Download MATH 329 - CSUSB Math Department

MATH 329-01: Transformation Geometry (82851)
JB-383, MoWe 6 - 7:50 PM
SYLLABUS Fall 2016
John Sarli
JB-328
[email protected]
661-341-1817
O¢ ce Hours: MoWeTuTh 4 - 5:30 PM, or by appt.
Text:
Brannan/Esplen/Gray (Cambridge)
Geometry (second edition)
ISBN 978-1-107-64783-1
Recommended references:
Coxeter/Greitzer, Geometry Revisited MAA New Mathematical Library (ISBN:
0883856190)
The Thirteen Books of Euclids Elements (3 vol., Dover).
Prerequisites: MATH 251, and high-school geometry or equivalent
This is an upper-division course in Euclidean geometry that serves two purposes:
1) develop the language of geometric transformations, classify those which establish congruence in the Euclidean plane, and apply them to some familiar
theorems; 2) explore some results that are not typically encountered in a …rst
geometry course. We will approach these goals concurrently through lectures,
reading, and your participation in presentations. The text listed above is primarily for reference. There is useful background in the …rst two chapters and
the remaining chapters develop material that is used in future courses such as
MATH 529 and MATH 614. However, I will structure this course from my own
notes in a way that …ts the above two purposes. Those notes will be posted on
my website (math.csusb.edu/faculty/sarli) as the course progresses. Occasionally I will suggest exercises and reading from Chapter 2 of the text.
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The midterm and …nal exams will be based on the material we develop for
geometric transformations. The seminar work will account for the remainder of
your course grade, as follows:
1)
2)
3)
4)
Presentation (10%)
Write-up of presentation (20%)
Midterm exam (30%)
Final exam (40%)
For the in-class exams you should bring your notes for the purpose of learning
how to use your own work when processing the material.
After computing your total scores weighted according to these percentages,
course grades will be assigned as follows:
A
A
B+
B
B
C+
C
C
D
F
91
86 90
81 85
76 80
71 75
66 70
61 65
51 60
45 50
< 45
2
Notes
1) Learning Outcomes: Upon successful completion of this course, students
will be able to:
1.2 make connections between mathematical ideas verbally, numerically, analytically, visually, and graphically;
2.1 correctly apply mathematical theorems, properties and de…nitions;
3.6 critique mathematical reasoning, both correct and ‡awed;
5.1 understand correct mathematical proofs.
2) The mid-term exam date is subject to change. Your presentation will be
scheduled based on the subject you choose. The write-up of that presentation
is due within two weeks after the presentation.
3) Please refer to the Academic Regulations and Policies section of your
current bulletin for information regarding add/drop procedures. Instances of
academic dishonesty will not be tolerated. Cheating on exams or plagiarism
(presenting the work of another as your own, or the use of another person’s
ideas without giving proper credit) will result in a failing grade and sanctions
by the University. For this class, all assignments are to be completed by the
individual student unless otherwise speci…ed.
4) If you are in need of an accommodation for a disability in order to participate
in this class, please let me know ASAP and also contact Services to Students
with Disabilities at UH-183, (909)537-5238.
Some important dates:
09-26: First day of class
10-12: Census Deadline
10-26: Midterm Exam
11-30: Last day of class
12-07: Final Exam (Wednesday 6-7:50)
3
Approximate Course Schedule
Week 1
History of transformations in geometry
Week 2
Re‡ection as a rigid motion
Week 3
Some theorems of Euclid
Week 4
Translations and Rotations
Week 5
Parallelism
Week 6
Some later theorems
Week 7
Classi…cation of rigid motions
Week 8
Inversion in a circle
Week 9
Representation of collineations
Week 10
Application to conic sections
4
Some Theorems from Euclids Elements
The …rst 28 propositions from Book I do not require Euclid’s Parallel Postulate.
I.1 Construct an equilateral triangle on a given segment.
I.2 Place at a given point a segment equal to a give segment.
I.3 Given two unequal segments, cut o¤ from the greater a segment equal
to the lesser.
I.4 SAS
I.5 Base angles of isosceles are equal (and exterior extensions).
I.6 Sides opposite equal base angles are equal.
I.7 Uniqueness of triangle with given base and sides in same half-plane relative to base.
I.8 SSS
I.9 Bisect an angle.
I.10 Bisect a segment.
I.11 Construct a ? at a given point on a line.
I.12 Drop a ? to a line from a given point.
I.13 Angles at intersection of two lines are supplementary.
I.14 Rays making supplementary angles at the vertex of a common ray
between them are collinear.
I.15 Vertical angles are equal.
I.16 An exterior angle of a triangle is greater than either of the interior and
opposite angles.
I.17 The sum of two interior angles in a triangle is less than two right angles.
I.18 In a triangle, the greater side subtends the greater angle.
I.19 Converse of I.18.
I.20 Strict triangle inequality.
I.21 If D is an interior point of ABC then BD + DC < BA + AC and
]BDC > ]BAC.
I.22 Construct a triangle from three segments of given lengths satisfying
I.20.
I.23 Construct an angle equal to a given angle at a given point on a given
line.
I.24 If in ABC and DEF we have AB = DE and AC = DF and
]BAC > ]EDF , then BC > EF .
I.25 Converse of I.24 (if BC > EF then ]BAC > ]EDF ).
I.26 ASA
I.27 Equal alternate angles implies parallel.
I.28 Equal opposite interior/exterior angles, or supplementary opposite interior angles, implies parallel.
I.29 Converses of I.27 and I.28 [existence of at least one parallel through
a given point].
I.30 Lines parallel to a given line are parallel to each other.
5
I.31 Construct a parallel to a given line through a given point.
I.32 Exterior angle of a triangle is equal to sum of opposite interior angles
and sum of interior angles is two right angles.
I.33 If AB = CD and AB k CD then AC = BD and AC k BD.
I.34 Opposite side and angles of a parallelogram are equal and a diagonal
divides the area in half.
I.35 Parallelograms with a common base and opposite side on the same line
have equal areas.
I.36 Parallelograms with equal and collinear bases and opposite side on the
same line have equal areas.
I.37 Triangles with a common base and third vertex on the line parallel to
the base have equal areas.
I.38 Triangles with equal and collinear bases and third vertex on the line
parallel to the base have equal areas.
I.39 If triangles ABC and DBC have equal areas then AD k BC.
I.40 If triangles ABC and CDE have equal areas with BC = CE and
B; C; E collinear, then AD k BC.
I.41 If a parallelogram and triangle have the same base and the third vertex
of the triangle is on the line through the opposite side of the parallelogram, then
the area of the parallelogram is twice that of the triangle.
I.42 Construct a parallelogram with a given angle equal in area to a given
triangle.
I.43 Given a parallelogram and a point on a diagonal, the two parallelograms
determined by this point and the other two vertices have
equal areas.
I.44 Construct a parallelogram with a given angle on a given segment as
side with area equal to that of a given triangle.
I.45 Construct a parallelogram with a given angle equal in area to a given
rectilineal …gure.
I.46 Construct a square on a given segment.
I.47 Pythagorean Theorem.
I.48 Converse of Pythagorean Theorem.
II.12 Law of Cosines (obtuse angle).
II.13 Law of Cosines (acute angle).
III.1 Construct the center of a given circle.
III.3 A diameter bisects a chord i¤ it is perpendicular to it.
III.4 Two chords that are not diameters cannot bisect each other.
III.10 Two distinct circles have at most two points in common.
III.11 The line through the centers of tangent circles contains the tangent
point (internal case).
III.12 The line through the centers of tangent circles contains the tangent
point (external case).
III.13 A circle cannot be tangent to another circle at more than one point.
III.14 Equal chords are equally distant from the center, and conversely.
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III.16 A perpendicular to a diameter at an endpoint falls outside the circle.
III.17 Construct a tangent to a circle from a given point.
III.18 A tangent is perpendicular to the diameter at the tangent point.
III.19 Converse of III.18.
III.20 A central angle with a given base arc is double a circumference angle
with the same base arc.
III.21 Corollary of III.20 (equality of circumference angles).
III.22 Opposite angles of a quadrilateral inscribed in a circle sum to two
right angles.
III.25 Construct the full circle from a given arc.
III.27 Generalization of III.21 to equal circles.
III.31 Thales’s Theorem and related angles.
III.32 The supplementary angles made by a tangent at the endpoint of a
chord are respectively equal to the circumference angles on the opposite sides
of the chord.
III.33 Given AB, construct a circle that circumscribes a triangle ABC,
where the angle at C has also been given.
III.34 Given a circle, inscribe a triangle having a given angle.
III.35 If chords AC and BD intersect at E then AE EC = BE ED.
III.36 If DB is tangent to a circle at B and DA cuts the circle at A and C
then DA DC = DB DB.
III.37 Converse of III.36 (equality implies DB is a tangent).
