and Q is not. P is fixed so P = PN. Then ÎQFP
–QNFP by SAS, and CPCF gives QP = QNP = QNPN.
Case 3: Neither P nor Q are on l,
but they are on the same side of l.
Let F and G be the feet of the
perpendiculars to l from Q and P,
respectively. By SASAS,
FQPG –FQNPNG. By CPCF,
PQ = PNQN.
Case 4: Neither P nor Q are on l, and
are on opposite sides of l. As in Case
3, let F and G be the feet of the
perpendiculars to l from Q and P,
respectively. Again by SASAS,
FQPNG –FQNPG. By CPCF, PQN =
PNQ and pQPNP – pQNPPN. Then SAS
gives us ÎQNPPN –ÎQPNP and CPCF
gives us PQ = PNQN. €
We are now going to prove an important result about isometries: Every
isometry is defined by what it does to three noncollinear points. We
make use of the ideas of fixed points and the identity transformation.
We need a couple of nice lemmas first:
Lemma: If P and Q are points on a line
BP = BQ, then P = Q.
such that AP = AQ and
~ Use the ruler postulate to establish a coordinate system. Using the
notation P[x} to denote that the coordinate of point P is x. Now let
A[0], B[b] with b > 0, P[p], and Q[q]. Then
, so p = ±q. If p = q, then P
= Q by the ruler postulate and we are done. Otherwise, suppose q = -p.
Now
so (p - b) = ± (q - b). If (p - b) =
(q - b), then p = q, and this combined with q = -p gives us p = -p so p =
0, and P = A = Q. If (p - b) = -(q - b), then p - b = - q + b. This,
together with q = -p, gives us -b = b, or b = 0, a contradiction. Thus in
all possible cases, P = Q.
Lemma: If an isometry f fixes two points A and B, it also fixes every
point on
.
~ Let P be on
. Because f is an isometry it will map
to
another line, and since A and B are fixed it will in fact map
unique line containing A and B, namely,
point of
, we need to show that
to the
itself. Now if P is a
. PN is on
and in fact APN = ANPN = AP and BPN =BNPN = BP since f is an
isometry. By the lemma above, P = PN, so P is a fixed point for f. €
Theorem: The only isometry to have three noncollinear fixed points is
the identity.
~ Let A, B, and C be noncollinear, and suppose f has A, B, and C as
fixed points. That is,
. We will show
that for any other point P of the plane P is also a fixed point. We note
first that by the previous lemma, every point on the triangle ªABC is
fixed. Now suppose P is a point not on ªABC. By picking a point D
in the interior of any side of ªABC, we can guarantee that
intersects the triangle exactly twice (either it contains a vertex, or
PASCH guarantees that is crosses another side); call the other
intersection point E. Since f fixes every point on the triangle, in
particular points E and D, it fixes all of
, including point P. Thus
f fixes every point, and must be the identity. €
Theorem: If f and g are isometries and A, B, and C are noncollinear
points such that
,
then f = g.
~ If
, then
. In other words,
is an isometry that fixes three noncollinear points. By the
previous theorem,
must be the identity, so
points P. But this implies that for all P,
for all
.
€
Theorem: Given two congruent triangles
a unique isometry that maps ÎABC onto ÎDEF.
, there is
~ First, note that it is sufficient to show that there is a unique isometry
that maps A to D, B to E, and C to F, because segments map to
segments. We build the isometry from reflections. We proceed in
stages:
Stage 1: Let l be the perpendicular bisector of
and let sl be the
reflection across line l. This maps A onto D. We now consider
, where
.
Stage 2: If sl mapped B onto E, proceed to Stage 3. Otherwise, let m
be the perpendicular bisector of
. We know that ED = BA =
BNAN = BND, so BN and E are equidistant from D. Thus m will contain
the point D (=AN) and so D will be a fixed point for the reflection sm.
Reflect
over line m. This will map BN onto E and leave A
mapped onto D.
Stage 3. If sm mapped CN onto F, we are done. The mapping smsl is the
desired mapping. Otherwise. AO = D and BO = E, so
mapped onto
. Let n be the line
has been
. We know that a reflection
through n will leave
fixed. Moreover, we know that
EF = BC = BOCO = ECO and DF = AC = AOCO = DCO so CO and F are
equidistant from both D and E. Thus
is the perpendicular
bisector of
. Thus sn will map CO onto F, while leaving AO = D
and BO= E fixed. The mapping snsmsl (the composition of sl followed
by sm followed by sn) is thus the desired mapping that takes
onto
.€
Theorem: Every isometry is the product of at most three reflections.
~ Let f be an isometry and A, B, and C be noncollinear points. Let
. Because f is an isometry and
preserves angles and distances,
. By the previous
theorem, we know that
is also mapped to
by a product
of at most three reflections. Because this product of reflections maps
the three noncollinear points A, B, and C to the same points as f, f is
equivalent to the product of reflections. €