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MATH 348 — TOPICS IN GEOMETRYCourse Outline, Lecture Notes, Further Reading, Assignments, Tests & Examinations

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Norman Do Mathematics and Statistics

May 3 to June 1 Faculty of Science

Summer 2010 McGill University

MATH 348 — TOPICS IN GEOMETRY CONTENTS

Contents

0 COURSE OUTLINE 3

1 EUCLIDEAN GEOMETRY 6

1.1 Euclid’s Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Triangles and Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.3 Triangle Centres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.4 Geometric Gems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2 SYMMETRY IN GEOMETRY 50

2.1 Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2.2 Symmetry and Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2.3 Symmetry in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

2.4 Crystals, Friezes and Wallpapers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

3 POLYHEDRA, GRAPHS AND SURFACES 93

3.1 From Polyhedra to Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

3.2 Platonic Solids and Beyond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

3.2 Surfaces and Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

3.4 The Classification of Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4 TILING AND DISSECTION 125

4.1 Tiling Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

4.2 Scissors Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

5 FURTHER READING 142

6 ASSIGNMENTS, TESTS AND EXAMINATIONS 146

Assignment 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

Assignment 1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

Assignment 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157


Assignment 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169


Sample Midterm Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

Sample Midterm Test Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

Midterm Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

Midterm Test Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

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MATH 348 — TOPICS IN GEOMETRY CONTENTS

Sample Final Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

Final Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

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0. COURSE OUTLINE

General Information

COURSE MATH 348 — Topics in GeometrySEMESTER Summer 2010 — Section 001PREREQUISITE MATH 133 or equivalent or permission of instructorCREDITS 3.000

INSTRUCTOR Norman DoOFFICE Burnside Hall 1125TELEPHONE 514 398 2998EMAIL [email protected]

OFFICE HOURS Tue 1400-1500, Thu 1400–1500 or by appointment

CLASSROOM Burnside Hall 1B24CLASS HOURS Mon 1105–1325, Tue 1105–1325, Wed 1105–1325, Thu 1105–1325

Monday 3 May to Tuesday 1 June except for Monday 24 MayWEBPAGE http://www.math.mcgill.ca/ndo/MATH-348.html

Learning Outcomes

I hope that, by the end of the course, students should

understand all of the topics in geometry described below;

know a little about some of the major players in mathematical history;

appreciate the mysterious notions of theorem and proof; and

think that maths is awesome.

Instructional Method

The course includes 16 sessions, each one lasting for an epic 140 minutes. These will generally consist of alecture in which I talk about the course material, a tutorial in which we talk about solving maths problems,and some much needed break in between.

Course Materials

There is no required textbook for the course, so you can save your hard-earned money. Instead, I will publishconcise notes of varying quality after each lecture on the course website, outlining the content of the lecture.However, these in no way constitute a replacement for going to the lectures. At various stages throughoutthe course, I might suggest books or websites which reinforce and/or complement the material delivered inlectures. The course website, which will also include electronic copies of any handouts, can be found at thefollowing URL.

http://www.math.mcgill.ca/ndo/MATH-348.html

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0. COURSE OUTLINE

Assignments and Evaluation

Your evaluation for this course will be based on assignments, a test, and an examination.

AssignmentsThere will be three assignments, each graded out of 20. You are not discouraged from talking aboutassignment problems with other students, but every solution that you hand in must be your ownwork.1 Every page submitted should clearly indicate your name, your student number, the coursenumber, and the assignment number.2 Late assignments will not be accepted, unless under particularlyextreme circumstances.

– Assignment 1 will be handed out at the end of class on Tuesday 4 May and is due at 11:00am onTuesday 11 May. It will test the material delivered in lectures 1.1 to 1.4.



TestThere will be a test graded out of 40. You will not be allowed to use calculators, computers, notes, orother aids. No provision will be made for a student who is absent on the day of the test.

– The test will occur in class on Thursday 20 May.It will test the material delivered in lectures 1.1 to 3.3.

ExaminationThe examination will be graded out of 100. You will not be allowed to use calculators, computers, notes,or other aids. There will not be a supplemental examination in this course.

– The examination will occur on Wednesday 2 June or Thursday 3 June.It will test the material delivered in lectures 1.1 to 4.4.

Your final mark will consist of

50% examination + 30% assignments + 20% test OR 100% examination,

whichever is greater. Your final grade will be a letter grade based on your final mark.

Course Content

We will be looking at the following four rather different topics in geometry.

Euclidean Geometry

Symmetry in Geometry

Polyhedra, Graphs and Surfaces

Tiling and Dissection

In an ideal world, we would be able to cover all of the course content outlined below. More realistically, wemight have to deviate from the plan a little, depending on time constraints and student response.

1McGill University values academic integrity. Therefore, all students must understand the meaning and consequences ofcheating, plagiarism and other academic offences under the Code of Student Conduct and Disciplinary Procedures. Please seewww.mcgill.ca/students/srr/honest/ for more information.

2In accord with McGill University’s Charter of Students’ Rights, students in this course have the right to submit in English or inFrench any written work that is to be graded.

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0. COURSE OUTLINE

1. EUCLIDEAN GEOMETRY

Mon 3 May 1.1. Euclid’s Elements. . . in which we start at the very beginning, where Euclid started,with the axioms and build the world of geometry

Tue 4 May 1.2. Triangles and Circles. . . in which we study two of the most basic and fundamentalobjects of geometry — namely, triangles and circles

Wed 5 May 1.3. Triangle Centres. . . in which we discover that the humble triangle has at least fourvery special points, each with very special properties

Thu 6 May 1.4. Geometric Gems. . . in which we use our geometric knowledge to discover wondrousthings before returning to the very beginning, where Euclid started, with the axioms

2. SYMMETRY IN GEOMETRY

Mon 10 May 2.1. Isometries. . . in which we learn the mathematics of flipping, sliding, turning andgliding, as well as what this all has to do with symmetry

Tue 11 May 2.2. Symmetry and Groups. . . in which we begin with the simple notion of symmetryand end with the abstract algebraic notion of a group

Wed 12 May 2.3. Symmetry in the Plane. . . in which we examine symmetry in the plane and prove atheorem of Leonardo da Vinci

Thu 13 May 2.4. Crystals, Friezes and Wallpapers. . . in which we consider the symmetry of crystals,friezes — those long decorations found where the wall meets the ceiling — and wallpapers

3. POLYHEDRA, GRAPHS AND SURFACES

Mon 17 May 3.1. From Polyhedra to Graphs. . . in which we glue together polygons to make polyhedraand introduce the notion of a graph

Tue 18 May 3.2. Platonic Solids and Beyond. . . in which we hunt down the Platonic solids and seethat there’s more to life than polyhedra

Wed 19 May 3.3. Surfaces and Topology. . . in which we define the notion of a hole, discover themysterious one-sided surfaces and play with the rubbery world of topology

Thu 20 May 3.4. The Classification of Surfaces. . . in which we hunt down all of the surfaces and thenconsider what it’s like to live in three dimensions

4. TILING AND DISSECTION

Tue 25 May 4.1. Tiling Rectangles. . . in which we decide whether or not a rectangle can be tiled withsmaller shapes, using coloured pencils and other tricks

Wed 26 May 4.2. Scissors Congruence. . . in which we explore the seemingly simple notions of areaand volume with the help of scissors and glue

Thu 27 May 4.3. Tiling the Plane. . . in which we find shapes which tile the plane and encounter somecrazy aperiodic tilings

Mon 31 May 4.4. My Favourite Problems. . . in which we see a handful of my favourite problems ingeometry, both solved and unsolved

REVIEW

Tue 1 June Review

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1. EUCLIDEAN GEOMETRY 1.1. Euclid’s Elements

Reductionism for Dummies

Today, Wikipedia told me that reductionism is “an approach to understand the nature of complex things byreducing them to the interactions of their parts, or to simpler or more fundamental things”. Here are twoexamples of reductionism at work which you may or may not be familiar with.

Prime factorisationThe Fundamental Theorem of Arithmetic says that every positive integer can be constructed bymultiplying prime numbers together. In fact, it even tells us that, for every positive integer, there isreally only one way to do this. So we can reduce every positive integer to its prime factorisation, whichacts like a fingerprint for that number. You probably know that one way to find the prime factorisationof a number is to write down its factor tree, like I’ve done below for 48 = 2× 2× 2× 2× 3.

48

8 6

4 2 3 2

22

So in this case, the “fundamental things” are the prime numbers and it turns out that there are infinitelymany of them. The oldest known proof of this fact is a beautiful piece of thinking by a guy namedEuclid who lived a really really long time ago.

PhysicsTo understand the universe, it makes sense to explore the “stuff” that we observe — this is calledmatter. Quite a while back, we discovered that matter is made up of molecules and, a little bit later, thatmolecules are made up of atoms. These atoms are, in turn, made up of smaller particles like electrons,neutrons and protons. And more recently, physicists have found that these particles are made up oftiny tiny things called quarks. There are six different flavours of quark known as up, down, charm,strange, top, and bottom.

In this overly simplistic view of physics, the “fundamental things” are the quarks and it turns out thatthere are only finitely many of them. I should mention that, even more recently still, some physicistsbelieve that quarks are actually just incredibly tiny pieces of vibrating string.

Where’s the Geometry?

So what does all this have to do with geometry? Well, suppose that you’re trying to convince me of thesimple geometric fact that all three angles in an equilateral triangle are 60◦. Your argument would probablylook something like this. . .

First, you might use the fact that the base angles in an isosceles triangle are equal. But theequilateral triangle is isosceles “all three ways”, so this means that all three angles are equal toeach other. And then you might invoke the well-known fact that the angles in a triangle alwaysadd up to 180◦. Since you now know that the three angles in an equilateral triangle are equal andadd up to 180◦, every single one of them must be 60◦, and you’re done.

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Now imagine that I’ve been listening really carefully to your argument, but that I’m not a particularlyintelligent nor knowledgeable individual.3 In this purely hypothetical world, I probably don’t know thatthe base angles in an isosceles triangle are equal. Nor would I believe the well-known fact that the anglesin a triangle always add up to 180◦. So you would have to prove these facts to me as well. And then in theprocess of proving these, you would probably use even more basic geometric facts and I would complainthat I have never heard of them before, and on and on it goes. But where does it all end?

Of course, the game ends once you’ve broken the proof down into geometric facts which are so simple that Imust know them to be true. What are these “fundamental things” in geometry? Do you need infinitely manyof them or only finitely many of them? And if the answer is finitely many, then just how many do you needexactly?

Euclid’s Elements

The “fundamental things”, the LEGO blocks, the basic truths from which we build up all of geometry — orany mathematical theory, in fact — are called axioms. The interesting thing about axioms is that you simplycannot prove them, so you just have to assume that they’re true.

The first person to successfully apply the reductionist approach to mathematics was a rather clever guy bythe name of Euclid, whom I’ve already mentioned. He was a Greek mathematician who lived around 300 BCand is most famous for his series of thirteen books known as the Elements. Even though you’ve probablynever heard of them, they’re some of the most successful and influential books ever written, even more sothan the Harry Potter or Twilight series. For example, Euclid’s Elements is supposed to be second only tothe Bible in the number of editions published. In fact, it was used as the basic text on geometry throughoutthe Western world for over two millennia, up until around one hundred years ago. In those days, if youdidn’t know Euclid’s Elements, then people didn’t think that you were well-educated. So to stop people fromthinking that we’re ignorant, let’s have a look at Book 1 of Euclid’s Elements, the most famous of them all.

Axioms and Common Notions

In Book 1, Euclid states ten axioms from which he deduces everything that he knows about geometry. Sothankfully, unlike the reductionist approach applied to prime factorisation, there are only finitely many“fundamental things”. If there had been infinitely many, then Euclid would never have been able to writethem all down.

The five main axioms of Euclidean geometry — which are often referred to as postulates — are stated below.I’ve taken some licence in paraphrasing Euclid’s old-fashioned Ancient Greek into more modern English.

A1. You can draw a unique line segment between any two given points.

A2. You can extend a line segment to produce a unique line.

A3. You can draw a unique circle with a given centre and a given radius.

A4. Any two right angles are equal to each other.

A5. Suppose that a line ` meets two other lines, making two interior angles on one side of ` which sum toless than 180◦. Then the two lines, when extended, will meet on that side of `.

3Of course, this is going to be a very difficult thing for you to imagine, but please try.

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`

a

b

these lines will meet on thisside of ` if a + b < 180◦

Euclid supplemented these with another set of five axioms of a slightly different nature, which we refer to ascommon notions.

C1. If A = C and B = C, then A = B.

C2. If A = X and B = Y, then A + B = X + Y.

C3. If A = X and B = Y, then A− B = X−Y.

C4. If A and B coincide, then A = B.

C5. The whole of something is greater than a part of something.

Surely you agree that each of these ten axioms is so obvious that it’s self-evident. Essentially, Euclid wasdeclaring his ten axioms to be the “rules of the game” from which he would build the world of Euclideangeometry. To really appreciate Euclid’s work, you have to remember that before he entered the picture,geometry consisted of a bunch of useful rules, like the fact that a triangle with side lengths 3, 4, 5 has a rightangle. You knew a fact like this was true because if you made a triangle with these side lengths out of ropeand used it as a protractor, then the home that you were building for your family would appear to be vertical.Euclid came along and said that if you believe my ten axioms — and I’m pretty sure that you all do — then Ican show by logic alone that you have to believe the more complicated things that I’m going to talk about. Inthis way, Euclid shows us what it means exactly for a theorem in mathematics to be true. In the remainder ofBook 1, Euclid proceeds to deduce, one by one, forty-eight propositions, the proof of each one dependingonly on the axioms and on previously proven propositions.

The First Few Propositions

In the timeless words of Maria von Trapp from The Sound of Music, “Let’s start at the very beginning, a verygood place to start”. To make digesting the proofs a little easier, I’ll write which axiom or proposition I’musing to deduce each statement. It’s generally a good habit when you’re constructing geometry proofs — orany proofs, as a matter of fact — to provide reasoning for every single statement that you write down.

Proposition (Proposition I). Given a line segment, you can draw an equilateral triangle on it.

A B

C

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Proof. Let AB be the given line segment.

Draw the circle with centre A and radius AB. [A3]

Now draw the circle with centre B and radius BA. [A3]

If the circles meet at a point C, then draw the line segments CA and CB. [A1]

Since A is the centre of one circle, AC = AB. And since B is the centre of the other circle, BC = BA. Butthese two statements together imply that AC = BC. [C1]

So the line segments AB, BC, CA are all equal which simply means that triangle ABC is equilateral.

Hopefully, you can appreciate that this is a fully rigorous proof and that it’s pretty hard to find fault with anysingle part of it. Because for every statement written down, Euclid can assert its truth, merely by pointing toone of his axioms. In some sense, Euclid’s axioms are like the ten commandments, and thou shalt not arguewith them. One down and forty-seven to go. . .

Proposition (Proposition II). Given a line segment and a point, you can draw a line segment from the given point,equal in length to the given line segment.

AB

C

D

G

E

F

H

Proof. Let A be the given point and BC the given line segment.

Draw the line segment AB. [A1]

Now draw the equilateral triangle DAB. [P1]

Extend the line segments DA and DB to obtain lines DE and DF. [A2]

Draw the circle with centre B and radius BC and let it meet DF at G. [A3]

Now draw the circle with centre D and radius DG and let it meet DE at H. [A3]

Since D is the centre of a circle, we know that DH = DG. And because we constructed DA = DB, wemay now subtract to obtain DH − DA = DG− DB, or equivalently, AH = BG. [C3]

Since B is the centre of a circle, we know that BC = BG. So we can deduce that AH = BC. [C1]

One thing you’ve probably noticed is that these propositions are ridiculously simple and seem pretty obvious.You’ve hopefully also noticed that it takes some ingenuity to prove them using only the axioms.

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Proposition (Proposition III). Given two line segments of unequal lengths, you can divide the longer one into twoparts, one of which is equal in length to the shorter one.

AB

C

D

E

F

Proof. Let AB and CD be the two given line segments, where the longer one is AB.

Draw the line segment AE equal to the line segment CD. [P2]

Draw the circle with centre A and radius AE and let it meet AB at F. [A3]

This means that AE = AF, but we also know that AE = CD, so we can deduce that AF = CD. [C1]

Therefore, we have divided AB into two parts using the point F, one of which is equal to CD.

Hopefully, you’re getting the hang of things now. . .

Proposition (Proposition IV). If two triangles have two pairs of equal sides and the angles formed by these sides areequal, then the two triangles are congruent — in other words, they have the same side lengths and the same angles.

A

B C

D

E F

◦ ◦

Proof. Let the two triangles be ABC and DEF, where AB = DE, AC = DF and ∠BAC = ∠EDF. Placetriangle ABC so that A coincides with D and the line AB coincides with the line DE. Since AB = DE, itmust be the case that B coincides with E. The equal angles ∠BAC = ∠EDF guarantee that the line AC willcoincide with the line DF while the equal lengths AC = DF guarantee that C will coincide with F.

The line segment BC must now coincide with the line segment EF. [A1]

Therefore, we know that BC = EF. [C4]

Furthermore, the angles of triangle ABC coincide with the angles of triangle DEF, so ∠ABC = ∠DEFand ∠ACB = ∠DFE. [C4]

We now know that triangles ABC and DEF are congruent.

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Pons Asinorum

We now come to Euclid’s fifth proposition from Book 1 of his Elements, a proposition which has historicallybeen given the nickname Pons Asinorum. If you’re well-versed in Latin, you’ll know that this means theBridge of the Asses. Why is it called this? One reason that has been proposed is that the diagram used inthe proof looks like a steep bridge which can be crossed by an ass but not by the fuller-figured horse. Themore commonly accepted reason is that this proposition is the first real test of intelligence. At this point,the intelligent people are able to cross over to the harder propositions while the unintelligent asses get leftbehind. So that we don’t feel like unintelligent asses, let’s now try to understand Euclid’s fifth proposition.

Proposition (Proposition V — Pons Asinorum). In an isosceles triangle, the base angles are equal and the anglesunder the base angles are equal.

A

B C

D E

F G

Proof. Let the isosceles triangle be ABC, where AB = AC.

Extend the sides AB and AC to produce the lines AD and AE, respectively. [A2]

Now choose a random point F on BD and let the point G divide the segment AE in such a way thatAF = AG. [P3]

Draw the line segments FC and GB. [A1]

Since AF = AG and AB = AC, the two line segments FA and AC are equal to the two line segmentsGA and AB, respectively. They also make a common angle ∠FAC = ∠GAB. So the two triangles AFCand AGB are congruent. [P4]

We know that AB = AC, but also that AF = AG, so we can subtract to obtain AF− AB = AG− AC.Of course, this is the same thing as BF = CG. [C3]

Remember that we already have FC = GB. So the two line segments BF and FC are equal to the twoline segments CG and GB, respectively. We’ve also shown that ∠CGB = ∠BFC, so triangle BFC iscongruent to triangle CGB. [P4]

Hence, we have the equal angles ∠ABG = ∠ACF as well as the equal angles ∠CBG = ∠BCF. Aftersubtracting, we obtain ∠ABG−∠CBG = ∠ACF−∠BCF or, equivalently, ∠ABC = ∠ACB, which arethe base angles. [C3]

Furthermore, we’ve already shown that ∠FBC = ∠GCB, which are precisely the angles under the baseangles.

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Pythagoras’ Theorem

Now that I’ve led you over the Pons Asinorum, it seems like a reasonably safe assumption that if you hadenough time and energy, you could understand the rest of Euclid’s Book 1. Rather than spend our timeworking through the remaining propositions, let’s fast forward to the second last proposition in Euclid’sBook 1, a theorem which you should already know and love.

Proposition (Proposition XLVII — Pythagoras’ Theorem). In a right-angled triangle, the square of the hypotenuseis equal to the sum of the squares of the other two sides.

A

B

C

D

E

F

G

H K

L

M

Proof. Let ABC be the right-angled triangle with ∠BAC = 90◦.

Draw the square BDEC on BC, the square AGFB on AB, and the square CKHA on CA. [P46]

Draw the line segment AL parallel to BD, where L lies on DE. [P31]

Draw the line segments AD and FC. [A1]

Since ∠BAC = ∠BAG = 90◦, the line segment CG passes through A and, similarly, the line segmentBH passes through A. [P14]

Note that ∠CBD = ∠FBA, so adding ∠ABC to both sides, we obtain ∠ABD = ∠FBC. [C2]

We also have BD = BC and AB = FB, so triangles ABD and FBC are congruent. [P4]

The parallelogram BDLM is twice the area of triangle ABD because they have the same base BD andlie between the same parallel lines BD and AL. [P41]

The square AGFB is twice the area of triangle FBC because they have the same base FB and are betweenthe same parallel lines FB and GC. [P41]

Thus, the parallelogram BDLM is equal in area to the square AGFB. In a similar fashion, we can deduce thatthe parallelogram CELM is equal in area to the square CKHA. Thus, the area of the square BDEC is equal tothe sum of the areas of the two squares AGFB and CKHA — exactly what we set out to prove.

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Of course, Euclid’s proof of Pythagoras’ Theorem relies on all sorts of propositions that we haven’t seenyet. So, for the sake of begin complete, here is a list of all forty-eight propositions from Book 1 of Euclid’sElements.

Euclid’s Propositions

P1. Given a line segment, you can draw an equilateral triangle on it.

P2. Given a line segment and a point, you can draw a line segment from the given point, equal in length tothe given line segment.

P3. Given two line segments of unequal lengths, you can divide the longer one into two parts, one of whichis equal in length to the shorter one.

P4. If two triangles have two pairs of equal sides and the angles formed by these sides are equal, then thetwo triangles are congruent — in other words, they have the same side lengths and the same angles.

P5. In an isosceles triangle, the base angles are equal and the angles under the base angles are equal.

P6. If a triangle has two equal angles, then the two sides opposite these angles are equal.

P7. Given a triangle ABC, there is no other point P on the same side of AB as C such that AC = AP andBC = BP.

P8. If two triangles have three pairs of equal sides, then they also have three pairs of equal angles.

P9. Given a right angle, you can draw a line which bisects it.

P10. Given a line segment, you can draw its midpoint.

P11. Given a line and a point on that line, you can draw another line perpendicular to the given line andpassing through the given point.

P12. Given a line and a point not on the line, you can draw another line perpendicular to the given line andpassing through the given point.

P13. If a line intersects another line, then it creates two angles which sum to 180◦.

P14. Given a point on a line, if two line segments drawn from the point lie on opposite sides of the line andform adjacent angles which sum to 180◦, then the line segments lie on a line.

P15. If two lines meet, then the two vertical angles are equal.

P16. In any triangle, if one of the sides is extended, then the exterior angle formed is greater than either ofthe two interior opposite angles.

P17. In any triangle, the sum of any two angles is less than 180◦.

P18. In any triangle, the angle opposite the longest side is the largest.

P19. In any triangle, the side opposite the largest angle is the longest.

P20. In any triangle, the sum of any two sides is longer than the remaining side.

P21. If P is a point inside triangle ABC, then AP + BP < AC + BC and ∠APB > ∠ACB.

P22. Given three line segments, it is possible to draw a triangle whose sides are equal in length to the givenline segments whenever the sum of any two is longer than the remaining one.

P23. Given an angle and a line with a point on it, you can draw a line passing through the given point whichcreates an angle with the given line equal to the given angle.

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P24. If two triangles have two pairs of equal sides and the angle formed by these two sides is larger in onetriangle, then the third side is longer in that triangle.

P25. If two triangles have two pairs of equal sides and the third side is longer in one triangle, then the angleformed by these two sides is larger in that triangle.

P26. If two triangles have two pairs of equal angles and one pair of corresponding equal sides, then the twotriangles are congruent.

P27. If a line meets two lines and forms equal alternate angles, then the two lines are parallel.

P28. If a line meets two lines and forms an exterior angle equal to the interior opposite angle on the sameside, or the sum of the interior angles on the same side is equal to 180◦, then the lines are parallel.

P29. If a line meets two parallel lines, then alternate angles are equal, the exterior angle is equal to theinterior opposite angle, and the interior angles on the same side sum to 180◦.

P30. Lines parallel to the same line are also parallel to each other.

P31. Given a line and a point, you can draw a line through the given point parallel to the given line.

P32. In any triangle, if one of the sides is extended, then the exterior angle formed equals the sum of the twointerior opposite angles, and the sum of the three interior angles of the triangle equals 180◦.

P33. Line segments which join the ends of equal parallel line segments are equal and parallel.

P34. In a parallelogram, opposite sides and angles are equal to each other, and the diagonals bisect the area.

P35. Parallelograms with the same base and equal heights have equal areas.

P36. Parallelograms with equal bases and equal heights have equal areas.

P37. Triangles with the same base and equal heights have equal areas.

P38. Triangles with equal bases and equal heights have equal areas.

P39. Triangles with equal areas and the same base have equal heights.

P40. Triangles with equal areas and equal bases have equal heights.

P41. If a parallelogram has the same base and equal height with a triangle, then the parallelogram is twicethe area of the triangle.

P42. Given an angle and a triangle, you can draw a parallelogram with area equal to the area of the giventriangle and with one angle equal to the given angle.

P43. Consider a point on the diagonal of a parallelogram. If lines are drawn through this point parallelto the sides of the parallelogram, then four triangles and two parallelograms are formed. The twoparallelograms have equal area.

P44. Given a triangle, a line segment and an angle, you can draw a parallelogram on the line segment withone angle equal to the given angle and with area equal to the area of the given triangle.

P45. Given a polygon and an angle, you can draw a parallelogram with one angle equal to the given angleand with area equal to the area of the given polygon.

P46. You can draw a square on a given line segment.

P47. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the othertwo sides.

P48. If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then thetriangle is right-angled.

14


The Problem with Reductionism

At this point, we’re going to leave Euclid’s Elements behind and make our own way through the world ofgeometry. We do this for three main reasons.

So far, we have followed Euclid’s reductionist approach and carefully proved the first five of hispropositions. At this rate, it would take us more than half the course to finish Book 1 of the Elementsand we’d still have twelve books left to go. As Willy Wonka from Willy Wonka and the Chocolate Factoryonce said, “So much time and so little to do! Strike that, reverse it.”

To be honest, I think that working through the remainder of Book 1 of Euclid’s Elements would notonly be incredibly laborious, but also excruciatingly boring.

Whenever you get too involved in the reductionist approach, you run the risk of missing the biggerpicture. For example, it is far from true that a physicist who unlocks the mysteries of the tiny tiny quarksand strings has a grasp of the universe around them. They would still have no further knowledgeabout topics such as beauty, music, linguistics, hockey and personal hygiene, to name a few. I think weshould take a broader overview of geometry, whereas if we continued with Euclid, we would not see theforest for the trees.

Still, there is great beauty in Euclid’s work, but it doesn’t lie in the individual propositions themselves. It liesin the fact that the reductionist approach lets us truly understand what it means for a mathematical theoremto be true, the fact that the Elements was the paradigm of mathematical rigour for centuries upon centuries,and the fact that Euclid has done a lot of the dirty work for us and we can use his results as a stepping stoneto look at some much more interesting facets of geometry.

Problems

At the end of each lecture, I’ll usually state a couple of example problems and solve them, to help you learnhow to approach problems in geometry. However, I’m going to end today’s lecture with the are-you-smarter-than-Euclid challenge.

Problem. As I mentioned earlier, Euclid’s fifth proposition — otherwise known as Pons Asinorum — is the first realtest of intelligence in Euclid’s Elements. This is because Euclid’s proof is relatively long and intricate. However, if youlook at the first four propositions very carefully, you might notice that these can be used to give a much slicker proof tothe Pons Asinorum. Try to find the proof that Euclid missed and write it out carefully.

15


Euclid

As I mentioned earlier, Euclid was a Greek mathemati-cian who lived a really really long time ago — around300 BC, to be precise. But we can’t be any more pre-cise than that, because there’s very little known aboutEuclid’s life. In fact, it’s been stated that any of thefollowing three possibilities may be the actual truthabout Euclid.

Euclid was a person who wrote the Elementsas well as various other works which are at-tributed to him.

Euclid led a team of mathematicians in Alexan-dria, who all contributed to writing the com-plete works of Euclid, even continuing after Eu-clid died.

Euclid was entirely fictional and his completeworks were written by a team of mathemati-cians in Alexandria who borrowed the nameof Euclid from an actual person — Euclid ofMegara — who had lived about 100 years ear-lier.

Most people who aren’t conspiracy theorists tend tobelieve the first possibility, in which case Euclid wasa very clever guy indeed. He has quite rightly beenreferred to as the “Father of Geometry”. You might

be wondering, if we don’t even know whether or notEuclid existed, then why is there a picture of himstudying geometry below? Well, this particular por-trait is just the product of an artist’s imagination.

16

1. EUCLIDEAN GEOMETRY 1.2. Triangles and Circles

Warming Up with Parallel Lines

Now that we’ve decided to leave Euclid’s Elements behind us, let’s embark on a much less-detailed, thoughfar more exciting, geometric journey. We’ll warm up with a fact about parallel lines — Euclid proved it, butwe’ll assume it.

Proposition. If a line meets two parallel lines and you label one angle x as shown in the diagram below, then you canlabel all eight angles as shown.

xx

180◦ − x180◦ − x

xx

180◦ − x180◦ − x

Basic Facts about Triangles

Now it’s time for some basic facts about triangles, facts which you should know like the back of your handand be able to use when solving geometry problems.

The angles in a triangle add up to 180◦.

If a triangle ABC satisfies AB = AC, then ∠ABC = ∠ACB. On the other hand, if a triangle ABCsatisfies ∠ABC = ∠ACB, then AB = AC. If this is true, then we say that the triangle is isosceles, a wordwhich comes from the Greek words “iso”, meaning same, and “skelos”, meaning leg.

The area of a triangle is given by 12 × b× h, where b denotes the length of the base and h denotes the

height of the triangle.

You might be wondering why we care so much about triangles. One reason is because a triangle is thesimplest shape which encloses a region that you can draw using only straight line segments. Such shapes— of which triangles, quadrilaterals and pentagons are examples — are usually called polygons. In fact, wecan consider triangles to be the “fundamental things” which we can glue together to build any polygon. Forexample, if you wanted to know what the sum of the angles are in a quadrilateral, then you could simplydraw one of its diagonals to split it up into two smaller triangles. Each triangle on its own has angles whichadd to 180◦ so together, their angles add to 360◦. Since the angles of the individual triangles account for all ofthe angles in the quadrilateral, we have shown that every quadrilateral has angles which add to 360◦. Thistrick doesn’t just work for quadrilaterals though, as we will now see.

Proposition. In a polygon with n sides, the angles add to (n− 2)× 180◦.

Proof. We already saw that a quadrilateral can be cut along a diagonal into two triangles. If you draw apentagon, you will notice that it can be cut along diagonals into three triangles. And if you draw a hexagon,you will notice that it can be cut along diagonals into four triangles. In general, a polygon with n sides can becut along diagonals into n− 2 triangles. This is not a particularly easy fact to prove, so we’ll just take it forgranted.

17


The figure above shows what you might get if you try this at home. Of course, each triangle on its own hasangles which add to 180◦ so together, their angles add to (n− 2)× 180◦. Since the angles of the individualtriangles account for all of the angles in the polygon, we now know that every polygon with n sides hasangles which add to (n− 2)× 180◦.

Congruence and Similarity

Congruence and similarity are two of the most important notions in geometry. We say that two shapes arecongruent if it is possible to pick one of them up and place it precisely on top of the other one. On the otherhand, we say that two shapes are similar if it is possible to pick one of them up, enlarge or shrink it by acertain factor, and then place it precisely on top of the other one. If the triangles ABC and XYZ are congruent,then we write this using the shorthand ABC ∼= XYZ and if they are similar, then we write this using theshorthand ABC ∼ XYZ. When we use this notation, we always mean that the first vertex from one trianglecorresponds to the first vertex from the other, the second vertex from one triangle corresponds to the secondvertex from the other, and the third vertex from one triangle corresponds to the third vertex from the other.

There are four simple rules to determine whether or not two triangles are congruent. Each one comes with acatchy TLA4 which should hopefully be self-explanatory.

SSS (side-side-side)If ABC and XYZ are two triangles such that AB = XY, BC = YZ and CA = ZX, then the two trianglesare congruent.

SAS (side-angle-side)If ABC and XYZ are two triangles such that AB = XY, BC = YZ and ∠ABC = ∠XYZ, then the twotriangles are congruent.

ASA (angle-side-angle)If ABC and XYZ are two triangles such that ∠ABC = ∠XYZ, ∠ACB = ∠XZY and BC = YZ, then thetwo triangles are congruent.

RHS (right-hypotenuse-side)If ABC and XYZ are two triangles such that AB = XY, BC = YZ and ∠BAC = ∠YXZ = 90◦, then thetwo triangles are congruent.

You have to be really careful with SAS, because the equal angles must be enclosed by the two pairs of equalsides for the rule to work. In other words, the fictitious rule SSA (side-side-angle) cannot be used to showthat two triangles are congruent.

4Three Letter Acronym

18


To see this, we should be able to find two triangles ABC and XYZ such that AB = XY, BC = YZ and∠BCA = ∠YZX such that the two triangles are not congruent. The following diagram should convince youthat such pairs of triangles certainly do exist.

A

B C

X

Y Z

There are also three simple rules to determine whether or not two triangles are similar. Each one comes witha catchy TLA which once again should hopefully be self-explanatory.

AAA (angle-angle-angle)If ABC and XYZ are two triangles such that ∠ABC = ∠XYZ, ∠BCA = ∠YZX and ∠CAB = ∠ZXY,then the two triangles are similar. In fact, since we know that the angles in a triangle add to 180◦, weonly need to know two of these equations and we get the third one for free.

PPP (proportion-proportion-proportion)If ABC and XYZ are two triangles such that the fractions AB

XY = BCYZ = CA

ZX are equal, then the twotriangles are similar.

PAP (proportion-angle-proportion)If ABC and XYZ are two triangles such that the fractions AB

XY = BCYZ are equal and ∠ABC = ∠XYZ,

then the two triangles are similar.

Once again, you have to be really careful with PAP, because the angle must be enclosed by the two propor-tional sides for the rule to work. In other words, the fictitious rule PPA (proportion-proportion-angle) cannotbe used to show that two triangles are similar. Let’s now apply our newfound knowledge about congruenceand similarity to prove the following simple, but extremely useful, theorem.

Theorem (Midpoint Theorem). Let ABC be a triangle where the midpoints of the sides BC, CA, AB are X, Y, Z,respectively. Then the four triangles AZY, ZBX, YXC and XYZ are all congruent to each other and similar to triangleABC.

A B

C

XY

Z

19


Proof. First, let’s prove that triangle ABC is similar to triangle AZY. It’s clear that ∠CAB = ∠YAZ becausethey actually coincide. We also have the equal fractions AB

AZ = ACAY = 2, so PAP tells us precisely that triangle

ABC is similar to triangle AZY and is twice the size.

An entirely similar argument — no pun intended — can be used to prove that triangle ABC is similar totriangles ZBX and YXC and is twice the size. So what we have deduced is that the triangles AZY, ZBX andYXC are all congruent to each other. In particular, we have the three equations XY = AZ, YZ = YZ andZX = AY which together imply that triangle XYZ is congruent to triangle AZY by SSS.

Pythagoras’ Theorem (Reprise)

We’ve already seen Euclid’s proof of Pythagoras’ Theorem — let’s reword the theorem a little differently nowand consider a much slicker proof.

Theorem (Pythagoras’ Theorem). Consider a right-angled triangle with side lengths a, b, c where c is the length ofthe hypotenuse. Then a2 + b2 = c2.

Proof. The entire proof is captured by the two diagrams below. On the left, we see a square of side lengtha + b. We’ve removed four right-angled triangles, each with side lengths a, b, c and the remaining shadedregions clearly have a combined area of a2 + b2.

a

a

a

a

b

b

b

b

a

a

a

a

b

b

b

b

On the right, we also see a square of side length a + b. We’ve once again removed four right-angled triangles,each with side lengths a, b, c, but in a slightly different way. The remaining shaded region clearly has area c2

and it follows that a2 + b2 = c2.