IV.4 Inscribe a circle in a given triangle.
IV.5 Circumscribe a circle about a given triangle.
IV.6 Inscribe a square in a given circle.
IV.7 Circumscribe a square about a given circle.
IV.8 Inscribe a circle in a given square.
IV.9 Circumscribe a circle about a given square.
IV.10 Construct an isosceles triangle with each base angle twice the vertex
angle.
IV.11 Inscribe a regular pentagon in a given circle.
IV.12 Circumscribe a regular pentagon about a given circle.
IV.13 Inscribe a circle in a regular pentagon.
IV.14 Circumscribe a circle about a regular pentagon.
IV.15 Inscribe a regular hexagon in a given circle.
IV.16 Inscribe a regular 15-gon in a given circle.
VI.1 Triangles and parallelograms with equal heights have areas in proportion to their bases.
VI.2 A segment between two sides of triangle is parallel to the third side i¤
it divides the sides proportionally.
VI.3 In a triangle, a cevian divides the opposite side in the same ratio as
the sides on the vertex i¤ it bisects the angle at that vertex.
VI.4 The corresponding sides of equiangular triangles are proportional.
VI.5 Converse of VI.4.
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VI.8 The altitude from a right angle creates two triangles equiangular to
the whole.
VI.13 Construct the mean proportional of two lengths.
VI.19 For equiangular triangles, the ratio of areas is the square of the ratio
of corresponding sides.
VI.24 Given parallelogram ABCD with diagonal AC, let P; Q; R; S be on
AB; BC; CD; DA respectively, such that T = P R \ QS is on AC with
P R k DA k BC and QS k AB k CD. Then the corresponding sides of
parallelograms ABCD; AP T S; T QCR are in proportion.
VI.30 Divide a segment so that the whole is to the longer as the longer is
to the shorter.
VI.31 Generalized Pythagorean Theorem for right triangles.
VI.33 Basis of radian measure.
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Greek Alphabet
A
alpha
B
beta
gamma
delta
" E
epsilon
Z
zeta
H eta
theta
I
iota
K kappa
lambda
M mu
N nu
ksi
o O
omicron
pi
R
rho
sigma
T
tau
upsilon
phi
X chi
psi
!
omega
9
Introduction.
It had for a long time been evident to me that geometry can in no
way be viewed, like arithmetic or combination theory, as a branch of
mathematics; instead geometry relates to something already given in
nature, namely, space.
(Hermann Grassmann, Die Ausdehnungslehre, 1844)
Geometric Transformations: Some History
Is Euclid I.4 (SAS) actually the proof of a theorem?
Pappus and other geometers didn’t think so, nor did they believe Euclid’s proof
of I.8 (SSS), because both proofs require that one triangle "be applied" to the
other. Pappus produced alternative proofs for many of Euclid’s propositions by
de…ning "congruence" to mean "relatable by rigid motion". His critiques were
part of an inquiry that lasted for centuries and culminated in David Hilbert’s
formulation (1899) of a complete and consistent set of axioms for Euclidean
geometry, and the formalization of congruence in terms of transformations. One
of Hilbert’s axioms (IV-6) is equivalent to SAS:
If triangles ABC and DEF are such that AB = DE, BC = EF
and ]B = ]E, then CA = F D, ]A = ]D and ]C = ]F .
Note that this statement is in terms of measurements .
Hilbert’s axioms are complete, but we can pare them down to a few basics.
For example, we accept without deeper justi…cation some of Euclid’s "common
notions" such as
If C is on segment AB but is neither A nor B then AB is greater
than AC.
In triangle ABC, if D is on segment AC but is neither A nor C then
!
]ABC is greater than ]ABE for any E on BD.
Euclid uses these as givens, together with SAS, to prove proposition I.16:
The exterior angle of a triangle is greater in measure than either of
its opposite interior angles.
!
Proof. In triangle ABC let D be on BC but not on BC. Let
!
E be the midpoint of AC and let F be on BE such that EF =
10
BE. Since \AEB and \CEF are vertical angles they are equal
in measure, and so triangles ABE and CF E correspond by SAS.
Therefore ]BAE = ]F CE, and since ]ECD > ]ECF it follows
that ]ACD > ]F CE = ]BAE = ]BAC. By a similar argument
(bisecting BC instead), ]ACD > ]ABC.
The …rst twenty-eight propositions in Book I of The Elements do not require
Euclid’s parallel postulate but many, such as I.16 and I.26 (ASA), use SAS.
Look at Euclid’s proof of I.26. It uses only SAS and the "common
notions" above. (He also uses I.16 in the proof but does not really
need it.)
Presentation 1*. Proof of Euclid I.26.
Later we will see that, using transformations, we can get SSS from SAS.
We add SAS to Euclid’s …ve "axioms":
Two distinct points determine a segment.
Any segment extends uniquely and continuously to a straight line.
The is a unique circle with given center and radius.
All right angles are equal.
SAS
The unique parallel postulate.
Hilbert unpacked the assumptions hidden in these axioms.
What, for example, did Euclid mean by the fourth postulate?
He was at once asserting that angles can be measured unambiguously, and that
this measurement does not change with location. (Pappus noted that Euclid
never de…nes congruence but refers to superposition of …gures on occasion. Thus
any right angle can be superimposed on any other right angle.) From this
postulate we obtain the notion of complementary and supplementary angles,
and …nd that vertical angles have equal measures.
11
α
γ
β
+
= 180 =
+
)
=
Euclid’s I.27 and I.28 are familiar from a …rst geometry course. As noted, they
do not require EPP (Euclid’s parallel postulate). However, their converses (also
familiar from a …rst course) do require it. Let’s sketch Euclid’s proof of the …rst
claim in I.29.
!
!
!
Theorem. Let a EF intersect the parallel lines AB and CD, as shown.
Then ]AGH = ]GHD.
E
A
G
B
H
C
D
F
12
Proof. Otherwise, one has greater measure, say, ]AGH > ]GHD. Then
180 = ]AGH + ]BGH > ]GHD + ]BGH
!
so the interior angles on the same side of EF sum to less than two right angles.
!
!
By EPP, AB and CD must intersect on that side, contradicting the hypothesis
! !
that ABkCD.
Presentation 2*. Determine whether Euclid’s I.30 requires EPP.
What about I.31?
Proposition I.32 includes the assertion that the measures of the interior angles
of a triangle sum to 180 . Later we will show that it is equivalent to EPP. The
remaining propositions in Book I that follow I.32 all require EPP.
Presentation 3*. Prove Euclid’s proposition I.33 and explain
why it depends on EPP.
Propositions I.34 through I.45 establish properties of triangles and parallelograms needed for Euclidean area formulas.
Presentation 4*. Prove Euclid’s proposition I.37 and explain
why it depends on EPP.
Presentation 5*. Proposition I.46 is a construction. Carry out
the construction and explain where it depends on EPP.
Book I concludes with the Pythagorean Theorem (proposition I.47, which uses
SAS, I.14, I.41 and I.46) and its converse (proposition I.48, which uses SSS
and I.47).
Hilbert’s work was motivated by the discovery of hyperbolic geometry, a type
of non-Euclidean geometry that results when we abandon EPP and postulate
that there is more than one parallel through a point not on a given line. For
the following presentations, assume Euclidean geometry.
13
Presentation 6*. Let ABC be isosceles with AB = AC, and let
measure the smaller angle between the ?-bisectors of AB and AC.
Express in terms of ]BAC.
Presentation 7*. Let ABC be isosceles with AB = AC, and let
measure the smaller angle between the bisectors of \ABC and
\ACB. Express in terms of ]BAC.
Presentation 8*. Let ABC be isosceles with AB = AC, and let
measure the smaller angle between the medians of AB and AC.
Express in terms of ]BAC.
Re‡ections.
Hilbert’s contemporary, Felix Klein, was motivated by the Pappus interpretation
of congruence by rigid motion, so he de…ned a geometry by the transformations
that preserve its properties.
A transformation (of the plane) is a 1-to-1 and onto mapping of the
points. Thus, for any transformation T there is an inverse transformation, which we denote T 1 .
A collineation is a transformation that takes lines to lines.
If a unit of measure has been de…ned then we can talk about the collineations
we need to establish congruence:
An isometry (rigid motion) is a collineation that preserves distance
between points.
We say two …gures are congruent provided there is a rigid motion
that maps one onto the other. If F1 and F2 are two …gures, T is a
rigid motion, and T (F1 ) = F2 then we write F1 ' F2 . It follows
that T 1 (F2 ) = F1 .
Re‡ection is a Rigid Motion
Pappus did not de…ne transformations as mappings but understood that re‡ection in a line is the most fundamental step in producing superposition.
We still use his de…nition:
The re‡ection in line l is the mapping that takes point P to the
point P 0 such that l is the perpendicular bisector of segment P P 0 .
Note that if P is on l then P 0 = P , whereas if P is not on l then P 0
and P are in opposite half-planes.
14
Any re‡ection R is a transformation, in fact, R 1 = R.
!
!
Further, R is a collineation: if C is on AB then C 0 is on A0 B 0 .
Proof. Without loss of generality we can assume C is on AB.
0
Let P be the midpoint of AA0 , Q be the midpoint of BB , and O be
0
the midpoint of CC . Then CP O and C 0 P O correspond by SAS,
0
so CP = C P and ]CP O = ]C 0 P O. But then the complements
of these angles are equal, ]AP C = ]A0 P C 0 , so A0 C 0 = AC, again
by SAS. Similarly, CQO and C 0 QO correspond by SAS, and so
B 0 C 0 = BC. Thus AC + CB = A0 C 0 + C 0 B 0 and so, by the triangle
inequality, C 0 is on A0 B 0 .
Eventually we will determine all possible isometries of the Euclidean plane by
showing that any rigid motion is a composition of re‡ections. The reasoning
in the above proof (re‡ections are collineations) also shows that re‡ections are
isometries.
Theorem. Let A0 and B 0 be the re‡ections of points A and B, respectively,
in a line l. Then AB = A0 B 0 .
Proof. Let P be the midpoint of AA0 and Q be the midpoint
of BB 0 . (So P and Q are on l.) Then triangles AP Q and A0 P Q
correspond by SAS because P Q is a common side, A0 P = AP , and
both angles at P are right angles. Thus AQ = A0 Q, ]P AQ =
]P A0 Q, and ]AQP = ]A0 QP . Then, triangles ABQ and A0 B 0 Q
correspond by SAS because AQ = A0 Q, BQ = B 0 Q, and ]AQB =
90
]AQP = 90
]AQP = ]A0 QB 0 . Thus AB = A0 B 0 (and
0 0
]ABQ = ]A B Q, ]BAQ = ]B 0 A0 Q).
This proof has also shown that if the vertex of an angle is on l (for example,
\AQB) then its re‡ection in l has the same measure. Thus:
Corollary. Re‡ection preserves the measure of angles.
Proof. Let \A0 B 0 C 0 be the re‡ection of \ABC in l. We can
assume A; B; C are not collinear, for then each of our angles is 180
and the proof follows from the fact that refection is a collineation.
!
We can also assume there is a point P = l \ BC (since at most
! ! !
one of the lines AB, BC, CA is parallel to l). Let D be the
!
!
intersection of AB with the perpendicular to l at P (use CA if
!
l k AB). Then ]BP D = ]B 0 P D0 because P is on l so, by SAS,
]DBP = ]D0 B 0 P since BP = B 0 P (re‡ection is an isometry) and
!
DP = D0 P (re‡ection is a collineation, so D0 is on A0 B 0 ). But then
]ABC = ]DBP = ]D0 B 0 P = ]A0 B 0 C 0 .
15
In summary, we have shown that re‡ection in a line is a rigid motion, and then,
as a corollary, that re‡ection is conformal (preserves angles). Any rigid motion
is conformal, but there are conformal collineations that are not isometries. We
will see, however, that any conformal collineation is the composition of a rigid
motion and a dilation, which we will de…ne later.
Pappus’s Proofs of Euclid I.5 and I.6
The proposition that the base angles of an isosceles triangle have equal measure
is usually proved in a …rst geometry course by constructing the bisector of
the vertex angle and then applying SAS. Remarkably, Euclid did not use this
simple proof, providing instead a proof that requires a relatively complicated
construction. The converse statement is also typically proved in a …rst course,
as an illustration of a proposition whose proof is not as straightforward as its
converse. Pappus preferred to use symmetry to show that both proofs can be
obtained by similar reasoning. We provide modern versions of his proofs using
re‡ection as a mapping (which we have shown, using SAS, to be a rigid motion).
I.5 The base angles of an isosceles triangle have equal measure.
Proof. Let ABC be a triangle with AB = AC and let l be the
angle bisector at A. Upon re‡ection in l, A 7! A, and B 7! B 0
!
with B 0 on ray AC because re‡ection preserves angle measure. But
it is given that AB = AC, so B 0 = C because re‡ection preserves
length. Similarly, C 0 = B, so \ABC re‡ects to \ACB. Therefore
these base angles have equal measure.
This proof shows that \ABC ' \ACB because the re‡ection in l
takes one to the other. The …gures \ABC and \ACB are congruent,
their measures ]ABC and ]ACB are equal.
I.6 If two angles in a triangle have equal measure then their opposite
sides have equal length.
Proof. Given that ]ABC = ]ACB, let l be the perpendicular bisector of segment BC. Upon re‡ection in l, ABC ! A0 CB
and so ]A0 CB = ]ABC = ]ACB. But then A and A0 are both on
!
ray CA. Similarly, ]A0 BC = ]ACB = ]ABC and so A and A0 are
!
both on ray BA. Thus A = A0 and so AB re‡ects to AC. Therefore
AB = AC.
This proof shows that AB ' AC because the re‡ection in l takes one
to the other. The …gures AB and AC are congruent, their measures
AB and AC are equal.
16
As a corollary we get the following special case of SSS:
For distinct points C and D, let triangles ABC and ABD be such
that AC = AD and BC = BD. Then the three pairs of corresponding angles are equal in measure.
Proof. If we construct segment CD we …nd, by I.5, that ]ACD =
]ADC and that ]BCD = ]BDC. Then ]ACB = ]ADB, and so,
by SAS the remaining pairs of corresponding angles are also equal.
!
We use SSS* for this special case of SSS. We have ABC ' ABD because AB
!
is the ?-bisector of CD, so the re‡ection in AB interchanges C and D.
Euclid III.3 can now be proved. It does not require the parallel postulate. Book
III of The Elements contains the theorems on circles typically encountered in a
basic geometry course.
Presentation 9*.
If a diameter of a circle bisects a nondiameter chord then it is perpendicular to it.
Presentation 10*. If a diameter of a circle is perpendicular
to a non-diameter chord then it bisects it.
The …rst nineteen propositions of Book III do not require EPP.
Presentation 11*. Prove Euclid III.20. Does it require EPP?
Presentation 12*. Prove Euclid III.22. Show where EPP is
required.
Presentation 13*. Prove the converse of Euclid III.22.
Presentation 14*. If a circle can be inscribed in a quadrilateral
show that the sums of the lengths of opposite sides of the quadrilateral are equal.
Presentation 15*. If the sums of the lengths of opposite sides
of the quadrilateral are equal show how to construct the inscribed
circle.
Presentation 16*. Prove Euclid III.32.
Presentation 17*. Prove Euclid III.35.
Presentation 18*. Prove Euclid III.36.
Presentation 19*. Prove Euclid III.37.
Presentation 20*. Let X be a point on the circumcircle of triangle
!
ABC. Let P be the closest point to X on BC, Q the closest point
!
!
to X on CA, and R be the closest point to X on AB. Show that
P; Q; R are collinear. (This is the Simson line for X.)
17
Translations and Rotations.
Using SAS we have shown that re‡ection is a rigid motion: Re‡ection preserves
length and, as a consequence, also preserves measure of angle.
We can now make new rigid motions by composing re‡ections.
A translation is a mapping that moves every point a given distance in
a given direction. (Thus, the translation is determined by P 7! P 0 ,
for any given point P . Any translation is a transformation.)
The rotation about point O through angle is the mapping P 7! P 0
such that
(1) if point P is di¤erent from O, then OP = OP 0 and ]P OP 0 =
; and
(2) if point P is the same as point O, then P 0 = P .
(Thus, P 0 is on the circle with center O and radius OP . Any rotation
is a transformation.)
Theorem. The composition of two re‡ections in parallel lines is a translation
perpendicular to the lines through twice the distance between the lines.
Theorem. The composition of two re‡ections in lines that intersect at point O
is a rotation about O through twice the angle between the lines.
We will prove these two theorems once we have discussed composition of transformations in more detail. We conclude from these theorems that translations
and rotations are rigid motions, and from this corollary we prove Euclid I.8:
SSS. If the sides of two triangles are correspondingly equal in pairs
then the pairs of corresponding angles are equal in measure.
Proof. Let ABC and DEF be the two triangles with correspondingly equal sides. Apply the translation A 7! A0 = D. Since
translation is a rigid motion, we have DB 0 = AB = DE, so we next
apply the rotation about D that takes B 0 7! B 00 = E.
Where is C 00 , the image of C after these two rigid motions?
On the one hand, it must be on the circle with center D that passes
through F , but it must also be on the circle with center E that
passes through F . So either C 00 = F (in which case A 7 ! D,
B 7! E, C 7! F ), or triangles DEF and DEC 00 correspond by
SSS* and (re‡ection in DE takes C 00 7! F ). In either case the
corresponding angles are equal in pairs.