Theorem (Converse of Pythagoras’ Theorem). Consider a triangle with side lengths a, b, c where a2 + b2 = c2.Then the triangle is right-angled and the hypotenuse has length c.

Proof. Construct a right-angled triangle whose legs have lengths a and b, and let its hypotenuse have length d.We can now invoke Pythagoras’ theorem since we proved it just a little bit earlier. It tells us that a2 + b2 = d2.But put this piece of information together with our assumption that a2 + b2 = c2 and you have the equationc2 = d2. This implies that c = d.

Therefore, the triangle with side lengths a, b, c that we were given possesses exactly the same side lengthsas the right-angled triangle that we have constructed. Because of SSS, this means that the two triangles are,in fact, congruent. So the given triangle with side lengths a, b, c is indeed right-angled, as we intended toprove.

20


Basic Facts about Circles

We now know lots about triangles, so let’s move on to circles — we start with some basic facts.

A circle is the set of points which are the samedistance r from some centre O. You should al-ready know by now that r is called the radius ofthe circle and O is called the centre of the circle.

A chord is a line segment which joins two pointson a circle.

An arc of a circle is the part of the circumferencecut off by a chord.

A diameter is a chord which passes through thecentre of the circle. Note that its length is twicethe radius.

diameter

chord

arc

This is all we need to know about circles in order to prove some pretty cool results, like the following.

Proposition. The diameter of a circle subtends an angle of 90◦. In other words, if AB is the diameter of a circle and Cis a point on the circle, then ∠ACB = 90◦.

Proof. Let O be the centre of the circle. The beauty of considering the centre is that we have the three equalradii OA = OB = OC. Equal lengths, for obvious reasons, often lead to isosceles triangles, and our diagramhappens to have two of them. There is the isosceles triangle OAC which means that we can label the equalangles ∠OAC = ∠OCA = x. There’s also the isosceles triangle OBC, which means that we can label theequal angles ∠OBC = ∠OCB = y. Labelling equal angles like this is an extremely common and extremelyuseful trick.

A B

C

O

x y

x y

Now it’s time for some angle chasing.5 In particular, let’s consider the sum of the angles in triangle ABC.

∠BAC + ∠ACB + ∠CBA = 180◦

5Angle chasing is the art of chasing angles. As Wikipedia puts it, the term is “used to describe a geometrical proof that involvesfinding relationships between the various angles in a diagram”.

21


We can replace all of these confusing angles with x’s and y’s in the following way.

x + (x + y) + y = 180◦

This equation is, of course, the same thing as 2(x + y) = 180◦ or, equivalently, x + y = 90◦. All we have to donow is recognise that ∠ACB = x + y so we have proven that ∠ACB = 90◦.

Proposition. The angle subtended by a chord at the centre is twice the angle subtended at the circumference, on thesame side. In other words, if AB is a chord of a circle with centre O and C is a point on the circle on the same side of ABas O, then ∠AOB = 2∠ACB.

Proof. The beauty of considering the centre is that we have the three equal radii OA = OB = OC. Equallengths, for obvious reasons, often lead to isosceles triangles, and our diagram happens to have three of them.There is the isosceles triangle OAB which means that we can label the equal angles ∠OAB = ∠OBA = x.There’s also the isosceles triangle OBC, which means that we can label the equal angles ∠OBC = ∠OCB = y.And there’s also the isosceles triangle OCA, which means that we can label the equal angles ∠OCA =∠OAC = z. Labelling equal angles like this is an extremely common and extremely useful trick.

A B

C

O

x x

y

yz

z180◦ − 2x

Now it’s time for some angle chasing. In particular, let’s consider the sum of the angles in triangle ABC.

∠BAC + ∠ACB + ∠CBA = 180◦

We can replace all of these confusing angles with x’s, y’s and z’s in the following way.

(x + z) + (z + y) + (y + x) = 180◦

This equation is, of course, the same thing as 2(x + y + z) = 180◦. Let’s keep this equation in the back ofour minds while we try and remember what it is exactly that we’re trying to do. We want to prove that∠AOB = 2×∠ACB. Using the angle sum in triangle OAB, we can write ∠AOB = 180◦ − 2x. We can alsowrite ∠ACB = y + z. So what we’re actually aiming for is the following equation.

180◦ − 2x = 2(y + z)

But after rearranging, this is just the same thing as 2(x + y + z) = 180◦, which we already proved.

22


Hopefully, you’ve managed to spot the similarity between these two proofs. They’re both indicative of thegeneral strategies that we’ll be using to solve tougher geometry problems. More specifically, I guess whatwe’ve used here is the old “find isosceles triangles–label equal angles–sum up the angles in a triangle–workout what you’re trying to prove–then prove it” trick.

I should probably mention that some of the proofs I’ve provided are a little incomplete. In particular, if youcheck the previous proof very carefully, you’ll notice that it only works when O lies inside triangle ABC. Itcould be possible that O lies on or even outside triangle ABC. However, the proofs in these other cases arequite similar, so I’ll leave it as a fun exercise for you to find them.

The Hockey Theorem

Suppose that a hockey coach wanted to see which player on their team had the best shot. They could lineall the players parallel to the goal and ask them to shoot to see who scores and who misses. But this wouldcertainly be unfair to the players on the ends who have to shoot further and have a smaller angle to aim for.The coach could try and fix the problem by placing everyone on a circle whose centre coincides with themiddle of the goal. Of course, this means that everyone is the same distance from the goal now, but someplayers have a much greater angle to aim at than others. In fact, due to the frictionless nature of the sport,distance is no problem when shooting at a hockey goal. What we would rather test is the accuracy of eachplayer. So it makes sense to place everyone somewhere where they all have the same angle to aim at. In thatcase, where should we put all the players? The following proposition tells us the answer — they should allstand on the arc of a circle whose end points are the goal posts.

goal goal goal

Proposition (Hockey Theorem). Angles subtended by a chord on the same side are equal. In other words, ifA, B, C, D are points on a circle with C and D lying on the same side of the chord AB, then ∠ACB = ∠ADB.

Proof. The proof to this is delightfully simple — we start by letting O be the centre of the circle. Wehave already proved that ∠AOB = 2∠ACB and also that ∠AOB = 2∠ADB. So it must be the case that∠ACB = ∠ADB.

A B

C

D

O

x

x

2x

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Cyclic Quadrilaterals

If I give you one point and ask you to draw a circle through it, then that’s a pretty easy thing to do, right?I could make life slightly more difficult for you by giving you two points and asking you to draw a circlethrough both of them. And even if I give you three points and ask you to draw a circle through all of them,then you could almost always do it, as long as they don’t lie on a line. To see this, consider a massive circle, somassive that all three points that I give you lie inside it. Now start shrinking the circle. Sooner or later, yourcircle has to hit one of the points. Now keep that point on the circle, but keep shrinking the circle. Sooner orlater, your circle has to hit another of the points. We are now in the happy situation of having the circle passthrough two of the given points. Now we either shrink or expand our circle, while keeping it in contact withthese two points, until it finally passes through the third given point. Hopefully, this should convince youthat there is actually only one circle which passes through any three points, as long as they don’t lie on a line.

So now I’m going to make life particularly difficult for you by giving you four points and asking you to drawa circle through all of them. The shrink/expand trick we used above shows that there is a unique circle whichpasses through three of them. So to be able to accomplish the task, the fourth point must already lie on thiscircle. The probability that I was nice enough to actually give you four points where this is true is incrediblysmall. What I’m trying to get at here is that if I give you a random quadrilateral, then it is extremely rare for acircle to pass through all four of its vertices. So if a circle does pass through all four of its vertices, then thequadrilateral must be very special indeed — so special that we should give it a special name. In fact, we referto such a quadrilateral as a cyclic quadrilateral. We’re now going to prove a very important fact about cyclicquadrilaterals.

Proposition. The opposite angles in a cyclic quadrilateral add up to 180◦. In other words, if ABCD is a cyclicquadrilateral, then ∠ABC + ∠CDA = 180◦ and ∠BCD + ∠DAB = 180◦.

Proof. If we draw in the diagonals AC and BD, the hockey theorem tells us that there are equal anglesgalore. For example, we can label ∠ACB = ∠ADB = w, ∠BDC = ∠BAC = x, ∠CAD = ∠CBD = y and∠DBA = ∠DCA = z. You can go crazy labelling equal angles like this whenever there’s a cyclic quadrilateralsomewhere in your diagram.

A B

C

D

x zy

wz

xw

y

Now we’re going to play a similar trick to one we played before — we’re going to add up all of the angles inthe quadrilateral and the answer should be 360◦.

∠DAB + ∠ABC + ∠BCD + ∠CDA = 360◦

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We can replace all of these confusing angles with w’s, x’s, y’s and z’s in the following way.

(x + y) + (y + z) + (z + w) + (w + x) = 360◦

This equation is, of course, the same thing as 2(w + x + y + z) = 360◦ or, equivalently, w + x + y + z = 180◦.All we have to do now is recognise that ∠ABC = y + z and ∠CDA = w + x, so that

∠ABC + ∠CDA = (y + z) + (w + x) = 180◦.

You could also have chosen to recognise that ∠BCD = z + w and ∠DAB = x + y, so that

∠BCD + ∠DAB = (z + w) + (x + y) = 180◦.

How to Find a Cyclic Quadrilateral

Something we’re going to learn is that it’s incredibly useful to keep your eyes open for cyclic quadrilateralswhen solving problems in Euclidean geometry. If the problem happens to mention a circle which has fourpoints on it, then of course, those four points form a cyclic quadrilateral. But quite often, cyclic quadrilateralscan be hidden somewhere in your diagram, even when there are no circles involved. In those cases, youwould probably use one of the two facts below to prove that the quadrilateral is cyclic.

We proved earlier that the opposite angles in a cyclic quadrilateral add up to 180◦. One thing you might bewondering is whether the converse is true — in other words, if someone gives you a quadrilateral where theopposite angles add up to 180◦, then is it necessarily true that the quadrilateral must have been cyclic? Thefollowing proposition tells you that the converse certainly is true.

Proposition. If the opposite angles in a quadrilateral add up to 180◦, then the quadrilateral is cyclic.

Another thing we proved earlier was the hockey theorem — that if A, B, C, D are points on a circle withC and D lying on the same side of the chord AB, then ∠ACB = ∠ADB. And another thing you might bewondering is whether the converse is true — in other words, if someone gives you a quadrilateral ABCDwhere ∠ACB = ∠ADB, then is it necessarily true that the quadrilateral must have been cyclic? The followingproposition tells you that the converse certainly is true, once again.

Proposition. If ABCD is a convex quadrilateral such that ∠ACB = ∠ADB, then the quadrilateral is cyclic.

A particularly useful case occurs when you are given four points A, B, C, D such that ∠ABC = ∠CDA = 90◦.If ABCD is a convex quadrilateral, then we can use the first proposition above to deduce that it must actuallybe a cyclic quadrilateral. On the other hand, if ABCD is a convex quadrilateral, then we can use the secondproposition above to deduce that it must actually be a cyclic quadrilateral. So either way, the four pointsA, B, C, D lie on a circle. This means that when right angles appear, you can often expect cyclic quadrilateralsto appear as well.

You can solve many many geometry problems by searching for cyclic quadrilaterals and using what youknow about them. The beauty of cyclic quadrilaterals is that if you find one pair of equal angles, then you getthree more for free. This is because if you find that ∠ACB = ∠ADB, then the quadrilateral ABCD is cyclic,in which case we can apply the hockey theorem to also obtain the equal angles

∠BDC = ∠BAC, ∠CAD = ∠CBD, and ∠ABD = ∠ACD.

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Tangents

A tangent is a line which touches a circle at preciselyone point. Given a point outside a circle, you candraw two tangents, thereby creating a diagram whichlooks very much like an ice cream cone.

Theorem (Ice Cream Cone Theorem). The pictureof the ice cream cone on the right is symmetric so thatAP = AQ and the line OA bisects the angle PAQ. A

P Q

O

The proof of the ice cream cone theorem is a very simple application of congruent triangles combined withthe fact that the radius drawn from the centre of a circle to a point on its circumference is perpendicular tothe tangent at that point. The following is another extremely useful fact about tangents.

Theorem (Alternate Segment Theorem). Suppose that AT is a chord of a circle and that PQ is a line tangent to thecircle at T. If B lies on the circle, on the opposite side of the chord AT from P, then ∠ABT = ∠ATP.

T

A

B

P Q

Proof. The hockey theorem guarantees that, as long as C lies on the circle, on the same side of AT as B,then ∠ACT = ∠ABT. So the trick here is to choose a particular point C on the circle and prove that∠ACT = ∠ATP. One nice way to choose the location of the point C is so that TC is a diameter of the circle.

T

O

A

B

C

P Q

x

26


Choosing C in this way is great because we’ve introduced two right angles into the diagram. We have∠CAT = 90◦, because it’s an angle subtended by the diameter TC and we also have ∠PTO = 90◦, becauseit’s an angle created by a radius and a tangent at the point T.

So if we label ∠ATP = x, then we have ∠OTA = 90◦ − x. Now use the fact that the angles in triangle CATadd to 180◦ and you find that ∠ACT = x. However, we already mentioned that ∠ABT = ∠ACT = x by thehockey theorem, so we can now conclude that ∠ABT = ∠ATP.

It turns out that the converse of the alternate segment theorem is also true — but I’ll let you try to write downexactly what it states.

Problems

Many geometry problems can be solved by angle chasing. This means labelling some — but not too many —well-chosen angles in your diagram and using what you know to determine other ones. Here are some verycommon ways to relate different angles in your diagram.

If two angles are next door to each other, then they add up to 180◦.

Parallel lines give you equal angles and angles which add up to 180◦.

The angles in a triangle add up to 180◦.

Congruent or similar triangles give you equal corresponding angles.

In any cyclic quadrilateral, the opposite angles add up to 180◦.

In any cyclic quadrilateral, you can apply the hockey theorem to give you four pairs of equal angles.

Apart from these tips, possibly one of the most useful things I can tell you is to always draw a very accurate,very large diagram, preferably in multiple colours. Below are two example problems and their solutions. Youshould try to read the solutions carefully to make sure that you understand them and then reread them untilyou think you can reconstruct the proofs on your own.

Problem. Two circles intersect at P and Q. A line through P meets the circles C1 and C2 at A and B, respectively. LetY be the midpoint of AB and suppose that the line QY meets the circles C1 and C2 at X and Z, respectively. Prove thattriangle XYA is congruent to triangle ZYB.

A

BP

Q

X

Y

Z

x

C1 C2

27


Proof. The first thing to notice is that the two triangles have the angles ∠XYA = ∠ZYB as well as the sidesAY = BY in common. So to prove that they’re congruent, we could try to use either ASA or SAS. It turns outthat the first choice is better for this problem, since the circles in the diagram will give us equal angles.

So let’s label ∠YAX = x and note that the problem is completely solved if we can only show that ∠YBZ = xas well. Since the angle ∠YAX = ∠PAX is subtended by the chord PX, we can use the hockey theorem todeduce that ∠PQX = ∠PAX = x. And since the angle PQZ = ∠PQX is subtended by the chord PZ, wecan use the hockey theorem again to deduce that ∠PBZ = ∠PQZ = x. However, ∠PBZ = ∠YBZ so we’veproved that ∠YBZ = x and the problem is solved.

Problem. Consider a semicircle with diameter AB. Suppose that D is a point such that AB = AD and AD intersectsthe semicircle at the point E. Let F be the point on the chord AE such that DE = EF and extend BF to meet thesemicircle at the point C. Prove that ∠BAE = 2∠EAC.

A B

D

E

FC

Proof. A good start would be to label ∠EAC = x and aim to prove that ∠BAE = 2x. Given that there is acircle — or at least half of one — there is probably a cyclic quadrilateral lurking about. Hopefully, you can seethat ABEC is a cyclic quadrilateral, so we can apply the hockey theorem to deduce that ∠EBC = ∠EAC = xas well.

Note that AB is a diameter and there is a certain fact that we know about diameters — namely, they subtendangles of 90◦. This means that ∠AEB = ∠ACB = 90◦. Since ∠EBC = x, considering the angle sum intriangle EFB yields the fact that ∠EFB = 90◦ − x.

Now if you’ve drawn a nice accurate picture, then you’ll see that the two triangles FEB and DEB looksuspiciously congruent. These suspicions can be confirmed by observing that that FE = ED, EB = EB,∠FEB = ∠DEB = 90◦ and using SAS. One consequence of this is the fact that ∠EDB = ∠EFB = 90◦ − x.

One piece of information we haven’t used yet is the fact that AB = AD or, in other words, that triangleABD is isosceles. The equal side lengths imply that there are equal angles in the diagram — namely,∠ABD = ∠ADB = ∠EDB = 90◦ − x.

Now we note that we have labelled two of the angles in triangle ABD, so we can definitely determine whatthe third angle is.

∠BAD = 180◦ −∠ABD−∠ADB = 180◦ − (90◦ − x)− (90◦ − x) = 2x

However, this tells us that ∠BAE = 2x as well, exactly what we set out to prove.

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Pythagoras

Pythagoras was a Greek mathematician who livedeven before Euclid was born — from around 570 BCto around 475 BC. Even though we know and lovehim for his theorem about right-angled triangles, itwas essentially known before Pythagoras arrived onthe scene and he never even managed to prove it. Ac-tually, Pythagoras was not a great mathematician atall, but was the first person to call himself a philoso-pher — a word which in ancient Greek literally meansa lover of wisdom.

Pythagoras founded a religious cult whose membersbelieved that everything was related to mathematicsand that numbers were the ultimate reality. Pythago-ras himself was pretty serious about this belief, asyou can tell from the following story. When his fellowcult member Hippasus of Metapontum managed toprove that the number

√2 was irrational — a rather

ingenious and important mathematical feat whichdidn’t sit well with the Pythagorean philosophy —Pythagoras supposedly could not accept the resultand sentenced Hippasus to death by drowning.

The Pythagorean cult also adhered to various rules,some mildly practical but most simply bizarre — thefollowing are examples.

Never eat beans.

Don’t pick up something that has fallen.

Don’t walk on highways.

When you get out of bed, take the sheets androll them all together.

Apart from these facts, there isn’t a great deal which isknown about Pythagoras because no written work ofhis has survived to this day. It’s commonly believedthat many of the accomplishments attributed to himwere simply accomplishments of people who were inhis cult.

29

1. EUCLIDEAN GEOMETRY 1.3. Triangle Centres

Triangle Centres

The humble triangle, with its three vertices and three sides, is actually a very remarkable object. For evidenceof this fact, just look at the Encyclopedia of Triangle Centers6 on the internet, which catalogues literally thousandsof special points that every triangle has. We’ll only be looking at the big four — namely, the circumcentre,the incentre, the orthocentre, and the centroid. While exploring these constructions, we’ll need all of ournewfound geometric knowledge from the previous lecture, so let’s have a quick recap.

Triangles

– Congruence : There are four simple rules to determine whether or not two triangles are congruent.They are SSS, SAS, ASA and RHS.

– Similarity : There are also three simple rules to determine whether or not two triangles are similar.They are AAA, PPP and PAP.

– Midpoint Theorem : Let ABC be a triangle where the midpoints of the sides BC, CA, AB are X, Y, Z,respectively. Then the four triangles AZY, ZBX, YXC and XYZ are all congruent to each otherand similar to triangle ABC.

Circles

– The diameter of a circle subtends an angle of 90◦. In other words, if AB is the diameter of a circleand C is a point on the circle, then ∠ACB = 90◦.

– The angle subtended by a chord at the centre is twice the angle subtended at the circumference, onthe same side. In other words, if AB is a chord of a circle with centre O and C is a point on thecircle on the same side of AB as O, then ∠AOB = 2∠ACB.

– Hockey Theorem : Angles subtended by a chord on the same side are equal. In other words, ifA, B, C, D are points on a circle with C and D lying on the same side of the chord AB, then∠ACB = ∠ADB.

Cyclic Quadrilaterals

– The opposite angles in a cyclic quadrilateral add up to 180◦. In other words, if ABCD is a cyclicquadrilateral, then ∠ABC + ∠CDA = 180◦ and ∠BCD + ∠DAB = 180◦.

– If the opposite angles in a quadrilateral add up to 180◦, then the quadrilateral is cyclic.

– Hockey Theorem : If ABCD is a convex quadrilateral such that ∠ACB = ∠ADB, then the quadrilat-eral is cyclic.

Tangents

– Ice Cream Cone Theorem : A picture of an ice cream cone is symmetric so that the two tangents havethe same length and the line joining the centre of the circle and the tip of the cone bisects the coneangle.

– Alternate Segment Theorem : Suppose that AT is a chord of a circle and that PQ is a line tangent to thecircle at T. If B lies on the circle, on the opposite side of the chord AT from P, then ∠ABT = ∠ATP.

These facts should all be burned into your memory, so that you can recall and use them, whenever youencounter a problem in Euclidean geometry.

6http://faculty.evansville.edu/ck6/encyclopedia/ETC.html

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http://faculty.evansville.edu/ck6/encyclopedia/ETC.html


The Circumcentre

A perpendicular bisector of a triangle is a line which passes through the midpoint of one side and is perpendic-ular to that side. Note that three randomly chosen lines will almost never ever meet at a point and yet, forany particular triangle we choose, we’ll see that its three perpendicular bisectors always do.

Proposition. The three perpendicular bisectors of a triangle meet at a point.

Proof. Our proof relies crucially on the following lemma, which can be proven using congruent triangles.

Lemma. A point P lies on the perpendicular bisector of AB if and only if AP = BP.

So take a triangle ABC and the perpendicular bisectors of the sides AB and BC. If we suppose that these twolines meet at a point O, then it must be the case that AO = BO and also that BO = CO. These two equationstogether imply that AO = CO, in which case O lies on the perpendicular bisector of the side CA as well.

In the previous proof, we noted that if the perpendicular bisectors of triangle ABC meet at O, then thedistances from O to the vertices are all equal. Another way to say this is that there’s a circle with centre Owhich passes through the vertices A, B, C. This circle is called the circumcircle of the triangle and the pointO is called the circumcentre.7 Furthermore, the radius of the circumcircle is known as the circumradius forobvious reasons. We now know that every triangle has exactly one circumcircle and that its centre lies on theperpendicular bisectors of the triangle.

Something interesting to note is that when triangle ABC is acute, O lies inside the triangle; when triangleABC is right-angled, O lies on the hypotenuse of the triangle, at its midpoint; and when triangle ABC isobtuse, O lies outside the triangle.

Let’s draw an acute triangle ABC and draw in the three perpendicular bisectors XO, YO, ZO, just like I’vedone below. There are three cyclic quadrilaterals lurking in the diagram — surely you can spot them.

A B

C

O

XY

Z

The cyclic quadrilaterals are

AZOY — the opposite angles ∠AZO and ∠OYA add to 180◦;

BXOZ — the opposite angles ∠BXO and ∠OZB add to 180◦; and

CYOX — the opposite angles ∠CYO and ∠OXC add to 180◦.7In Latin, the word “circum” means around and this makes sense because the circumcentre goes around the triangle.

31


Let’s now draw in the three line segments AO, BO, CO as well as the triangle XYZ. An extremely usefulexercise is to label all of the thirty-six angles in the diagram in terms of a = ∠CAB, b = ∠ABC and c = ∠BCA.

A B

C

OXY

Z

The midpoint theorem states that the triangles AZY and ABC are similar, so we have ∠YZA = b. Youcan use this strategy to label six of the angles in the diagram.

Since the quadrilateral AZOY is cyclic, the hockey theorem tells us that ∠YOA = ∠YZA = b. You canuse this strategy to label six more of the angles in the diagram.

Since the sum of the angles in triangle YOA is 180◦, we must have ∠YAO = 90◦ −∠YOA = 90◦ − b.You can use this strategy to label six more of the angles in the diagram.

Since the quadrilateral AZOY is cyclic, the hockey theorem tells us that ∠YZO = ∠YAO = 90◦ − b.You can use this strategy to label six more of the angles in the diagram.

The remaining twelve angles can be labelled by using the fact that the angles in a triangle add to 180◦.

The Incentre

An angle bisector of a triangle is a line which passes through a vertex and bisects the angle at that vertex. Notethat three randomly chosen lines will almost never ever meet at a point and yet, for any particular trianglewe choose, we’ll see that its three angle bisectors always do.

Proposition. The three angle bisectors of a triangle meet at a point.

Proof. Our proof relies crucially on the following lemma, which can be proven using congruent triangles.

Lemma. A point P lies on the angle bisector of ∠ABC if and only if the distance from P to AB isequal to the distance from P to CB.

So take a triangle ABC and the angle bisectors at the vertices A and B. If we suppose that these two linesmeet at a point I, then it must be the case that the distance from I to CA equals the distance from I to AB andthe distance from I to AB equals the distance from I to BC. These two statements together imply that thedistance from I to CA equals the distance from I to BC, in which case I lies on the angle bisector at the vertexC as well.

In the previous prof, we noted that if the angle bisectors of triangle ABC meet at I, then the distances from Ito the three sides are all equal. Another way to say this is that there’s a circle with centre I which touchesthe sides AB, BC, CA. This circle is called the incircle of the triangle and the point I is called the incentre.8

8In Latin, the word “in” means inside and this makes sense because the incentre goes inside the triangle.

32


Furthermore, the radius of the incircle is known as the inradius for obvious reasons. We now know that everytriangle has exactly one incircle and that its centre lies on the angle bisectors of the triangle.

Let’s draw a triangle ABC and draw in the three radii of the incircle PI, QI, RI, just like I’ve done below.There are three cyclic quadrilaterals lurking in the diagram — surely you can spot them.

A B

C

I

PQ

R


ARIQ — the opposite angles ∠ARI and ∠IQA add to 180◦;

BPIR — the opposite angles ∠BPI and ∠IRB add to 180◦; and

CQIP — the opposite angles ∠CQI and ∠IPC add to 180◦.

Let’s now draw in the three line segments AI, BI, CI as well as the triangle PQR. An extremely useful exerciseis to label all of the thirty-six angles in the diagram in terms of a = ∠CAB, b = ∠ABC and c = ∠BCA.

A B

C

I

PQ

R

The ice cream cone theorem states that the angles QAI and RAI are equal, so we have ∠QAI =∠RAI = a

2 . You can use this strategy to label six of the angles in the diagram.

Since the quadrilateral ARIQ is cyclic, the hockey theorem tells us that ∠QRI = ∠QAI = a2 . You can

use this strategy to label six more of the angles in the diagram.

Since the sum of the angles in triangle QAI is 180◦, we must have ∠QIA = 90◦ −∠QAI = 90◦ − a2 .

You can use this strategy to label six more of the angles in the diagram.

Since the quadrilateral ARIQ is cyclic, the hockey theorem tells us that ∠QRA = ∠QIA = 90◦ − a2 .

You can use this strategy to label six more of the angles in the diagram.

The remaining twelve angles can be labelled by using the fact that the angles in a triangle add to 180◦

and, in fact, they are all equal to 90◦.

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The Orthocentre

An altitude of a triangle is a line which passes through a vertex and is perpendicular to the opposite side. Ifthe triangle happens to have an angle greater than 90◦, then you will need to extend the sides in order todraw all three altitudes. Note that three randomly chosen lines will never ever meet at a point yet, for anyparticular triangle we choose, we’ll see that its three altitudes always do.

Proposition. The three altitudes of a triangle meet at a point.

Proof. Draw the triangle A′B′C′ such that A′B′ is parallel to AB and C lies on A′B′, B′C′ is parallel to BC andA lies on B′C′, and C′A′ is parallel to CA and B lies on C′A′.

A B

C

D

E

F

A′B′

C′

This creates the parallelograms ABCB′ and ABA′C, so that we have the equal lengths B′C = AB = CA′.Similarly, we have the equations C′A = BC = AB′ and A′B = CA = BC′. So the points A, B, C are simplythe midpoints of the sides B′C′, C′A′, A′B′, respectively. This is precisely the setup for the midpoint theorem.

Since CF is perpendicular to AB, it’s also parallel to B′A′ — in other words, CF is the perpendicular bisectorof A′B′. Similarly, we know that AD is the perpendicular bisector of B′C′ and BE is the perpendicular bisectorof C′A′. But we proved earlier that the three perpendicular bisectors of a triangle meet at a point. Therefore,the three altitudes AD, BE, CF of triangle ABC meet at a point.

The three altitudes AD, BE, CF of triangle ABC meet at a single point H called the orthocentre.9 Somethinginteresting to note is that when triangle ABC is acute, H lies inside the triangle; when triangle ABC isright-angled, H lies at a vertex of the triangle; and when triangle ABC is obtuse, H lies outside the triangle.

Let’s draw an acute triangle ABC and draw in the three altitudes AD, BE, CF, just like I’ve done below.Amazingly, there are six cyclic quadrilaterals lurking in the diagram — can you find them all?

9In ancient Greek, the word “ortho” means vertical and this makes sense because the altitude of a triangle is vertical with respect tothe base.

34


A B

C

D

E

F

H


AFHE — the opposite angles ∠AFH and ∠HEA add to 180◦;

BDHF — the opposite angles ∠BDH and ∠HFB add to 180◦;

CEHD — the opposite angles ∠CEH and ∠HDC add to 180◦;

ABDE — the angles ∠ADB and ∠AEB are equal;

BCEF — the angles ∠BEC and ∠BFC are equal; and

CAFD — the angles ∠CFA and ∠CDA are equal.

Let’s now draw in the triangle DEF. An extremely useful exercise is to label all of the thirty-six angles in thediagram in terms of a = ∠CAB, b = ∠ABC and c = ∠BCA.

A B

C

D

E

F

H

Since the sum of the angles in triangle ABD is 180◦, we must have ∠BAD = 90◦ −∠DBA = 90◦ − b.You can use this strategy to label six of the angles in the diagram.

Since the quadrilateral AFHE is cyclic, the hockey theorem tells us that ∠FEH = ∠FAH = 90◦ − b.You can use this strategy to label six more of the angles in the diagram.

Since ∠HEA is a right angle, we must have ∠FEA = 90◦ −∠FEH = b. You can use this strategy tolabel six more of the angles in the diagram.

Since the quadrilateral AFHE is cyclic, the hockey theorem tells us that ∠FHA = ∠FEA = b. You canuse this strategy to label six more of the angles in the diagram.

The remaining twelve angles can be labelled by using the fact that the angles in a triangle add to 180◦.

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The Centroid

A median of a triangle is a line which passes through a vertex and the midpoint of the opposite side.

Proposition. The three medians of a triangle meet at a point.

Proof. Suppose that the medians AX and BY meet at G1. By the midpoint theorem, we know that XY isparallel to AB which implies that the triangles ABG1 and XYG1 are similar by AAA. We also know by themidpoint theorem that AB

XY = 2, so the constant of proportionality is 2. This means that AG1G1X = 2. So the

median BY cuts the median AX at a point G1 such that AG1 is twice as long as G1X.

A B

C

G1

XY

A B

C

G2

X

Z

Now suppose that the medians AX and CZ meet at G2. By the midpoint theorem, we know that XZ isparallel to AC which implies that the triangles ACG2 and XZG2 are similar by AAA. We also know by themidpoint theorem that AC

XZ = 2, so the constant of proportionality is 2. This means that AG2G2X = 2. So the

median CZ cuts the median AX at a point G2 such that AG2 is twice as long as G2X.

Putting these two pieces of information together, we deduce that the points G1 and G2 must actually be thesame point. In other words, the medians BY and CZ meet the median AX at the same point.

The three medians AX, BY, CZ of triangle ABC meet at a single point G called the centroid. It is the centre ofgravity of ABC in the sense that if you cut the triangle out of cardboard, then it should theoretically balanceon the tip of a pencil placed at the point G. One consequence of our proof above is the fact that we have theequal fractions

AGGX

=BGGY

=CGGZ

= 2.

More Fun with Triangle Centres

There is so much more to triangle centres than we have mentioned. Let’s write down some interesting factshere, which you can try to prove on your own.

Proposition. If the orthocentre of triangle ABC is H, then the orthocentre of triangle HBC is A, the orthocentre oftriangle HCA is B and the orthocentre of triangle HAB is C.

Proposition. If you take a triangle ABC and draw in the three medians AX, BY, CZ, then the six resulting trianglesall have equal area.

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The following proposition shows that the four triangle centres we have looked at are related in variouspeculiar ways — you should try to prove all of these statements.

Proposition. Given a triangle ABC, let the midpoints of the sides be X, Y, Z, let the incircle touch the sides at P, Q, R,and let the feet of the altitudes be D, E, F.

The circumcentre O of triangle ABC is the orthocentre of triangle XYZ.

The incentre I of triangle ABC is the circumcentre of triangle PQR.

The orthocentre H of triangle ABC is the incentre of triangle DEF.

The centroid G of triangle ABC is the centroid of triangle XYZ.

Problems

When solving difficult geometry problems, here are a few things to always keep a look out for. If you becomewell practised at spotting these objects in your geometry diagrams, then you are well on the way to becominga geometry guru.

cyclic quadrilaterals

isosceles triangles

equal angles and lengths

similar or congruent triangles

right angles

ice cream cones

Problem. Let ABC be a triangle with incentre I and extend AI until it meets the circumcircle of triangle ABC at X.Prove that X is the circumcentre of triangle BIC.

A

B C

X

I

a a

bb

cc

Proof. When the incentre of triangle ABC is involved, I like to let the angles at A, B, C be 2a, 2b, 2c, respectively.This is because I lies on the angle bisectors so I can label

∠BAI = ∠CAI = a, ∠CBI = ∠ABI = b, ∠ACI = ∠BCI = c.

We want to prove that X is the circumcentre of triangle BIC or equivalently, that the lengths BX, IX, CX areall equal. Hopefully you can see that the problem is solved if we can prove that triangle BXI is isosceles withBX = IX. This is because the same reasoning will tell us that triangle CXI is isosceles with CX = IX.

37


So let’s focus on proving that BX = IX. Since there are many relationships between angles in circles, it makessense to try to instead prove that ∠IBX = ∠BIX. Using our notation, we obtain that

∠IBX = ∠CBX + ∠IBC = ∠CAX + ∠IBC = a + b.

Here, we’ve used the hockey theorem on chord CX to deduce that ∠CBX = ∠CAX = a.

Now observe that ∠BIX + ∠BIA = 180◦, since they form a straight line. And if we sum up the angles intriangle ABI, we obtain the equation ∠BAI + ∠ABI + ∠BIA = 180◦ or equivalently, a + b + ∠BIA = 180◦.These two facts imply that ∠BIX = a + b, so we have deduced that ∠IBX = ∠BIX. As we mentionedearlier, it follows that BX = IX and similar reasoning will lead to CX = IX as well. Hence, the three lengthsBX, IX, CX are all equal and X is the circumcentre of triangle BIC.

Problem. If ABC is a triangle with a right angle at C, prove that the angle bisector from C bisects the angle formed bythe altitude from C and the median from C.

Proof. Let’s call the altitude, the angle bisector and the median CF, CT and CZ, respectively. Then what we’dlike to prove can be rephrased as ∠FCT = ∠TCZ. However, we already know that CT bisects the right angle,so that ∠TCA = ∠TCB. What this means is that we can rephrase the problem once again as ∠FCA = ∠ZCB.With this in mind, let’s write a = ∠CAB and try to determine the angles ∠FCA and ∠ZCB in terms of a.

A B

C

F ZTa

The first of these angles is easy — since triangle CFA is right-angled, we can write ∠FCA = 90◦ − a. Nowif we consider the sum of the angles in triangle ABC, we obtain ∠ABC = 90◦ − a which is equivalent to∠ZBC = 90◦ − a. Remember that our goal is to prove that ∠ZCB = ∠FCA = 90◦ − a. So what we shouldtry to prove now is that triangle ZBC is isosceles, with ZB = ZC. But remember that the diameter of a circlesubtends an angle of 90◦, so a circle with diameter AB passes through C. As Z is the midpoint of AB, thismeans that Z is the circumcentre of triangle ABC and ZB = ZC, as desired.