18
Overview of Transformations
General to Speci…c
Mappings - assign an image point to a given point
Transformations - invertible (one-to-one and onto) mappings
Collineations - transformations that take lines to lines
Similarities - conformal collineations (preserve absolute angle measurement)
Isometries (Rigid Motions) - collineations that preserve distance between points
19
Equivalents of Euclid’s Fifth Postulate
There are many propositions in Euclidean plane geometry that are equivalent
to EPP. The equivalence of EPP to the axiom of Proclus/Playfair/Hilbert
(PPH) and to the result that the sum of interior angles in any triangle is two
right angles (
) is shown below. We say that two lines are parallel (l k m)
if there is no point in the plane that lies on both lines. (Euclid uses this word
in his list of "common notions" but not in his postulates.) We will make use
of Euclid’s propositions I.5, I.23, and I.27. We also require Pasch’s Axiom,
which in one form states that a line through a vertex of a triangle, properly
within the interior angle at that vertex, must intersect the segment opposite the
vertex.
EPP implies PPH
Assuming EPP we prove there is a unique line through a given point X, not
!
!
on a given line CD, that is parallel to CD. Let Y be any point on CD and let
!
!
!
AB be a line through X such that ]DY X = ]AXY . Then AB k CD by I.27.
!
!
If A0 B 0 is any line through X distinct from AB, with A0 in the same half-plane
!
as A (determined by XY ) and B 0 in the same half-plane as B, then one of the
following must be true:
]A0 XY
< ]AXY
]B XY
< ]BXY
0
In the …rst case, since ]DY X =
]CY X it follows that ]A0 XY + ]CY X <
. In the second case, since ]CY X =
]DY X it follows that ]B 0 XY +
!
!
]DY X < . In either case, EPP implies that A0 B 0 must intersect CD. Thus
!
!
AB is the unique line through X that is parallel to CD.
PPH implies
!
!
Given triangle ABC, let DE be the unique line through A parallel to BC, with
D and E chosen such that \BAD is alternate to \ABC and \EAC is alternate
to \BCA. Then ]BAD = ]ABC because any choice of point D0 in the same
!
half-plane as D (determined by AB) such that ]BAD0 = ]ABC (which exists
!
!
!
!
by I.23) implies AD0 k BC (by I.27) and so, by PPH, AD0 and DE are
the same line. Similarly, ]EAC = ]BCA. It follows that ]ABC + ]BCA =
]BAD +]EAC. Since ]BAD +]EAC +]CAB = (straight angle) it follows
that the sum of the interior angles of ABC also equals .
20
implies EPP
This last step requires Mathematical Induction, which helps explain why the
independence of EPP was disputed for so many years. Let l be a transversal
!
!
that intersects AB at A and CD at C. Let ]ACD = and ]CAB = and
assume that + < . We will inductively de…ne a sequence of points fDn g:
!
Let D1 = D, let D2 be the point on CD such that D1 is on CD2
!
with AD1 = D1 D2 , and in general let Dn+1 be the point on CD
such that Dn is on CDn+1 with ADn = Dn Dn+1 .
By construction, triangle ADn Dn+1 is isosceles with vertex Dn . Let n =
]ADn C. Then, for n > 1, we have n 1 =
]ADn 1 Dn and, by Euclid
I.5, n = ]Dn ADn 1 . By hypothesis, the sum of interior angles of any triangle
equals , so n 1 =
(
2 n ) = 2 n . By induction we have
n
Now let
n
1
2n
=
1 1
= ]CADn . Again, by hypothesis,
n
=
n
and so
lim
n!1
n
=
because
lim
n!1
n
=0
But
> by construction, so for some large enough value of n we must have
!
!
>
.
Then
AB is properly within \CADn and so, by Pasch’s Axiom, AB
n
!
and CD intersect in the half-plane of l where and are measured.
21
Isometries of the Euclidean Plane
Theorem. An isometry is determined by its action on any three non-collinear
points.
Proof. Let T be a rigid motion and let A; B; C be non-collinear
points, T (A) = A0 ; T (B) = B 0 ; T (C) = C 0 . Given any point X
we must show that T (X) = X 0 is determined by A0 ; B 0 ; C 0 . Let
a = AX; b = BX; c = CX. Let C(P; r) be the circle with center P
and radius r. Then, by Euclid III.3, C(A; a), C(B; b) and C(C; c)
have only X in common because A; B; C are non-collinear. Since T
is an isometry X 0 must be on C(A0 ; a), C(B 0 ; b), and C(C 0 ; c). Since
A; B; C are non-collinear so are A0 ; B 0 ; C 0 (otherwise T 1 would not
be a collineation). Thus X 0 is the only point common to C(A0 ; a),
C(B 0 ; b) and C(C 0 ; c). It follows that X 0 = T (X).
In fact, any collineation is determined by its action on any three non-collinear
points. This result is known as the Fundamental Theorem of A¢ ne Geometry
(FTAG), which is equivalent to the following representation theorem:
Theorem. Let T be a collineation and represent any point P by the
ordered pair (x; y) in R2 . Then the coordinates of T (P ) are given
by
a b
x
p
+
c d
y
q
for some choice of a; b; c; d; p; q with ad
bc 6= 0.
We will not prove this theorem but it implies that the isometries of the plane
are represented by the collineations with
a
c
b
d
a
b
c
d
=
1
0
0
1
We will determine all such rigid motions without using matrices or coordinates.
Transformations generated by two re‡ections
We have already seen that a re‡ection T in a line l is an isometry. Thus the
composition of any number of re‡ections is also an isometry. To keep notation
as simple as possible we identify T with the line l. If m is another line we write
m l
22
for the isometry obtained by …rst re‡ecting in l and then in m. Thus, if l = m
(same line, where equality refers to equality as transformations) we have
m l=l m=I
where I is the identity transformation, that is, a re‡ection is an involution
(transformation equal to its inverse). Now suppose that l and m are distinct
lines.
Case 1): l k m
First look at T = m l. For any point P we want to …nd T (P ). Let d be the
oriented distance from l to m measured along their common perpendicular n
through P . Thus d is the distance from L to M , where L is l \ n and M is
m \ n. (The distance from M to L is d.) Let p be the distance from P to L.
Then the distance from P to l(P ) is 2p, whether p is positive, negative, or zero.
We know that T (P ) is on n and so its location will be determined provided we
can specify its distance from P .
Note that the distance from P to M is d+p, whether p is positive,
negative, or zero.
Next, the distance from P to l(P ) plus the distance from l(P ) to M is the
distance from P to M , and so
the distance from l(P ) to M is (d + p)
2p = d
p.
Finally, just as the distance from P to l(P ) is 2p, the distance from l(P ) to
m (l(P )) is 2(d p). Thus,
the distance from P to m (l(P )) = T (P ) is 2p + 2(d
p) = 2d.
Conclusion. The re‡ection in l followed by the re‡ection in m takes any
point P to the point P 0 on n such that the distance from P to P 0 is twice the
distance from to l to m. The composition m l de…nes a translation T (naturally
associated with the vector of length jdj in the direction along n from l to m).
The translation T factors as l2 l1 , where l1 and l2 are any two lines in the
parallel pencil of l such that the distance from l1 to l2 is equal to d.
It follows that T
1
= l m.
The identity transformation I can be associated with the zero vector, whereby
composition of translations corresponds to addition of vectors. In particular,
composition of translations is commutative, even though composition of re‡ections is not. Note that a non-identity translation has no …xed point.
23
Case 2): l , m
Let O be the intersection of l and m. We want to …nd T (P ) for T = m l,
relative to O. By analogy to the distance between parallel lines we let be the
measure of the angle from l to m, with the convention that > 0 if measured
in the counterclockwise direction. Given P di¤erent from O let C be the circle
through P centered at O and let L be the midpoint of the segment P l(P ). We
will combine central angles of C, observing the sign convention. First, ]P OL =
]LOl(P ), and so ]P Ol(P ) = 2]P OL. Next, let M be the midpoint of the
segment l(P )T (P ). Then ]l(P )OM =
]LOl(P ) = ]M OT (P ). Thus
]P OT (P ) = ]P Ol(P )+]l(P )OT (P ) = 2]P OL+2]l(P )OM =
2 .
Conclusion. The re‡ection in l followed by the re‡ection in m takes any point
P to the point P 0 on C such ]P OP 0 = 2 . The composition de…nes a rotation
about the point O with through the angle 2 . This rotation factors as l2 l1 ,
where l1 and l2 are any two lines in the pencil of lines concurrent at O such
that the measure of the angle from l1 to l2 is equal to .
It follows that T
1
= l m.
A rotation about any point through angle = 0 is the identity transformation
I. Note that a non-identity rotation has a unique …xed point, which is
the intersection of l1 and l2 . Any two rotations about O commute with each
other, though, as we will see, rotations about distinct centers do not.
24
Suggested Exercises for Midterm Study
Draw two parallel lines l and m, and a third line n perpendicular to them. Let
L = n \ l and let M = n \ m, and assume the distance from L to M is 2 units.
Let P be the midpoint of the segment LM . Let P 0 = (m l) (P ).
a) Locate P 0 .
b) Determine the signed distance from M to P 0 .
c) Determine the signed distance from L to P 0 .
Let l and m be lines through point O such that the angle from l to m is
T = l m.
a) Determine the angle from l to m (l).
b) Determine the angle from l to T (l).
4.
Let
Draw two lines l and m through the point O such that the angle from l to m is
3 . Locate a point P on the bisector of this angle such that the distance from
O to P is 4 units. Let P 0 = (m l) (P ).
a) Locate P 0 .
b) Determine the angle from m to the line through OP 0 .
c) Determine the angle from l to the line through OP 0 .
Let ABCD be a square. Label the vertices in this order counter-clockwise. Let
!
!
T1 be the re‡ection in line AB followed by the re‡ection in line AC, and let
!
!
T2 be the re‡ection in line BD followed by the rezre‡ection in line CD. Let
T = T2 T1 .
a) Determine T1 (B).
b) Determine T2 (A).
c) Determine T (X), where X = AC \ BD.
d) Determine T (A), T (B), T (C), and T (D).
e) Let P be an arbitrary point in the plane. Explain how to …nd T (P ).
Let ABC be an isosceles triangle with angle ]ABC = ]ACB. Let b be the
altitude cevian through B, let c be the altitude cevian through C, and let
H = b \ c. Find ]AHA0 in terms of ]BAC , where A0 = (c b) (A).
Let ABC be an isosceles triangle with angle ]ABC = ]ACB. Let b be the
internal angle bisector at B, let c be the internal angle bisector at C, and let
I = b \ c. Find ]AIA0 in terms of ]BAC , where A0 = (c b) (A).
Let ABC be an isosceles triangle with angle ]ABC = ]ACB. Let b be the
?-bisector of AC, let c be the ?-bisector of AB, and let O = b \ c. Find ]AOA0
in terms of ]BAC , where A0 = (c b) (A).
Let ABC be an equilateral triangle. Describe the isometry that takes
a) triangle ABC to triangle ACB
25
b) triangle ABC to triangle BCA
c) triangle ABC to triangle CAB
d) triangle ABC to triangle CBA
e) triangle ABC to triangle BAC
f ) triangle ABC to triangle ABC
Let ABCD be a square. Label the vertices in this order counter-clockwise.
For which permutations of the vertices does there exist an isometry that takes
ABCD to the permuted vertices? Describe the isometry in each of these cases.
How many are there?
Let ABCD be a rhombus. Label the vertices in this order counter-clockwise
and let P = AC \ BD. Describe the isometry that takes
a) triangle ABC to triangle ADC
b) triangle ABD to triangle CBD.
c) triangle AP B to triangle CP B.
d) triangle AP B to triangle AP D.
e) triangle AP B to triangle CP D.
f ) If the rhombus is not a square, for which permutations of its vertices does
there exist an isometry that takes ABCD to the permuted vertices? Describe
the isometry in each of these cases. How many are there?
In the coordinate plane, let l be the line y = x and let m be the line y = x. Let
AOB be the triangle with A = (1; 0), O = (0; 0), and B = (0; 1). Let T = m l,
and for any point P let P 0 = T (P ).
a) Find the vertices of triangle A0 O0 B 0 .
b) Find the image of AOB under T 1 .
c) Let n be the line y = x + 2, and let S = n l . Find the image of AOB
under S.
d) Find the image of AOB under S 1 .
26
Ceva’s Theorem (1678)
Let ABC be any triangle and let a; b; c be lines through A; B; C respectively,
! !
!
each distinct from AB, BC, or CA. Such lines are called cevians. Ceva’s
Theorem provides a necessary and su¢ cient condition, in terms of the ratios
in which they divide the opposite sides of the triangle, for these lines to be
concurrent.
!
!
!
Let P = a \ BC, Q = b \ AC, R = c \ AB, and suppose a; b; c are concurrent
at X. For any triangle U V W let (U V W ) denote its area. Consider triangles
P AB and P CA. We have
(P AB)
BP
=
PC
(P CA)
because both triangles have the same height when we consider both bases to be
!
on BC. (This step requires EPP.) For the same reason we also have
BP
(P XB)
=
PC
(P CX)
and so
BP
(P AB)
=
PC
(P CA)
(P XB)
(XAB)
=
(P CX)
(XCA)
Here we have used the fact that if two ratios
then each is equal to p+tr
q+ts , for any t.
p
q
and
r
s
are equal
Similarly,
AR
RB
CQ
QA
=
=
(XCA)
(XBC)
(XBC)
(XAB)
Consequently
AR BP CQ
(XCA) (XAB) (XBC)
=
=1
RB P C QA
(XBC) (XCA) (XAB)
AR BP CQ
Conversely, if RB
P C QA = 1 then the cevians must be concurrent because if X
is the intersection of any two of them, say a \ b = X , with
!
!
CX \ AB = R0
then
and so
AR0 BP CQ
AR BP CQ
=1=
0
R B P C QA
RB P C QA
AR
AR0
=
0
RB
RB
27
But the ratio of distances to two given points uniquely determines
the location of any point on the line through the given points. This
is the concept of a¢ ne coordinates.
Thus, R = R0 .
Corollary. The medians of a triangle are concurrent (because
CQ
BP
AR
RB = P C = QA = 1), the medians divide ABC into six triangles of
equal area, and the medians trisect each other.
CQ
AR
Proof. First, RB
= BP
P C = QA = 1, so let G be the intersection
of the medians (centroid ). Next, let p = (BP G) = (GP C) ; q =
(CQG) = (GQA) ; r = (ARG) = (GRB). Then 2p + r = (CRB) =
(CAR) = 2q + r and so p = q. Similarly, q = r. Finally, (GAB) =
2 (GBP ) so, using a common altitude from B, AG = 2GP , and
similarly, BG = 2GQ and CG = 2GR.
By considering the ratios to be signed the notation is consistent with signed
areas, so Ceva’s Theorem is completely general in that it applies to cevians
concurrent at a point exterior to the triangle as well.
Corollary. The altitudes of a triangle are concurrent.
Proof. Using the six right triangles, with BC = a; CA =
b; AB = c, we have the following equal ratios
AR
b
RB
a
PC
b
=
=
=
QA
c
BP
c
CQ
a
Then
AR
QA
BP
RB
CQ
PC
and so
=
=
=
b
c
c
a
a
b
AR BP CQ
AR BP CQ
=
=1
RB P C QA
QA RB P C
Exercise. The ?-bisectors of a triangle are concurrent.
28
Presentation 21*. Prove Euclid VI.3.
Presentation 22*. Use Ceva’s Theorem, along with Euclid VI.3,
to prove that the bisectors of the interior angles of a triangle are
concurrent.
Presentation 23*. Use Ceva’s Theorem to prove that the cevians
through the contact points of the circle inscribed in a triangle are
concurrent.
Our proof of Ceva’s Theorem uses the ’adjacent/hypotenuse’ratio that we now
call cosine of the included angle measure. Euclid II.12 and II.13 are equivalent
to what we now call the Law of Cosines.
Presentation 24*. Prove Euclid II.12.
Presentation 25*. Prove Euclid II.13.
29
Classi…cation of Euclidean Isometries
We have encountered three types of rigid motion: re‡ection, translation, rotation. The last two were obtained by composing two re‡ections, so to see if
there are others we should experiment with composing additional re‡ections. It
will turn out not to be necessary to consider the composition of more than three
re‡ections.
Transformations generated by three re‡ections
Let l; m; k be three distinct lines. We …rst consider two special cases.
Case 1: l; m; k are members of the same parallel pencil.
For any point P , we need to determine T (P ) where T = m l k. By associativity,
we can interpret T in two ways: the re‡ection k followed by the translation
m l, or the translation l k followed by the re‡ection m. Choosing the …rst
way, T = (m l) k, let n be the common perpendicular through P and let
K = k \ n. Let L = l \ n and M = m \ n, with d the distance from L to M . Let
P 0 = k(P ) and let P 00 = T (P ). If the distance from P to K is p then P P 0 = 2p
and
P P 00
=
P P 0 + P 0 P 00
=
2p + 2d
=
2 (p + d)
It follows that any point P moves along n through the distance 2(p + d). However, if J is the midpoint of P P 00 then
KJ = KP + P J =
p + (p + d) = d
so if j is the line in this pencil obtained by translating k through the distance
d, then j is the ?-bisector of P P 00 . Thus, T (P ) = j(P ).
Conclusion. The composition of re‡ections in three distinct parallel lines is a re‡ection in another line of their parallel pencil; in
particular, that line is the translation of the …rst line of re‡ection
through the distance from the second line to the third line.