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Fermat

Pierre de Fermat was actually a lawyer by day atthe Parlement of Toulouse and an amateur mathe-matician by night. He lived in the early seventeenthcentury from 1601 to 1665 and is often credited withthe development of a very early form of what we nowcall calculus. However, Fermat is probably most fa-mous for his work in number theory. One theorem ofhis — Fermat’s Little Theorem — says that if you pickyour favourite positive integer a and your favouriteprime number p, then the number ap− a will be divis-ible by p. The most famous story about Fermat tellsof how he pencilled in the margin of a mathematicsbook the following problem.

It is impossible to find two perfect cubeswhich sum to a perfect cube, two per-fect fourth powers which sum to a perfectfourth power, or in general, two perfectnth powers which sum to a perfect nthpower, if n is an integer greater than 2.

Rather tantalisingly, Fermat also wrote that he hada marvellous proof which was too small to fit intothe margin. Of course, many people tried to recreatethe supposedly unwritten proof, but to no avail. Infact, Fermat’s Last Theorem — as the result is com-monly called — was not proved until 1995, the proofbeing over one hundred pages long and using ex-tremely technical mathematical tools which haven’tbeen around for very long. Almost every mathemati-cian believes that Fermat must have been either mis-taken or lying.

In geometry, there is a triangle centre known as theFermat point. Given a triangle ABC, it is the point Pwhich makes the sum of the distances PA + PB + PCas small as possible. If no angles of the triangleare greater than or equal to 120◦, the point P willbe the unique point inside the triangle such that∠APB = ∠BPC = ∠CPA = 120◦.

39

1. EUCLIDEAN GEOMETRY 1.4. Geometric Gems

The Nine Point Circle

Now we come to one of the real gems of geometry, a theorem which was discovered more than two thousandyears after Euclid’s day. Remember that for every three points which do not lie on a line, there is a uniquecircle which passes through them. For four points, there is very rarely going to be a circle which passesthrough them all, and if there is, then those four points form a very special type of quadrilateral called acyclic quadrilateral. Given that it’s already quite difficult for four points to lie on a circle, then it must benear impossible to find nine points which lie on a circle. And yet, this is precisely what the nine point circletheorem tells us — we can find nine points which lie on a circle, associated to any particular triangle wechoose to think about.

Theorem (Nine Point Circle Theorem). Let ABC be a triangle with altitudes AD, BE, CF, medians AX, BY, CZ,and orthocentre H. If A′, B′, C′ are the midpoints of AH, BH, CH, respectively, then the nine points A′, B′, C′, D, E, F,X, Y, Z all lie on a circle which is — for obvious reasons — called the nine point circle of triangle ABC.

A B

C

XY

Z

D

E

F

A′ B′

C′

H

Proof. There are midpoints galore in this problem — in fact, six of the nine points that we are interested in aredefined as midpoints. So it seems like a prime opportunity to use the midpoint theorem. Applied in triangleABH, we obtain that A′B′ is parallel to AB. Applied in triangle ABC, we obtain that XY is also parallel toAB. Applied in triangle ACH, we obtain that A′Y is parallel to CH. Applied in triangle BCH, we obtain thatB′X is also parallel to CH.

In summary, XY and A′B′ are parallel to each other and to AB. Furthermore, A′Y and B′X are parallel to eachother and to CH. However, since CH is perpendicular to AB by assumption, we know that A′B′XY must bea rectangle. The exact same argument can be used to show that B′C′YZ and C′A′ZX are also rectangles.

Now let N be the midpoint of A′X and note that this means that N is the centre of the circle passing throughthe vertices of rectangle A′B′XY as well as the centre of the circle passing through the vertices of C′A′ZX,both of which have A′X as diameters. Therefore, the six points A′, B′, C′, X, Y, Z all lie on a circle with centreN. Six points down and three to go. . .

Since A′B′XY is a rectangle, we have ∠A′YX = 90◦. But ∠A′DX = 90◦ as well, so we know that thequadrilateral A′XDY is cyclic. In particular, D lies on the circumcircle of triangle A′XY, which we havealready seen actually passes through the six points A′, B′, C′, X, Y, Z. The same argument shows that E and Flie on this circle as well. So we can now conclude that the nine points A′, B′, C′, D, E, F, X, Y, Z all lie on acircle.

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The Euler Line and Other Gems

This next theorem tells us that three of the four triangle centres that we’ve already considered are very closelyrelated — in fact, they always lie on a line. It’s amazing that such a simple fact seems to have escapedeverybody’s notice until Euler, a pretty amazing mathematician whom we’ll encounter again later, arrived onthe scene in the eighteenth century.

Theorem. The orthocentre H, the centroid G and the circumcentre O of any triangle lie on a line known as the Eulerline. Furthermore, we have the equation HG = 2GO.

A

B C

H G O

There are many, many, many, many, many more geometric gems out there and we’ve really only scratchedthe surface of Euclidean geometry. For example, there is the following fact which adds the nine point circlecentre to the list of points lying on the Euler line.

Theorem. The orthocentre H, the nine point circle centre N, the centroid G and the circumcentre O of any triangle lieon a line known as the Euler line. Furthermore, we have the equation HN = NO.

To be honest, Euclidean geometry is not a thriving area of research mathematics. This is probably becausenew theorems in Euclidean geometry have limited uses in other areas. But that doesn’t mean they aren’tinteresting, and it doesn’t mean that people aren’t discovering new geometric gems all the time. Here’s apair of related facts, the first which was discovered just over a century ago, while the second was discoveredabout ten years ago.

Theorem.

Observe that if you take any four points A, B, C, D, then you can draw the nine point circles of the trianglesABC, BCD, CDA and DAB. Amazingly, all four of these circles meet at a single point — let’s call this point P.

Now suppose that we consider the feet of the perpendiculars from A to the sides of triangle BCD, where you mayhave to extend the sides. This will give us three points whose circumcircle we call the A-circle. Similarly, there isa B-circle, a C-circle and a D-circle. Amazingly, all four of these circles — you guessed it — meet at a singlepoint, which happens to coincide precisely with the point P.

It’s amazing — or at least I think it’s amazing — that you can build this incredibly intricate world of geometrywith just pen, paper and ten little axioms.

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Non-Euclidean Geometry

We’re now going to go back to where our journey began, way back to Euclid and his axioms. The commonnotions are pretty much trivial statements with no real geometric content. So let’s look at the other fiveaxioms, which are commonly referred to as postulates.

A1. You can draw a unique line segment between any two given points.

A2. You can extend a line segment to produce a unique line.

A3. You can draw a unique circle with a given centre and a given radius.

A4. Any two right angles are equal to each other.

A5. Suppose that a line ` meets two other lines, making two interior angles on one side of ` which sum toless than 180◦. Then the two lines, when extended indefinitely, will meet on that side of `.

It didn’t take very long at all for mathematicians to notice that one of the postulates sticks out like a sorethumb. Can you guess which one it is? Of course you can. . . it’s the only one which takes more than oneline to write down. At a deeper level, it just seems to be conceptually so much more complicated than therest. So for centuries upon centuries after Euclid announced his axioms, one of the burning questions inmathematics was whether or not the fifth axiom was needed at all, whether or not it could be proved fromthe other axioms.

Many mathematicians throughout the ages tackled the problem, but to no avail. A small advance was madeby showing that you could write down the parallel postulate in the following simpler looking, but actuallyequivalent, form. In fact, when Euclid’s fifth axiom is written in this way, it’s often referred to as the parallelpostulate.

A5′. Given a line and a point not on the line, there exists a unique line through the given point, parallel tothe given line.

One common approach was to to assume the opposite of the parallel postulate and keep deducing anddeducing, with the aim of finding a contradiction. If a contradiction could be found, then the opposite ofthe parallel postulate was false, thereby proving that the parallel postulate was true. Unfortunately — orpossibly, fortunately — no one was successful in doing so. What they should have done is stop looking fora contradiction, because there are none. If they had done this, then they would have realised that all thestatements that they were deducing were theorems in a new type of geometry. Since this geometry comesfrom taking the opposite of one of Euclid’s axioms, it’s commonly known as non-Euclidean geometry.

There were various people who contributed to the discovery of non-Euclidean geometry, all of whom livedaround the turn of the nineteenth century. The mathematicians Bolyai and Lobachevsky probably deservethe most credit, but their ideas are related to work by Saccheri, Gauss and various other people. Note thatthere are actually two ways to change the parallel postulate so as to create a non-Euclidean geometry.

Given a line and a point not on the line, there exists no line through the given point, parallel to the givenline.

Given a line and a point not on the line, there exists more than one line through the given point, parallelto the given line.

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Spherical Geometry

Euclidean geometry, for thousands of years, seemed to reflect the world around us and was the only geometrystudied by mathematicians. But why should mathematics follow the rules of the real world? Why can’tyou invent your own rules? And this is precisely what mathematicians decided to do. Nowadays, there aremany different types of geometry which people study. And usually, after studying these, people discoveredthat they turned out to be both interesting and surprisingly useful to the real world. So what makes theseseemingly crazy-sounding theories deserve to be called geometry? Loosely speaking, you can think of ageometry as anything where you have objects called points and lines, where points can lie on lines and linescan intersect lines, and so on.

The simplest non-Euclidean geometry is actually rather simple. It’s spherical geometry, which is the geometrywe are used to when we talk about flying around the Earth. Take a sphere — let’s say it has radius 1 to bedefinite — and consider the paths of shortest distance between two points. These are just arcs of great circles,those circles whose centre is the centre of the sphere. We can now define the points in spherical geometry tobe the points on the unit sphere, and we also define the lines in spherical geometry to be the great circles.We now have a new type of geometry with points and lines and angles and circles and so forth, but whichsatisfies the following statement.

Given a line and a point not on the line, there exists no line through the given point, parallel to the givenline.

You can see that two lines in spherical geometry can never be parallel, because they always meet at twopoints on opposite sides of the sphere.

It’s possible to prove lots of interesting theorems in spherical geometry — here’s just one interesting example.It shows that you can calculate the area of a spherical triangle from its three angles alone, something which isimpossible to do in Euclidean geometry.

Theorem. A spherical triangle whose angles measured in radians10 are a, b, c has area a + b + c− π.

Proof. Let’s think of our sphere as an orange with radius 1 and recall that its surface area is simply 4π.Consider what happens if you slice through the orange twice, with the knife passing through the centre ofthe orange, such that the two cuts make an angle of a radians with each other. This will create two slices, onopposite sides of the orange, whose total surface area is 2a

2π times the total surface area of the orange. So thesetwo slices will have surface area 2a

2π × 4π = 4a.

Now take a spherical triangle whose angles measured in radians are a, b, c. If we extend the sides, we obtaina diagram very much like the one below, where the blue spherical triangle is the one whose area we wishto calculate. Note that there is the same triangle, which we’ve shaded in grey, appearing on the back of thesphere as well.

Cutting along the two lines which form the angle a will create two orange slices with area 4a. Similarly,cutting along the two lines which form the angle b will create two orange slices with area 4b. And similarlyagain, cutting along the two lines which form the angle c will create two orange slices with area 4c.

10I’m hoping that you already know how to measure angles in radians, but if not, then it’s easy to learn. All you need to know is thatπ radians is the same thing as 180◦. Now you’re probably wondering why on earth someone would decide to measure angles in thisway. A better thing to wonder is why on earth someone would decide to split a straight angle into 180 parts and call each one a degree.Radians are very natural, because if you take a slice of pizza of a given angle, then the angle measured in radians is simply the length ofthe crust divided by the length of one of the cuts — in other words, the length of the arc divided by the radius of the circle. Because thisis so natural, many formulas in mathematics look a whole lot nicer when you use radians to measure angles.

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a b

c

Now let’s see what happens when we add up all of these areas. Most of the sphere is accounted for exactlyonce, but the blue spherical triangle has been accounted for three times and the grey spherical triangle hasalso been accounted for three times. So if we let A denote the area of the blue triangle, or equivalently, thearea of the grey triangle, we have the equation

4a + 4b + 4c = 4π + 4A.

Here, the number 4π represents the total surface area of the sphere with radius 1. This equation rearranges togive the desired formula A = a + b + c− π.

One very interesting consequence of this theorem is the fact that the angles in a spherical triangle must addto more than 180◦. This should seem intuitively true, because the sides of a spherical triangle seem to “bulgeoutwards”, creating larger angles than a Euclidean triangle. This is one of the major differences betweenspherical and Euclidean geometry.

Remember that one of our goals is to show that you can have a geometry in which all of Euclid’s axiomsare true except for the parallel postulate. Certainly, the parallel postulate fails to hold in spherical geometry,but unfortunately it’s not the only one which fails. For example, it’s not true in spherical geometry that youcan draw a unique line segment between any two given points. That’s because between the north pole andthe south pole of the sphere, there are infinitely many line segments. However, knowing a bit about howspherical geometry works will help us with hyperbolic geometry, a case where all of Euclid’s axioms do holdexcept for the parallel postulate.

Hyperbolic Geometry

We now consider a type of geometry which satisfies the following statement.

Given a line and a point not on the line, there exists more than one line through the given point, parallelto the given line.

This geometry will be far more difficult to visualise and is conceptually more removed from everydayexperience. The geometry we will talk about is called hyperbolic geometry and there are many ways to describeit. The particular way that we’re going to use is called the Poincare disk model.

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In the Poincare disk model, the hyperbolic plane is an open disk — let’s say it has radius 1 to be definite —or, in other words, all of the points which are inside, but not on the circumference of, a circle. Very intuitively,you can think of the hyperbolic plane as a circular pond of quicksand, one where it’s easy to walk aroundwhen you’re close to the middle, but which gets infinitely difficult to travel through when you’re close to theedge. So the effort that it takes you to get from point A to point B or equivalently, the hyperbolic distance frompoint A to point B, is warped and not like the normal Euclidean distance between two points. This meansthat if you were told to walk from point A to point B, you would probably take a curved path, which bendstoward the middle of the circular pond of quicksand. Anyway, it’s possible to make all of this mumbo jumboprecise and the end result is that hyperbolic geometry works as follows.

A hyperbolic point is just a normal point inside the hyperbolic plane.

A hyperbolic line is the arc of a circle which is perpendicular to the boundary of the disk or a diameterof the disk.

You can see some examples of hyperbolic lines drawn in the Poincare disk model below. Remember thatlines are said to be parallel if they never meet. You should now be able to convince yourself that, given a linein hyperbolic geometry and a point not on the line, there exists more than one line through the given point,parallel to the given line.

Remember that in spherical geometry, the area of a triangle with angles a, b, c is simply a + b + c− π. Notonly does this mean that you can calculate the area of a triangle from its angles alone, but also that in everyspherical triangle, the angles add up to more than 180◦. Hyperbolic geometry is, in many ways, the exactopposite of spherical geometry. For example, in every hyperbolic triangle, the angles add up to less than180◦. This should seem intuitively true, because the sides of a hyperbolic triangle seem to “curve inwards”,creating smaller angles than a Euclidean triangle. Furthermore, you can calculate the area of a hyperbolictriangle from its angles alone in the following way.

Theorem. A hyperbolic triangle whose angles measured in radians are a, b, c has area π − a− b− c.

Note that there are theorems which will be the same in Euclidean, spherical or hyperbolic geometry. This issimply because there are results which you can prove which don’t rely on the parallel postulate at all. Eventhough hyperbolic geometry seems the most far removed from our everyday experience, it is, in some sense,the most mathematically important and is used in various areas of pure mathematics and theoretical physics.

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Problems

Problem. Consider a point P on the circle which passes through the vertices of triangle ABC. Let D be the point onBC such that PD is perpendicular to BC. Let E be the point on CA such that PE is perpendicular to CA. Let F be thepoint on AB such that PF is perpendicular to CA. (You may have to extend the lines AB, BC and CA for this to bepossible.) Prove that the points D, E and F lie on a line.11

A

B CD

E

F

P

Proof. The first thing to do, as always, is to draw a large accurate diagram. With a big fat circle in the picture,it is hard to miss the fact that the quadrilateral ABCP must be cyclic. This is a very useful piece of informationto note down because we’re going to use it later.

In general, right angles will often lead to cyclic quadrilaterals, and there are actually three of them that wehaven’t yet mentioned. One of the easiest ones to see is the quadrilateral PFAE, which is cyclic because itsopposite angles add up to 180◦.

∠PFA + ∠AEP = 90◦ + 90◦ = 180◦.

Another one which is not too difficult to spot is the quadrilateral PFBD, which is also cyclic because itsopposite angles add up to 180◦.

∠PFB + ∠BDP = 90◦ + 90◦ = 180◦

Finally, we consider quadrilateral PEDC, which is cyclic because of the hockey theorem applied to the equalangles

∠PEC = ∠PDC = 90◦.11The line passing through the points D, E and F is actually known as the Simson line of the point P and the triangle ABC.

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Let’s remember exactly what we’re trying to prove — namely, that the points D, E and F lie on a line. Oneway to prove this would be to show that ∠PEF + ∠PED = 180◦. If we label the angle ∠PEF = x, then thehockey theorem applied to the cyclic quadrilateral PFAE tells us that ∠PAF = x as well. The angle next doorto this must then be ∠PAB = 180◦ − x. Since the opposite angles in the cyclic quadrilateral ABCP must addup to 180◦, this in turn yields the fact that ∠BCP = x. And now we use the fact that the opposite angles inthe cyclic quadrilateral PEDC must add up to 180◦ to tell us that

∠PED = 180◦ −∠DCP = 180◦ −∠BCP = 180◦ − x.

If you’ve labelled all of these angles on your diagram, then you will have noticed that we have now labelledboth ∠PEF and ∠PED. This is most useful, because we now have

∠PEF + ∠PED = x + (180◦ − x) = 180◦.

And this is exactly what we’re aiming to prove, because it implies that the points D, E and F lie on a line.

Problem. In the hyperbolic plane, there exist quadrilaterals all of whose angles are equal to 45◦. Sketch one example ofsuch a quadrilateral in the hyperbolic plane using the Poincare disk model. What is the area of this quadrilateral?

Proof. The large circle below represents the Poincare disk model of the hyperbolic plane. The four dottedcurves represent four hyperbolic lines. As you can see from the diagram, they form a quadrilateral all ofwhose angles are equal to 45◦.

45◦45◦

45◦45◦

Now consider the following schematic diagram for the quadrilateral, where the hyperbolic lines are repre-sented by normal straight lines. We have divided the quadrilateral into two triangles — numbered 1 and 2 —and labelled every single angle in the diagram.

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a

b cd

ef

1

2

Since each angle of the quadrilateral is equal to 45◦, we have four equations that these labelled angles mustsatisfy. Here, we have labelled the angles using radians rather than degrees.

a + f =π

4, b =

π

4, c + d =

π

4, e =

π

4

One of the advantages of using radians is that the area of a hyperbolic triangle is much easier to determine.In fact, we know that the area of triangle 1 is π− a− b− c and the area of triangle 2 is π− d− e− f . Addingup the areas of these two triangles, we deduce that the area of the quadrilateral is

(π − a− b− c) + (π − d− e− f )

= 2π − (a + b + c + d + e + f )

= 2π − (a + f )− (b)− (c + d)− (e)

= 2π − π

4− π

4− π

4− π

4= π.

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Gauss

Johann Carl Friedrich Gauss — who was born in 1777and died in 1855 — was a German mathematicianwho contributed significantly to a variety of subjects.The book Men of Mathematics by Eric Temple Bell andpublished in 1937 — which apart from the sexist ti-tle, is a decent read — has a chapter on Gauss whichbegins “Archimedes, Newton and Gauss, these three,are in a class by themselves among the great mathe-maticians, and it’s not for ordinary mortals to attemptto range them in order of merit.” Hopefully this con-vinces you that Gauss is regarded among the bestmathematicians who ever walked the earth.

Gauss was known to be a child prodigy and thereis a famous story which tells of his primary schoolteacher asking the class to add up the numbers from1 up to 100. The very young Gauss produced thecorrect answer in seconds, to the astonishment of histeacher. Presumably, his method was to pair the num-bers 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and soon, obtaining fifty pairs, each with sum 101. Hence,the answer is 50× 101 = 5050.

It seems that Gauss may have discovered the possibil-ity of non-Euclidean geometry but, for some strangereason, decided not to publish it. In fact, when ayounger mathematician by the name of Janos Bolyaidiscovered non-Euclidean geometry and publishedhis work in 1832, Gauss wrote that “To praise it would

amount to praising myself. For the entire content ofthe work. . . coincides almost exactly with my ownmeditations which have occupied my mind for thepast thirty or thirty-five years.”

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2. SYMMETRY IN GEOMETRY 2.1. Isometries

What is Symmetry?

You probably have an intuitive idea of what symme-try means and can recognise it in various guises. Forexample, you can hopefully see that the letters N, O

and M are symmetric, while the letter R is not. Butprobably, you can’t define in a mathematically pre-cise way what it means to be symmetric — this issomething we’re going to address.

Symmetry occurs very often in nature, a particularexample being in the human body, as demonstratedby Leonardo da Vinci’s drawing of the Vitruvian Manon the right.12 The symmetry in the picture arisessince it looks essentially the same when we flip itover. A particularly important observation about thedrawing is that the distance from the Vitruvian Man’sleft index finger to his left elbow is the same as thedistance from his right index finger to his right elbow.Similarly, the distance from the Vitruvian Man’s leftknee to his left eye is the same as the distance fromhis right knee to his right eye. We could go on andon writing down statements like this, but the pointI’m trying to get at is that our intuitive notion of sym-metry — at least in geometry — is somehow tied upwith the notion of distance.

What is an Isometry?

The flip in the previous discussion was a particular function which took points in the picture to other pointsin the picture. In particular, it did this in such a way that two points which were a certain distance apartwould get mapped to two points which were the same distance apart. This motivates us to consider functionsf which map points in the plane to points in the plane such that the distance from f (P) to f (Q) is the sameas the distance from P to Q for any choice of points P and Q. Any function which satisfies this property iscalled an isometry. This comes from the ancient Greek words “isos”, meaning equal, and “metron”, meaningmeasure.

Example. The best way to get a feeling for what an isometry looks like is to consider some examples.

Identity : The identity is the function which simply takes a point P in the plane to the same point P. Inother words, it does nothing, so hopefully you can see that it’s an example of an isometry — in fact, thesimplest example of an isometry. We’ll sometimes denote the identity map by I.

Translation : A translation is a function which takes every point in the plane and slides it in a certaindirection by a certain distance. In other words, if f is a translation such that f (A) = B and f (X) = Y,then the quadrilateral ABYX will always be a parallelogram. This means that if you want to specify a

12Leonardo da Vinci was very interested in symmetry and later on we’re going to see a theorem which bears his name.

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translation, then you only have to specify a point A and the point f (A) = B. We’ll sometimes denotethe translation which takes the point A to the point B by TAB.

A

B

X

Y

f

f

Rotation : A rotation is a function which takes the whole plane and turns it about a certain point Othrough a certain angle a. In other words, if f is a rotation such that f (A) = B, then the centre ofrotation O and the angle of rotation a satisfy OA = OB and ∠AOB = a. This means that if you want tospecify a rotation, then you only have to specify the centre of rotation and the angle of rotation. We’llsometimes denote the rotation about the point O through an angle a by RO,a.

O A

B

a

Reflection : A reflection is a function which takes the whole plane and flips it over a certain line ` whichwe also refer to as a mirror. It’s important to observe that if a reflection maps the point A to the point B,then ` will be the perpendicular bisector of the line segment AB. This means that if you want to specifya rotation, then you only have to specify the mirror. We’ll sometimes denote the reflection through theline ` by M`.

A

B

`

Suppose that P and Q are two points in the plane which are distance d apart from each other. If we apply anisometry to both points, then the result will be two points which are still distance d apart from each other.If we apply another isometry to the resulting points, then the new result will be two points which are stilldistance d apart from each other. The application of two or more isometries in a row is called composition ofisometries. From what we’ve just said, you can see that the composition of two or more isometries will alwaysbe an isometry. This means that we can compose the examples we’ve listed above in an attempt to discovernew isometries.

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Interesting Facts about Isometries

One of our guiding questions will be to determine what sorts of isometries there are. For example, is it truethat the examples we discussed earlier — identity, translation, rotation and reflection — account for everysingle possible isometry? Before we answer this question, we need to prove some basic facts.

Proposition. If f is an isometry such that f (A) = A′, f (B) = B′, f (C) = C′, then triangles ABC and A′B′C′ arecongruent.

A

B C

A′

B′ C′

Proof. By the very definition of an isometry, we have the equal lengths AB = A′B′, BC = B′C′ and CA = CA′.So it follows by SSS that the triangle ABC and the triangle A′B′C′ are congruent.

What if I give you two triangles ABC and A′B′C′ and ask you whether you can find an isometry such thatf (A) = A′, f (B) = B′ and f (C) = C′? The previous proposition says that you couldn’t possibly be able todo this unless the triangles ABC and A′B′C′ are congruent. So if we suppose that they actually are congruent,then can you definitely find an isometry satisfying the conditions? If so, then how many can you find exactly?The next proposition states that you can always find one and only one.

Theorem. If triangles ABC and A′B′C′ are congruent, then there is a unique isometry such that f (A) = A′,f (B) = B′ and f (C) = C′.

Proof. To show that there is at least one such isometry, we can construct it explicitly. First, we translate triangleABC until A lines up with A′. Then we rotate triangle ABC around the point A until B lines up with B′. Nowif C and C′ don’t already line up, then we reflect through the line AB so that they do. Since the compositionof these two or three isometries is also an isometry, we have constructed an isometry satisfying the conditionsof the problem. Note that we need triangles ABC and A′B′C′ to be congruent so that everything lines upexactly as we have described.

Now we need to show that there is at most one such isometry. Take any point P in the plane — we will showthat there is only one possible point P′ for f (P). This is because the definition of an isometry forces the pointP′ to satisfy the conditions A′P′ = AP, B′P′ = BP, and C′P′ = CP.

Another way to say this is that P′ has to

lie on the circle with centre A′ and radius AP;

lie on the circle with centre B′ and radius BP; and

lie on the circle with centre C′ and radius CP.

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A

B C

P

A′

B′ C′

P′

Q′

But hopefully you can see that the latter two circles can only meet at two points which we’ve labelled P′ andQ′, while the first circle can only pass through one of these two. And this single point where the three circlesmeet is the only possible location for the point f (P). So we have shown that for every point P in the plane,there is only one possible location for the point f (P). This means that there is at most one isometry f suchthat f (A) = A′, f (B) = B′ and f (C) = C′.

Since we’ve proven that there is at least one isometry satisfying the given conditions as well as the fact thatthere is at most one isometry satisfying the given conditions, we can put these two statements together todeduce that there’s a unique isometry satisfying the given conditions.

One very important consequence of this theorem is the fact that two isometries f and g must be the same ifthey satisfy f (A) = g(A), f (B) = g(B) and f (C) = g(C) for some triangle ABC.

Composition of Isometries

We’ve already mentioned that you can take isometry f followed by an isometry g and that the result isanother isometry called the composition of f and g. It’s going to be very useful to have some notation for thisidea. So let’s denote the result of “doing f followed by doing g” as g ◦ f . Yes, that’s right — even though wedo f first and g second, we write the result with g to the left of f .13

Pf−→ f (P)

g−→ g( f (P)) = g ◦ f (P)

13The reason for this is because in function notation, we would normally write the result of doing f followed by doing g to a point Pas g( f (P)) — note that even though we do f first and g second, we write the result with g to the left of f .

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As our first application of composition, I’m going to introduce a new isometry which didn’t appear in ouroriginal list of examples. It’s the isometry which you get when you do a reflection in a line ` followed by atranslation parallel to `. Using our earlier notation, we could write such a function as TAB ◦M`, where AB isa line segment parallel to `. An isometry which takes this form is called a glide reflection and we’ll sometimesdenote it by GAB. The process of repeatedly applying a glide reflection is something you’ve no doubt alldone before when walking along the beach. Each successive footprint that you leave in the sand is a glidereflection applied to the previous footprint.

Since you can compose different isometries, a natural question to ask is what the result is. For example, thefact below follows immediately from the definition of a translation and doesn’t really need a proof.

Proposition. The composition of a translation with another translation is always a translation.14 In fact, we have therather obvious formula TBC ◦ TAB = TAC.

To write down a similar fact concerning the composition of two reflections is a more difficult matter — butwe can still do it.

Proposition. Let Mk denote a reflection in the line k and M` denote a reflection in the line `. Then the compositionM` ◦Mk of a reflection in k followed by a reflection in ` is

the identity if k and ` are the same line;

a translation if k and ` are parallel lines — the direction of translation is perpendicular to k and ` while thedistance of translation is twice the distance between k and `; or

a rotation if the two lines meet — the centre of rotation is the intersection of k and ` while the angle of rotation istwice the angle from k to `.

Proof.

This one’s obvious because if you reflect twice through the same line, you end up where you started.

In the following diagram, you can see that the result of applying M` ◦Mk to P is to move it by a distanceof 2a + 2b, where a + b is the distance between k and `. Furthermore, the direction of translation isperpendicular to both k and `.

k `

P Mk(P) M` ◦ Mk(P)

a a b b

14You might think that we haven’t accounted for the fact that the composition of two translations could possibly be the identity.However, you can and should consider the identity as a translation which moves every point by zero distance.

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In the following diagram, you can see that the result of applying M` ◦Mk to P is to rotate by an angle2a + 2b, where a + b is the angle from k to `. Furthermore, the centre of rotation is the intersection of kand `.

k

`

P

Mk(P)M` ◦ Mk(P)

aab b

Although we’ve demonstrated these facts for only one particular point P, the argument would have workedfor several different choices of P. By the theorem we proved earlier, we only need to demonstrate these factsfor three points which form a triangle to know that they hold for all points.

We’ll also state a couple more results concerning composition of isometries. We won’t provide the proofs,since they aren’t particularly interesting and they use ideas very similar to those in the previous proof.However, you should definitely draw a few diagrams to convince yourself that they are true.

Proposition. Let R1 be a rotation by angle a1 and R2 be a rotation by angle a2. Then the composition R2 ◦ R1 is

a rotation by angle a1 + a2 if a1 + a2 6= 360◦; or

a translation if a1 + a2 = 360◦.

Proposition. A reflection followed by a translation is a reflection or a glide reflection. A reflection followed by arotation is a reflection or a glide reflection.

Classification of Isometries

Since the composition of two isometries is again an isometry, you can try to build every possible isometryout of some simple building blocks.15 The next result tells us that we can take the reflections as our buildingblocks.

Proposition. Every isometry is the composition of at most three reflections.

Proof. Consider any isometry f , any triangle ABC, and let f (A) = A′, f (B) = B′ and f (C) = C′. All weneed to prove is that it takes at most three reflections to send A to A′, B to B′ and C to C′. We simply reflectin the perpendicular bisector of AA′ so that A ends up coinciding with A′. Now we simply reflect in theperpendicular bisector of BB′ so that B ends up coinciding with B′. The congruence of triangles ABC andA′B′C′ ensures that A still coincides with A′. Now either the two triangles coincide and the job took tworeflections, or we need one more reflection through A′B′ to finish off the job, so that C coincides with C′.

15This reeks of the reductionist approach that we used to begin our journey into Euclidean geometry.

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Theorem. Every isometry is a translation, a reflection, a rotation or a glide reflection.

Proof. We just proved that every isometry is the product of at most three reflections, so the proof can bedivided into the following cases.

The only isometry which is the product of zero reflections is, of course, the identity isometry. As usual,we can think of the identity isometry as a translation which moves every point by zero distance.

The only isometries which are the product of one reflection are, of course, reflections themselves.

The proposition which we proved above tells us that the product of two reflections is the identity, atranslation or a rotation.

Now let’s consider the product of three reflections as a reflection followed by the product of tworeflections. Thus, the product of three reflections can be the product of a reflection and the identity, theproduct of a reflection and a translation, or the product of a reflection and a rotation. In the first casewe obtain a reflection, in the second case we obtain a reflection or a glide reflection, and in the thirdcase we obtain a reflection or a glide reflection.

So what you can hopefully see is that products of up to three reflections always result in a translation, areflection, a rotation or a glide reflection. Furthermore, since every isometry is the composition of at mostthree reflections, this accounts for all possible isometries.

Problems

Problem. Let ABC be a triangle with the vertices labelled clockwise such that AC = BC and ∠ACB = 90◦. Let MAB

be the reflection in the line AB, MAC be the reflection in the line AC, and R be the rotation by 90◦ counterclockwisearound B. Identify the composition R ◦MAB ◦MAC.

Proof. The idea is to use the theorem we proved earlier which states that

if triangles ABC and A′B′C′ are congruent, then there is a unique isometry such that f (A) = A′,f (B) = B′ and f (C) = C′.

What this means is that we can solve problems likethis one using the following simple strategy. Findthree points which form a triangle and see where thecomposition of isometries takes them. Next, it’s timeto guess what the isometry is. If your guess is cor-rect for the three vertices of the triangle, then it mustbe correct. And this is because the theorem aboveguarantees that if you know what an isometry doesto three corners of a triangle, then you know what theisometry does to every point in the plane.

A

B

CP

Q

R S

The diagram above shows triangle ABC drawn on a grid of squares. Since we want to choose three pointswhich form a triangle, we may as well choose the points A, B and C.

It’s easy to check that MAC(A) = A, MAB(A) = A and R(A) = P.In other words, R ◦MAB ◦MAC(A) = P.

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It’s easy to check that MAC(B) = Q, MAB(Q) = R and R(R) = Q.In other words, R ◦MAB ◦MAC(B) = Q.

It’s easy to check that MAC(C) = C, MAB(C) = S and R(S) = C.In other words, R ◦MAB ◦MAC(C) = C.

So can you think of an isometry which takes A to P, B to Q and C to C? If you think hard enough, youshould realise that it’s just a rotation by 180◦ around C. So we’ve managed to deduce that the compositionR ◦MAB ◦MAC is a rotation by 180◦ around C.

Problem. Let ABCD be a rectangle with the vertices labelled counterclockwise such that BC = 2AB. Let

MAB be the reflection in the line AB;

RB be the counterclockwise rotation by 90◦ about B;

TDB be the translation which takes D to B; and

GCD be the glide reflection in the line CD which takes C to D.

Identify the composition MAB ◦ RB ◦ TDB ◦ GCD.

Proof. The diagram below shows rectangle ABCD drawn on a grid of squares. Since we want to choose threepoints which form a triangle, we may as well choose the points A, B and C.

It’s easy to check that GCD(A) = E, TDB(E) = D, RB(D) = F and MAB(F) = G.In other words, MAB ◦ RB ◦ TDB ◦ GCD(A) = G.

It’s easy to check that GCD(B) = H, TDB(H) = C, RB(C) = I, and MAB(I) = I.In other words, MAB ◦ RB ◦ TDB ◦ GCD(B) = I.

It’s easy to check that GCD(C) = D, TDB(D) = B, RB(B) = B, and MAB(B) = B.In other words, MAB ◦ RB ◦ TDB ◦ GCD(C) = B.

A

B C

D

EF G

H

I

O

So can you think of an isometry which takes A to G, B to I and C to B? If you think hard enough, you shouldrealise that it’s a rotation, although you might not be sure of where the centre lies. However, we can use thefact that if a rotation takes X to Y, then the centre of rotation must lie on the perpendicular bisector of XY. Inparticular, the centre of the rotation that we’re interested in must lie on the perpendicular bisector of AG aswell as the perpendicular bisector of BI. And there’s only one point which does that — namely, the point Olabelled in the diagram above. It’s now easy to deduce that the composition must be a rotation about O by∠AOG = 90◦ in the clockwise direction.

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Archimedes

Archimedes of Syracuse was a Greek mathematicianwho is widely held to be the greatest mathematicianof antiquity and one of the greatest of all time. Liv-ing from about 287 BC to 212 BC, his thoughts onmathematics were far ahead of his time. He used atechnique called the method of exhaustion to calcu-late areas under parabolas and give a very accurateapproximation of the number π. He also applied thistechnique to prove that a sphere which fits perfectlyinside a cylinder has two-thirds of its surface area aswell as two-thirds of its volume. He regarded this asthe greatest of his mathematical achievements andasked for a diagram of a sphere inside a cylinder tobe placed on his tombstone. Archimedes’ method ofexhaustion was a precursor to the modern day dif-ferential and integral calculus which was discoverednearly two thousand years later.