Exercise. Show that j is also the translation of the third line of
re‡ection through the distance from the second line to the …rst line.
Suggestion: Write T = m (l k).
Presentation 26*. Is it possible for j to be m, l, or k ? If so,
how? If not, why not?
30
When the …rst and second re‡ections (or the second and third re‡ections) are
grouped together to produce a translation, their locations in the pencil are no
longer relevant. All that matters is that they are members of the pencil with
the given oriented distance d between them. This makes it easy to remember
which line to move in order to obtain j: Translate the remaining line of the
pencil.
Case 2: l; m; k are concurrent.
For any point P , we need to determine T (P ) where T = m l k. Let l\m\k = O.
As in Case 1, we can write T = (m l) k, the re‡ection k followed by the
rotation m l. Given P , let C be the circle through P centered at O, and let
be the measure of the angle from to l to m. Let P 0 = k(P ) and let K be
the midpoint of P P 0 . Let P 00 = T (P ). If ]P OK = then ]P OP 0 = 2 and
]P 0 OP 00 = 2 . Thus,
]P OP 00
= ]P OP 0 + ]P 0 OP 00
=
2( + )
However, if J is the midpoint of P P 00 then ]KOJ =
]KOJ = ]KOP + ]P OJ =
+
and
+( + )=
so if j is the line in this pencil obtained by rotating k through the angle , then
j is the bisector of \P OP 00 . Thus, T (P ) = j(P ).
Conclusion. The composition of re‡ections in three distinct concurrent lines is a re‡ection in another line of their concurrent pencil;
in particular, that line is the rotation of the …rst line of re‡ection
through the angle from the second line to the third line.
Exercise. Show that j is also the rotation of the third line of re‡ection through the angle from the second line to the …rst line. Suggestion: Write T = m (l k).
Presentation 27*. Is it possible for j to be m, l, or k ? If so,
how? If not, why not?
When the …rst and second re‡ections (or the second and third re‡ections) are
grouped together to produce a rotation, their locations in the pencil are no longer
relevant. All that matters is that they are members of the pencil through O
with the given oriented angle between them. This makes it easy to remember
which line to move in order to obtain j: Rotate the remaining line of the pencil.
31
We say that three lines are in general position if their intersections are the
vertices of a triangle. Suppose that l \ m = O and that O is not on k. Let
T = m l k and let be the measure of the angle from l to m. Since the rotation
m l is the composition of re‡ections in any two lines through O where the angle
from the …rst to the second is , there is no loss of generality in assuming l k k.
Then, when = 2 the isometry T is called a glide. A glide has no …xed points
but has a single invariant line (m in this case), called the axis of the glide.
Apparently this is a new type of rigid motion.
Exercise. Suppose T = m
T = l k m.
l
k is a glide with l k k. Show that
The idea now is to show that the remaining case for the composition of three
re‡ections always results in a glide.
Case 3: T = m l k with l k k and m a transversal.
The strategy is to express T = c b a with b k c and a a perpendicular
transversal. Let K be the intersection of k with its perpendicular through O.
Let a be the line through K that is parallel to m, let b be the line through K
that is perpendicular to a, and let c be the line through O that is perpendicular
to a. If d = KO then the distance from a to m is d cos and the distance from b
to c is d sin . Let L = b \ l. If we can show that T 0 = c b a takes the vertices
of the triangle KOL to the same images as T does then we have proved that
T = T 0 , a glide. First, note that
T (K)
0
T (K)
=
(m l) (K)
= c (K)
But the circle through K centered at O intersects a at K and its re‡ection in c
because c is a diameter line of this circle. Thus T 0 (K) = T (K).
Next,
k (O) = (b a) (O)
because b a is a half-turn about K. But the circle through k (O) centered at
O intersects the line through a (O) that is perpendicular to c at the re‡ection
of k (O) in c, so this intersection must also be the the rotation of k (O) through
2 about O. Thus T 0 (O) = T (O).
Finally, note that LK = d sec so the distance from (l k) (L) to m is
d (sec + cos )
But then the points K, (b a) (L), T (L), and c (K) are the vertices of a rectangle
with length d sec and width 2d sin . Since the distance from (b a) (L) to its
re‡ection in c is also 2d sin , it follows that T (L) = T 0 (L).
32
Conclusion. For composition of re‡ections T = m l k in three
lines in general position we can always assume l k k and that m is a
transversal. Let O = l\m and let K be the foot of the perpendicular
through O to k. Let a be the line through K parallel to m, let b
be the line through K perpendicular to a, and let c be the line
through O parallel to b. Then T = (c b) a = a (c b), so T is a
glide, a re‡ection followed by a non-identity translation in a direction
parallel to the line of the re‡ection: the re‡ection component of
the glide commutes with the translation component. A glide has no
…xed point and has a single invariant line, the axis of the refection
component.
Notes.
1) Let be the measure of the angle from l to m, and let M =
m \ b. Then the translation component of the glide moves every
!
point in the direction M O through the distance 2d sin , where d is
the length of KO .
2) T = c (b a) = (a c) b, so we can also interpret T as the
half-turn about K followed by the re‡ection in c, which is the same
as the re‡ection in b followed by the half-turn about J = a \ c.
3) T 1 = a b c = b c a so the inverse of the glide is obtained
by reversing the direction of the translation component; equivalently,
re‡ect in c and then perform the half-turn about K; equivalently,
perform the half-turn about J and then re‡ect in b.
Presentation 28*. Let k; l; m be the lines through the sides of an
equilateral triangle ABC. Let T = m l k and let P 0 = T (P ).
Find A0 B 0 C 0 .
33
Composition of …nitely many re‡ections
We now show that increasing the number of re‡ections results in no new type of
transformation, so we have already found all rigid motions that can be produced
by composing re‡ections. Suppose T is a re‡ection, translation, rotation, or
glide then we have seen that T 1 is of the same type Since T l and l T 1 are
inverses of each other so it su¢ ces to consider l T when we add a new re‡ection
l. Now, if T is a re‡ection then l T is either a translation or rotation, whereas,
if T is a rotation or translation, we have seen that T is either a re‡ection or a
glide. Therefore, it su¢ ces to suppose that T is a glide and to analyze l T .
We assume T = c b a with a ? b and b k c.
Case1: l = a or l = b
If l = a then l T = a (c b a) = a
component of the glide T .
(a c b) = c
b, the translation
If l = b then l T = (b c b) a. We have seen that b c b = k where k
is the line obtained by translating b through the distance from c to b. Thus
l T = k a and k ? a, so l T is the half-turn about k \ a.
Case 2: l k a
Let O = a \ b and let Q = c \ l. let H1 be the half-turn about O and let H2 be
the half-turn about Q. Then l T = (l c) (b a) = H2 H1 . If P is any point
of the plane then O is the midpoint of segment P H1 (P ) and Q is the midpoint
of segment H1 (P ) P 0 , where P 0 = (l T ) (P ). Consider triangle P H1 (P ) P 0 .
!
!
We have OQ k P P 0 and P P 0 = 2OQ. It follows that l T is the translation in
!
a direction parallel to OQ through twice the distance from O to Q.
Conclusion. The composition of a glide followed by re‡ection in a
line parallel to the axis of the glide is the composition of half-turns
about two distinct points resulting in a translation through twice the
distance between these points in the direction from the …rst point to
the second.
Note that this conclusion agrees with Case 1 if we take l = a.
Case 3: l , a
34
Let Q = l \ a. Since c b is a translation we can assume that c \ a = Q. Now
l c b a = l c a b because b a is a half-turn. We have seen that l c a = j,
where j is the line obtained by rotating l about Q through the angle from c to
a, so j is the line through Q that is perpendicular to l. Thus T = j b is a
rotation about J = j \ b. Note that the angle from a to l has the same measure
as the angle from b to j.
Exercise. Let Q0 = T (Q) and let d = 12 QQ0 . Show that the length
of QJ is jd csc j, where is the measure of the angle from a to l.
Note that j]Q0 QJj = 2 + j j.
Conclusion. The composition of a glide T followed by re‡ection in
a line l that intersects the axis a of the glide in a single point Q is
a rotation. The amount of rotation is 2 , where is the measure of
the angle from a to l; if Q0 = T (Q) and d = 12 QQ0 then the center
J of the rotation is on the line through Q that is perpendicular to l
whose distance from Q is jd csc j and such that j]Q0 QJj = 2 + j j.
Note that the composition is a half-turn if and only if l ? a.
Presentation 29*. Let l1 ; l2 ; l3 be the lines through the sides of
an equilateral triangle ABC, and let l be one of the altitudes of the
triangle. Let T = l3 l2 l1 and let P 0 = S(P ), where S = l T .
Find A0 B 0 C 0 . What type of rigid motion is S ?
Theorem. The composition of …nitely many re‡ections is either a re‡ection,
translation, rotation, or glide. If the number of re‡ections is even (direct isometry) the result is either a translation or rotation; if the number of re‡ections
is odd (opposite isometry) the result is either a re‡ection or glide.
Classi…cation of isometries
The above theorem classi…es the rigid motions of the plane that can be produced by a sequence of re‡ections. It remains to show that this accounts for all
isometries of the plane. We have seen that any isometry is determined by its
action on any three non-collinear points. Let’s review the proof of this fact:
Let ABC be a triangle and T be an isometry with T (A) = A0 ; T (B) =
B 0 ; T (C) = C 0 . We have seen that if two circles intersect in two distinct points then the ?-bisector of their common chord contains the
centers of both circles. Thus, for any point P , the circle centered at
A of radius AP , the circle centered at B of radius BP , and the circle
35
centered at C of radius CP intersect only at P (otherwise A; B; C
would be collinear). Similarly, the circle centered at A0 of radius AP ,
the circle centered at B 0 of radius BP , and the circle centered at C 0
of radius CP intersect at a single point, which must be P 0 = T (P )
since AP = A0 P 0 ; BP = B 0 P 0 ; CP = C 0 P 0 .
Main Theorem. Any isometry T of the plane is either a re‡ection, translation,
rotation, or glide. In particular, T can be produced by a sequence of three or
fewer re‡ections.
Proof. Let T be an isometry and let ABC be a triangle with T (A) =
A0 ; T (B) = B 0 ; T (C) = C 0 .
Case 1: Suppose triangles ABC and A0 B 0 C 0 have the same orientation. Let T1 be the translation A 7! A0 and let T2 be the rotation
about A0 such that (T2 T1 ) (B) = B 0 . The circle centered at A0 of
radius AC and circle centered
at B 0 of radius BC intersect at C 0
!
0 0
and its re‡ection in A B . Since translation and rotation preserve
orientation we must have (T2 T1 ) (C) = C 0 . Now T is determined
by the images A0 ; B 0 ; C 0 and so T = T2 T1 , which is itself either a
translation or rotation.
Case 2: Suppose triangles ABC and A0 B 0 C 0 have opposite orien!
tation. Let T3 be the re‡ection in A0 B 0 . With T1 and T2 as in Case 1
we note that T3 T2 T1 reverses orientation, so (T3 T2 T1 ) (C) =
0
C . Thus T = T3 T2 T1 and T is either a re‡ection or a glide.
Presentation 30*. Exhibit triangles ABC and A0 B 0 C 0 with six
distinct vertices and corresponding sides of equal length such that
i) T : ABC ! A0 B 0 C 0 is a translation
ii) T : ABC ! A0 B 0 C 0 is a rotation
iii) T : ABC ! A0 B 0 C 0 is a re‡ection
iv) T : ABC ! A0 B 0 C 0 is a glide
36
Suggested Exercises for Final Exam Study
i) Draw two parallel lines l and m and a third line k perpendicular to them.
Let L = k \ l and let M = k \ m , and let LM = 2. Let P be the midpoint of
LM .
a) Let P 0 = (m l) (P ). Locate P 0 and determine the signed distance M P 0 .
b) Let P 00 = (l m l) (P ). Locate P 00 and determine the signed distance
00
P P.
c) Find the line j such that P 00 = j(P ).
ii) Draw two lines l and m through the point O such that the measure of the
angle from l to m is 3 . Locate a point P on the bisector of this angle such that
OP = 2.
a) Let P 0 = (m l) (P ). Locate P 0 and determine the measure of the angle
!
from m to OP 0 .
b) Let P 00 = (l m l) (P ). Locate P 00 and …nd the measure of the angle
!
!
from OP to OP 00 .
c) Find the line j such that P 00 = j(P ).
iii) Let ABC be an equilateral triangle with side length 2, and let a; b; c be
! ! !
BC; CA; AB, respectively.
a) Find the invariant line of the isometry T = c b a.
b) Represent ABC in the coordinate plane so that the origin is the foot
of the perpendicular from A to a and C is on the positive x-axis. Find the
coordinates of T (A); T (B); T (C).
iv) In the coordinate plane, let k; l; m; n be the lines y = 0, x = 0, x = 1, y = 1,
respectively. Determine each of the following transformations T , and …nd the
image of the triangle AOB, where A = (1; 0), O = (0; 0), and B = (0; 1).
a) T = k l m n
b) T = m n k l
c) T = m l k n
d) T = n k l m
v) Let P and Q be distinct points. Let T1 be the rotation about P through
and let T2 be the translation from P to Q.
a) Show that the isometry T = T2 T1 is a rotation.
b) Where is the center of the rotation T ?
c) What happens if T1 is the rotation about P through ?
vi) Let P be a point on line l and let T be the rotation through
a) Determine the isometry T l.
b) Determine the isometry l T l.
37
about P .
4
vii) Label the vertices of triangle ABC counter-clockwise and measure the inter! ! !
nal angles at A; B; C counter-clockwise. Let a; b; c be BC; CA; AB, respectively.
For each vertex V , Let TV be the rotation about V through twice the internal
angle measure at V .
a) Describe each of TA ; TB ; TC in terms of the re‡ections a, b, and c.
b) Determine the isometry TB TC TA .
! ! !
viii) For triangle ABC let a; b; c be the ?-bisectors of BC; CA; AB, respectively.
Determine the lines k; l; m such that
a) k = c b a
b) l = b a c
c) m = a c b
d) There are six ways to order the re‡ections a; b; c. What happens with the
other three orders of composition?
ix) Let P and Q be distinct points on the line l. Let HP be the half-turn about
P and let HQ be the half-turn about Q.
a) Describe the isometry T = HQ l HP .
b) Find the invariant line of T .
c) Locate T (M ), where M is the midpoint of P Q.
x) The composition of re‡ections in the three angle bisectors of a triangle is a
re‡ection in some line through the incenter of the triangle. Determine this line.
xi) Let ABC be an isosceles triangle with AB = AC and ]BAC = 6 . Let
a; b; c be the altitudes from A; B; C, respectively.
a) Find l such that l = c b a.
!
b) Let P = l \ BC. Find ]HCP , where H is the orthocenter of ABC.
!
c) For what value of = ]BAC would be l parallel to BC ?
xii) Let C1 and C2 be circles that i ntersect at two points X and Y . Let HX be
the half-turn about X. Then C2 intersects HX (C1 ) at X and another point Z.
a) Locate U = HX (Z).
b) Locate V = HX (Y ).
c) What type of quadrilateral is Y U V Z ?
!
xiii) For any triangle ABC, choose a point P on BC. Let Q be the point on
!
!
!
!
!
!
CA such that P Q k AB, and let R be the point on AB such that QR k BC.
! ! !
For what choice of P will the cevians AP ; BQ; CR be concurrent?
xiv) In the coordinate plane, let O = (0; 0) ; A = (1; 0) ; B = (0; 1). Let T be
the collineation P 7! P 0 such that O0 = (1; 1) ; A0 = (2; 2) ; B 0 = (0; 2). Express
T as a rotation followed by a dilation, and then as a dilation followed by a
rotation. Show how to express each dilation as the composition of inversions in
concentric circles.
38
3
4
x
5
+
is the com4 3
y
10
position of a rotation and dilation about the same center. Find the center, the
dilation factor k, and the amount of rotation.
xv) The transformation T (x; y) =
39
Similarity Transformations
We have de…ned two …gures to be congruent provided there is an isometry that
relates one to the other. In Euclidean geometry we can also de…ne what it means
for two …gures to be similar.
Definition. Two …gures are similar provided there is a conformal
collineation that takes one to the other.
Recall that a transformation is conformal if it preserves absolute angle measure. Since rigid motions are conformal they are special cases of conformal
collineations. Just as isometries are called rigid motions, conformal collineations
are called similarities. The following theorem is a corollary of the Fundamental
Theorem of A¢ ne Geometry.
Theorem. Any similarity of the Euclidean plane is the composition
of a rigid motion and a central dilation.
Definition. The (central ) dilation about point O with factor k 6= 0
!
!
is the transformation P 7! P 0 such that OP = OP 0 and OP 0 = kOP .
Notes.
1) If k < 0 the dilation is the composition of the dilation with
factor jkj followed by the half-turn about O. For this reason the
de…nition is sometimes stated for k > 0.
2) Dilation is a conformal collineation that preserves orientation,
and the image of any line l is a line parallel to l.
Dilation and Inversion
Dilation is closely related to translation in that the image of any line l is a line
parallel to l. In fact, just as a translation can be factored into two re‡ections,
a dilation can be factored into two inversions.
Definition. The inversion in the circle
center O and radius r
! with
!
is the mapping P 7! P 0 such that OP 0 = OP and (OP 0 ) (OP ) = r2 .