It’s quite common for a mathematician to be multi-talented and Archimedes was certainly no exception.He is also renowned as a physicist, engineer, inventor,and astronomer. Among his great discoveries andinventions are the foundations of hydrostatics, theexplanation of the principle of the lever, machines tobe used in siege warfare, and many many more. He isthe one who is said to have cried “Eureka!” and ranthrough the streets of Syracuse naked upon discov-ering the principle of buoyancy while in the bathtub.Another story about Archimedes is that his servantsneeded to take him against his will to the baths. Andwhile they bathed him and anointed him with oils,he would be drawing diagrams on his body with theoils, such was his enthusiasm for geometry.

The mathematical writings of Archimedes were notparticularly well known throughout antiquity. How-ever, the few copies which survived through to theMiddle Ages became an influential source of ideas forscientists. Amazingly, previously unknown works ofArchimedes were discovered in 1906. These writings,now known as the Archimedes Palimpsest, providenew insights into how he obtained mathematical re-sults. We are quite lucky to have them because thiscopy of his writings had been made in the tenth cen-

tury AD and the pages subsequently erased, foldedin half, and reused for a Christian text. Fortunately,the erasure was incomplete and we can now readArchimedes’ work after scientific and scholarly workover the past ten years involving digital image pro-cessing using ultraviolet, infrared and X-ray technol-ogy.

Although Archimedes had built many machines ofwar to keep the Romans out, they finally managed tocapture his home town of Syracuse. Apparently, oneRoman soldier happened to find him hard at workon a geometry problem. Archimedes was so trans-fixed that he never noticed the soldier nor even thefact that the city had been taken. When Archimedesrefused to follow the soldier until he had finishedsolving the problem, the soldier decided to run hissword through him, despite orders to keep him alive.And that was the end of Archimedes.

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2. SYMMETRY IN GEOMETRY 2.2. Symmetry and Groups

Direct and Opposite Isometries

Consider a triangle ABC in the plane such that the vertices A, B, C occur counterclockwise around theboundary of the triangle. If you apply an isometry to the triangle, then the result will be a triangle where thevertices A, B, C can occur clockwise or anticlockwise. If the orientation stays the same, then we say that theisometry is direct but if the orientation changes, then we say that the isometry is opposite.

A

B C

A′

B′ C′

A′

C′ B′

direct

opposite

Remember that we classified the isometries into four types — translations, rotations, reflections and glidereflections. It’s easy to see which of these are direct and which are opposite.

Every single translation is a direct isometry.

Every single rotation is a direct isometry.

Every single reflection is an opposite isometry.

Every single glide reflection is an opposite isometry.

One of the nice things about composition of direct and opposite isometries is that they behave very much likemultiplication of positive and negative numbers. This should be obvious when you compare the followingtwo “multiplication tables” which have the same underlying structure. We’re going to be looking at many“multiplication tables” like this and examining their underlying structure, so keep this example in mind.

◦ dir opp

dir dir oppopp opp dir

× pos neg

pos pos negneg neg pos

Fixed Points of Isometries

A fixed point of an isometry f is a point P such that f (P) = P — in other words, a point which does not getmoved by the isometry. Remember that we classified the isometries into four types — translations, rotations,reflections and glide reflections. It’s easy to see which of these have fixed points and which of these don’t.

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Every single translation which is not the identity has no fixed points.

Every single rotation has one fixed point — namely, the centre of rotation.

Every single reflection has infinitely many fixed points — namely, the points on the mirror.

Every single glide reflection which is not a reflection has no fixed points.

Putting this information together with our knowledge of direct and opposite isometries, we have thefollowing table. As long as the isometry we’re interested in is not the identity, this table allows us to deducethe type of an isometry just by knowing whether it’s direct or opposite and whether or not it has fixed points.

isometry direct or opposite fixed points

translation direct norotation direct yes

reflection opposite yesglide reflection opposite no

Symmetry in the Plane

So we now know something about isometries — but what does this all have to do with symmetry? Well,we’re now in a position where we can define what we mean by a symmetry, at least in the realm of Euclideangeometry. Informally, a symmetry of a geometric shape will be something we can do to the Euclidean planewhile someone’s back is turned so that when they turn around again, the shape will look exactly the same.More precisely, given a set X of points in the plane — it could be finite or infinite — a symmetry of X is anisometry which leaves the set X unchanged. You should think of X as a black and white picture, where thepoints in the plane coloured black are those that belong to X while the points in the plane coloured white arethose that don’t belong to X.

Note that the isometry doesn’t have to leave every point of X exactly where it is — that would be way toorestrictive — but only has to leave X as a whole exactly where it is. By this precise mathematical definition,every single subset of the Euclidean plane, no matter how crazy it looks, has at least one symmetry — namely,the identity isometry. Our intuitive notion of a shape being symmetric corresponds to the mathematicallyprecise fact that it has a symmetry which is not the identity.

Example. The following diagram lists the letters of the alphabet and below it, the number of symmetriesthat it has. You should check to see that all the numbers are correct and, for each letter, determine what theisometries are which leave the letter exactly where it is.

A B C D E F G H I J K L M2 2 2 2 2 1 1 4 4 1 1 1 2

N O P Q R S T U V W X Y Z2 4 1 1 1 2 2 2 2 2 4 2 2

Example. As another example, consider the symmetries of the square ABCD. We can prove that there are atmost eight symmetries, since any symmetry must take the triangle ABC to one of the triangles ABC, BCD,CDA or DAB. Each of these triangles is isosceles, so there are two ways to map the triangle ABC to each ofthem. Tally all these up and, as promised, you see that there can be at most eight symmetries of the square.

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A

B C

D

To see that there are, in fact, exactly eight symmetries of the square, we simply need to write them all down.Below, we give each symmetry a symbol, describe the isometry geometrically, and describe what the isometrydoes to the vertices of the square.

I : the identity isometryA→ A, B→ B, C → C, D → D

R1 : rotation by 90◦ anticlockwise about the centre of the squareA→ B, B→ C, C → D, D → A

R2 : rotation by 180◦ anticlockwise about the centre of the squareA→ C, B→ D, C → A, D → B

R3 : rotation by 270◦ anticlockwise about the centre of the squareA→ D, B→ A, C → B, D → C

Mh : reflection across the horizontal line passing through the centre of the squareA→ B, B→ A, C → D, D → C

Mv : reflection across the vertical line passing through the centre of the squareA→ D, B→ C, C → B, D → A

MAC : reflection across the line ACA→ A, B→ D, C → C, D → A

MBD : reflection across the line BDA→ C, B→ B, C → A, D → D

Qualifying Symmetry in the Plane

So we’re now in a position where we can quantify — in other words, count — the symmetries of a shape.However, it will be much more interesting to qualify — in other words, examine the structure of — thesymmetries of a shape.16

Earlier, we stated that the letters H and X each have four symmetries. Not only do they have the samenumber, but their symmetries also seem to have a similar structure — there is the identity, rotation by 180◦,reflection in a horizontal axis, and reflection in a vertical axis. So in some sense, the letters H and X not onlyhave the same quantity of symmetries, but also the same quality of symmetries, whatever that might mean.

16People who don’t know what mathematics is about seem to think that it is about quantifying — in other words, counting — things,when it is really about qualifying — in other words, examining the structure of — things.

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A slightly more interesting example is to compare the symmetries of the letters A and B, both of which havetwo symmetries. This time, however, the symmetries of the letter A are the identity and reflection in a verticalaxis, while the symmetries of the letter B are the identity and reflection in a horizontal axis. So it seems thatthe letters A and B have the same quantity of symmetries, but not the same quality — or do they? On furtherthought, if we consider the letter B to be simply made up of a set of points in the plane, who cares which wayis up, down, left or right? As a mathematical object, it is essentially the same thing as B, which like the letterA, has the identity and reflection in a vertical axis as its symmetries.

One crucial observation about the symmetries of an object is that if you compose two of them, then the resultis always a symmetry. This means that if you find some symmetries of an object, then you can try to findother ones by composing the ones you already have. The composition of symmetries captures their structurein a way that can be represented by a sort of “multiplication table”.

Example. Let’s continue with our example from earlier, which involved the symmetries of a square. We wereable to verify that there are eight symmetries, each of which we gave a name. We may now use these to fillout a table which describes precisely how these symmetries compose with each other. Note that if you wantto work out the entry corresponding to the row labelled A and the column labelled B, then the entry shouldbe A ◦ B. Remember that this is the composition B followed by A, because we always apply the isometry onthe right before the one on the left. You should carefully check the following table to make sure that youunderstand exactly how to construct it on your own.

◦ I R1 R2 R3 Mh Mv MAC MBD

I I R1 R2 R3 Mh Mv MAC MBD

R1 R1 R2 R3 I MBD MAC Mh Mv

R2 R2 R3 I R1 Mv Mh MBD MAC

R3 R3 I R1 R2 MAC MBD Mv Mh

Mh Mh MAC Mv MBD I R2 R1 R3

Mv Mv MBD Mh MAC R2 I R3 R1

MAC MAC Mv MBD Mh R3 R1 I R2

MBD MBD Mh MAC Mv R1 R3 R2 I

The set of symmetries of a subset X of the plane is called the symmetry group of X. The “multiplication table”describing how the symmetries of X compose with each other is called the Cayley table of the symmetrygroup. We should note the following things about the Cayley table we have just written down.

It is not true that A ◦ B = B ◦ A for all choices of A and B. In particular, you can see in the tablethat MBD ◦Mv = R3 while Mv ◦MBD = R1. This means that Cayley tables are not all that similar tomultiplication tables — the entries are not symmetric when you flip along the diagonal, a propertywhich multiplication tables obey.

Every row and column contains every element of the symmetry group exactly once. We will restateand prove this property — which I will refer to as the sudoku property — later on.

As we mentioned earlier, the whole table of entries is not symmetric when you flip along the diagonal.However, the location of the entries which are I is symmetric when you flip along the diagonal. Anotherway to say this is that if A ◦ B = I, then B ◦ A = I as well.

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Properties of Symmetry Groups

Symmetry is a very far-reaching idea in mathematics and extends way beyond the notion of symmetry whichwe have defined. Indeed, we have only defined symmetries for subsets of the Euclidean plane, while thenotion of symmetry applies to many, many other things. As a simple example, consider the expressionsx + 2y and x + y. To many a mathematician, the first expression does not seem symmetric, because swappingx and y changes it. On the other hand, the second expression does seem symmetric, because swapping xand y results in y + x which, although it may look different, is exactly the same expression. This example isso far removed from geometry that, to widen our definition of symmetry to incorporate it, we have to dosomething rather drastic.

The idea we will use is a relatively modern one in mathematics. Take the set of objects that you are studying— in our case, the symmetries of a geometric shape — write down the most important properties that theyobey, and then consider anything at all which obeys those properties. In some cases, this will give a veryuseful and interesting set of objects which is far more general than the set of objects that you started with.This probably makes no sense to you whatsoever, so the best thing to do is probably just to forge ahead. Thefollowing are four very important properties which all symmetry groups obey.

(Closure) If A and B are symmetries, then the composition A ◦ B is also a symmetry.

(Identity) There always exists the identity symmetry I such that, for each symmetry A, the compositionI ◦ A and the composition A ◦ I are both equal to A.

(Inverse) For each symmetry A, there exists a symmetry B such that the composition A ◦ B and thecomposition B ◦ A are both equal to I.

(Associative) For all symmetries A, B, C, the symmetry obtained by composing them as (A ◦ B) ◦ C isthe same as the symmetry obtained by composing them as A ◦ (B ◦ C).

The first property states that composition of symmetries is a symmetry, the second that doing nothing isalways a symmetry, and the third that doing the reverse of a symmetry is again a symmetry. Hopefully theseare all obvious statements which you believe are true. The fourth statement is a little different, possibly tooobvious to seem important. It merely says that when you are calculating the composition of three or moresymmetries, you never need to use brackets. So an example like (A ◦ (B ◦ C)) ◦ ((D ◦ E) ◦ F) is just the samething as A ◦ B ◦ C ◦ D ◦ E ◦ F. Anyway, if you think that this just seems silly, then you may be right, but itcertainly is important mathematically.

The Definition of a Group

So the idea now is to take these four properties and use them as rules to define an object as follows. Whateverobject we obtain is going to behave very similarly to a symmetry group but will capture a notion of symmetrythat is much broader than the geometric symmetries that we’ve been discussing.

A group is a set G with a “multiplication table” such that the following four properties hold.

(Closure) For all g and h in G, the expression g · h is also in G.

(Identity17) There is a special element e in G such that if g is in G, we have e · g = g · e = g.

(Inverses) For every g in G, there is an element h in G such that g · h = h · g = e.

(Associative) For all g, h, k in G, we have (g · h) · k = g · (h · k).

17For some reason, when you deal with groups, it’s common to call the identity element e — hopefully, this won’t be too confusing.

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Because we are simply assuming that these are the rules of the game and we can never prove them, thefour properties above are often called the group axioms. Hopefully you can see that the definition mimicsthe four properties of symmetry groups which we discussed earlier. So, in some sense, you can think of theelements of G as the symmetries of some object, although that object may or may not be geometric and maybe something you can’t even imagine.

Some Examples of Groups

It’s a good habit, whenever anyone throws a mathematical definition at you, to think of as many examples asyou can. So here are some simple examples of groups. You should take each one and verify in your ownmind why it obeys the group axioms listed above.

The set of symmetries of any subset of the plane, where · represents composition, is a group. We knowthat this has to be true, because we constructed the definition of a group so that the set of symmetriesof any subset of the plane would be an example.

The set of all isometries, where · represents composition, is a group. Actually, this is just a particularcase of the example above, where the subset of the plane that we take is the empty set consisting of nopoints.

The real numbers, where · represents addition, is a group. In this case, the identity is the number 0 andthe inverse of x is the number −x. The fact that addition of numbers is associative is just something wealways take for granted.

The set of integers, where · represents addition, is a group. If we call this group Z and the group in theprevious example R, note that Z is a group which lives inside of R, so we say that Z is a subgroup of R.

The positive real numbers, where · represents multiplication, is a group. In this case, the identity isthe number 1 and the inverse of x is the number 1

x . If · represents multiplication, then you know thatthe number 0 cannot be a part of the group, because it has no inverse. In fact, you can just remove thenumber 0 from the real numbers to obtain another example of a group where · represents multiplication.

The set of all m× n matrices, where · represents addition, is a group. In this case, the identity is the zeromatrix and the inverse of a matrix M is the matrix −M. The fact that addition of matrices is associativefollows directly from the fact that addition of numbers is associative.

The set of all n× n matrices with determinant 1, where · represents multiplication, is a group. You needthe condition that the determinant is 1 — or at least, something similar to it — to get rid of exampleslike the zero matrix which have no inverse. In this case, the fact that multiplication of matrices isassociative is something you may have taken for granted but is definitely not immediately obvious. Tryto prove it — even just for 2× 2 matrices — and you’ll see what I mean.

The set of two numbers {1,−1}, where · represents multiplication, is a group. In this case, the identityis the number 1 and the inverse of each element is itself. Note that the Cayley table for this group looksremarkably similar to the two tables we wrote down when talking about direct and opposite isometries,just with some of the names changed.

In fact, it’s a good habit, whenever anyone throw a mathematical definition at you, to also think of coun-terexamples. Obviously, examples of things which aren’t groups are easy to come up with — a banana, ahippopotamus, your index finger, and so on. However, can you think of something which obeys the groupaxioms except for the identity property, something which obeys the group axioms except for the inverseproperty, and something which obeys the group axioms except for the associative property?

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Three Special Types of Group

The examples of groups that we have seen, as special as they are, will not feature very much in this course.But there are three types of groups which will, and they are known as dihedral groups, cyclic groups andsymmetric groups.

Dihedral Groups Dn

We can describe the dihedral group Dn as the symmetries of the regular polygon with n sides.18 Thecase n = 4 amounts to studying the symmetries of a square, which we considered earlier. We notedthen that there are four rotations (including the identity) and four reflections which are symmetries ofthe square. In the general case, there are n rotations (including the identity) and n reflections which aresymmetries of the regular polygon with n sides. Therefore, the group Dn contains 2n elements.

The diagram below shows a regular polygon with eight sides and the mirrors corresponding to theeight reflective symmetries.

Cyclic Groups Cn

We can describe the cyclic group Cn as the symmetries of the “decorated” regular polygon with n sides.The diagram below shows a decorated regular polygon with eight sides and hopefully you’ll be able todraw a decorated regular polygon with any number of sides. The extra decoration removes all of thereflections as symmetries, but keeps all of the rotations. Another way to describe the cyclic group Cn isas the set of direct symmetries of a regular polygon with n sides. Either description you decide to use,you see that the group Cn contains n elements.

18A regular polygon is just a polygon whose side lengths are all equal and whose angles are all equal.

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Symmetric Groups Sn

The elements of the group Sn are simply the permutations of the numbers from 1 up to n. For example,the group S3 contains the following six elements — these are the ways to write the numbers 1, 2, 3 insome order.

123 132 213 231 312 321

You should think of the element abc as the function which takes a number from the set {1, 2, 3} andspits out a number from the set {1, 2, 3} in the following way.

1→ a 2→ b 3→ c

In this group, the · stands for composition of these permutations — maybe an example is the best wayto illustrate this. Suppose that we want to determine which permutation corresponds to the product132 · 231. The permutation 231 takes

1→ 2 2→ 3 3→ 1

while the permutation 132 takes1→ 1 2→ 3 3→ 2.

Let’s see where the permutation 132 · 231 takes the number 1. We always start from the rightmostpermutation which, in this case, is 231. This takes 1 to 2 and this 2 is then fed into the next permutationalong which, in this case, is 132. This takes 2 to 3, so the composition 132 · 231 takes 1 to 3.

Now let’s see where the permutation 132 · 231 takes the number 2. We always start from the rightmostpermutation which, in this case, is 231. This takes 2 to 3 and this 3 is then fed into the next permutationalong which, in this case, is 132. This takes 3 to 2, so the composition 132 · 231 takes 2 to 2.

Finally, let’s see where the permutation 132 · 231 takes the number 3. We always start from the rightmostpermutation which, in this case, is 231. This takes 3 to 1 and this 1 is then fed into the next permutationalong which, in this case, is 132. This takes 1 to 1, so the composition 132 · 231 takes 3 to 1.

Putting all of these facts together, we see that the composition 132 · 231 is a permutation which takes 1to 3, 2 to 2 and 3 to 1 and this resulting permutation we have called 321. Therefore, we can write thecomposition as

132 · 231 = 321.

Hopefully, you can see that when you compose permutations in this way, the identity of S3 will be theelement we’ve described as 123. That’s because this permutation does nothing — it takes 1 to 1, 2 to 2and 3 to 3.

Problems

Problem. Let ABC be a triangle with the vertices labelled clockwise such that AC = BC and ∠ACB = 90◦. Let MAB

be the reflection in the line AB, MAC be the reflection in the line AC, and R be the rotation by 90◦ counterclockwisearound B. Identify the composition R ◦MAB ◦MAC.

If X denotes the composition R ◦MAB ◦MAC, let n be the minimum number of reflections whose composition is equalto X. Determine the value of n and carefully describe n reflections whose composition is equal to X.

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Proof. We solved this problem in the previous lecture, but today we’ll give a slightly different solution. Thediagram below shows triangle ABC drawn on a grid of two squares. The composition R ◦MAB ◦MAC mustbe direct because it’s the composition of two opposite isometries and one direct isometry. It’s easy to checkthat MAC(C) = C, MAB(C) = Y and R(Y) = C — in other words, R ◦MAB ◦MAC(C) = C and C is a fixedpoint of the composition. Since the composition R ◦MAB ◦MAC is a direct isometry with C as a fixed point,it must be the identity or a rotation about C.

A

B

C X

Y

To determine which of these cases applies, we consider the location of R ◦MAB ◦MAC(A). It should be clearthat MAC(A) = A, MAB(A) = A and R(A) = X — in other words, R ◦MAB ◦MAC(A) = X and A is not afixed point of the composition. Hence, the composition R ◦MAB ◦MAC must be a rotation by ∠ACX = 180◦

about C.

As we’ve already learnt with these sorts of problems, it’s useful to draw the diagram on a grid of squares.We’ve also learnt that it’s useful to keep in mind the table which characterises isometries by whether they aredirect or opposite and whether they have fixed points or not.

Problem. Write down the Cayley table for the group S3.

Proof. We’ve already seen an example of how the composition of permutations works, but it’s probably agood idea to see another example. Makes sure you understand very carefully how the notation and theprocess works. Suppose that we want to determine which permutation corresponds to the product 213 · 321.The permutation 213 takes

1→ 2 2→ 1 3→ 3

while the permutation 132 takes1→ 3 2→ 2 3→ 1.

Let’s see where the permutation 213 · 321 takes the number 1. We always start from the rightmost permutationwhich, in this case, is 321. This takes 1 to 3 and this 3 is then fed into the next permutation along which, inthis case, is 213. This takes 3 to 3, so the composition 132 · 231 takes 1 to 3.

Now let’s see where the permutation 213 · 321 takes the number 2. We always start from the rightmostpermutation which, in this case, is 321. This takes 2 to 2 and this 2 is then fed into the next permutation alongwhich, in this case, is 213. This takes 2 to 1, so the composition 132 · 231 takes 2 to 1.

Finally, let’s see where the permutation 213 · 321 takes the number 3. We always start from the rightmostpermutation which, in this case, is 321. This takes 3 to 1 and this 1 is then fed into the next permutation alongwhich, in this case, is 213. This takes 1 to 2, so the composition 132 · 231 takes 3 to 2.

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Putting all of these facts together, we see that the composition 132 · 231 is a permutation which takes 1 to 3, 2to 1 and 3 to 2 and this resulting permutation we have called 312. Therefore, we can write the composition as

213 · 321 = 312.

This means that we can fill in the entry in the Cayley table whose row is labelled by 213 and whose column islabelled by 321 with the permutation 312. After a while, you can get pretty quick are doing these computations,and I’m sure it won’t take you long to check that the resulting Cayley table looks like the following.

◦ 123 132 213 231 312 321

123 123 132 213 231 312 321132 132 123 312 321 213 231213 213 231 123 132 321 312231 231 213 321 312 123 132312 312 321 132 123 231 213321 321 312 231 213 132 123

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Galois

I’m going to tell you about one of my favourite math-ematicians ever, the French mathematician EvaristeGalois. It seems that Galois was a very passionateman, particularly when it came to mathematics, pol-itics and women — the first made him famous, thesecond got him locked up in prison and the thirdcaused his death. In fact, after being born in 1811,Galois didn’t even make it to his 21st birthday beforedying, all because of a woman.

Apparently, Galois wasn’t particularly fond of school— you can’t blame him — and his teachers didn’trecognise his talents at all. In fact, the story goes thatit wasn’t until Galois was a teenager and confined tobed with illness that he really discovered mathemat-ics. Once hooked, it seems that Galois devoted muchof his time to mathematics in the few remaining yearsof his life.

After school, Galois attempted the entrance exam tothe Ecole Polytechnique in Paris, but failed due to hislack of explanation in the oral exam. This was due toGalois’ unusual upbringing in mathematics, learningeverything on his own from advanced textbooks. Soinstead, he went to the Ecole Preparatoire, a far infe-rior institution where he found some professors whowere sympathetic to him. The following year, Galoistried, for his second and last time, the entrance examto the Ecole Polytechnique and, for reasons we aren’ttoo sure about, failed again despite being more thanqualified. Some say that he thought the exercise givento him was boring and rather than solve it, decided tothrow the blackboard cleaner at the examiner’s head.Another possible reason is that Galois’ logical leapswere far too advanced for the incompetent examiner,which angered Galois. Yet another explanation couldbe Galois’ emotional state, since his father had com-mitted suicide two days earlier.

There were other ways to get ahead in the mathemati-cal world in those days. So Galois decided to write uphis thoughts and send them to the very prestigiousAcademy of Sciences. For very mysterious reasons,the famous mathematician Augustin Louis Cauchyread these papers but refused to publish them. When

he tried the following year to submit them, but thistime to another famous mathematician by the nameof Joseph Fourier, it turns out that Fourier suddenlydied and the paper was lost. Later on, Galois triedonce again, this time turning to the famous mathe-matician Simeon Poisson. However, Poisson declaredthat Galois’ work was incomprehensible, saying thathis ”argument is neither sufficiently clear nor suffi-ciently developed to allow us to judge its rigour”.

Meanwhile, Galois was also involved in the politicalturmoil going on in France. He was expelled from hisuniversity for a particularly heated letter concerningthe political situation and then joined the staunchlyRepublican artillery unit of the National Guard. Ata large and rather riotous banquet, Galois made atoast to King Louis–Philippe with a dagger above hiscup, which was interpreted as a threat against theking’s life. Although he was arrested, he was lateracquitted. But on the following Bastille Day, Galoisheaded a protest, while wearing the uniform of theNational Guard, carrying pistols, a rifle and a dagger.For this he was arrested and sentenced to six monthsin prison — but this quiet time allowed him to returnto mathematics and further develop his ideas.

Only one month after Galois’ release from prison, itappears that he was involved in a duel. Althoughthe reasons behind the duel are not particularly clear,we know that it was the result of a love affair. Someconspiracy theorists believe that the whole situationwas orchestrated by the government in order to getrid of the troublemaking Galois, but this doesn’t seemto be too likely. Whatever the reasons behind the duel,Galois was so convinced of his impending death thathe stayed up all night writing letters and composingwhat would become his mathematical testament, a fa-mous letter to Auguste Chevalier outlining his ideas.Hermann Weyl, one of the greatest mathematicians ofthe twentieth century, said that “this letter, if judgedby the novelty and profundity of ideas it contains, isperhaps the most substantial piece of writing in thewhole literature of mankind.” Weyl’s statement isundoubtedly an exaggeration, but you get the point.

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Unfortunately for the world of mathematics, in thevery early morning on 30 May 1832, Galois was shotin the abdomen and died the following day with hisbrother at his side, at the tender age of twenty. I al-ways wonder whether Galois would have done betterin the duel had he not stayed up all night writingdown his mathematical ideas.

Galois laid the foundations for a whole new areaof mathematics which we now call Galois theory.One of the main applications was the problem ofsolving the quintic equation. So we all know thata quadratic equation is something which looks likeax2 + bx + c = 0 and that its solution is given by thequadratic formula

x =−b±√b2 − 4ac

2a.

It’s much less commonly known that there is a for-mula for the solutions of the cubic equation ax3 +bx2 + cx + d = 0 which was discovered in the seven-teenth century. This formula would probably take meabout a third of this page to write down, but it pales incomparison to the formula for for the solutions of thequartic equation ax4 + bx3 + cx2 + dx + e = 0. Thisbeast of a formula would probably take me pages,but it was discovered not long after its cubic cousin.Then, there was a dry spell, when mathematicianstried to look for a formula for the solutions to thequintic equation ax5 + bx4 + cx3 + dx2 + ex + f = 0.They tried and tried but to no avail.

The reason why they weren’t successful is becausethere is no formula, and we know that because Ga-lois proved it.19 To understand the significance ofthis, you have to realise that to find a formula is easybecause you just have to write it down and demon-strate that it works. But how do you prove that thereis no formula for the quintic equation? Galois’ didit by finding some “hidden symmetry” among thesolutions of polynomials, which led to him using theconcept of groups in his work. In fact, we have Galoisto thank for the word group in mathematics.

19Actually, the Norwegian mathematician Niels Henrik Abel proved the same thing at around the same time, although the ideas in hisproof are not as far-reaching. Unfortunately, Abel also reached an untimely end, dying at the tender age of twenty-six.

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2. SYMMETRY IN GEOMETRY 2.3. Symmetry in the Plane

Recap

Now is probably a good time for a quick recap of our exploration into symmetry so far. Philosophically, ourbroad goal is to give a mathematically precise definition of symmetry and see where this definition can takeus. When we began, I tried to convince you that the notion of symmetry in geometry is somehow tied to thenotion of distance. In particular, this observation motivated us to define isometries, functions which takepoints in the plane to points in the plane and which preserve distances. Mathematicians are always tryingto classify things, so it made sense to try to classify all of the possible isometries. We discovered that therewere essentially four types — namely, translations, rotations, reflections and glide reflections. The definitionof an isometry allowed us to mathematically define a symmetry of a subset of the Euclidean plane as anisometry which leaves the subset exactly where it is. We observed that the symmetries of a given shape canbe composed with each other and are collectively known as the symmetry group of the shape. Then wefound that the structure of a symmetry group is encapsulated in its “multiplication table”, which is moreaccurately known as its Cayley table.

At this point, we turned to more abstract matters, using four very simple properties of symmetry groups todefine the notion of a group. A group is a set G with a “multiplication table” such that the following fourproperties hold.

(Closure) For all g and h in G, the expression g · h is also in G.

(Identity) There is a special element e in G such that if g is in G, we have e · g = g · e = g.

(Inverses) For every g in G, there is an element h in G such that g · h = h · g = e.

(Associative) For all g, h, k in G, we have (g · h) · k = g · (h · k).

So, from the seeds of our intuition about symmetry, we have developed the abstract notion of a group. Andnow we’re in a position to delve a little deeper into the mysterious world of group theory.

When are Two Groups the Same?

Quite a while ago, we noted that the letters H and X have the same symmetry group structure — the identity,a rotation by 180◦, a reflection in a horizontal mirror, and a reflection in a vertical mirror. We also decidedthat the letters A and B have the same symmetry group structure, even though they superficially seem tobe different. The crucial point is that, in both cases, there is the identity as well as one reflection, and thatreflection composed with itself gives back the identity. This all sounded like mumbo jumbo back then and itprobably sounds like mumbo jumbo now — but it’s time now to make all this mumbo jumbo mathematicallyprecise.

Remember that we looked at the symmetry group ofthe square and named its elements

I, R1, R2, R3, Mh, Mv, MAC, MBD.

In fact, we actually wrote out the whole Cayley ta-ble for this group, which is an example of a dihedralgroup — the dihedral group D4, to be precise.

A

B C

D

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If you don’t actually remember, then that’s fine because here’s the Cayley table once again.










Now suppose that, due to failing the course, you were forced to take it again next summer.20 And supposethat I was the lecturer once again and decided to write out the Cayley table for the symmetry group of thesquare. If I had instead named the elements

i, r1, r2, r3, mh, mv, mAC, mBD,

would you think that I had made some sort of mistake? Would you think that this new Cayley table isdifferent from its capitalised version? No, of course you wouldn’t, and rightly so. And that’s because thetwo Cayley tables have essentially the same structure, even though we are using different symbols for eachelement of the group. In fact, by this reasoning, I could even have changed the names of the symmetries ofthe square to

A, B, C, D, E, F, G, H,

and you’d still have to agree that the resulting Cayley table and the resulting symmetry group have essentiallythe same structure. It would certainly be silly to treat two groups differently just because you’ve written theirCayley tables using different symbols. This idea is most aptly described by Shakespeare himself, in his playRomeo and Juliet.

What’s in a name? That which we call a roseBy any other name would smell as sweet.

Translating this couplet from the world of Shakespearean tragedy to its group theoretic equivalent, we havethe following.

Taking a Cayley table and renaming the elementsGives a group with the same structure.

So we consider two groups to be the same if the entries in the Cayley table of one can be renamed to give theCayley table of the other. If this is the case, then we say that the two groups are isomorphic, which in ancientGreek means “same structure”. An even more mathematically precise way to express this is as follows. Twogroups G and H are isomorphic if there exists a bijection — that is, a one-to-one dictionary correspondence —f : G → H such that

f (g1 · g2) = f (g1) · f (g2).

20Hopefully none of you will fail the course, so this story is purely hypothetical.

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Here, the function f is called an isomorphism and simply describes the relabelling of the elements of G intoelements of H. The equation above simply encapsulates the idea that the elements of G should be relabelledinto elements of H in such a way that respects the structure of the two Cayley tables.21 If two groups G andH are isomorphic, then we usually write this using the shorthand notation G ∼= H.

Examples of Isomorphisms

The notion of isomorphism is a very powerful one indeed in mathematics, appearing in all sorts of areasapart from group theory. The definition is truly very simple, but it will be useful to see some small examplesof isomorphisms.

Example. We’ve actually already considered an isomorphism between two groups a long time ago, when wediscussed direct and opposite isometries. Back then, we observed that the following two tables seem to havea very similar structure.

◦ dir opp

dir dir oppopp opp dir

× pos neg

pos pos negneg neg pos

More formally, you can verify that they both correspond to Cayley tables of groups and that the two groupsare isomorphic. In fact, it’s easy to describe the isomorphism between them as

f (dir) = pos and f (opp) = neg.

To verify that this is indeed an isomorphism, all you need to do is check that the following statements aretrue, which is quite easy to do.

f (dir ◦ dir) = f (dir)× f (dir)

f (dir ◦ opp) = f (dir)× f (opp)

f (opp ◦ dir) = f (opp)× f (dir)

f (opp ◦ opp) = f (opp)× f (opp)

Example. Now consider the group — let’s call it G — whose Cayley table looks like the table below left.

· a b c

a c a bb a b cc b c a

◦ I R1 R2

I I R1 R2

R1 R1 R2 IR2 R2 I R1

We know about a group with three elements already — namely, the cyclic group C3. Recall that this consists ofthe rotational symmetries of an equilateral triangle or, if you prefer, the symmetries of a decorated equilateraltriangle. Above right is the Cayley table of C3, where we denote the identity isometry by I, the rotation

21Note that the · on the left hand side of the equation corresponds to composition using the Cayley table of G while the · on the righthand side corresponds to composition using the Cayley table of H, even though I’ve used the same symbol for both.

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by 120◦ about the centre of the equilateral triangle by R1 and the rotation by 240◦ about the centre of theequilateral triangle by R2. With this notation, an isomorphism f : G → C3 is given by

f (a) = R2 f (b) = I f (c) = R1.

I was careful to say “an isomorphism” because it’s not a priori clear that there will only be one of them. And,in fact, there happen to be two possible isomorphisms in this case and the other one is given by

f (a) = R1 f (b) = I f (c) = R2.

Example. At one stage, we wrote out the Cayley table for the symmetric group S3, which consists of thepermutations of the numbers 1, 2, 3. Another group with six elements is the group D3, which consists of thesymmetries of an equilateral triangle ABC. The Cayley tables for these two groups are listed below, where Iis the identity isometry, R1 is a rotation by 120◦ about the centre of ABC, R2 is a rotation by 120◦ about thecentre of ABC, MA is a reflection through a mirror passing through A, MB is a reflection through a mirrorpassing through B, and MC is a reflection through a mirror passing through C.

◦ 123 132 213 231 312 321

123 123 132 213 231 312 321132 132 123 312 321 213 231213 213 231 123 132 321 312231 231 213 321 312 123 132312 312 321 132 123 231 213321 321 312 231 213 132 123

◦ I R1 R2 MA MB MC

I I R1 R2 MA MB MC

R1 R1 R2 I MC MA MB

R2 R2 I R1 MB MC MA

MA MA MB MC I R1 R2

MB MB MC MA R2 I R1

MC MC MA MB R1 R2 I

It turns out that these two groups are isomorphic, but an isomorphism is a little tricky to find. And onceyou’ve found it, there would still be thirty-six things to check to make sure that it’s an isomorphism, at leastif you try to do it the naive way.

We can actually describe the isomorphism quite easily in such a way that it should be reasonably clear that it’san isomorphism, without having to check all thirty-six entries of the Cayley table. Simply label the vertices ofthe equilateral triangle 1, 2, 3 rather than A, B, C. Then any symmetry of the equilateral triangle permutes thevertices and hence, corresponds to a permutation of the numbers 1, 2, 3 — in other words, an element of thegroup S3. Since composition of symmetries will behave in the same way as composition of permutations, thisis an isomorphism between D3 and S3. If you really want to, you can write out the isomorphism explicitly,and this is what you’d get.

f (I) = 123 f (R1) = 231 f (R2) = 312 f (MA) = 132 f (MB) = 321 f (MC) = 213

When are Two Groups Different?