Notes. Inversion is a transformation of the plane with the point
O removed. It is an involution on this so-called ‘punctured plane’.
Note that P = P 0 if and only if P is on the circle.
40
Since circles may have di¤erent centers it is not convenient to work with various
punctured planes. To avoid this situation we consider all lines to be circles
that are concurrent at a point that is not in the Euclidean plane. When this
point, commonly denoted by the symbol 1, is added to the plane we obtain the
inversive plane.
Theorem. Inversion is a conformal involution of the inversive plane
that reverses orientation. (If O is the center of the circle of inversion
then O0 = 1.) Any line through the center of the circle of inversion
is invariant. The image of any other line is a circle through O, and
the image of any circle not through O is another circle.
The proof of this theorem is not di¢ cult. It is usually given in a course on
inversive geometry. Instead we will show how inversions can be combined to
produce dilations.
Theorem. The composition of inversions in two concentric circles
is a dilation about their center.
Proof. Let C1 with radius r1 and C2 with radius r2 be centered
at O. Let P be any point of the Euclidean plane and let P 0 be the
!
inversion of P in C1 . Then P 0 is on OP with (OP ) (OP 0 ) = r12 . (If
P = O then P 0 = 1.) Let P 00 be the inversion of P 0 in C2 . Then P 00
!
!
is on OP 0 = OP with (OP 0 ) (OP 00 ) = r22 . (If P 0 = 1 then P 00 = O.)
Thus, if P = O inversion of P in C1 followed by inversion in C2 is
just O. Otherwise
r2
OP 00
= 22
OP
r1
so in any case
OP 00 =
r2
r1
2
OP
Therefore, inversion in C1 followed by inversion in C2 is the dilation
about O with factor k =
r2
r1
2
.
Exercise. The power of the point P with respect to the circle C, with center
O and radius r, is
p = d2 r2
where d = OP . In the coordinate plane, let C be the circle with center (a; b)
and radius r. Let P = (x; y). Show that P 0 , the inversion of P in C, is
1
d2
d2
p x + pa; d2
41
p y + pb
Thus if C is the unit circle then the coordinates of P 0 are
y
x
x2 +y 2 ; x2 +y 2
.
Fundamental Theorem of A¢ ne Geometry
FTAG. Given two triangles ABC and A0 B 0 C 0 , the correspondence
ABC ! A0 B 0 C 0 determines a collineation. Any collineation can be
described by such a correspondence.
A constructive proof of FTAG is not as easy to achieve as the proof we gave
for rigid motions. Such a proof requires the theory of the a¢ ne plane, which
depends on the following assignment of a¢ ne coordinates to a given line l.
Given two distinct points A; B on the line l there is a unique point
P on l with the ratio PAP
B.
These are signed ratios as usual, so every real number corresponds to a point on
l, but if P = B then we must assign it a new coordinate, 1. The assignment of
coordinates on l is independent of the assignment on another line, so the a¢ ne
plane does not work with absolute measurements, just ratios. The following
proposition is now equivalent to FTAG:
Any collineation of the a¢ ne plane preserves ratios on any given
line.
With this proposition the proof of FTAG is easy:
Proof. Given triangle ABC, any line l must intersect at least two
of the lines through its sides (because it can be parallel to at most
!
!
one side). Suppose l \ AB = R and l \ AC = Q. Now R0 is on
!
!
0 0
AR
AR
0
0 0
A0 B 0 and we must have R
0 B 0 = RB , so the location of R on A B
!
is determined. Similarly, the location of Q0 on A0 C 0 is determined
0 0
Q
CQ
0
0
0
because C
Q0 A0 = QA . Thus l is is the line through Q and R .
FTAG implies that not all collineations are similarity transformations, that is,
there are collineations that are not conformal. As an example, let ABC be a
triangle with AB ? BC and AB = BC, and let T be the collineation P 7! P 0
!
!
such that A0 = A, B 0 = B, and C 0 is the point with C 0 A ? AB and C 0 A = CA
such that the image triangle has the same orientation as ABC.
42
Given P , where is P 0 ?
!
The image point P 0 is on the line through P parallel to AB such that the
distance from P to P 0 is equal to the length of P D, where D is the foot of
!
the perpendicular to AB through P , in the direction from B to A. (Such a
!
collineation is called a shear. It follows that any line parallel to ABis invariant
!
and the …xed points are precisely those on AB.) How do we know this? Again,
we can use the preservation of ratios but for our purposes the representation
theorem for collineations is more straightforward.
Let T be a collineation and represent any point P by the ordered
pair (x; y) in R2 . Then the coordinates of T (P ) are given by
a
c
b
d
x
y
p
q
+
for some choice of a; b; c; d; p; q with ad
entation if ad
bc 6= 0. T preserves ori-
bc > 0 and reverses orientation if ad
bc < 0.
For the shear T in our example we can let A = (1; 0) ; B = (0; 0) ; C = (0; 1).
Then C 0 = (1; 1), and since B 0 = B we must have p = q = 0, so
T (x; y) =
1
0
1
1
x
y
x+y
y
=
from which we see that every point on the line y = 0 is …xed and every line
parallel to y = 0 is invariant. Remember that the representation of T depends on
the choice of coordinate system. For example, if we had chosen A = ( 1; 1) ; B =
( 2; 1) ; C = ( 2; 2) with A0 = A and B 0 = B then C 0 = ( 1; 2), so
a+b+p
=
c+d+q
=
2a + b + p
=
2c + d + q
=
2a + 2b + p
=
2c + 2d + q
=
1
1
2
1
1
2
which has the unique solution a = 1; b = 1; c = 0; d = 1; p =
T (x; y) =
1
0
1
1
x
y
+
1
0
=
x+y
y
1; q = 0. Thus
1
!
Now the …xed points comprise the line y = 1 which is, of course, the line AB,
and every point P is sheared relative to this line as expected.
43
This analytic description allows us to describe any collineation in terms of the
chosen coordinates, even when the geometric properties of the motion are not
obvious.
2 1
x
2
+
. This is a
1 1
y
1
collineation because ad bc 6= 0, and if O is the origin then O0 has
coordinates ( 2; 1). Does T have any invariant lines? It is easy
to see that no vertical line is invariant (why?), and any other line l
in the plane consists of points with coordinates (x; mx + t) for some
m and t.
Exercise. Show that l is invariant if and only if
Example. Let T (x; y) =
t =
m
1
1
2
=
m
p
5
1
Thus, there are exactly two invariant lines
p
p
5 1 x 2y =
5 3
p
p
5 + 1 x + 2y =
5+3
These lines intersect at (1; 1), which must be a …xed point since its
image is on both lines.
Exercise. Show that P = (1; 1) is the only point …xed by T .
y
4
3
2
1
-4
-3
-2
-1
1
-1
2
3
4
x
-2
-3
-4
T has exactly two invariant lines and a unique …xed point
44
Representation of Similarities
The collineations used most frequently in Euclidean geometry are those that are
conformal. What do they look like in matrix/vector form?
a
c
Theorem. Let T (x; y) =
b
d
x
y
a
b
represents a similarity if and only if
a
c
b
d
The matrix
p
q
+
c
d
. Then T
is the adjugate of
.
a
b
c
d
a
c
is the transpose of
b
d
. The adjugate of a matrix
is its inverse multiplied by its determinant, so the adjugate of
a
c
b
d
is
d
b
. The theorem says that T represents a similarity precisely when
c a
the transpose and adjugate of the matrix part of T coincide. For example,
1
1
T (x; y) =
1
1
x
y
+
=
2
0
2
3
is a similarity because
1
1
1
1
1
1
1
1
0
2
Since
1
1
1
1
x
y
=
p
p1 x
2
p1 x
2
2
+
p1 y
2
p1 y
2
!
the matrix part of T is a rotation through 4 about the origin followed by dilation
p
about the origin with factor 2. So T is the composition of a rotation, dilation,
and then translation by the vector 2i + 3j. In fact, T is rotation (through 4 )
p
and dilation (with factor k = 2) about its unique …xed point ( 3; 2).
Now, if
a
b
c
d
=
d
c
b
a
then
a
= d
b
=
45
c
so we can always …nd
such that
a
c
b
d
=
k cos
k sin
k sin
k cos
a
c
b
d
=
k cos
k sin
k sin
k cos
or
The …rst matrix represents dilation and rotation about the origin, the second
represents dilation about the origin and re‡ection in a line through the origin.
With k = 1 we obtain the following result.
Theorem. If T is a rigid motion then for some ; p; q its representation is either
cos
sin
sin
cos
x
y
+
p
q
x
y
+
p
q
if T preserves orientation, or
cos
sin
sin
cos
if T reverses orientation.
Exercise. The matrix
cos
sin
sin
cos
represents the rotation
cos
sin
represin
cos
sents the re‡ection in a line through the origin. Find the slope of
this line.
about the origin through . The matrix
46