If you know what it means for two groups to be the same, then you must also know what it means for twogroups to be different. To prove that two groups are the same — remember the technical term is isomorphic— you can just go ahead and find the isomorphism and check that it is indeed an isomorphism. On theother hand, how do you prove that two groups are different – in other words, that there exists no possibleisomorphism? Well, there are various tricks, but here are two very simple ones.

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Example. Two groups of different sizes cannot be isomorphic. This is simply because if two groups havedifferent numbers of elements, then there cannot possibly exist a bijection — in other words, a one-to-onedictionary correspondence — between them. Therefore, we can say things like C3 and C4 are not isomorphic.

Example. Two groups cannot be isomorphic if one is abelian while the other is not. We say that a group G isabelian if for all g and h in G, it is true that g · h = h · g. So in an abelian group, it doesn’t matter in whichorder you compose elements. In terms of the Cayley table, an abelian group is one where the entries aresymmetric when you flip over the main diagonal.22 Therefore, we can say things like D4 and C8 are notisomorphic, even though they have the same number of elements. This is because we know that D4 is notabelian since the two Cayley table entries Mh ◦MBD and MBD ◦Mh aren’t equal. On the other hand, you cansee from the Cayley table for C8 — which I’ve written below — that it’s abelian.

◦ I R1 R2 R3 R4 R5 R6 R7

I I R1 R2 R3 R4 R5 R6 R7

R1 R1 R2 R3 R4 R5 R6 R7 IR2 R2 R3 R4 R5 R6 R7 I R1

R3 R3 R4 R5 R6 R7 I R1 R2

R4 R4 R5 R6 R7 I R1 R2 R3

R5 R5 R6 R7 I R1 R2 R3 R4

R6 R6 R7 I R1 R2 R3 R4 R5

R7 R7 I R1 R2 R3 R4 R5 R6

Actually, it’s useful to know that the cyclic group Cn is abelian for all n ≥ 1. This is essentially becauseit consists of n rotations all with the same centre, and it doesn’t matter in which order you compose suchrotations. It’s also useful to know that the dihedral group Dn is not abelian for n ≥ 3, a fact that you can andshould try to prove on your own.

Properties of Cayley Tables

From a group, we obtain a Cayley table and from a Cayley table, we obtain a group. So any property thatapplies to all Cayley tables is really a property that applies to all groups. Here are two important facts thatapply to all Cayley tables and which we’ll prove right now.

Sudoku property : In any row or column of a Cayley table, no element of the group appears twice.If this were not true, then there might be a row labelled r in which there are two equal entries. So let’ssuppose that these two equal entries happen to be in the column labelled c1 and the column labelled c2.Of course, part of this setup is the assumption that the columns c1 and c2 are distinct. Now the fact thatthese two entries are equal implies the equation r ◦ c1 = r ◦ c2. We’ll use the group axioms to deduce acontradiction from this equation in the following way.

r−1 ◦ (r ◦ c1) = r−1 ◦ (r ◦ c2)

(r−1 ◦ r) ◦ c1 = (r−1 ◦ r) ◦ c2

e ◦ c1 = e ◦ c2

c1 = c2

22The main diagonal is the one which runs from top left to bottom right.

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To obtain the first line, we’ve used the inverse property to multiply both sides of the equation by theinverse of r on the left.23 To get from the first line to the second, we’ve used the associative property. Toget from the second line to the third, we’ve used the inverse property. And to get from the third line tothe fourth, we’ve used the identity property. So we have proved that c1 = c2, which is in clear violationof the assumption that the columns c1 and c2 are distinct. This contradiction means that it’s not possiblefor two entries of the same row to be equal in a Cayley table. And by an analogous argument, we alsoknow that it’s not possible for two entries of the same column to be equal in a Cayley table.

There are two important consequences of our proof. The first is that if G is a finite group, then everyrow and every column of the Cayley table of G must contain every element of G exactly once. Thesecond is that you can always “cancel group elements from an equation”.

Symmetric identity property : The entries of the Cayley table in which the identity appears are symmetric whenyou flip over the main diagonal.Another way to say this is that if the identity e appears in row r and column c, then it also appears inrow c and column r. So our goal is to show that the equation r ◦ c = e implies that c ◦ r = e.

r−1 ◦ (r ◦ c) = r−1 ◦ e

(r−1 ◦ r) ◦ c = r−1

e ◦ c = r−1

c = r−1

c ◦ r = r−1 ◦ r

c ◦ r = e

To obtain the first line, we’ve used the inverse property to multiply both sides of the equation by theinverse of r on the left. To get the from the first line to the second, we’ve used the associative property.To get from the second line to the third, we’ve used the inverse property. To get from the third line tothe fourth, we’ve used the identity property. To get from the fourth line to the fifth, we’ve multipliedboth sides of the equation by r on the right. And to get from the fifth line to the sixth, we’ve used theinverse property. And this completes the proof of the symmetric identity property.

I’ve been really meticulous here and broken down these proofs into very basic steps, each one involving atmost one of the group axioms. It’s good for you to see all of the gory details of the proof now since these areour first real proofs in group theory. However, once you get the hang of working with groups, you can take alot of shortcuts and not go into quite so much detail.

Finite Symmetry Groups

A while ago, we managed to define the symmetery group of a subset of the Euclidean plane. We used certainproperties which these symmetry groups obey to broaden our definition of symmetry. The resulting object isa group, an abstract algebraic object which was constructed to behave a lot like a symmetry group. This justbegs the question. . . which groups arise as symmetry groups of subsets of the Euclidean plane?

One phenomenon which occurs in group theory — the area of mathematics dealing with groups — is thefact that finite groups have certain qualitative differences to infinite groups. Of course, for infinite groups,

23It is very important here to say “on the left” because you would get a different answer if you multiplied both sides of the equationon the right. And you are definitely not allowed to multiply one side of the equation on the left and the other side of the equation on theright — you must do the same thing to both sides.

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you’ll find it pretty hard to write down the Cayley table, but there are various other fundamental differencesbetween the two. So let’s start with the following simpler question. . . which finite groups arise as symmetrygroups of subsets of the Euclidean plane? This question was raised way back in the fifteenth century byLeonardo da Vinci who was interested in the notion of symmetry in art, particularly in architecture.

We already know that the cyclic group Cn and the dihedral group Dn are finite groups which arise as thesymmetry group of a subset of the Euclidean plane. In fact, we know it because that’s precisely how theywere defined. The following result states that these are actually the only finite groups which arise as thesymmetry group of a subset of the Euclidean plane.

Theorem (Leonardo’s Theorem). If a subset of the Euclidean plane has finitely many symmetries, then its symmetrygroup must be the cyclic group Cn or the dihedral group Dn for some positive integer n.

At this point, we should point out that our previous definition of cyclic and dihedral groups only reallyworked for n ≥ 3. However, it’s easy enough to define the cyclic and dihedral groups for n = 1 and n = 2 bywriting down their Cayley tables. The Cayley tables of C1 and C2 are as follows. Note that C1 is the symmetrygroup for the letter R while C2 is the symmetry group for the letter N.

◦ I

I I

◦ I R

I I RR R I

The Cayley tables of D1 and D2 are as follows. Note that D1 is the symmetry group for the letter M while D2

is the symmetry group for the letter O.

◦ I M

I I MM M I

◦ I R1 Mv Mh

I I R1 Mh Mv

R1 R1 I Mv Mh

M1 Mh Mv I R1

M2 Mv Mh R1 I

You can hopefully see from the Cayley tables for C2 and D1 that they’re isomorphic. However, the groupsC2n and Dn — despite having the same number of elements — are certainly not isomorphic for any n ≥ 2.For n = 2, you can prove this by nothing that every element of D2 composed with itself gives the identity, afact which doesn’t hold for C4. For n ≥ 3, you can prove this using our earlier observation that C2n is abelianwhile Dn is not.

Dihedral and Cyclic Symmetry

Since Leonardo’s theorem claims that every subset of the Euclidean plane has either cyclic or dihedralsymmetry, we should be able to find lots of examples of each and recognise the difference between the twotypes. Just keep in mind that the main difference between dihedral and cyclic symmetry is that the formerincludes reflective and rotational symmetries while the latter only includes rotational symmetries.

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Example. As examples of dihedral symmetry, we have an apple, the logo for Mercedes–Benz and ThePentagon.24

Example. As examples of cyclic symmetry, we have the blades of a windmill, the Isle of Man flag, theperiwinkle flower, the Penrose triangle and the logo for Sun Microsystems.25

Example (Examples of ambigrams). An ambigram is a design or artform that may be read as one or morewords not only in its form as presented, but also from another viewpoint, direction or orientation. Thefollowing shows some interesting examples of ambigrams — you should easily be able to determine whichones have dihedral symmetry and which ones have cyclic symmetry.

24As with any real object, the symmetry is not quite exact — no apple is going to be perfectly circular and have five equally spacedseeds and The Pentagon doesn’t have exactly the same rooms and furniture on all five sides. Also, if you look carefully, you’ll see thatthe drawn logo for Mercedes–Benz does not really have dihedral symmetry — it’s the three-dimensional object which the drawingrepresents which has dihedral symmetry.

25The Penrose triangle is an “impossible object” named after the well-known mathematical physicist Sir Roger Penrose. The logo forSun Microsystems is one of my favourites.

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Ambigrams have been recently popularised through Dan Brown’s book Angels and Demons. They alsoappear on the front cover of the twentieth anniversary collector’s edition of The Princess Bride.26

Facts about Finite Symmetry Groups

So how does one go about proving Leonardo’s theorem? Well, you need a couple of useful little lemmasunder your belt before you can start. In the following, when we say finite symmetry group, we are referringto the symmetry group of a subset of the Euclidean plane.

Lemma. A finite symmetry group cannot contain a translation or a glide reflection.

Proof. Of course, when we say translation here, we mean a translation which is not the identity. And whenwe say glide reflection, we mean a glide reflection which is not itself a reflection. That’s because we alreadyknow that every group of isometries necessarily contains the identity and we’ve already seen examples ofsymmetry groups which contain reflections.

If a symmetry group contains the translation T, then the group must also contain T ◦ T, T ◦ T ◦ T, T ◦ T ◦ T ◦ T,and so on. These isometries cannot possibly be the same, because they are translations by different distances.Since these infinitely many isometries can’t possibly fit into a finite group, a finite symmetry group cannotcontain a translation.

If a symmetry group contains the glide reflection G, then the group must also contain the translationT = G ◦ G. However, we’ve already deduced that translations cannot occur in a finite symmetry group, sonor can glide reflections.

Lemma. In any finite symmetry group, either every isometry is direct or there is an equal number of direct and oppositeisometries.

Proof. Let’s call the direct isometries R1, R2, . . . , Rm. If there is at least one opposite isometry, then let’s callthe opposite isometries M1, M2, . . . , Mn. So our finite symmetry group has m direct isometries and n oppositeisometries, and our goal is to prove that m = n.

26I’ve never read anything by Dan Brown, so I can’t comment on this book, but I can say that The Princess Bride is an excellent film, atleast in my opinion.

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Suppose that M is any one of the opposite isometries. Then M ◦ R1, M ◦ R2, . . . , M ◦ Rm are all oppositeisometries since they are compositions of direct and opposite isometries. Furthermore, they must all be distinctby the sudoku property of groups. Therefore, {M ◦ R1, M ◦ R2, . . . , M ◦ Rm} is a subset of {M1, M2, . . . , Mn}and this implies that m ≤ n.

Again suppose that M is any one of the opposite isometries. Then M ◦ M1, M ◦ M2, . . . , M ◦ Mm are alldirect isometries since they are compositions of two opposite isometries. Furthermore, they must all bedistinct because by the sudoku property of groups. Therefore, {M ◦M1, M ◦M2, . . . , M ◦Mn} is a subset of{R1, R2, . . . , Rm} and this implies that n ≤ m.

The two inequalities m ≤ n and n ≤ m obviously lead to m = n, and we’re done.

Problems

Problem. Show that any group with three elements must be isomorphic to C3.

Proof. Let’s call the elements of the group e, a, b, where e is the identity element. Given that e is the identity,we can fill in most of the Cayley table, as shown below left. In fact, if we use the sudoku property, there isonly one way in which we can complete the table, as shown below right.

· e a b

e e a ba a * *b b * *

· e a b

e e a ba a b eb b e a

We can see that this group is isomorphic to C3 via the isomorphism

f (e) = I f (a) = R1 f (b) = R2,

where I is the identity isometry, R1 is rotation by 120◦ and R2 is rotation by 240◦.

Problem. The following is a groupoku puzzle — a Cayley table for the group G, where some of the entries are missing.Use the properties which you know about Cayley tables to fill in all of the missing entries. Give full reasoning only forthe first entry of the table that you manage to fill in.

· a b c d e f

a e d f * * *b d f e * * *c * * * * * bd b * * * * *e * * * * e *f c a b * * *

In the lectures, we have seen two groups with six elements — the cyclic group C6 and the dihedral group D3. Prove thatG is isomorphic to one of these groups by writing down an explicit isomorphism. Prove that the cyclic group C6 and thedihedral group D3 are not isomorphic to each other.

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Proof.

The first thing to notice from the Cayley table is that acannot be the identity since a · a 6= a. Similarly, b, c, dand f cannot be the identity since b · a 6= a, c · f 6= f ,d · a 6= a and f · a 6= a. But every group has to havean identity so in this case, it must be e. This allowsus to fill in all of the entries of the Cayley table in therow and column labelled by e.

· a b c d e f

a e d f * a *b d f e * b *c * * * * c bd b * * * d *e a b c d e ff c a b * f *

Now we use the sudoku property to fill out the restof the table. As an example, consider the c · b entry.Since it’s in the same row as entries equal to b andc and in the same column as entries equal to a, b, dand f , the only possibility left is that c · b = e. Usingthis strategy, we can fill in all of the missing entries togive the complete table.

· a b c d e f

a e d f b a cb d f e c b ac f e d a c bd b c a f d ee a b c d e ff c a b e f d

The group G is isomorphic to C6, which is the symmetry group formed by the rotational isometries of aregular hexagon. If we write the elements of C6 as follows, then its Cayley table will look like the one below.

I : the identity isometry

R1 : rotation by 60◦ counterclockwise





· I R1 R2 R3 R4 R5

I I R1 R2 R3 R4 R5

R1 R1 R2 R3 R4 R5 IR2 R2 R3 R4 R5 I R1

R3 R3 R4 R5 I R1 R2

R4 R4 R5 I R1 R2 R3

R5 R5 I R1 R2 R3 R4

An explicit isomorphism F : G → C6 is given by the equations

F(a) = R3 F(b) = R1 F(c) = R5 F(d) = R4 F(e) = I F( f ) = R2.

This is certainly not the only isomorphism possible. To find one, you can use the fact that the identity in Gmust map to the identity in D2. After this, the problem can be finished with a little trial and error.

Note that C6 is an abelian group and, in fact, so are all cyclic groups. On the other hand, D3 is not abelian, soC6 cannot be isomorphic to D3.

Problem. Prove that a group G can only have one identity. In other words, prove that there is only one e ∈ G suchthat e ◦ g = g ◦ e = g for all g ∈ G.

Proof. This proof is so short that it can be hard to find. The idea is to argue by contradiction — so supposethat there is a group with two elements which could be the identity and call them e and f . What we’ll donow is show that they must actually be the same element. Since e is an identity, it follows that e ◦ f = f andsince f is an identity, it follows that e ◦ f = e. And there’s the proof, since this means that e = f .

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Cayley

Arthur Cayley, between his birth in 1821 and hisdeath in 1895, was a British mathematician who alsoworked as a lawyer. As a child, Cayley enjoyedsolving math problems for amusement and when heentered Trinity College, Cambridge, he excelled inGreek, French, German, and Italian, as well as mathe-matics.

Cayley only became a lawyer because he had a prettylimited fellowship, but this didn’t seem to slow hismathematics down. Another well-known Britishmathematician by the name of James Joseph Sylvester,was in a similar position to Cayley and became anactuary in London. The two would walk togetheraround the courts, discussing mathematics. It wasduring this fourteen-year span of his life that Cayleyproduced over two hundred mathematical papers. Atthe age of 42, Cayley was offered a prestigious pro-fessorship at Cambridge. He never regretted givingup his lucrative practice for a modest salary becauseit enabled him to end the divided allegiance betweenlaw and mathematics, and to devote his energies tothe pursuit which he liked best.

If you’ve studied some linear algebra, then you mayalready have come across a result of Cayley’s knownas the Cayley–Hamilton theorem. He was very inter-

ested in symmetries, which is why the multiplicationtables for groups are usually known as Cayley tables.

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2. SYMMETRY IN GEOMETRY 2.4. Crystals, Friezes and Wallpapers

More Facts about Finite Symmetry Groups

Last time we proved that a finite symmetry group cannot contain a translation or a glide reflection. We alsodiscovered that in any finite symmetry group, either every isometry is direct or there is an equal number ofdirect and opposite isometries. Now we just need a few more facts before we can put the pieces together toobtain Leonardo’s Theorem.

Lemma. In a finite symmetry group, every rotation must have the same centre.

Proof. Suppose that our finite symmetry group has a rotation R1 with centre O1 and a rotation R2 with centreO2. Obviously, our goal is to prove that O1 = O2. Consider the composition R−1

2 ◦ R−11 ◦ R2 ◦ R1 — it must

be a direct isometry and, due to the way that rotations compose together, it must actually be a translation.27

But we already know that our group can’t contain translations which aren’t the identity — obviously thismeans that the composition R−1

2 ◦ R−11 ◦ R2 ◦ R1 must in fact be the identity.

Therefore, we know that R−12 ◦ R−1

1 ◦ R2 ◦ R1(O1) = O1 from which it follows that R−12 ◦ R−1

1 ◦ R2(O1) = O1.If we apply R2 to both sides, the R2 and the R−1

2 on the left hand side knock each other out and the equationsimplifies to R−1

1 ◦ R2(O1) = R2(O1). Now we apply R1 to both sides so that the R1 and the R−11 on the left

hand side knock each other out and the equation simplifies yet again to

R1 ◦ R2(O1) = R2(O1).

Another way to say the same thing is that R2(O1) is a fixed point of R1. However, since R1 is a rotation,we know that it has a unique fixed point, which we have called O1. So we can deduce that R2(O1) = O1.Another way to say the same thing is that O1 is a fixed point of R2. However, since R2 is a rotation, we knowthat it has a unique fixed point, which we have called O2. So we’ve deduced that O1 = O2.

A simple consequence of the previous result is the following.

Lemma. In a finite symmetry group, every mirror of a reflection passes through the same point.

Proof. If there is only one reflection in the finite symmetry group, then there is nothing to prove. If there areat least two mirrors, then you can compose the corresponding reflections to obtain rotations. By the previouslemma, we know that all of these rotations have the same centre O.

Now suppose that two mirrors meet at a point P. Then the composition of the corresponding reflections is arotation about P. But we’ve already stated that this rotation, since it belongs to our finite symmetry group,must have centre O. Therefore, the two mirrors must have met at O and it follows that every mirror passesthrough O.

One final result that we’ll need to use is the following little lemma. We’ll leave the proof of this as an exercisefor the enthusiastic reader.

Lemma. In a finite symmetry group, the rotations are of the form I, R, R2, R3, . . . , Rn−1 for some rotation R.

27Expressions like R−12 ◦ R−1

1 ◦ R2 ◦ R1 which take the form a · b · a−1 · b−1 are known as commutators and are incredibly useful ingroup theory and other areas of mathematics.

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The Proof of Leonardo’s Theorem

Recall that Leonardo’s theorem states that if a subset of the Euclidean plane has finitely many symmetries,then its symmetry group must be the cyclic group Cn or the dihedral group Dn for some positive integer n.Most of the mathematical content in the proof of Leonardo’s theorem is in the lemmas we proved above. Allwe need to do now is fit the pieces of the puzzle together as follows.

Proof.

Suppose that someone gives you a finite symmetry group. By the lemma proved earlier, we knowthat there are no translations nor are there glide reflections. In other words, every element of the finitesymmetry group must be a rotation or a reflection.

If there are only rotations, then the last lemma above guarantees that the finite symmetry group mustbe Cn for some positive integer n. If there are only reflections as well as the identity, then the finitesymmetry group must be D1. This is because there must be an equal number of direct and oppositeisometries.

So we’re left with the case that there are both rotations and reflections, in which case there must beequal numbers of each.28 The first lemma above states that all of the rotations must share the samecentre O while the second lemma above states that all of the mirrors of reflections must pass through O.

We know that the rotations are “evenly spaced” by the previous lemma, so the rotations must be by theangles

0◦, 360◦

n, 2× 360◦

n, . . . , (n− 1)× 360◦

n,

for some positive integer n. If the mirrors were not evenly spaced as well, then there must be two ofthem which create an angle strictly less than 180◦

n . The composition of the reflections through these twomirrors will then yield a rotation of strictly less than 360◦

n — a contradiction.

Therefore, we can conclude that the rotations are equally spaced with centre O and that the mirrors areequally spaced and pass through O. But this just means that the finite symmetry group is dihedral.

Frieze Patterns

Let’s turn our attention now to infinite symmetry groups. Since we’ve already seen that a finite group ofsymmetries can’t contain a translation, an easy way to create an infinite symmetry group is to consider asingle translation T. A symmetry group which contains T necessarily contains the infinitely many elements29

. . . , T−3, T−2, T−1, I, T, T2, T3, . . . .

Keeping this in mind, we define a frieze pattern to be a subset of the Euclidean plane whose symmetry groupcontains a horizontal translation T along with

. . . , T−3, T−2, T−1, I, T, T2, T3, . . . ,

and no other translations.28Here, we’re considering the identity as a rotation through an angle of zero.29Here, the notation Tn stands for the composition of n copies of T, while T−n stands for the composition of n copies of T−1.

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A frieze pattern can certainly have other symmetries which aren’t translations, but there aren’t too manypossibilities for such other symmetries. Actually, it shouldn’t be too difficult to see that they can only come infour types.

H = a reflection in a horizontal mirror

V = a reflection in a vertical mirror

R = a 180◦ rotation

G = a glide reflection along a horizontal axis

So for each frieze pattern, we can give it an HVRG symbol, depending on which of these four symmetries itpossesses. For example, a frieze pattern with symbol VRG would have symmetry group which contains areflection in a vertical mirror, a 180◦ rotation and a glide reflection along a horizontal axis, but not a reflectionin a horizontal mirror. If we decide to classify frieze patterns in this way, then we obtain the following result.

Theorem. There are exactly seven types of frieze pattern.

Proof. There are at most sixteen possible HVRG symbols.

none, H, V, R, G, HV, HR, HG, VR, VG, RG,

VRG, HRG, HVG, HVR, HVRG

However, the following four observations allows us to exclude some of these examples.

If you have H, then you have G.

If you have V and R, then you have G.

If you have R and G, then you have V.

If you have G and V, then you have R.

You can prove the first of these observations, for example, by composing H and T to obtain G. The otherthree facts can be proved in a similar way. If you go ahead and use these observations to eliminate some ofthe possibilities, then you should find that there are only seven remaining.

none, V, R, G, HG, VRG HVRG

Of course, this only shows that there are at most seven types of frieze. We still have to demonstrate that eachof these possibilities actually arises, which we accomplish in the following diagrams.

none

V

R

G

HG

VRG

HVRG

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The mathematician John Conway — who often has his own spin on certain mathematical theorems andproofs — has coined his own set of names for the types of frieze patterns.

HOP (none)

SIDLE (V)

SPINNING HOP (R)

STEP (G)

JUMP (HG)

SPINNING SIDLE (VRG)

SPINNING JUMP (HVRG)

The following diagrams should hopefully explain Conway’s rather strange nomenclature for the friezepatterns.

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The Crystallographic Restriction

Consider applying two translations S and T in differ-ent directions to a point P. By taking various combi-nations of S and T, we obtain a set of points in theplane. Any set of points in the plane which you canobtain in this way is called a lattice. You can think ofa lattice as an equally spaced orchard of apple trees.

Of course, the translations S and T are symmetries of the lattice, as is any combination involving S andT. Another symmetry of the lattice is a rotation by 180◦ about one of the lattice points. However, it couldbe possible that there are other rotational symmetries, although there is some restriction as to what thoserotational symmetries could be. In order to state the theorem more explicitly, let’s define the order of arotation R to be the smallest positive integer n such that Rn = I.

Theorem (Crystallographic Restriction). If the symmetry group of a lattice contains a rotation, then that rotationmust have order 2, 3, 4 or 6.

Proof. First, we state a simple fact, which the enthusiastic reader is encouraged to prove on their own.

Any group which contains a rotation of order n also contains a rotation by angle 360◦n .

Let’s call a point in the plane special if it’s the centre of a rotational symmetry of the lattice of order n. Pickany special point O and let P be the closest special point to it. By the fact stated above, a rotation by 360◦

nabout any special point must be a symmetry of the lattice. But, furthermore, any such rotation must be asymmetry of the set of special points. So if we rotate O about P by 360◦

n , we obtain a point Q which is special.For n ≥ 7, this point Q is closer to O than P, which contradicts our assumption. So we can deduce that thereis no rotational symmetry of the lattice of order greater than or equal to 7.

OP

Q

360◦n

OP

Q

R

360◦5

360◦5

If n = 5, then we rotate P about Q by 360◦n to obtain another special point R. However, this point R is now

closer to O than P, which contradicts our assumption. So we can deduce that there is no rotational symmetryof the lattice of order 5. Hence, we may conclude that any rotational symmetry of the lattice must have order2, 3, 4 or 6.

This result derives its name from the fact that it also holds for three-dimensional lattices. In that case, thestatement is important in crystallography — the study of crystals — and guarantees that any rotationalsymmetry of a crystal must have order 2, 3, 4 or 6. It would be nice if the theorem also held in greaterthan three dimensions, but it just isn’t true. This is because rotations in higher dimensions behave in a verydifferent way than in two or three dimensions.

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Wallpaper Patterns

If a subset of the Euclidean plane has a symmetry group whose translations form a lattice, then we callthat subset a wallpaper pattern. Simply put, they’re patterns that cover the whole plane and are repetitive intwo different directions. Just as we did for friezes, we can try to classify the types of wallpaper patterns —however, the game is much harder this time. It turns out that there are exactly seventeen different types ofwallpaper patterns, although we won’t provide a proof here. We’ll merely content ourselves with knowinghow to identify them. To each wallpaper pattern, we associate an RMG symbol consisting of the followingthree numbers R, M and G.

R = the maximum order of a rotational symmetry

M = the maximum number of mirrors which pass through a point

G = the maximum number of proper glide axes which pass through a point

By a proper glide axis, we mean the axis of a glide reflection which is not itself a mirror. The seventeenpossible types of wallpaper patterns are pictured below. For each wallpaper pattern, make sure that you canfind a centre of a rotational symmetry of order R, a point through which M mirrors pass, and a point throughwhich G proper glide axes pass.

Unfortunately, the RMG symbol doesn’t quite distinguish all seventeen wallpaper patterns — there are twowhich are described by 332. So let’s call them 332A and 332B and note that they can be distinguished usingthe following observation.

For the wallpaper pattern 332A, it’s possible to find a centre of rotation which doesn’t lie on a mirror.

For the wallpaper pattern 332B, every centre of rotation lies on a mirror.

100 101

110 111

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200 202

211 220

222 300

332A 332B

89


400 423

442 600

664

Problems

Problem. Prove that if a symmetry group of a frieze pattern contains a reflection in a vertical mirror and a rotation by180◦, then it must also contain a glide reflection.

Proof. If we denote the reflection in a vertical mirror by V and the rotation by 180◦ by R, then the symmetrygroup of the frieze pattern must also contain R ◦V. This composition is an opposite isometry and it is easy tosee that it is either reflection in a horizontal mirror or a glide reflection along a horizontal axis. (Rememberthat the centre of rotation for R does not have to lie on the mirror for V.) Since the symmetry group of afrieze pattern contains a horizontal translation by definition, in either case, the symmetry group of the friezepattern must contain a glide reflection along a horizontal axis.

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So how do you know that the composition R ◦V is a reflection in a horizontal mirror or a glide reflectionalong a horizontal axis? One way to see this is to consider what happens to the picture of a left footprintwalking right, above the centre of rotation of R. After applying the reflection V, this becomes a right footprintwalking left, above the centre of rotation of R. And then after applying R, this becomes a right footprintwalking right below the centre of rotation of R. The only way to start with a left footprint walking right andend up wit ha right footprint walking right is via a reflection in a horizontal mirror or a glide reflection alonga horizontal axis.

Problem. For each wallpaper pattern pictured above, if it has symbol RMG, find on the diagram a point which is thecentre of a rotational symmetry of order R, a point where M mirrors meet, and a point where G proper glide axes meet.

Proof. For example, the diagram below shows the wallpaper pattern with RMG symbol equal to 423 witha point which is the centre of a rotational symmetry of order 4, a point which 2 mirrors pass through, anda point which 3 proper glide axes pass through. You can do something similar for the other wallpaperpatterns.

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Abel

Earlier we spoke about Galois, a mathematician whoproved that it was impossibile to write down a for-mula to solve the quintic equation, before dying atthe tender age of twenty. In fact, the first person togive a complete proof of this beat Galois by severalyears. He was a Norwegian mathematician by thename of Niels Henrik Abel and he was born in 1802before dying in 1829, living a whole six years longerthan Galois.

Abel’s proof of the impossibility of solving the quinticequation was not as deep and far-reaching as Galois’proof, but was nonetheless extremely novel. Unfor-tunately, his work was extremely difficult to read,partly because he had to cut out all of the details tosave money on printing. His success in mathematicsgave him some finances to travel around Europe andmeet some of the more well-known mathematicians.Unfortunately, he contracted tuberculosis while inParis and became quite ill. During this time, a friendof his had found a prestigious professorship for Abelin Berlin and wrote him a letter to tell him the goodnews. Unfortunately, the letter arrived two days afterAbel died.

The early death of this extremely talented mathemati-cian cut short a career of extraordinary brilliance andpromise. Abel had managed to clear some of theprevailing obscurities of mathematics and paved theway for several new fields. His complete works wereedited and eventually published by the Norwegiangovernment. The adjective “abelian” is derived fromAbel’s name and is so commonplace in mathematicsthat mathematicians don’t even bother to capitalise itany more.

Abel is still a relatively big deal in Norway — therehave been stamps, banknotes and coins bearing hisportrait; there is a crater on the Moon named afterhim; and there is the prestigious Abel prize for math-ematicians, presented by the King of Norway andworth about a million dollars in prize money.

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3. POLYHEDRA, GRAPHS AND SURFACES 3.1. From Polyhedra to Graphs

What is a Polyhedron?

Now that we’ve covered lots of geometry in two dimensions, let’s make things just a little more difficult.We’re going to consider geometric objects in three dimensions which can be made from two-dimensionalpieces. For example, you can take six squares all the same size and glue them together to produce the shapewhich we call a cube.

More generally, if you take a bunch of polygons and glue them together so that no side gets left unglued,then the resulting object is usually called a polyhedron.30 The corners of the polygons are called vertices, thesides of the polygons are called edges and the polygons themselves are called faces. So, for example, the cubehas 8 vertices, 12 edges and 6 faces.

Different people seem to define polyhedra in very slightly different ways. For our purposes, we will needto add one little extra condition — that the volume bound by a polyhedron “has no holes”. For example,consider the shape obtained by drilling a square hole straight through the centre of a cube. Even though thesurface of such a shape can be constructed by gluing together polygons, we don’t consider this shape to be apolyhedron, because of the hole.

We say that a polyhedron is convex if, for each planewhich lies along a face, the polyhedron lies on oneside of that plane. So, for example, the cube is aconvex polyhedron while the more complicated spec-imen of a polyhedron pictured on the right is certainlynot convex. Note that this definition is just a gener-alisation of the definition of a convex polygon in theplane. One very important thing to keep in mind isthe fact that we usually think of a polyhedron as justthe outside, the surface of the shape, and not as asolid object carved out of wood.

I’m sure you’ve probably seen and played with polyhedra many times before. For example, if you’ve everpicked up a standard soccer ball, you might have noticed that they’re usually made by sewing togetherpatches in the shape of pentagons and hexagons. Other examples include the shape formed by the buckmin-sterfullerene carbon molecule C60 which is found in soot, the Montreal Biodome at Parc Jean Drapeau andthe pyramids of Egypt.

30Although it’s all right to say “polyhedrons”, the more common plural to use is “polyhedra”. That’s because the word polyhedroncomes from the ancient Greek words which mean many faces.

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What is a Graph?

It’s quite unfortunate that the word graph means at least two completely different things in mathematics.One type of graph that you are no doubt already aware of is the graph of a function or equation, plotted on aset of axes. You probably know that it’s possible to plot the graph of an equation like y = x2 and you obtain ashape called a parabola which looks like a large smiley face. We won’t talk about these types of graph here atall — so forget you ever heard about them.

For us, a graph will always mean a set of points called vertices, connected in pairs by lines or curves callededges. We make no assumptions about whether the graph is in one connected piece or not. However, weusually won’t allow loops — an edge whose endpoints are the same vertex — or multiple edges — morethan one edge whose endpoints are the same pair of vertices.

We’ll call this a graph. We won’t call this a graph.

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For our purposes, it’s a good idea to think of a graph as a diagram whose vertices represent people at a partyand whose edges represent friendship. So we should consider two graphs to be the same if they represent thesame party. In particular, the important thing about a graph is not the way that you draw it on paper, but therelationships that the edges describe. Keeping this in mind, we say that two graphs G and H are isomorphicif there is a one-to-one correspondence between their vertices such that two vertices are connected by anedge in G if and only if their two corresponding vertices are connected by an edge in H. For example, thefollowing two graphs are isomorphic and the labelling on the vertices tells you exactly why.

1

2

3 4

5

1

23

4

5

Every Polyhedron is a Graph

The fact that every polyhedron is a graph is a rather simple statement. This is because if you have apolyhedron and simply ignore the faces, then what you have left over is just a bunch of vertices connected inpairs by edges — in other words, a graph.

A more interesting statement is the following fact. Every polyhedron corresponds to a planar graph — inother words, a graph which can be drawn in the plane without any of its edges crossing. So why is this true?Well, suppose that your polyhedron is made from some sort of rubbery material, like a balloon. If you popthe balloon by removing one of the faces, then what remains is a rubbery sheet with the vertices and edgesstill drawn on it. Now just stretch this out flat onto a table and there you have your planar graph. Note thatwe needed to define our polyhedra to “not have holes” for this to trick to work — donut-shaped and otherhole-possessing balloons just don’t do the job.

The Handshaking Lemma

The degree of a vertex in a graph corresponds to a person’s popularity index at a party — it’s basically howmany friends they have. So the degree of a vertex in a graph is simply the number of ends of edges whichmeet around that vertex. The following diagram shows a graph with each vertex labelled by its degree.

3 2

2 1

0

1

1

If you give me some numbers and ask me to find a graph with those numbers as its degrees, then the taskis often not possible. Obviously, the numbers have to be non-negative integers, but it turns out that otherthings have to be true as well. The most important relation which these numbers must obey is the following.

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Lemma (Handshaking lemma). At any party, if you ask everyone in the room (including yourself) how many handsthey shook, and add up all of the answers, then you will get an even number. In fact, the number you get will be twicethe number of handshakes that have occurred during the party. In graph theory terminology, this translates into thefollowing fact — in any graph, the sum of the degrees of all the vertices is equal to twice the number of edges.

Proof. At the start of the party, everyone has shaken zero hands because no handshakes have taken place.Each time a handshake takes place, the number of hands shaken increases by two, once for each personshaking hands. Therefore, after all handshakes have taken place, the sum of all of the answers must be equalto twice the number of handshakes. In graph theory terms, this means that the sum of the degrees of all thevertices is equal to twice the number of edges in the graph.

The Bridges of Konigsberg

Actually, graph theory began with the mathematician Leonhard Euler in the early eighteenth century. It’spretty amazing that such a simple mathematical construction took so long to be developed. It’s now athriving area of mathematics and, due to its very simplicity, has lots and lots of applications. In fact, thereason why Euler invented graph theory to begin with was to solve a very simple problem, known as theSeven Bridges of Konigsberg problem.

The above map shows what Konigsberg — now known as Kaliningrad — looked like back in Euler’s day.When the weather was nice, the locals liked to stroll about the town. One thing that many of them tried todo was to walk around Konigsberg passing over every bridge exactly once, with no swimming allowed.Since people tried and tried without success, the problem became known as the Seven Bridges of Konigsbergproblem.

Proposition. It’s impossible to walk around Konigsberg, passing over every bridge exactly once.

Proof. Euler’s approach was to pick out only the relevant aspects of the diagram. For example, it didn’t reallymatter that there was a church on the corner of Third Avenue and Fifth Street, nor that there was a troll-likehobo standing on one of the bridges. So if we remove a lot of the irrelevant details, we end up with a picturesuch as the one below left. However, Euler realised that you could simplify the diagram further by creatingone point for every land mass on the map and joining two points each time there is a bridge joining the twoland masses. Of course, the result is just the graph pictured below right.31

31OK, so I said earlier that a graph can’t have multiple edges — but we should be allowed to change the rules whenever we like, tosuit our application. So for the purposes of solving this problem, let’s temporarily allow graphs with multiple edges.

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Let’s suppose now that it’s possible to walk around Konigsberg, passing over every bridge exactly once. We’llstart with the obvious fact that this walk must start at some vertex of the graph and end at some vertex of thegraph. Between the start and end, each time we visit a vertex, we must have walked along two edges incidentto it — one going in and one coming out. Even though we can visit a vertex many times, this reasoningtells us that every vertex other than the start and end must have even degree. So to be able to walk aroundKonigsberg, passing over every bridge exactly once, the Konigsberg graph above must have at most twovertices of odd degree. But you can see for yourself that all four vertices of the graph have odd degree, anobvious contradiction.

In honour of Euler, we say that a graph is Eulerian if it’s possible to walk around the graph, passing over eachedge exactly once. The following theorem tells us exactly when a graph is Eulerian.

Theorem. A connected graph is Eulerian if and only if it has zero or two vertices with odd degree.

Note that if a graph is Eulerian, then it should consist of one piece and such graphs are called connected.Our argument used to solve the Seven Bridges of Konigsberg problem tells us that if a connected graphis Eulerian, then it must have at most two vertices with odd degree. However, the handshaking lemmaimplies that no graph can possibly have exactly one vertex with odd degree. Therefore, if a connected graphis Eulerian, then it must have zero or two vertices with odd degree.

Conversely, it’s always possible in such graphs to find a walk which traverses each edge exactly once. Thegeneral idea behind the proof is to imagine a drunk person stumbling randomly around the graph, and neverpassing over an edge twice. Eventually, they have to get stuck. If they’ve passed over each edge exactly onceafter they get stuck, then that’s great. If not, then they they can always alter their route slightly to pass overmore edges — to do this, we need to use the fact that the graph is connected and has zero or two verticeswith odd degree. They can keep altering their route in this way until they finally have a route which passesover each edge exactly once.

Six People at a Party

There is a very famous result in graph theory which states that, at any party with six people, there must existthree people who all know each other or three people who all don’t know each other. One way to representthis problem is to draw a graph with six vertices, one for each person, and then draw a red edge between twopeople who know each other or a blue edge between two people who don’t know each other. So the theoremcan be rephrased in graph theoretic terms in the following way. Prove that if, in a graph with six vertices,every pair of vertices is connected by an edge coloured red or blue, then there must exist a red triangle or ablue triangle. In fact, to make life easier, let’s define the complete graph Kn to be the graph with n vertices such

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that every pair of vertices is connected by an edge. Then the six people at a party problem can be stated inthe following way.

Proposition. If every edge of the complete graph K6 is coloured red or blue, then there must exist a red triangle or ablue triangle.

Proof. Consider a random partygoer A — of the five edges which are connected to A, note that at least threeof them have to be the same colour. So let’s just go ahead and assume that there are three edges connected toA, all of which are red. The same argument will apply for the case that there are three edges connected to A,all of which are blue. Suppose that these red edges connect A to the party people B, C and D. If BC is red,then triangle ABC is red; if CD is red, then triangle ACD is red; and if DB is red, then triangle ADB is red.So to avoid a red triangle, the edges BC, CD and DB must be all be blue, which forces triangle BCD to beblue. Therefore, there must exist a red triangle or a blue triangle in the graph.

Euler’s Formula

If you ask me to build a polyhedron which has V vertices, E edges and F faces, then sometimes I won’t beable to do it — not because I’m incompetent, but because such a polyhedron may not exist. In fact, Eulernoticed that the equation V − E + F = 2 holds for all polyhedra. For example, you can check that it’s true forthe following three examples, where a tetrahedron is just a fancy name for a triangular pyramid.

polyhedron V E F

tetrahedron 4 6 4cube 8 12 6

soccer ball 60 90 32

In fact, since polyhedra correspond to planar graphs, you might expect Euler’s formula to work for planargraphs as well — and you’d be right. We have to be a little careful here, because Euler’s formula will onlywork if you include the region outside the graph as a face. This is due to the fact that when we construct aplanar graph from a polyhedron by flattening it, we had to remove a face first. Another subtle point is thatyour planar graph must be connected — otherwise, Euler’s formula just cannot hold, and you should checkthis for yourself.

Theorem (Euler’s Formula). For a connected planar graph with V vertices, E edges and F faces, V − E + F = 2.

Proof. I’m just going to give a sketch proof here, some-thing which is incredibly convincing, but which isn’tactually a complete proof. Suppose that we wouldlike to calculate V − E + F for the graph on the right,but we’re just too lazy to count the number of vertices,edges and faces. Consider an edge, like the one la-belled a, which connects one vertex to the remainderof the graph. If you remove this edge and the vertexattached to it, then you obtain a new graph with oneless vertex and one less edge. So this new graph hasexactly the same value of V − E + F.

a

b

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Now consider an edge, like the one labelled b, which separates two distinct faces of the graph. If you removethis edge, then you obtain a new graph with one less edge and one less face. So this new graph also hasexactly the same value of V − E + F. So we can continue removing edges and vertices, without ever changingthe value of V − E + F, until we are left with the simplest graph possible, consisting of only one vertex andhence, only one face. Now we can calculate the value of V − E + F for this graph and the answer is simply1− 0 + 1 = 2. Therefore, in our original planar graph, we know that V − E + F = 2.

Dual Graphs

Given a planar graph, we can construct a new graph with a vertex for each of the faces of the original graphand an edge between two vertices if the corresponding faces meet along an edge. This new graph is calledthe dual graph. You should draw several examples to convince yourself of the following facts.

The dual graph may have loops and multiple edges, but is always planar.

The number of vertices of the dual graph is equal to the number of faces of the original graph. Thenumber of edges of the dual graph is equal to the number of edges of the original graph. The numberof faces of the dual graph is equal to the number of vertices of the original graph.

The dual graph of the dual graph is the original graph.

One extremely useful thing to do with a polyhedron or a planar graph is apply the handshaking lemma to itsdual. To do this, you need to know what the degree of a vertex in the dual graph is. However, a vertex in thedual graph corresponds to a face in the original graph and the degree of a vertex in the dual graph is exactlythe number of edges around the corresponding face in the original graph. Therefore, we obtain the followingresult.

Proposition. In a planar graph or a polyhedron, the sum of the numbers of edges around each face is equal to twice thenumber of edges.

More on Planar Graphs

As fascinating as graph theory is, we’re actually here to learn about geometry. And as we saw earlier, oneway that geometry makes contact with graph theory is via polyhedra. In particular, we showed that everypolyhedron corresponds to a graph and, not just any old graph, but a planar graph. We have to take a littlecare with the definition of a planar graph, because we say that a graph is planar if it can be drawn in theplane without any of its edges crossing, not if it is drawn in the plane without any of its edges crossing. So,for example, although the following graph has been drawn with two edges crossing, it actually is planar, asyou can check for yourself.

Some graphs just aren’t planar, no matter how hard you try to draw them without edges crossing. The onewith the smallest number of vertices happens to be the complete graph K5.

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Proposition. The complete graph K5 is not planar.

Proof. To obtain a contradiction, let’s suppose that K5 is planar. Given that V = 5 and E = 10, Euler’s formulatells us that, if we could draw K5 in the plane without edges crossing, then we would have F = 7. But itshould be clear that if we could draw K5 in the plane without edges crossing, then every face must have atleast three edges around it — this is because K5 has no multiple edges or loops. This means that the sum ofthe numbers of edges around each face is at least 7× 3 = 21. However, by the proposition above concerningthe handshaking lemma in the dual graph, we know that the sum of the number of edges around each facecoincides with twice the number of edges. In other words, we obtain the inequality 2E ≥ 21 which impliesthat 20 ≥ 21. Contradictions don’t come more blatant than that, so we may now conclude that there is noway to draw K5 in the plane without edges crossing — in short, K5 is not planar.

You can try and prove in an entirely analogous way that K3,3 is not planar either, although the proof is a littlebit more difficult. A very difficult theorem to prove which is, in my opinion, quite amazing, is the followingresult which lets you decide whether or not a graph is planar.

Theorem (Kuratowski’s Theorem). A graph is planar if and only if it does not contain a a smaller graph which lookslike K5 or K3,3.

When I say a graph which looks like K5, I mean a graph which you can make out of K5 by adding in someessentially useless vertices along its edges. Here’s another amazing theorem which involves planar graphsand is also quite difficult to prove.

Theorem (Fary’s Theorem). Every planar graph can be drawn in the plane without edges crossing so that each edgeis represented by a line segment.

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Euler

Leonhard Paul Euler — pronounced “oiler”, not“yooler” — was a Swiss mathematician who livedfrom 1707 to 1783. He is widely regarded as the pre-eminent mathematician of the eighteenth century andone of the greatest of all time. He is the most prolificmathematician to ever have lived, with his collectedworks filling about 75 quarto volumes. He was alsoquite prolific in other ways as evidenced by the factthat he fathered at least thirteen children.

I could go on all day about Euler’s contributions tomathematics and science. For example, he createdgraph theory when he solved the Seven Bridges ofKonigsberg problem; he discovered the Euler linein Euclidean geometry; he discovered the formulaV − E + F = 2 for polyhedra; he introduced the no-tation e for the natural base of logarithms, f (x) for afunction, ∑ for a summation, and i for the square rootof negative one; he was the master of the branch ofmathematics known as analysis; he discovered manytheorems in number theory including a generalisationof Fermat’s Little Theorem; his influence was integralin the birth of analytic number theory, a very impor-tant branch of mathematics; and he is renowned forwork in mechanics, fluid dynamics, optics and astron-omy.

Euler discovered many remarkable formulas in hisday — for example, he showed that the exponentialfunction ex could be represented as an infinite poly-nomial, or Taylor series.

ex =x0

0!+

x1

1!+

x2

2!+

x3

3!+ · · ·

He solved the famous Basel problem which asked forthe sum of the reciprocals of the perfect squares —quite remarkably, the answer is

112 +

122 +

132 +

142 + · · · = π2

6.

Euler also established the following formula concern-ing complex numbers, the exponential function, andtrigonometric functions.

eiθ = cos θ + i sin θ

This was called “the most remarkable formula inmathematics” by the famous physicist Richard Feyn-man. In 1988, readers of the Mathematical Intelli-gencer voted it the most beautiful formula ever andtheir poll also included two other formulas of Eulerin the top five.

Euler’s eyesight worsened throughout his life andhe eventually became nearly blind in his right eye.His eyesight worsened so much over time that heearned the nickname Cyclops. Later in life, Euler suf-fered a cataract in his good left eye, rendering himalmost totally blind in 1766. Even so, his conditionappeared to have little effect on his productivity, as hecompensated for it with his mental calculation skillsand photographic memory. For example, Euler couldrepeat the Aeneid of Virgil from beginning to endwithout hesitation, and for every page in the editionhe could indicate which line was the first and whichthe last. With the aid of his scribes, Euler’s productiv-ity on many areas of study actually increased. In fact,he amazingly produced one mathematical paper perweek in the year 1775.

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3. POLYHEDRA, GRAPHS AND SURFACES 3.2. Platonic Solids and Beyond

Classifying the Platonic Solids

A Platonic solid is a convex polyhedron whose faces are all congruent regular polygons, with the same numberof faces meeting at each vertex. In some sense, these are the most regular and most symmetric polyhedra thatyou can find. Our goal now will be to classify the Platonic solids — in other words, hunt them all down.

Theorem (Classification of Platonic solids). There are exactly five Platonic solids.

Proof. The following geometric argument is very similar to the one given by Euclid in the Elements.

Let’s say that the regular polygons have n sides and that d of them meet at every vertex. It should beclear that to form a vertex, you need d ≥ 3.

Because the Platonic solids are convex by definition, at each vertex of the solid, the sum of the anglesformed by the faces meeting there must be less than 360◦. If the sum was equal to 360◦, then the faceswhich meet at the vertex would all lie in the same plane and there wouldn’t be a vertex at all. And ifthe sum was more than 360◦, then it would be impossible for the vertex to be convex. By this reasoning,the angle in the regular polygon with n sides must be less than 360◦ ÷ 3 = 120◦.

Note that a regular hexagon has angles of 120◦ so that a regular polygon with more than six sides hasangles which are greater than 120◦. Therefore, we only need to consider the cases when n is equal to 3,4 or 5.

– When n = 3, we have faces which are equilateral triangles, all of whose angles are 60◦. Using thefact that the sum of the angles formed by the faces at a vertex must be less than 360◦, we obtainthat d must be equal to 3, 4 or 5.

– When n = 4, we have faces which are squares, all of whose angles are 90◦. Using the fact that thesum of the angles formed by the faces at a vertex must be less than 360◦, we obtain that d must beequal to 3.

– When n = 5, we have faces which are regular pentagons, all of whose angles are 108◦. Using thefact that the sum of the angles formed by the faces at a vertex must be less than 360◦, we obtainthat d must be equal to 3.

So, in summary, we know that there are only five possibilities for the pair of integers (n, d) — namely,(3, 3), (3, 4), (3, 5), (4, 3) and (5, 3). We will simply state the fact here that for each of these possibilities,there is exactly one Platonic solid. To see that there is at least one Platonic solid corresponding to a pair(n, d), all you need to do is construct it out of regular polygons with n sides, with d of them meeting atevery vertex. On the other hand, to see that there is at most one Platonic solid corresponding to a pair(n, d) is a more difficult matter, but should seem believable. This is because if you start to glue togetherregular polygons with n sides, with d of them meeting at every vertex, then you don’t have any choicein what the resulting shape will look like.

Constructing the Platonic Solids

Of course, if you want to build a Platonic solid — the one corresponding to (n, d) = (5, 3), for example —then it helps to know how many vertices, edges and faces are required. So let’s have a look at how you candetermine these numbers. If you want to work out the three unknown quantities V, E and F, then it makessense to look for three relations that these numbers obey. One relation will come from using the handshakinglemma, another will come from using the handshaking lemma on the dual, and another is given to us byEuler’s formula.

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Since three faces meet at every vertex, we know that every vertex in the polyhedron must have degreethree. The handshaking lemma asserts that the sum of the degrees is equal to twice the number ofedges, so we have the equation 3V = 2E or equivalently, V = 2

3 E. This means that if we know thevalue of E, then we also know the value of V.

Since every face is a pentagon, we know that every vertex in the dual must have degree five. Thehandshaking lemma on the dual asserts that the sum of the numbers of edges around each face is equalto twice the number of edges, so we have the equation 5F = 2E or equivalently, F = 2

5 E. This meansthat if we know the value of E, then we also know the value of F.

Now we can use Euler’s formula V − E + F = 2 and substitute for V and F. If you do this properly,you should obtain

23

E− E +25

E = 2 ⇒ E = 30.

From this, we can easily deduce that V = 23 E = 20 and F = 2

5 E = 12.

You can and should use this method to determine the number of vertices, edges and faces for each of thePlatonic solids and, if you do so, you’ll end up with a table like the following.

polyhedron n d V E F

tetrahedron 3 3 4 6 4cube 4 3 8 12 6

octahedron 3 4 6 12 8dodecahedron 5 3 20 30 12icosahedron 3 5 12 30 20

The following diagram shows the five Platonic solids — they are called the tetrahedron, the hexahedron, theoctahedron, the dodecahedron and the icosahedron.32 The first three are easier to imagine, because they aresimply the triangular pyramid, the cube, and the shape obtained from gluing two square pyramids together.

32These names come from the ancient Greek and simply mean four faces, six faces, eight faces, twelve faces and twenty faces,respectively. Of course, we almost always refer to the hexahedron more affectionately as the cube.

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Those of you who are fans of role-playing games such as Dungeons & Dragons — you know who you are —may be able to visualise these more easily. This is because such games involve d4’s, d6’s, d8’s, d12’s andd20’s, where a dn is simply an n-sided die. Another way to visualise a polyhedron is via a net, a figure whichyou can cut out of cardboard and then fold to create the polyhedron. The following diagram shows nets forall of the Platonic solids.

Platonic Solids in Nature

The aesthetic beauty and symmetry of the Platonic solids have made them a favourite subject of geometers forthousands of years. They are named for the ancient Greek philosopher Plato who thought that the classicalelements — earth, water, air, fire and ether — might be constructed from Platonic solids. In the sixteenthcentury, the German astronomer Johannes Kepler attempted to find a relation between the five known planetsat that time — excluding Earth — and the five Platonic solids. In the end, Kepler’s original idea had tobe abandoned, but out of his research came the realization that the orbits of planets are not circles and thediscovery of Kepler’s laws of planetary motion.

Each of the five Platonic solids occurs regularly innature, in one form or another. The tetrahedron, cube,and octahedron all occur in crystals, as does a slightlywarped version of the dodecahedron. In the earlytwentieth century, it was observed that certain speciesof amoeba known as radiolaria possessed skeletonsshaped like Platonic solids — the picture on the rightgives an icosahedral example. Also, the outer proteinshells of many viruses form regular polyhedra — forexample, the HIV virus is enclosed in an icosahedron.

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Scientists have also discovered new types of carbon molecule, known as fullerenes, which have very symmet-ric polyhedral shapes. The most common is C60, which has the shape of a soccer ball, though there are otherswhich possess the shape of Platonic solids.

Symmetries of Platonic Solids

It’s definitely worth mentioning that the whole symmetry game we played for subsets of the Euclideanplane can also be played for subsets of Euclidean space. We can define three-dimensional isometries andsymmetries of figures in space in an entirely analogous way, although some details are slightly more difficult.For example, the result that any isometry in the plane is a product of three reflections becomes the result thatany isometry in space is a product of four reflections. Furthermore, there are more types of isometry than justtranslation, rotation, reflection and glide reflection — they are called twists and rotatory reflections — andyou can try to imagine what these might do.

There is also a three-dimensional analogue of Leonardo’s Theorem, a result which classifies all of the possiblesymmetry groups in Euclidean space. Essentially, the result says something like the only symmetry groupsyou can get are cyclic groups, dihedral groups, certain slightly more complicated versions of the cyclic anddihedral groups, and symmetry groups of Platonic solids.

As an example, let me tell you about the symmetry group of the tetrahedron. An isometry has to take aparticular face of the tetrahedron to one of the four faces of the tetrahedron. And once you decide where oneof the equilateral faces ends up, there are still six possibilities, one for each element of D3. This means that thesymmetry group of the tetrahedron has 24 elements. But we just happen to know a group with 24 elementsand that group is the symmetric group S4. And if you guess that the symmetry group of the tetrahedronis isomorphic to S4, then you’d be right. Now you should try to determine how many elements are in thesymmetry groups of the remaining four Platonic solids using the same idea.

What’s Wrong with Euler’s Formula?

We’re now going to leave Platonic solids behind and embark on a new adventure. Recall that when wedefined what a polyhedron was, we were very careful to say that it “had no holes”. One of the reasons fordoing this is because it doesn’t seem that Euler’s formula works when there are holes. For example, considerthe following geometric object which is made by gluing together polygons.

If you count the number of vertices, edges and faces, then you’ll find that V − E + F does not equal 2 at all.Remember to be very careful if you do this, because the regions shown on the top and bottom in the diagram

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are not actually faces, since they have holes in the middle of them. All you need to do is divide these regionsinto bona fide faces with the help of some extra edges.

Since Euler’s formula doesn’t work for this shape, there are two things we can do — start to cry or try tomake it work. Mathematicians would generally prefer the latter approach. To make Euler’s formula workfor more general shapes, you simply need to note that V − E + F = 0 in this particular case and, in fact,V − E + F = 0 for any shape which has one hole through the middle of it. In fact, if you try the same thingfor shapes with g holes, you’ll eventually discover that

V − E + F = 2− 2g.

So V − E + F seems to change when you talk about very different geometric objects — for example, objectswith different numbers of holes — but seems to be the same when you talk about similar geometric objects —for example, objects with the same number of holes. Another observation is that if we take the geometricobject pictured above and draw it so that it looks a bit curvier, then that doesn’t change the number of vertices,edges and faces, so V − E + F doesn’t change at all. In fact, we can bend, stretch, warp, morph or deform itand the value of V − E + F wouldn’t change.

The Earth

Suppose that you lived a really really long time ago.Then you would probably believe, as did most peo-ple, that the surface of the Earth is a big flat plane.And what makes you think that? Well, it’s simplydue to the fact that everywhere you stand, you no-tice that there’s a pretty flat piece of earth immedi-ately surrounding your feet. And since a big flatplane seems to have this same property, you’ve sim-ply jumped to the conclusion that the Earth must bebig flat plane. But, as you know, this is rather fool-ish thinking. There are many geometric objects apartfrom the plane which have this property. The sphereis just one more example, and we’re going to exploreother shapes that can arise. This idea leads us to studythings called surfaces which we’ll soon talk about.

Topology

The two stories above — one about Euler’s formula and one about the Earth — motivate us to considertopology, which very roughly studies intrinsic fundamental properties of geometric objects and which doesn’tcare about length, size, angle, and so forth. When you study objects in mathematics, you always need tohave some notion of when two of those objects are the same. For example, in Euclidean geometry we havecongruence, in group theory we have isomorphism, in graph theory we have isomorphism, and in topologywe have the notion of homeomorphism. Intuitively speaking, two geometric objects are homeomorphic if it’spossible to bend, stretch, warp, morph or deform one so that it becomes the other one. Note that you are notallowed to cut and glue here.

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A more mathematically precise definition of homeomorphism is as follows. Two geometric objects arehomeomorphic if there exists a bijection — in other words, a one-to-one correspondence — from one to theother which is continuous and has a continuous inverse. If two geometric objects A and B are homeomorphic,then we write A ∼= B, and we call a bijection from A to B which is continuous and has a continuous inverse ahomeomorphism.33

Example. The sphere and the cube are certainly homeomorphic to each other. This is because you can takethe sphere and squash it into a box until it looks like a cube. Or, on the other hand, you could take a cubeand blow it up like a balloon until it looks like a sphere. These sorts of deformations are certainly allowedand show that the two shapes are homeomorphic. For the same sort of reason, a disk and a square are alsohomeomorphic to each other. You can probably see why topology is sometimes informally called rubbersheet geometry.

An explicit homeomorphism in this case isn’t too difficult to describe and, if you were really keen, you couldeven write down an equation for one. The idea is to stick the sphere inside of the cube and suppose that thesphere is a light bulb with a source of light at its centre. Any point on the sphere now casts a shadow on thecube, and this gives a map from the sphere to the cube which is a continuous bijection with a continuousinverse.

Example. Consider an unknotted piece of string like the one shown below left and a knotted piece of a stringlike the one shown below right.

Believe it or not, these two are homeomorphic to each other. Sure, you need to cut the knot open and glue theends together again to make the unknotted loop — but that’s only if you happen to live in three dimensions.Topology, unlike we mere mortals, doesn’t live in any number of dimensions. Intuitively, you can deformthe unknotted loop into a knot if you make those deformations in four-dimensional space. The reasonbeing that lines can just move past each other in four-dimensional space without hitting. It’s just the higherdimensional analogue of the fact that two points on the line can’t move past each other without hitting yet intwo dimensions, they can do so with ease.

If this doesn’t convince you, then you can always go back to the mathematical definition of homeomorphism.Suppose that you wrote the numbers from 1 to 100 along the unknotted loop, in order. You could also dothe same thing around the knotted loop and convince yourself that there’s a function from one to the otherwhich matches up the numbers. It’s easy to see that such a function can be made to be a bijection which iscontinuous and has a continuous inverse — hence, the two are homeomorphic.

33The word homeomorphism comes from the ancient Greek words meaning similar shape.

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Example. A sphere and a torus — the surface of a donut — are not homeomorphic, and this is a little difficultto see. And that’s because if you want to show that two geometric objects are homeomorphic, then youjust go ahead and find the deformation or homeomorphism that makes them so. On the other hand, if youwant to show that two geometric objects are not homeomorphic, then you need some tricks. This should bereminiscent of the fact that if you want to show that two groups are isomorphic, then you just go ahead andfind the isomorphism but if you want to show that two geometric objects are not isomorphic, then you needsome tricks. We’re going to learn some of these tricks later on.

Problems

Problem. Show that at any party, there are always at least two people with exactly the same number of friends at theparty.

Proof. For this problem, we need to assume that if A is friends with B, then B is friends with A.34 Theproblem can obviously be translated into graph theoretic terms as follows.

Show that in any graph — without loops or multiple edges — there are always two vertices withthe same degree.

Just to be concrete, let’s suppose that we’re dealing with a graph which has seven vertices. The degree of avertex in such a graph can only be 0, 1, 2, 3, 4, 5 or 6 since there are no multiple edges or loops, by assumption.Note that there are seven vertices in total as well as seven possibilities for their degrees. This means that ifthere aren’t two vertices with the same degree, then there must be exactly one vertex with degree 0, exactlyone vertex with degree 1, exactly one vertex with degree 2, and so on, up to exactly one vertex with degree 6.However, this situation simply cannot arise, since a graph with seven vertices can’t have a degree 0 vertexas well as a degree 6 vertex. This is because a degree 0 vertex is connected to no others by an edge, while adegree 6 vertex is connected to all others by an edge. In party terms, you can’t have a loner who is friendswith nobody as well as a social butterfly who is friends with everybody. This contradiction means that thereare always two vertices with the same degree. Of course, our argument can be generalised to graphs withany number of vertices.

Problem. Does there exist a polyhedron with exactly thirteen faces, all of which are triangles?

Proof. The first thing you should do is try to draw such a polyhedron as a planar graph, keeping in mind thatthe outside face has to be a triangle. You can try doing this all day, but what you’ll find is that the task isimpossible. And hopefully, you’ll also find that you can have such polyhedra with exactly ten faces or twelvefaces or fourteen faces. What this tells you is that there is something to do with oddness and evenness goingon in this problem and that suggests that we are going to use the handshaking lemma

So now let’s suppose that there does exist a polyhedron with exactly thirteen faces, all of which are triangles.The trick here is to use the handshaking lemma on the dual graph. This asserts that the sum of the numbersof edges around each face is equal to twice the number of edges. In our case, this means that 13× 3 is equalto twice the number of edges or, in other words, that the number of edges must be 19 1

2 . Of course, this isa contradiction so we can deduce that there does not exist a polyhedron with exactly thirteen faces, all ofwhich are triangles.

34As you probably know, such an assumption isn’t always true in the real world.

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Problem. If a planar graph with E edges divides the plane into F faces, prove that F ≤ 2E3 .

Proof. For this problem, we need to assume — as is usual — that the graph contains no loops or multipleedges. Note that the dual graph has F vertices and, since every face of the original graph has at least threesides, every vertex of the dual graph has degree at least three. So the sum of the degrees of the vertices inthe dual graph is at least 3F. However, the number of edges in the dual graph is equal to E. So we can nowinvoke the handshaking lemma to deduce that 2E ≥ 3F, which rearranges to give the desired result.

Problem. Consider a polyhedron all of whose faces are triangles such that four faces meet at every vertex. Determinethe number of vertices, edges and faces of the polyhedron.

Proof. If you want to work out the three unknown quantities V, E and F, then it makes sense to look for threerelations that these numbers obeys. One relation will come from using the handshaking lemma, another willcome from using the handshaking lemma on the dual, and another is given to us by Euler’s formula.

Since four faces meet at every vertex, we know that every vertex in the polyhedron must have degreefour. The handshaking lemma asserts that the sum of the degrees is equal to twice the number of edges,so we have the equation 4V = 2E or equivalently, V = 1

2 E. This means that if we know the value of E,then we also know the value of V.

Since every face is a triangle, we know that every vertex in the dual must have degree three. Thehandshaking lemma on the dual asserts that the sum of the numbers of edges around each face is equalto twice the number of edges, so we have the equation 3F = 2E or equivalently, F = 2

3 E. This meansthat if we know the value of E, then we also know the value of F.

Now we can use Euler’s formula V − E + F = 2 and substitute for V and F. If you do this properly,you should obtain

12

E− E +23

E = 2 ⇒ E = 12.

From this, we can easily deduce that V = 12 E = 6 and F = 2

3 E = 8.

In fact, an example of such a polyhedron is given by the Platonic solid known as the octahedron.

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Newton

Sir Isaac Newton was born on Christmas Day wayback in 1642 and lived a most productive eighty-fouryears until his death in 1727. He was an Englishmathematician who was also a great physicist, ac-complished natural philosopher, prolific theologianand dedicated alchemist. His Philosophiæ NaturalisPrincipia Mathematica is considered to be among themost influential books in the history of science. Inthis work, Newton described how gravity works andproduced three laws of motion, which gave the mostaccurate description of how the universe works forthe following three centuries until around the timeof Einstein. Famously, Newton formed his theory ofgravitation after seeing an apple fall from a tree. TheFrench mathematician Joseph–Louis Lagrange oftensaid that Newton was the greatest genius who everlived, and once added that he was also “the most for-tunate, for we cannot find more than once a systemof the world to establish”.

Newton’s contributions to mathematics and scienceare incredibly diverse. In mathematics, he is mostfamously remembered for his discovery of calculus,independently to the German mathematician Leibniz,but at about the same time. According to Newton’sfriends, he had worked out his method years beforeLeibniz, but published almost nothing about calculusuntil 1693. Meanwhile, Leibniz began publishing hisaccount of calculus up to fifteen years earlier. It didn’ttake long for scientists from the Royal Society — ofwhich Newton was a member — to accuse Leibnizof plagiarism. Thus began an ugly, bitter and contro-versial dispute between Newton and Leibniz, whichmarred the lives of both until the latter’s death.

Some people believe that Newton was quite modestabout his own achievements, writing that “If I haveseen a little further, it is by standing on the shouldersof giants”. However, it is more likely that Newtonwas not modest at all and that this quote was an at-

tack on the scientist Robert Hooke, who was shortand hunchbacked. Modest or not, Newton is consid-ered by many to be the greatest scientist who everlived. A survey of scientists from Britain’s Royal Soci-ety deemed that Newton has had more impact on thehistory of science than Einstein.

To be a great mathematician or scientist, I think thatyou have to be motivated by curiosity, which is whyI like the following quote from Newton himself: “Ido not know what I may appear to the world, butto myself I seem to have been only like a boy play-ing on the sea-shore, and diverting myself in nowand then finding a smoother pebble or a prettier shellthan ordinary, whilst the great ocean of truth lay allundiscovered before me.”

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3. POLYHEDRA, GRAPHS AND SURFACES 3.3. Surfaces and Topology

What is a Surface?

For our purposes, a surface refers to a geometric object which obeys the following conditions.

If you pick a point on a surface and look at all of the points close to it, then they should form somethingwhich looks like a disk. (More precisely, every point on the surface has a neighbourhood which ishomeomorphic to a disk.)

A surface must be finite in the sense that you can put it inside your house. (More precisely, a surfaceshould be bounded.)

A surface must be in one piece, so that you can’t have some of it inside your house, some of it outsideyour house, and be able to close the door. (More precisely, a surface should be connected.)

A surface should not have boundaries. (More precisely, a surface should not have edges which you canwalk off.)

We’ve already seen some examples of surfaces, such as the sphere and the torus. You should convinceyourself that they obey all of the properties that a surface should obey. Note that the plane is not a surface,because it obeys all of the required properties except for the finiteness condition. A disk is not a surfacebecause it obeys all of the required properties except for the boundary condition. And a sphere sitting next toa torus is not a surface because it obeys all of the required properties except for the connectedness condition.

Our goal is to classify all surfaces — in other words, hunt them all down. This is going to be a two stepprocess — first, we have to be able to list all of the surfaces possible, and second, we have to make sure thatno two surfaces in our list are homeomorphic to each other. As a start, we have the following examples ofsurfaces — the sphere which has no holes, the torus which has one hole, a surface which has two holes, andso on. The technical name for the number of holes on a surface is the genus.

genus = 0 genus = 1 genus = 2

Pac-Man World

Let’s change topic a little and look at Pac-Man world, the world in which Pac-Man lives.35 Pac-Man lives in aworld which appears to us as a rectangle on the computer screen.

35For those of you who are too young to know, Pac-Man is a classic arcade game in which you control Pac-Man through a maze sothat he eats lots of pac-dots. It’s universally considered one of the classics of the medium, virtually synonymous with video games, andan icon of 1980s popular culture. To be gender neutral, I should also mention Ms Pac-Man, a spin-off arcade game involving a femaleversion of Pac-Man who looks exactly the same as Pac-Man, but with a bow on her head.

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But his world is much more interesting than just any old rectangle, because when he moves off the right edgeof the screen, he reemerges on the left edge of the screen.

And if you try to move Pac-Man off the bottom edge of the screen, he’ll reemerge on the top edge of thescreen.

In fact, even if Pac-Man is very naughty and you decide to put him in the corner, you can’t hide him becausehe’ll now appear at every corner of the screen.

What this tells us is that Pac-Man world is, in fact, an example of a surface. To see this, all you need to do ischeck that the four properties required of a surface are true. It should be clear that no matter where Pac-Manstands, there is a little disk around him. This is particularly easy to see when he is in the middle of the screen.But even if he is on the right edge of the screen, there is a patch around him which will appear as a semicircleon the right edge and a semicircle on the left edge — these together make up a disk. Of course, this verysame argument applies no matter which edge Pac-Man decides to stand on. The most tricky case is whenPac-Man is in the corner of the screen, but we’ve already seen that he will be surrounded by a disk whichwill appear to us as four quarter circles.

Pac-Man world also happens to be finite, because it appears on our finite computer screen. It has no boundarybecause Pac-Man can never run off the edge of the world. And it is certainly connected, because Pac-Mancan visit every part of it just by running around eating pac-dots.

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So the obvious question to ask is what surface does Pac-Man live on? Is it one we’ve seen before, or is itsomething brand new? To see what the answer is, we use the following strategy. Note that a point on thebottom edge of the screen is the same as a point on the top edge of the screen, even though it seems to appearin two different places. So the obvious thing to do is simply glue these two points together — in fact, we’llend up gluing the whole bottom edge to the whole top edge. And similarly, we’ll end up gluing the wholeleft edge to the whole right edge. So we can think of Pac-Man world as a rectangular piece of fabric whichcomes to us with gluing instructions, indicated by arrows. Edges with matching arrows get glued togetherwith the arrow telling you how the edges should line up with each other.

The diagrams below show what happens first when you glue together the top and bottom edges, and thesecond when you glue together the remaining two edges. The result, as you can plainly see, is simply thetorus. Pac-Man lives on a torus.36

Connected Sums

In many branches of mathematics, it’s useful to take a reductionist approach. For example, when we studiedEuclidean geometry, we reduced all theorems to just a handful of axioms. In group theory, you can slice anddice groups into pieces called simple groups. When talking about surfaces, there is a very useful way toconstruct a new surface out of two old ones which will help us apply the reductionist approach. To find theconnected sum of two surfaces, you simply cut out a little hole from each and glue the two surfaces togetheralong these holes. Probably the best thing to do is look at the following sequence of diagrams to see exactlyhow this works. We denote the connected sum of two surfaces A and B as A # B.

36This is all a bit of a lie, because Pac-Man doesn’t exactly work this way, but I hope you understand the point. Actually, it’s possibleto take many games on rectangles and turn them into games on tori by gluing the edges of the board together. For example, toroidalchess is a variant of chess which uses such a board. Of course, the pieces have to start in different locations and the rules have to bealtered very slightly.

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#

cut

glue

Of course, if you have a complicated surface, then you should be able to cut it up into smaller pieces whoseconnected sum is the original surface. The obvious question we’d like to answer is the following — is everysurface the connected sum of just a few surfaces?

One interesting observation is the fact that if you take the connected sum of a surface and a sphere, youalways get the original surface back again. So, in some sense, the sphere is to connected summation as zero isto addition or one is to multiplication or the identity is to group composition, and so on.

We’re now going to ask a very subtle question — are there different ways to produce the connected sum oftwo surfaces? In other words, if I have two surfaces A and B and you have two surfaces A and B, and weboth try to form the connected sum A # B, then will the results always be homeomorphic? The answer is thatthey definitely will be homeomorphic, although there are two issues at play here.

First, we may choose different locations to cut our holes on A and B. It’s not too hard to see thatthis makes no difference, because you can slide the holes around the surfaces and this corresponds tobending, stretching, warping, morphing or deforming the resulting shape. In other words, sliding holesaround before forming the connected sum will give surfaces which are homeomorphic to each other.

Second, and a lot harder to see, is the fact that there are always two ways to glue two holes together.For example, suppose that you have a very thin wall with a small hole in it and a tube like what youmight find on a vacuum cleaner. If someone tells you to plug the vacuum cleaner into the hole, thenyou could do it from two different sides of the wall. We won’t say too much about why these give thesame result when you form the connected sum. However, it follows from the fact that you can turn ashape inside out merely by deforming it, especially if you do it in four or more dimensions.

The Mobius Strip

Here is something that you can do in the comfort of your own home, and all you need is some paper, somescissors, some glue and three dimensions. Cut a long strip of paper and glue the ends together — but beforeyou glue the ends together, give the strip a half twist. The resulting object is called a Mobius strip. Notethat it isn’t a surface by our definition, since it has a boundary. In fact, one of the interesting things abouta Mobius strip is that it has only one edge, unlike its relative the cylinder, which is formed by gluing theends of a long strip of paper together without any twists at all. Furthermore, a Mobius strip only has oneside, unlike the cylinder. This means that if you start painting one side of the Mobius strip and come back towhere you started painting, then the whole Mobius strip will be painted over. Don’t just take my word for it— try this at home for yourself.

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The following woodcut by M. C. Escher entitled Mobius Strip II shows some ants walking around a Mobiusstrip. The Mobius strip also appears on the universal recycling symbol. In fact, the Mobius strip appears invarious places, such as the shape of some conveyor belts, since this allows the belt to wear evenly on bothsides, rather than on just one. They also appear in the design of computer printer and typewriter ribbons.

Unfortunately, as interesting as Mobius strips are, they aren’t surfaces — and we’re interested in surfaces.Fortunately, we can easily make a surface out of a Mobius strip simply by observing that it has precisely oneedge. Now take a patch in the shape of a disk which also has only one edge. And then glue the two edgestogether, so that there are no spare edges remaining. Try as you might, this just isn’t possible to do in threedimensions. However, that’s just fine, since we don’t always have to draw something which looks like asurface for it to be a surface. For example, Pac-Man world didn’t look a surface and yet we discovered that itwas one, the torus in fact.

One way to think of this surface obtained from patching a Mobius strip with a disk is as follows. Supposethat all along the edge of the Mobius strip are numbers — say from 1 to 100 — and that all along the edge ofthe disk are numbers — also from 1 to 100. Suppose that you are an ant walking around the Mobius stripand, as soon as you hit the edge of the strip — say at the number 73 — you suddenly get teleported over tothe number 73 on the disk and so you just continue walking along the disk. If you try and leave the disk, youget teleported back to the edge of the Mobius strip — precisely where you get teleported back corresponds towhere you left the disk. This is now obviously a surface because every point on the edge of the Mobius stripor the patch has a little disk around it made up of one semicircle on the strip and one semicircle on the patch.

The resulting surface is called the projective plane and it’s definitely not homeomorphic to any of the surfaceswe’ve previously encountered. This is because all of the surfaces we’ve seen before have two sides whereasthis new beast only has one. We say that a surface with two sides is orientable and that a surface with onlyone side is non-orientable. It seems that the world of surfaces isn’t quite so simple after all.

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Orientability

Remember that we’re trying to classify surfaces, and that classification is always a two step process. First, weneed to be able to list out all of the possible surfaces that exist and second, we need to make sure that wedon’t have two things in our list which are homeomorphic. We’ve already mentioned that to show that twosurfaces are homeomorphic is often an easy task — you simply find the homeomorphism between them. Onthe other hand, to show that two surfaces are not homeomorphic is often a difficult task — you need tricks.We’ll see that when it comes to surfaces, there are really only two tricks that you need.

The first trick is to consider the orientability of a surface — in other words, whether it’s one-sided or two-sided. All of the simple examples of surfaces that we could think of, such as the sphere or the torus, seemedto have two sides so we called them orientable. On the other hand, we now have the projective plane, a surfacewhich results from gluing the one edge of a Mobius strip to the one edge of a disk. This surface has only oneside and surfaces like this are called non-orientable. As another example of a non-orientable surface, you couldalso take two copies of the Mobius strip and glue them together along their edges. The resulting surface isknown as the Klein bottle.

Another way to think of orientability is to consider whether or not there is a notion of clockwise andcounterclockwise. If you draw a counterclockwise arrow on an orientable surface and move it around, it willalways appear to be a counterclockwise arrow. On the other hand, consider a counterclockwise arrow on theMobius strip — if you take it for a walk all the way around the strip, then it will come back looking like aclockwise arrow. From this definition of orientability, it’s easy to prove the following fact.

Proposition. If S1 and S2 are orientable surfaces, then the connected sum S1 # S2 is orientable. If S1 is non-orientableand S2 is any surface, then the connected sum S1 # S2 is non-orientable.

Example.

The sphere is orientable, as is any connected sum of tori.

The projective plane is non-orientable, as is any connected sum of projective planes.

Euler Characteristic

Euler’s formula states that in any polyhedron, the number of vertices, edges and faces satisfies the equationV − E + F = 2. Actually, we shouldn’t treat this as a statement about polyhedra, but a statement about mapson the sphere. We define a map to be a way to divide a surface into vertices, edges and faces, where the edgesno longer have to be straight lines, just any old curves on the surface. What Euler’s formula is really tellingus is that for every single map you can draw on the sphere — because all polyhedra are homeomorphic tothe sphere — the number of vertices, edges and faces satisfies the equation V − E + F = 2. We define thenumber V − E + F to be the Euler characteristic of a surface, so what Euler’s formula tells us is that the Eulercharacteristic of the sphere is 2.

Using the same strategy as for the sphere, we can calculate the Euler characteristic of any surface — but thisbegs the following question. How do we know that the Euler characteristic is an intrinsic property of thesurface and doesn’t depend on the map that we draw? In other words, if I draw a map on a surface andyou draw a map on the same surface, will they always have the same value of V − E + F? This question isanswered by the following result.

Proposition. Two maps on the same surface will always have the same value of V − E + F.

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Proof. We proved this fact for planar graphs by deleting the number of vertices, edges and faces and showingthat V − E + F remained the same after each deletion. Unfortunately, this proof wasn’t very precise and youwould have difficulty repeating it for a general surface. This is because we reduced planar graphs to a singlevertex and a single face on the sphere, but there is no analogous “simplest map” on other surfaces.

So instead, we’re going to see that adding vertices, edges and faces doesn’t change the value of V − E + F.For example, if we simply take an edge and break it into two by adding a vertex in the middle, then thenumber of vertices has increased by one, the number of edges has increased by one, while the number offaces is the same. So V − E + F is the same. Similarly if we divide a face into two by an edge, or if we decideto add a new edge which has one endpoint a vertex of degree one. There are various cases to check, but thecrux of the argument is that you can add vertices, edges and faces to your map without changing the value ofV − E + F.

Now I simply take my map and you take your map and we know that the union of these two maps — thatis, the map obtained by overlapping them on the surface — can be obtained from mine by adding vertices,edges and faces. By our previous argument, this means that V − E + F for my map is the same as V − E + Ffor the union of our maps. And by the same reasoning, V − E + F for your map is the same as V − E + F forthe union of our maps. In short, our maps have the same value of V − E + F.

This result means that the Euler characteristic of a surface S — usually denoted by χ(S) — is an intrinsicnumber associated to S and doesn’t depend on which map you decide to draw on S. For example, Euler’sformula asserts that χ(sphere) = 2 and you can calculate that χ(torus) = 0. A more difficult example is tocalculate the Euler characteristic of the projective plane — it’s equal to χ(projective plane) = 1. We’ll seesome techniques later for computing this and essentially the Euler characteristic of any given surface.

Proposition. If S1 and S2 are two surfaces, then the Euler characteristic of their connected sum is

χ(S1 # S2) = χ(S1) + χ(S2)− 2.

Proof. Suppose that we have a map on S1 which contains at least one triangular face and a map on S2 whichcontains at least one triangular face. Since we are allowed to remove any hole from the surfaces in order tocreate the connected sum, let’s decide to remove the triangular faces. After gluing the three edges from onehole to the three edges from the other, we can see what happened to the total number of vertices, edges andfaces. Clearly, the number of vertices decreased by three, because the three around one hole and the threearound the other hole were glued together to result in three vertices. Similarly, the number of edges decreasedby three, because the three around one hole and the three around the other hole were glued together to resultin three edges. And finally, the number of faces simply decreased by two, since we removed two triangularfaces to make the holes and never added any back in. Therefore, the value of V − E + F simply decreased bytwo and we have the equation χ(S1 # S2) = χ(S1) + χ(S2)− 2, as desired.

This theorem allows us to calculate various Euler characteristics using the following result, which you shouldtry to prove.

Corollary. The Euler characteristic of the connected sum of g tori is 2− 2g and the Euler characteristic of theconnected sum of n projective planes is 2− n.

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Poincare

Jules Henri Poincare was a French mathematician,theoretical physicist and philosopher who lived from1854 to 1912. He is often described as The Last Uni-versalist, since he excelled in every branch of mathe-matics known in his day. Poincare is considered oneof the founders of topology, his work on the famousthree-body problem led to the birth of chaos theory,and he is also renowned for introducing the mod-ern principle of relativity, although Einstein seems tohave received most of the credit.

At school, Poincare was a pretty clever student,achieving the top marks in almost every topic hestudied. Poincare went on to study mathematics andmining engineering at university. Although he wassoon offered a post as a lecturer in mathematics, henever fully abandoned his mining career for mathe-matics. Poincare went on to become chief engineerof the Corps de Mines, held various chairs at the Sor-bonne, became president of the French Academy ofSciences and was elected to the Academie francaise.

One of the most famous problems in mathematics —the Poincare conjecture — was formulated by him.To understand just how important this problem is inmathematics, note that when Stephen Smale provedthe conjecture in dimensions five and up in 1961, hewas awarded a Fields Medal, the highest honour inmathematics. When Michael Freedman proved theconjecture in dimension four in 1982, he was awardeda Fields Medal as well. And when Grigori Perelmanproved the conjecture in dimension three in 2002, hewas also awarded a Fields Medal.37

Poincare’s work habits have been compared to a beeflying from flower to flower. He didn’t care aboutbeing rigorous and disliked logic. He believed thatlogic was not a way to invent but a way to struc-ture ideas and that logic limits ideas. A psychologistby the name of Toulouse wrote the following aboutPoincare.

He worked for short periods of time, doingmathematical research for only four hours a day.

His normal work habit was to solve a problemcompletely in his head, before committing thecompleted solution to paper.

His ability to visualise what he heard provedparticularly useful when he attended lectures,since his eyesight was so poor.

He was physically clumsy and artistically inept.

He was always in a rush and disliked goingback for changes or corrections.

His method of thinking is well summarised by thefollowing quotation. “Accustomed to neglecting de-tails and to looking only at mountain tops, he wentfrom one peak to another with surprising rapidity,and the facts he discovered, clustering around theircenter, were instantly and automatically pigeonholedin his memory.”

37Interestingly, Perelman declined the Fields Medal and seems to no longer be working in mathematics. In fact, as of 2003, he gave uphis job at the Steklov Institute and is currently unemployed, living with his mother in Saint Petersburg.

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3. POLYHEDRA, GRAPHS AND SURFACES 3.4. The Classification of Surfaces

Polygon Models

We’re going to look at how to very simply draw a surface on a piece of paper, and the idea is inspired byPac-Man world. Recall that when you glue up the sides of Pac-Man world, you obtain a torus. Furthermore,if you look at the seams of your gluing, they form a map on the torus and, most importantly, this map hasexactly one face. This face corresponds precisely to the rectangle that made up Pac-Man world in the firstplace.

So, if you could find a map on a surface which has one face, then you could cut the surface open along theedges of the map to give a polygon. The surface can be reconstructed from this polygon as long as you havethe right gluing instructions. When you can do this, the resulting polygon and its gluing instructions arereferred to as a polygon model for the surface. Note that a polygon model must always have an even numberof sides which are glued together in pairs, without any edges left unglued. Fortunately, every surface can beturned into a polygon model, as the following result shows.

Theorem.

Every surface corresponds to a polygon model.

Every polygon model corresponds to a surface.

Proof.

The idea here is to use the simple sounding though difficult to prove theorem which states that everysurface can be cut up into triangles. So suppose you have a surface and that you cut it up into trianglesand lay them down on a table. We can label each side of a triangle with a gluing instruction which tellsyou how the triangles can be glued back together to recreate the surface. Now just take two triangleswhich are supposed to be glued together and glue them together. Find another triangle which can beglued onto these two and just go ahead and glue it to them. If you keep gluing more triangles onto theshape that you have, they will all end up making one piece which will form a polygon — in fact, thegluing instructions on the unglued edges of the polygon will ensure that it’s a polygon model for thesurface.

A polygon model is really just a polygon with 2n sides which are glued together in n pairs. You cancheck that after performing such a gluing, the resulting object is indeed a surface. The main thingto check is that around every point, there is a little disk — this is certainly true for points inside thepolygon. And for points on the edges, there will be half of a disk protruding from one edge and half ofa disk protruding from the edge that it’s glued to — these two halves of disks glue together to make awhole disk. Finally, you need to check that if you take a point which is a vertex of the polygon model,then it’s surrounded by a disk. But this is easy to check, since all of the vertices which glue togetherwill have a little pizza slice of a disk and these pizza slices glue together to make a whole disk.

Edge Words

Imagine that someone calls you up on the phone and asks you to describe your favourite surface — howcan you do it? One way is to cut up your surface into triangles, glue the triangles together until they form apolygon, and then describe the resulting polygon and its gluing instructions — remember that this is called apolygon model. These gluing instructions are usually given by writing the same letter on each pair of sideswhich get glued together and placing arrows on the sides to show in which direction they get glued.

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So you can describe a surface by reading around its polygon model — usually in a counterclockwise fashion— calling out x if the side is labelled x and the arrow is facing in the right direction and calling out x−1 if theside is labelled x and the arrow is facing in the wrong direction. This sequence of letters, some with inversesand some without, is called an edge word for a surface. Note that one surface can be described by many manydifferent edge words. Hopefully it’s clear that once you tell someone an edge word to your favourite surface,then they can reconstruct the surface that you’re talking about.

Example. Let’s consider the Pac-Man world example from earlier, which we know to be the torus. Startingin the bottom-left corner and reading around counterclockwise, the first letter is a, the second is b, the third isa−1 and the fourth is b−1, which takes us back to where we started. So an edge word for the torus is aba−1b−1.

a

b

a

b

Example. Some other useful edge words to keep in mind are aa, which represents the projective plane, andaba−1b which represents the Klein bottle.

Playing with Edge Words

What we’d like to do is translate some of the constructions to do with surfaces into the world of edge words.This will allow us to play with surfaces by just playing with edge words — and that is a much simpler game,because edge words don’t involve any crazy pictures, imagining things in four-dimensional space, and so on.

First, let’s consider what happens to edge words when we take connected sums. We earlier discussed thatwhen you take connected sums, it doesn’t matter in the slightest where you decide to cut out your holes fromeach surface. So if we want to take the connected sum of two tori, for example, we could describe one byaba−1b−1, the other by cdc−1d−1, and decide to cut out holes from each one as shown in the diagram below.

a

b

a

b

c

d

c

d

If you open each of these holes up, the rectangles look more like pentagons, where four of the edges arelabelled and the unlabelled edges in each pentagon will get glued together. The result will be an octagonwhose edge word is simply aba−1b−1cdc−1d−1.

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a

a

b

b c

c

d

d

It is a simple matter to generalise this reasoning and the final result is the following fact.

Proposition. Given polygon models for the surfaces S1 and S2 with edge words W1 and W2, an edge word for theconnected sum S1 # S2 is W1W2, where the words W1 and W2 have simply been written one after the other.

If you analyse what happens when you slide a counterclockwise arrow around a polygon model, then you’llfind that nothing happens to the direction of the arrow unless you happen to slide over an edge and emergefrom the edge that it is glued to. In fact, the direction of the arrow doesn’t change unless these two edges arefacing in the same direction around the polygon. In other words, we have the following fact which allows usto deduce the orientability of a surface very simply from its edge word.

Proposition. In an edge word for an orientable surface, every letter appears with its inverse. In an edge word for anon-orientable surface, there exists at least one letter which does not appear with its inverse.

Finally, let’s consider how to calculate the Euler characteristic from an edge word, for example the edge wordabcdc−1b−1a−1d−1. Once you’ve seen how to calculate the Euler characteristic for this edge word, you’ll nodoubt be able to calculate the Euler characteristic for just about any edge word I give you.

d

a

b

c

d

c

b

a

1

12

3

1

1 3

2

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Above we have a diagram of the polygon model corresponding to the edge word. Of course, to work outthe Euler characteristic of the corresponding surface, we’d like to determine the number of vertices, edgesand faces. The easiest of these is the number of faces, since we always have F = 1 for a polygon model. Thenext easiest is the number of edges, since the eight edges will get glued into four pairs, so the resulting mapon the surface will have E = 4. And finally we can calculate the number of vertices as V = 3. We do thisby determining which vertices get matched up with each other after the gluing occurs. In the diagram, thethree different vertices are labelled 1, 2 and 3. The strategy for finding this labelling is to make sure that thetwo vertices at the tip of the edges labelled a correspond to the same vertex and the two vertices at the tailof the edges labelled a correspond to the same vertex. If you do this for every edge label, then you shouldfind that the vertices of the polygon divide into the three groups indicated, each group corresponding toone of the three vertices of the resulting map. Now that we have the number of vertices, edges and faces,it’s a simple matter to calculate χ = V − E + F = 3− 4 + 1 = 0. Furthermore, we know that this surface isorientable because, in the edge word abcdc−1b−1a−1d−1, every letter appears with its inverse.

The Classification of Surfaces

Finally, we’re in a position where we can classify the surfaces — that is, write a list of every different surfacepossible. A lot of effort by mathematicians goes into classifying mathematical objects. However, surfaces isone of the few examples where we’ve been successful and achieved a complete classification. In fact, this wasall worked out back in the early twentieth century.

Theorem (The classification of surfaces). Every surface is homeomorphic to the sphere, a connected sum of tori or aconnected sum of projective planes. Furthermore, no two of these surfaces are homeomorphic to each other.

Sketch of the proof. The proof of this theorem, although reasonably elementary, is rather involved. Rather thanget bogged down in all the gory details, I’ll just give an overview of the proof, which relies heavily on thefact that surfaces can be described by edge words and that every edge word describes a surface.

Take any edge word — our goal is to show that that it corresponds to a sphere, a connected sum of torior a connected sum of projective planes.

The idea is to apply simplifying moves to the edge word which do not change the surface. Such movescan be applied until the edge word can be recognised as the sphere or a connected sum of tori andprojective planes. The only moves you require to perform this task are the following — you shouldcheck that they do not change the surface.

– σ(aXa−1Y) ∼= σ(b−1XbY)

– σ(XY) ∼= σ(YX)

– σ(X) ∼= σ(X−1)

– σ(aa−1X) ∼= σ(X)

– σ(aXaY) ∼= P # σ(XY−1)

– σ(aWbXa−1Yb−1Z) ∼= T # σ(WZYX)

The idea is that a sequence of such moves allows you to break an edge word into a piece which is aprojective plane or a torus connected sum with a surface whose edge word is shorter. Here, I am usinga and b to represent letters, and W, X, Y, Z to represent words. Furthermore, P and T represent theprojective plane and the torus, respectively. And finally, the word X−1 is what you get if you read theword X backwards. For example, if X was aba−1cdc−1, then reading it backwards would give youcd−1c−1ab−1a−1. Here, we are using the notation σ(W) to denote the surface which corresponds to theedge word W.

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So you can use the aforementioned simplifying moves to reduce the surface to a connected sum of toriand projective planes. However, connected sums of tori and projective planes do not actually appear inour classification of surfaces. But that’s all right, because when you have a mix of the two appearing ina connected sum, you can replace the torus with two projective planes. Another way to say this is thatT # P ∼= P # P # P. The proof of this statement is something that you can try in the comfort of your ownhome — it’s simply equivalent to the statement σ(aba−1b−1cc) = σ(aabbcc).

These tricks can be used to reduce any edge word until it looks like it represents the sphere, a connectedsum of tori or a connected sum of projective planes. All that remains is to show that these are alldifferent from each other. However, this is quite easy, because we can use orientability and the Eulercharacteristic to tell all of these surfaces apart. For more details, you can consult the following usefultable.

SURFACE ORIENTABILITY EULER CHARACTERISTIC

sphere orientable 2connected sum of g tori orientable 2− 2gconnected sum of n projective planes non-orientable 2− n

So we now have a very useful corollary — it took a lot of work to get to this point in time, but we can nowsay that we essentially understand everything there is to understand about surfaces.

Corollary. In order to identify a surface, all you need are its orientability and its Euler characteristic.

Problems

Problem. Identify the surface corresponding to the edge word abcdc−1b−1a−1d−1.

Proof. We earlier calculated that this surface has Euler characteristic equal to zero and is orientable. Now wecan consult the table above and identify the surface as a connected sum of one torus — namely, a torus.

Problem. Identify the surface corresponding to the edge word abcdec−1d−1bae−1.

Proof. The following is a diagram of the polygonmodel, with the vertices of the polygon labelled 1and 2. All of the vertices labelled 1 will coincideafter the gluing is performed and all of the ver-tices labelled 2 will coincide after the gluing is per-formed. Therefore, the Euler characteristic is simplyχ = V − E + F = 2− 5 + 1 = −2.

It is easy to see from the edge word that the surface isnon-orientable, since the letter a does not appear withits inverse. Therefore, the classification of surfacestells us that it must be a connected sum of projectiveplanes. In fact, the Euler characteristic tells us that itmust be a connected sum of four projective planes,which we can write as P # P # P # P.

b

cd

e

c

d

b a

e

a

1

2

1

2

1

2

1

2

1

2

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Noether

Amalie Emmy Noether was a German mathemati-cian who was born into a Jewish family in 1882 anddied from surgery complications in 1935 at the ageof fifty-three. She is known for her groundbreakingcontributions to abstract algebra and to theoreticalphysics, despite many obstacles in her life. For exam-ple, her decision to attend the University of Erlangenwas unconventional since the Academic Senate of theuniversity had declared that allowing coeducationwould “overthrow all academic order”. She was oneof only two female students in a university of nearlyone thousand and was only allowed to audit classeswith the permission of each individual professor.

After completing her dissertation, Noether workedat the Mathematical Institute of Erlangen, but had todo so without pay for seven years. Finally in 1915,she was invited by the great mathematicians DavidHilbert and Felix Klein to join the mathematics de-partment at the University of Gottingen, a world-renowned centre of mathematical research. However,when the Nazis came to power in 1933, Noether wasforced to leave her job due to her Jewish background.She accepted the decision calmly and was fortunatelyoffered a job at Bryn Mawr College near Philadelphia,which provided a welcoming home for her duringthe last two years of her life.

Noether was highly respected for her teaching. Ap-parently, she did not follow a lesson plan for her lec-tures, which frustrated some students. Instead, sheused her lectures as a spontaneous discussion time,to think through and clarify important cutting-edgeproblems in mathematics. Once, when the mathemat-ics department was closed for a state holiday, shegathered the class on the steps outside, led themthrough the woods, and lectured at a local coffeehouse. Later, after she had been dismissed by theThird Reich, she invited students into her home todiscuss their future plans and mathematical concepts.In addition to her own publications, Noether was

generous with her ideas and is credited with severallines of research published by other mathematicians,even in fields far removed from her main work.

Noether was described by Albert Einstein and oth-ers as the most important woman in the history ofmathematics. She revolutionised abstract algebra anddiscovered Noether’s theorem in physics, which ex-plains the fundamental connection between symme-try and conservation laws — this tells us why wecan expect conversation of physical properties suchas energy, momentum and angular momentum. Hertheorem has been called “one of the most importantmathematical theorems ever proved in guiding thedevelopment of modern physics”.

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4. TILING AND DISSECTION 4.1. Tiling Rectangles

The Mutilated Chessboard

One of the most famous of tiling conundrums is the following, a problem which almost every mathematicianmust have encountered at one time or another.

Consider an 8× 8 chessboard, where the top-right and bottom-left squares have been removed. Isit possible to tile this mutilated chessboard with 2× 1 dominoes?

The first thing you should do is take out some pen and paper, draw a mutilated chessboard, and try to tile itwith 2× 1 dominoes. However, I can tell you right now that you’ll fail — not because your tiling skills areinadequate, but because the task is impossible. The answer to this problem should seem surprising to anunsuspecting audience.

Prior to removing the two squares, there is a myriad of ways to perform such a domino tiling — actually,36042 = 12988816 ways to be precise, but that’s another story. So why should such a trivial alteration of theboard reduce this number to zero? The argument is stunning in its simplicity and the key to the solution liesin the seemingly unimportant colouring of the chessboard into black and white squares. This colouring issuch that the placement of any domino on the board will cover exactly one square of each colour. So if it’spossible to tile the board with dominoes, then it must be the case that there is an equal number of black andwhite squares. However, a quick count reveals that the mutilated chessboard has 30 black squares and 32white squares. A slicker way to see that there are unequal numbers of black and white squares is to noticethat we removed two squares of the same colour from a board that previously had equal numbers of each.From this disparity, we are led to the conclusion that the mutilated chessboard cannot be tiled by dominoes,no matter how hard one might try.

Colouring Arguments

We were lucky with the mutilated chessboard problem, because the standard 8× 8 chessboard came with acolouring which helped our cause, free of charge. But sometimes, as in the next problem, you have to inventyour own colouring.

Is it possible to tile a 10× 10 square with 4× 1 rectangles?

Once again, try as you might, you’ll find that it’s impossible to tile a 10× 10 board with 4× 1 rectangles. So itseems like a good idea to generalise the colouring trick which worked for the mutilated chessboard so that itworks for this problem too. And if it works for this problem too, then who knows how many other problemsthis colouring trick will work for?

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The crucial aspect of the chessboard colouring that we used is the fact that any domino placed on the boardoccupied one square of each colour. So the idea here is to find a colouring of the 10× 10 square such thatany 4× 1 rectangle placed on the board occupies one square of each colour. Of course, this means that werequire four colours, which we will call 0, 1, 2 and 3. Working along the top row of the board, we may as welllabel the first four squares 0, 1, 2 and 3, in that order. In order to satisfy the property that any 4× 1 rectangleplaced on the board occupies one square of each colour, the next square along must be labelled 0. And thenext square along must be labelled 1, and the next square along must be labelled 2, and the next square alongmust be labelled 3, and so on. So we see that every square in the first row will be coloured according to therepeating pattern 0, 1, 2, 3, 0, 1, 2, 3, . . .. Since this seems to work along the first row, we can use the same trickto fill the first column. And after a little trial and error, you should find the following very pretty lookingcolouring of the 10× 10 square.

0 1 2 3 0 1 2 3 0 1

1 2 3 0 1 2 3 0 1 2

2 3 0 1 2 3 0 1 2 3

3 0 1 2 3 0 1 2 3 0

0 1 2 3 0 1 2 3 0 1

1 2 3 0 1 2 3 0 1 2

2 3 0 1 2 3 0 1 2 3

3 0 1 2 3 0 1 2 3 0

0 1 2 3 0 1 2 3 0 1

1 2 3 0 1 2 3 0 1 2

I like to call this a modulo 4 colouring, because if we label the rows and columns 0, 1, 2, . . ., then the square inrow i and column j is coloured i + j modulo 4. If you have no idea what I’m talking about, then that’s fine,because the colouring is easy to describe without all of this jargon. You simply cycle through the colours 0, 1,2, 3 along the first row, and every other row is the same as the previous one, but shifted to the left by one.Hopefully, I don’t need to tell you that for other problems, you might need to use a modulo k colouring forsome positive integer k.

This colouring certainly obeys the rule that a 4× 1 rectangle on the board always occupies one square of eachcolour. You can easily check this for a 4× 1 rectangle placed in the first row which means that it’s also truefor a 4× 1 rectangle placed in any other row or any column. Of course, we’re hoping that there are not thesame number of squares of each colour, so that we can deduce that it’s impossible to tile the board. One wayto verify this is to simply count them and you would indeed find that this is true — there are twenty-fivesquares coloured 0, twenty-six squares coloured 1, twenty-five squares coloured 2, and twenty-four squarescoloured 3. However, that’s rather pedestrian, so let’s use a slicker, more stylish, approach.

We simply note that it’s possible — and quite easy to demonstrate — a tiling of the entire board except for the2× 2 square in the top-left corner. This is because any 4× n rectangle is very easy to tile with 4× 1 rectangles.So you can tile the bottom four rows of the grid, leaving a 6× 10 rectangle. Then you can tile the bottom fourrows of this grid, leaving a 2× 10 rectangle. Now you can tile the rightmost four columns of this grid, leavinga 2× 6 rectangle. And then you can tile the rightmost four columns of this grid, leaving a 2× 2 square.

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The part of the board which we’ve covered in tiles must certainly contain the same number of squares ofeach colour, otherwise we wouldn’t have been able to tile it. Since the remaining part of the board does not —there is one square coloured 0, two squares coloured 1, one square coloured 2 and zero squares coloured 3 —there cannot be the same number of squares of each colour on the entire board. We conclude that a 10× 10square cannot be tiled with 4× 1 rectangles.

These colouring arguments are extremely useful and are most commonly applied with a modulo k colouringfor some positive integer k. Hopefully you can imagine what such a colouring might look like and how suchan argument might work, but if not, then we’ll see an example very soon. One thing to keep in mind is that acolouring argument can be used to prove that a tiling is impossible, but it can never ever be used to provethat a tiling is possible. Anyway, to prove to someone that a particular tiling is possible is usually easy —you just have to demonstrate it to them.

Tiling Rectangles with Skinny Rectangles

Since a colouring argument was so successful for the previous problem, we may as well try to solve thefollowing far more general problem.

For which values of m, n and k is it possible to tile an m× n rectangle with k× 1 rectangles?

If someone gives you a problem like this, the very first thing you should do with it is experiment with variousvalues of m, n and k. One thing you should realise very quickly is that if m is a multiple of k, then the tilingis really easy to find. And that’s because the first column consists of m squares and can be tiled with m

krectangles. Once you can tile the first column this way, then you can tile every column on the board in thisway. Similarly, the tiling is very easy to find if n is a multiple of k. But what happens if neither m nor n isa multiple of k? Well, you should find that the task is impossible, and that’s precisely what we’re going toprove.

Theorem. It’s possible to tile an m× n rectangle with k× 1 rectangles if and only if m is a multiple of k or n is amultiple of k.38

Proof. We’ve already shown that if m is a multiple of k or n is a multiple of k, then the tiling is easy to find. Solet’s now assume that m and n are not multiples of k and prove that the tiling is impossible.

This is where our coloured pencils come to the rescue. Hopefully, you haven’t forgotten the problem wesolved earlier about tiling with 4× 1 rectangles. We solved that one by using a colouring which repeats everyfour squares. For this problem, we simply use a colouring which repeats every k squares. The strategy hereis the same — fill out the first row by cycling through the colours 0, 1, 2, . . . , k− 1 and let every other rowbe the same as the previous row one, but shifted to the left by one. I just happen to be using the colours0, 1, 2, . . . , k− 1 because that’s what I’m used to. You’re more than welcome to use the colours 1, 2, 3, . . . , k oreven real colours — whatever takes your fancy.

Now if m is not a multiple of k, then it must leave some remainder a after you divide by k. Similarly, if n isnot a multiple of k, then it must leave some remainder b after you divide by k. And these numbers a and bcan’t be any old numbers — they must be positive integers which lie strictly between 0 and k.

38In mathematics, when we say A or B, we allow the possibility that both A and B could happen. It’s not like when you go to a friend’shouse and they ask you whether you want tea or coffee, in which case they are usually excluding the fact that you might want both.

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We’ll also make the assumption that a ≤ b which we certainly can do, because if it wasn’t true, then we couldjust switch the names of m and n and the names of a and b to make it true.39

Note that it’s possible — and quite easy to demon-strate — a tiling of the entire board except for thea × b square in the top-left corner. This is becauseyou can tile the bottom k rows of the grid, leaving an(m− k)× n rectangle. Then you can tile the bottomk rows of this grid, leaving an (m − 2k) × n rectan-gle. And you can keep tiling the bottom k rows ofthe grid, until you are left with an a × n rectangle.Now you can tile the rightmost k columns of thisgrid, leaving an a× (n− k) rectangle. Then you cantile the rightmost k columns of this grid, leaving ana× (n− 2k) rectangle. And you can keep tiling therightmost k rows of the grid, until you are left withan a× b rectangle.

a

k

k

b k k

m

n

......

...

· · ·

Remember that our goal is to show that there aren’t equal numbers of squares of each colour on the entireboard. The trick we’ve used here is to tile a large part of the board, which tells us that the tiled part definitelydoes have equal numbers of squares of each colour. In other words, we’ve reduced the problem to showingthat there aren’t equal numbers of squares of each colour in the a× b rectangle in the top-left corner.

All we have to do now is note the following two things.

Note that the colour a− 1 appears in the bottom-left corner. In fact, it has to appear in every row of thea× b rectangle. This is because it also appears in the square one up and one right of the bottom-leftcorner, and in the square one up and one right from that one, and in the square one up and one rightfrom that one, and so on. Since a ≤ b, this means that the colour a− 1 appears at least once in everyrow. The fact that b < k tells us that no colour can appear more than once in a row. So, in summary,colour a− 1 actually appears exactly a times, once in each row.

0 1 2 · · · b − 2 b − 1

1 2 3 · · · b − 1 b

2 3 4 · · · b b + 1

......

......

...

a − 2 a − 1 a · · ·

a − 1 a a + 1 · · ·

39Often, mathematicians will say that we can make an assumption of this sort without loss of generality. It basically means that theassumption is allowed, and that you are still covering all possible cases, even though it might not at first appear to be so. In fact, sucharguments are often referred to as WLOG arguments.

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Note that the colour 0 appears in the top-left corner but it definitely does not appear in the secondrow — and this is because the second row looks like 1, 2, 3, . . .. So if it did appear, then the second rowwould have at least k squares, clearly in contradiction of our assumption that b < k. This means thatthe colour 0 definitely does not appear in the second row and yet, we already discussed the fact that nocolour can appear more than once in a row. So, in summary, the colour 0 appears fewer than a times.

This tells us that the colour 0 and the colour a− 1 do not appear the same number of times on the entire board.And this is precisely what we wanted to prove, because we can now deduce that the tiling is impossible.

Tiling Rectangles with Rectangles

Thus far, we’ve considered only the case of tiling with skinny rectangles — in other words, those of the formk× 1. Let’s now broaden our horizons and consider the more general case of tiling with a× b rectangles,where a and b are positive integers. Of course, we can start by making the simplifying assumption that a andb have no common factors greater than 1, since other cases reduce to this after scaling the size of the tiles andthe board down. For example, if a and b were both even and we wanted to know whether an m× n rectanglecan be tiled with a× b rectangles, then this is the same problem as determining whether an m

2 × n2 rectangle

can be tiled with a2 × b

2 rectangles. Obviously, the question we would like to answer is the following.

For which values of m, n, a and b is it possible to tile an m× n rectangle with a× b rectangles?

Before we state the answer, let’s consider three instructive cases.

Can you tile a 12× 15 rectangle with 4× 7 rectangles?No, of course not, since the area of each tile does not divide the area of the board.

Can you tile a 17× 28 rectangle with 4× 7 rectangles?The answer is again in the negative, although for a more subtle reason. It turns out that 4× 7 rectanglescannot even be used to cover the first column of a 17× 28 rectangle. For if such a tiling is possible, wemust certainly be able to write the number 17 as a sum of 4’s and 7’s. A quick check shows that this isnot the case.

Can you tile an 18× 42 rectangle with 4× 7 rectangles?It is not actually possible to carry out this task. If you could, then you could certainly tile the 18× 42rectangle with 4× 1 rectangles, by tiling each 4× 7 rectangle with seven 4× 1 rectangles. But ourearlier result — which we proved using a colouring argument — tells us that you can’t tile an 18× 42rectangle with 4× 1 rectangles because neither 18 nor 42 are multiples of 4.

These arguments can be generalised to prove the following theorem, which gives a complete answer to ouroriginal problem.

Theorem. Let a and b be positive integers with no common factors greater than 1. A tiling of an m× n rectangle witha× b rectangles exists if and only if

both m and n can be written as a sum of a’s and b’s; and

either m or n is a multiple of a and either m or n is a multiple of b.

Faulty Tilings

Let’s now turn our attention to the following beautiful tiling problem.

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A 6× 6 square is tiled with 2× 1 dominoes. Prove that it’s possible to cut the board into twosmaller rectangles with a straight line which doesn’t pass through any of the dominoes.

Given a tiling, let’s call a line which cuts the board into two pieces and yet does not pass through any of thetiles a fault line. For example, the diagram below shows two tilings of a 5× 6 rectangle with dominoes, onewhich has a fault line and one which doesnt. This particular problem asserts that every possible dominotiling of the 6× 6 square must have a fault line.

In order to obtain a contradiction, let’s suppose that we have a domino tiling of the 6× 6 square which has nofault line. Consider any one of the ten potential fault lines — five horizontal and five vertical — and, withoutloss of generality, we may assume that it is vertical. Since our tiling has no fault line, at least one dominomust cross this vertical. However, it cannot be the only such domino, since otherwise, an odd number ofsquares would remain to the left of the line and this part of the board cannot be tiled with dominoes. So atleast two dominoes must cross the given vertical line. The same argument applies for all ten potential faultlines, so at least two dominoes must cross each of the ten potential fault lines. Since a domino may cross onlyone such line, we conclude that the tiling must involve at least 10× 2 = 20 dominoes. However, 20 dominoescover an area of 40 squares, more than the area of the board in question. This contradiction implies that everytiling of the 6× 6 square with dominoes must have a fault line.

Having solved this question, it’s only natural to ask the following more general question.

When can an m× n rectangle be tiled with a× b rectangles without any fault lines?

Despite first appearances, there is a natural answer to this problem as described by the following result.Interestingly enough, the case of tiling a 6× 6 rectangle with dominoes which had such an elegant proof, isthe only exception to the rule.

Theorem. Let a and b be positive integers with no common factors greater than 1. A faultless tiling of an m× nrectangle with a× b rectangles exists if and only if

both m and n can be written as a sum of a’s and b’s in such a way that at least one a and one b is used;

either m or n is a multiple of a, and either m or n is a multiple of b; and

for the case where the tiles are dominoes, the rectangle is not 6× 6.

More Mutilated Chessboards

We started with the mutilated chessboard problem — from such humble beginnings, we began our journeyinto the amazing world of tiling. The mutilated chessboard problem spawns a further interesting questionwhose answer is not quite so well-known.

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Which pairs of squares may be removed from the regular 8× 8 chessboard so that the remainingboard can be tiled with dominoes?

Of course, the colouring argument we used to solve the mutilated chessboard problem implies that any suchpair of squares must be of opposing colours. But if we remove two such squares, is it always possible to tilethe remaining board with dominoes? The answer is in the affirmative and the simplest proof requires us toconsider the chessboard as a labyrinth, as pictured below. This labyrinth is hardly the design that might beused for a hedge maze, since it not only has no entrance and exit, but also consists simply of a tour whichtraverses all of the 64 squares. All that’s required now is to note that the removal of two squares of oppositecolours divides the path now into two shorter paths, one of which may be empty. Furthermore, these twopaths are of even length, so it’s easy to tile them both.

Trominoes, Tetrominoes and Polyominoes

We’ll end with some interesting questions which can be solved with the help of colouring arguments andother tiling tricks. But first, we have to introduce trominoes, shapes which can be made by gluing togetherthree unit squares edge to edge. You should be able to see that there are essentially two distinct trominoes —one looks like a 3× 1 rectangle while the other looks like an L-shape. We already know which rectangles canbe tiled by 3× 1 rectangles, so it’s natural to ask which rectangles can be tiled by L-trominoes.

And once you’ve mastered trominoes, of course, you would move on to tetrominoes — shapes which can bemade by gluing together four unit squares edge to edge. Anyone who’s played the excellent computer gameTetris before will be well-acquainted with them, but even if you haven’t, you should be able to see that thereare essentially the five different types as shown in the diagram below. With these simple shapes, you have amyriad of tiling problems that you can try, such as the following.

Can the five tetrominoes tile a rectangle of area 20?

Can two copies of each of the five tetrominoes tile a rectangle of area 40?

For each tetromino, determine which rectangles can be tiled by them.

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And, of course, once you’ve mastered tetrominoes, there are many other shapes you can play with. In general,any shape which can be made by gluing together unit squares edge to edge is called a polyomino. For anypolyomino, you can try to determine which rectangles can be tiled by them.

Problems

Problem. The 8× 8 chessboard can be tiled with twenty-one 3× 1 rectangles and one 1× 1 square. Determine allpossible locations for the 1× 1 square and prove that these are the only ones possible.

Proof. Our approach will, of course, use a modulo 3 colouring like the one pictured in the diagram below.Recall that the great thing about this colouring is the fact that any 3× 1 rectangle placed on the board willcover precisely one square of each colour. This means that twenty-one 3× 1 rectangles must cover exactlytwenty-one squares of colour 0, twenty-one squares of colour 1 and twenty-one squares of colour 2. But youcan plainly see from the diagram that there are actually twenty-one squares of colour 0, twenty-two squaresof colour 1 and twenty-one squares of colour 2. What this means is that the 1× 1 square must definitely beon a square which has the colour 1. Unfortunately, it’s not true that if you put the 1× 1 square on a squarewhich has the colour 1 that you can actually tile the remainder of the board with twenty-one 3× 1 rectangles.

0 1 2 0 1 2 0 1

1 2 0 1 2 0 1 2

2 0 1 2 0 1 2 0

0 1 2 0 1 2 0 1

1 2 0 1 2 0 1 2

2 0 1 2 0 1 2 0

0 1 2 0 1 2 0 1

1 2 0 1 2 0 1 2

The trick here is to observe that there are actually two different modulo 3 colourings that we could have tried.In the previous colouring, each row is equal to the previous row, but shifted by one square to the left. Now

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we simply use a colouring where each row is equal to the previous row, but shifted by one square to theright. Again, the great thing about this colouring is the fact that any 3× 1 rectangle placed on the board willcover precisely one square of each colour. This means that twenty-one 3× 1 rectangles must cover exactlytwenty-one squares of colour 0, twenty-one squares of colour 1 and twenty-one squares of colour 2. But youcan plainly see from the diagram that there are actually twenty-two squares of colour 0, twenty-one squaresof colour 1 and twenty-one squares of colour 2. What this means is that the 1× 1 square must definitely beon a square which has the colour 0.

0 1 2 0 1 2 0 1

2 0 1 2 0 1 2 0

1 2 0 1 2 0 1 2

0 1 2 0 1 2 0 1

2 0 1 2 0 1 2 0

1 2 0 1 2 0 1 2

0 1 2 0 1 2 0 1

2 0 1 2 0 1 2 0

So what we’ve deduced is that the 1× 1 square must be on a square with the colour 1 in the first colouringand on a square with the colour 0 in the second colouring. And there are only four such squares on the 8× 8chessboard — the ones indicated in the diagram below.

I’ll leave it as an exercise for you to show that, whichever one of these squares you decide to put the 1× 1square on, you can tile the remainder of the chessboard with 3× 1 rectangles.

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Erdos

Paul Erdos was a Hungarian mathematician famousfor being incredibly prolific but also incredibly eccen-tric. He has published more mathematics papers thananyone else in history, even more so than Euler, al-though Euler published more pages. He wrote nearly1500 articles in his lifetime, in collaboration with over500 different people. This is due to Erdos’ philosophythat mathematics is a social activity.

Erdos was born in 1913 to Jewish parents who wereboth mathematics teachers. He learnt much fromthem as a child and supposedly, at the age of three,could calculate how many seconds his friends hadlived for. After receiving a doctorate in mathematicsat the age of twenty-one, he moved first to Englandand then to the United States, to escape the growinganti-Semitic sentiment in Europe. At this time, hebegan to develop the habit of travelling from campusto campus and staying with friends. He would typ-ically show up at a colleague’s doorstep, announcethat “my brain is open”, and stay long enough to col-laborate on a few papers before moving on a few dayslater. Possessions meant very little to Erdos and mostof his belongings would fit in a suitcase. Awards andother earnings were generally donated to people inneed and various worthy causes. He kept up thisvagabond lifestyle until his death in 1996.

His colleague Alfred Renyi once said that “a math-ematician is a machine for turning coffee into theo-rems”. Erdos certainly drank copious amounts of cof-fee but later in life also started to take amphetamines.At one stage, a friend and colleague bet him $500 thathe couldn’t stop taking the drug for a month. Erdoswon the bet, but complained during his abstinencethat mathematics had been set back by a month: “Be-fore, when I looked at a piece of blank paper my mindwas filled with ideas. Now all I see is a blank pieceof paper.” Needless to say, after he won the bet, hepromptly resumed his amphetamine habit.

Erdos had his own idiosyncratic vocabulary — hespoke of “The Book”, an imaginary book in whichGod had written down the most elegant proofs forevery mathematical theorem. Children were referred

to as “epsilons”, women were “bosses”, men were“slaves”, people who stopped doing math had “died”,people who physically died had “left”, alcoholicdrinks were “poison”, music was “noise”, peoplewho had married were “captured”, people who haddivorced were “liberated”, to give a mathematicallecture was “to preach”, and to give an oral exam toa student was “to torture”. For his epitaph he sug-gested, “I’ve finally stopped getting dumber.”

He contributed to many areas of mathematics — mostnotably combinatorics, graph theory, number theory,analysis, approximation theory, set theory, and proba-bility theory. As a teenager, Erdos managed to give avery nice proof of Bertrand’s postulate, which statesthat there is always a prime number between n and2n. He discovered the first elementary proof of theprime number theorem, which states that the numberof primes less than n is approximately n

log n . Erdosalso proved new results in several fields which wereof little interest to him, such as topology.

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Erdos is perhaps most well-known for his applicationof the probabilistic method to extremal combinatorics,particularly Ramsey theory. Ramsey theory is thebranch of mathematics concerned with problems ofthe following type.

How many people do you need at a partyto guarantee that there exist at least n peo-ple who all know each other or n peoplewho all don’t know each other?

We’ve already seen that the answer in the case n = 3is six, and we write this as R(3) = 6. It turns out thatR(4) = 18 and that the value of R(5) is known onlyto be between 43 and 49 inclusive. These so-calledRamsey numbers are incredibly difficult compute, asevidenced by the following story that Erdos used totell. Imagine that a large alien force, vastly more pow-erful than us, lands on Earth and asks for the value ofR(5) within a year or they will destroy our planet. Inthat case, we should gather together all of our human-

power and technology to try and find the value. Onthe other hand, if they ask for R(6) instead, then weshould simply gather together all of our humanpowerand technology to try and launch a preemptive attackon the aliens.

Erdos’ friends created a humorous tribute for him,defining the Erdos number of a mathematician. Erdoshimself is the only mathematician with Erdos numberzero. Anyone who has written a paper with him hasErdos number one, anyone who has written a paperwith someone who has written a paper with him hasErdos number two, and so on. Some have estimatedthat ninety percent of the world’s active mathemati-cians have an Erdos number smaller than eight. Atleast twice, there have been eBay auctions offeringthe chance to collaborate on a paper with someone inorder to gain a small Erdos number. I have an Erdosnumber of two and, unfortunately, it will never everdecrease.

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4. TILING AND DISSECTION 4.2. Scissors Congruence

What is Area?

Despite being such a fundamental notion of geometry, the concept of area is very difficult to define. Ashuman beings, we have an intuitive grasp of the idea which suffices for our everyday lives. However, asmathematicians, we’ve only relatively recently been able to identify what we actually mean by the word.Efforts to define and generalise the notions of length, area and volume have led to the development of thebranch of mathematics known today as measure theory.

Part of the difficulty in defining area lies in the fact that subsets of the plane can be quite wild in comparisonto the objects we encounter in the physical world. In particular, there exist certain sets for which we cannotascribe an area without disturbing one or more of the fundamental premises that govern measure theory.These so-called non-measurable sets make any efforts to characterise area rather complicated. To avoid suchintricacies, let’s propose the more modest task of defining area for polygons in the plane.

A seemingly elementary approach to the problem is to use the method taught to children in school. Thisinvolves drawing the shape on graph paper and then counting the number of squares which lie within thefigure. This, of course, gives a lower bound for the area, while counting the squares which contain some partof the shape in question will yield an upper bound. The idea now is to consider graph paper with smallerand smaller squares, thus giving more and more accurate lower and upper bounds on the area in the hopethat they will converge to the same number. It’s this number which is defined to be the area. It turns out thatthis definition of area, although unsatisfactory for more complicated subsets of the plane, is well-defined forevery polygon in the plane. In this way, we have constructed the area function which, given a polygon P,returns a real number a(P).

It’s quite unfortunate that this definition of area, even for polygons, should involve infinite processes andcontinuity arguments. Could we perhaps find an alternative definition, one which requires purely elementarytechniques? A common trend in contemporary mathematics is to define an object by the properties which itsatisfies. For example, we may like to extract enough properties of the area function defined above that therecan be only one such function which satisfies them all. In fact, the four properties listed below perform sucha task and we present them as the area axioms.

Non-negative. If P is a polygon, then a(P) ≥ 0.

Additive. If P1 and P2 are polygons with no interior points in common, then a(P1 ∪ P2) = a(P1) + a(P2).

Invariant. If P is a polygon and f is an isometry, then a(P) = a( f (P)).

Normalised. If S is the square with sides of length one, then a(S) = 1.

I think you’d have to agree that any notion of area must satisfy these four axioms. A careful look at themleads us naturally to the following definition. We say that two polygons are scissors congruent if one can becut into finitely many polygons which can be rearranged to give the other. In other words, scissors congruentpolygons are common solutions to a jigsaw puzzle with the same set of pieces. Let’s denote the fact thatpolygons P and Q are scissors congruent by P ∼ Q. Also, note that we can extend the definition to unions ofpolygons with no interior points in common. In this case, we denote the union by P1 + P2 + · · ·+ Pn, whereP1, P2, . . . , Pn denote the individual polygons. The notion of scissors congruence, first for polygons, and thenfor polyhedra, will be our main topic of investigation.

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Scissors Congruence in the Plane

Our exploration into the areas of polygons has naturally led us to the notion of scissors congruence. It’sa simple consequence of the area axioms that if two polygons are scissors congruent, then they have thesame area. Now it’s only natural to ask the converse — if two polygons have the same area, then are theynecessarily scissors congruent?

This question was answered in the affirmative by Bolyai in 1832 and independently by Gerwien one year later.Before embarking on the proof, let’s begin by presenting a few important lemmas on scissors congruence.

Lemma. Scissors congruence is an equivalence relation, which means that

for all polygons P, P ∼ P;

if P ∼ Q, then Q ∼ P; and

if P ∼ Q and Q ∼ R, then P ∼ R.

Proof. The first two statements are immediately evident from the definition of scissors congruence. For thethird statement, suppose that we trace out the cuts required to decompose Q into pieces which rearrange togive P as well as the cuts required to decompose Q into pieces which rearrange to give R. Then cutting alongall of these lines will yield a finite set of pieces which can be rearranged to produce either P or R. It followsthat P and R are scissors congruent.

Lemma. Every triangle is scissors congruent with somerectangle.

Proof. Cut along the altitude to the longest side of thetriangle as well as along the perpendicular bisectorof this altitude. This will decompose the triangle intotwo triangles and two quadrilaterals which can berearranged to give a rectangle with the same base andhalf the height of the triangle.

Lemma. Any two rectangles with the same area are scissors congruent.

Proof. Place the two rectangles in the plane so that they have one vertex in common, with two adjacent sidesaligned as shown in the diagram. Cutting one of the rectangles along the lines shown produces a pentagon, aquadrilateral, a large triangle and a small triangle which can be rearranged to give the other rectangle.

The astute reader may have noticed that this particu-lar decomposition may not always work if one of therectangles is “too long”. More explicitly, suppose thatthe rectangles have dimensions `1 × h1 and `2 × h2

where we may assume without loss of generality thath1 < h2 ≤ `2 < `1. Then the construction works onlywhen `1 ≤ 2`2. However, this can be arranged byrepeatedly cutting the longer rectangle in half andstacking the two pieces on top of each other until thecondition `1 ≤ 2`2 is satisfied.

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With these three lemmas under our metaphorical belts, we’re now ready to prove the Bolyai–GerwienTheorem.

Theorem. Two polygons are scissors congruent if and only if they have the same area.

Proof. First, note that any polygon P can be cut into finitely many triangles. Furthermore, each of thesetriangles is scissors congruent to some rectangle and each of these rectangles is scissors congruent to arectangle, one of whose sides has length one. Therefore, we can write

P ∼ R1 + R2 + · · ·+ Rn,

where each Ri is a rectangle with one side of length one.

Concatenating these rectangles together produces a single rectangle R which has one side of length one. Ofnecessity, the other side of R must have length equal to the area of P. Therefore, if P and Q have the samearea, they are both scissors congruent to the rectangle R, so they are scissors congruent to each other.

Philosophically, the Bolyai–Gerwien Theorem tells us that it’s possible to see whether two polygons have thesame area or not using a finite process of cutting and pasting. This is mathematically more elementary —and hence, more satisfactory — than the usual definition of area which requires an infinite process involvinggraph paper with smaller and smaller squares.

What is Volume?

It’s simple enough to develop a theory of volume for polyhedra in an analogous manner to the theory ofarea for polygons. The area axioms transform naturally into volume axioms to give a rigorous definitionof the volume of a polyhedron. However, it had been noted by Gauss that proofs for the volume of atetrahedron had all used in some way or another infinite processes and continuity arguments, rather thanentirely elementary methods. Such an elementary proof would require that polyhedra with the same volumebe scissors congruent — if two polyhedra have the same volume, then are they necessarily scissors congruent?

Hilbert considered this question of such importance that he included it in his famous address to the In-ternational Congress of Mathematicians at Paris in 1900. As is well known, his address included a listof twenty-three unsolved problems in mathematics which he considered to be of great significance. Thisquestion of scissors congruence in three dimensions was the third on his list.

It’s clear from Hilbert’s exposition on the matter that he didn’t expect the Bolyai–Gerwien theorem to carryover from polygons in the plane to polyhedra in space — and he was exactly right. Hilbert’s third problemwas answered by his own student Max Dehn in 1900, the very year in which Hilbert had announced his listof problems. The crux of the proof lies in constructing ingenious invariants which have since been namedafter Dehn. Using these, Dehn managed to prove that the cube and the regular tetrahedron of unit volumeare not scissors congruent. Philosophically, Dehn’s result tells us that it’s impossible to see whether twopolyhedra have the same volume or not using a finite process of cutting and pasting.

Cantor and Cardinality

Naively, it might seem that counting and measuring are two of the simplest and most fundamental conceptsin mathematics. On the other hand, if you examine them closely, you’ll find that exploring these two areas

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can throw up a lot of seemingly nonsensical but mathematically true statements.

When you count the number of fingers of your left hand, you’re just finding a bijection — that is, a one-to-onecorrespondence — from the fingers to the set {1, 2, 3, 4, 5}. If you wanted to convince someone that you hadthe same number of fingers on each hand, then you could simply count the number of fingers on your lefthand and your right hand and find that the two answers are the same. Of course, this means that you wouldessentially be finding a bijection from the fingers of your left hand to the set {1, 2, 3, 4, 5} and a bijection fromthe fingers of your right hand to the same set. Obviously, a far more economical way to perform the sametask is to simply line the fingers from your hands together, so that every finger from one hand matches upwith a unique finger from the other. In other words, you can convince someone that you have the samenumber of fingers on each hand by finding a bijection between the two sets. Therefore, we say that two setshave the same cardinality — in other words, the same size — if there exists a bijection from one to the other. Itturns out that not only is this a sensible definition, but it’s essentially the only sensible one possible.

This definition for cardinality was first proposed by Georg Cantor in the nineteenth century and used todemonstrate a number of pretty counter-intuitive facts.

The set of positive even integers has the same size as the set of positive integers.

The set of integers has the same size as the set of positive integers.

The set of rational numbers has the same size as the set of positive integers.

The set of numbers in the interval [0, 1] has size greater than the set of positive integers. There is afamous and beautiful proof of this fact which uses a trick known as Cantor’s diagonal argument.

The set of numbers in the interval [0, 1] has the same size as the set of points in a square with sides oflength one.

More Nonsense

Peano curves.Cantor’s work shows that the set of numbers in the interval [0, 1] has the same size as the set of pointsin a square with sides of length one. However, the usual bijection used to prove this fact is a rathercrazy function with no nice properties. In the late nineteenth century, Giuseppe Peano managed tofind a function from the interval [0, 1] to the set of points in a square with sides of length one which issurjective — in other words, contained every point of the square in its range — and also continuous.This is an amazing fact, because it means that it’s possible to draw a curve which passes through everypoint in the square at least once. One of the easiest ways to describe such a curve is as the limit of thefollowing sequence of diagrams.

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Continuum hypothesis.Cantor managed to find many infinite sets which have the same size as the set of positive integers. Wesay that these sets have cardinality ℵ0, which is pronounced “aleph-nought”. He also managed to findmany infinite sets which have the same size as the set of numbers in the interval [0, 1]. We say that thesesets have cardinality c, which stands for “continuum”. In fact, Cantor found an infinity of infinities byshowing that for any set with cardinality x, there is a set with cardinality greater than x. On the otherhand, his search didn’t turn up any sets whose cardinality was between ℵ0 and c, which led Cantor toconjecture that none existed. This conjecture is now known as the continuum hypothesis.

Rather amazingly, due to work by Kurt Godel in 1940 and Paul Cohen in 1963, we now know thatthe continuum hypothesis can neither be proved nor disproved using the usual axioms — known asZermelo–Fraenkel set theory — which form the foundation of modern mathematics. In other words, wemay be able to add the continuum hypothesis or its opposite to our axioms and obtain a consistentsystem of mathematics. For this piece of work, Cohen was awarded the Fields Medal, often viewed asthe top honour a mathematician can receive.

Banach–Tarski paradox.In 1924, Stefan Banach and Alfred Tarski showed that it’s possible to split a solid sphere into a finitenumber of non-overlapping pieces which can then be put back together in a different way to yield twoidentical copies of the original sphere. The reassembly process involves only moving the pieces aroundusing isometries. However, the pieces themselves are complicated — they are not usual solid figuresbut look more like infinite scatterings of points.

The Banach–Tarski theorem is often referred to as a paradox because it contradicts basic geometricintuition. Doubling the sphere by dividing it into parts and moving them around by isometries —without any stretching, bending, or adding new points — seems to be impossible, since all theseoperations preserve volume. This paradoxical decomposition shows that the notion of volume certainlycan’t apply to arbitrary sets of points in space.

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Hilbert

David Hilbert was a German mathematician wholived from 1862 to 1943 and is recognised as one ofthe most influential and universal mathematicians ofhis day. He discovered and developed a broad rangeof fundamental ideas in mathematics and establishedrigour in much of the mathematics used in modernphysics. Hilbert famuosly presented a list of twenty-three unsolved problems at the 1900 InternationalCongress of Mathematicians in Paris. This is gener-ally considered to be the most successful and deeplyconsidered compilation of open problems ever to beproduced by an individual mathematician.

Hilbert managed to put geometry on far more rig-orous foundations than Euclid did. He proposed amuch larger set of axioms which avoided the weak-nesses which had been identified in the work of Eu-clid. His approach signalled the shift to the mod-ern axiomatic method which is now considered fun-damental to mathematics. The main idea is that al-though geometry may treat things which we havepowerful intuitions about, it’s not necessary to as-sign any explicit meaning to the undefined concepts.For example, points, lines, planes and other conceptscould be substituted, as Hilbert says, by tables, chairs,glasses of beer and other such objects.

In 1920, Hilbert proposed a grand research project toreformulate all of mathematics on a complete logicalfoundation. Hilbert dreamed that all of mathematics

could follow mechanically from a finite set of axiomsand that one could prove that this system containedno contradictions. Unfortunately, this program wasdoomed to fail because Kurt Godel remarkably man-aged to prove that Hilbert’s dream was impossible.

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5. FURTHER READING

Euclidean Geometry

The Elements by EuclidThis is one of the most published and most influential works in the history of humankind. In our veryfirst lecture, we looked at a small part of Book I from Euclid’s Elements, with the main goal being tounderstand the philosophy behind Euclid’s work. In my opinion, The Elements is an incredibly boringread, and there are many better ways to learn mathematics. Still, you might like to look at a copy to geta bit more of a feel for what Euclid was doing. There are many decent translations around, but I like touse the bilingual edition by Richard Fitzpatrick which is available for free on the internet.

http://farside.ph.utexas.edu/euclid.html

Encyclopedia of Triangle CentersThis website catalogues over 3500 points associated to a triangle. Of course, the circumcentre, orthocen-tre, incentre and centroid are the most important, but some of these others are quite meritorious as well.The website doesn’t make for a particularly interesting read, but should definitely convince you thatthere’s more than meets the eye when it comes to triangles.


The Geometer’s SketchpadThis is a program for drawing Euclidean geometry diagrams. I haven’t used it very much, but fromwhat I’ve seen, it does a fine job. You can find some information about it at the first link below and theprogram itself at the second link below.

http://www.keypress.com/x5521.xml

http://www.themathlab.com/toolbox/geometrystuff/geosketch.htm

GeoGebraThis is another program for drawing Euclidean geometry diagrams which I quite like. Unlike TheGeometer’s Sketchpad, the full program is available for free from the following website. Programs likethis may be particularly useful for those who want to teach mathematics.

http://www.geogebra.org

Geometry Revisited by H. S. M. Coxeter and Samuel L. GreitzerThis is supposedly a classic book which touches many different topics in Euclidean geometry. Peoplewho love Euclidean geometry seem to love this book, although I’m not a particular fan.

Episodes in Nineteenth and Twentieth Century Euclidean Geometry by Ross HonsbergerEuclidean geometry is, in some sense, a lost art. There are very few people who seem to care about itthese days, although the few who do seem to keep on coming up with new gems. This book collectstogether many of these, accompanied with beautiful proofs.

Symmetry in Geometry

Groups and Symmetry by M. A. ArmstrongIt was such a long time ago when I learnt group theory that I don’t remember whether I learnt it from abook or not. Anyway, this is a nice, relatively simple, introduction to group theory that you might liketo look at. It’ll show you a lot more about group theory than we covered in the course. It’s part of theUndergraduate Texts in Mathematics series which generally has books which are easy to digest.

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http://farside.ph.utexas.edu/euclid.html


http://www.keypress.com/x5521.xml

http://www.themathlab.com/toolbox/geometry stuff/geosketch.htm

http://www.geogebra.org

5. FURTHER READING

Group Theory in the Bedroom, and Other Mathematical Diversions by Brian HayesA nice little taste of group theory can be found in a very easy-to-read article Group Theory in the Bedroomwhich appears in American Scientist. The article can be found at the following website.

http://www.americanscientist.org/issues/pub/group-theory-in-the-bedroom

However, it also appears in this collection of articles, all by the same author. I haven’t had a chance tolook at the book, but I’m hoping that it’s pretty good.

Polyhedra, Graphs and Surfaces

Graphs and Their Uses by Oystein OreFor things which are as simple as dots and lines, graphs can be very complicated things to study. Thereis a wealth of literature on graph theory, but one of the simplest books to read is this one. I haven’tlooked at it in a while, but I remember it being quite a good read, despite being a bit old-fashioned instyle.

Introduction to Graph Theory by Robin J. WilsonAlthough an introduction to graph theory, this book is more technical than Graphs and Their Uses. Iquite like it though, and think that it’s a good book to learn graph theory from.

Flatland: A Romance of Many Dimensions by A. Square (otherwise known as Edwin A. Abbott)Looking for a book which offers pointed observations on the social hierarchy of Victorian culture whileexploring the concept of living in a two-dimensional world? Well, this is the book for you! This is abrief, though very entertaining, novella and, in my opinion, one of the best books that you can buy for$2.75.

The Shape of Space by Jeffrey R. WeeksThis book deals with some very interesting, crazy, modern geometry. It talks about surfaces, which youshould already know about, but also about higher-dimensional versions of surfaces. There are manypretty pictures and the book should help you to visualise geometry much better. I highly recommendthis for anyone who is interested in continuing with pure mathematics or theoretical physics.

Tiling and Dissection

Tilings and Patterns by Branko Grunbaum and G.C. ShephardAt a whopping 720 pages, this is a rather lengthy tome dedicated almost completely to tiling problems.You might like to pick it up and have a browse through it. But I really just wanted to point it out toconvince you that there is a lot to say about tiling, a lot more than we’ve learnt about.

Polyominoes: Puzzles, Patterns, Problems, and Packings by Solomon W. Golomb and Warren LushbaughDominoes, trominoes, tetrominoes — these are all just simple examples of what are generally calledpolyominoes. This book talks all about them and shouldn’t be too difficult to read.

Polyominoes: A Guide to Puzzles and Problems in Tiling by George MartinHere’s another book which deals solely with polyominoes. I’ve only browsed this book briefly and itseems to contain some interesting information.

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http://www.americanscientist.org/issues/pub/group-theory-in-the-bedroom

5. FURTHER READING

Mathematicians

The Man Who Loved Only Numbers: The Story of Paul Erdos and the Search for Mathematical Truth by PaulHoffmanIn this book, you’ll learn about the life, the mathematics and the eccentricities of Paul Erdos. Actually, thebook is very light on mathematics, but it does give you some idea of what the world of mathematiciansis like. Erdos is a pretty crazy character, so anything you read about him is likely to be entertaining.

N is a Number: A Portrait of Paul Erdos by George Paul CsicseryThis one hour documentary was made not too long before Erdos passed away. It’s pretty entertainingto watch, if you can get your hands on it.

The Man Who Knew Infinity: A Life of the Genius Ramanujan by Robert KanigelRamanujan was one of the most remarkable mathematicians who ever lived. In his brief thirty-twoyears on earth, he left us with thousands of results, mostly without proof. People are still workingtoday on deciphering his work and trying to understand how he could possibly have come up with it.This book tells the story of his life and his mathematics.

Surely You’re Joking, Mr. Feynman by Richard P. FeynmanI think that everyone should have this book. It’s essentially a very readable collection of anecdotesabout the physicist and Nobel laureate Richard Feynman. Mathematicians and scientists should bedriven by curiosity and reading this book, you’ll see what a crazily curious character Feynman is.

The MacTutor History of Mathematics ArchiveIf you want to know more about mathematicians and the history of mathematics, then this is thewebsite for you. If you head into the Biographies Index part of the website, then you’ll see a long list ofmathematicians in alaphebtical order, each of which is accompanied by a short biography.

http://www.gap-system.org/~history

Men of Mathematics by E. T. BellThis book is terribly outdated, as you can tell from the rather sexist title. However, it is a classic book toread about the lives of great mathematicians. Many spurious stories about mathematicians have beenpropagated by this book, but we can forgive the author for wanting to add a few embellishments.

Mathematics Genealogy ProjectThis website is dedicated to keeping a register of people’s mathematical ancestry. Here, I mean that amathematician is the parent of another if they were their graduate advisor. You can do things like tracemy lineage all the way back to such mathematical greats as Gauss and Euler.

http://www.genealogy.ams.org

Other Stuff

Godel, Escher, Bach: An Eternal Golden Braid by Douglas R. HofstadterThis is a fantastic book, dealing with mathematics, music, art, computer science, cognitive science,language, symmetry and so on. The book’s chapters alternate between non-fictional prose and fictionaldialogues which are used to demonstrate various ideas. For an example, read the Crab Canon, whichcan be found at the following website.

http://www.evl.uic.edu/swami/crabcanon

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http://www.gap-system.org/~history

http://www.genealogy.ams.org

http://www.evl.uic.edu/swami/crabcanon

5. FURTHER READING

QED: The Strange Theory of Light and Matter by Richard P. FeynmanIf you think you are interested in physics at all — even if you’ve studied it and found it incrediblyboring — then you should read this book. Feynman was one of the greatest and most entertainingscientists of the twentieth century. This book is based on four lectures delivered to the public andcaptures what physics is about more than any other book I’ve read. It’s amazing how much physicsFeynman can explain to a layperson in such a short book. If you actually want to see the lecturesthemselves, then you can find them online at the following website.

http://vega.org.uk/video/subseries/8

LATEXThis is the mathematical typesetting software which has become standard in the mathematical worldand is spreading to other areas of science and engineering. If you’ve ever wondered how I’ve mademy notes so pretty, or why Microsoft Word is so annoying for mathematics, then you should definitelycheck this out. If you want to continue in mathematics or physics, then you will definitely need it andif you enter into mathematics education, then I would definitely encourage you to use it. LATEXwascreated by Donald Knuth, one of the most famous computer scientists alive today. Two of the hardestthings about learning to use LATEXare installing it on your computer and getting started. There’s a heapof information on the internet which will help you with both, but if in doubt, I’m more than happy tohelp you.

The On-Line Encyclopedia of Integer SequencesSay you come across a sequence of integers and want to know some more information about it. Thenthe first thing you should do is go to this website and type it into the search engine.

http://www.research.att.com/~njas/sequences/

Or say you want to know how many graphs there are with 11 vertices? Just go to this website and it’lltell you the answer.

The Colossal Book of Mathematics: Classic Puzzles, Paradoxes, and Problems by Martin GardnerFrom 1956 to 1981, Martin Gardner wrote a column entitled Mathematical Games in Scientific American.His efforts played a great part in renewing and sustaining interest in recreational mathematics, whateverthat might mean. This book is a collection of some of his columns and you should find it prettyinteresting.

Mathellaneous (The Gazette of the Australian Mathematical Society) by Norman DoI have a collection of my own articles, slightly less recreational in nature than those of Martin Gardner,which you might like to peruse. These were all published as part of the Mathellaneous column whichfeatured in The Gazette of the Australian Mathematical Society.

http://www.math.mcgill.ca/ndo/articles.html

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http://vega.org.uk/video/subseries/8

http://www.research.att.com/~njas/sequences/

http://www.math.mcgill.ca/ndo/articles.html

6. ASSIGNMENTS, TESTS AND EXAMINATIONS Assignment 1

Release date: 1:30pm on Tuesday 4 May, 2010Due date: 11:00am on Tuesday 11 May, 2010

You are not discouraged from talking about assignment problems with other students, but every solutionthat you hand in must be your own work. Every page submitted should clearly indicate your name, studentnumber, the course number, and the assignment number. Late assignments will not be accepted, unlessunder particularly extreme circumstances.

Problems

1. Let ABC be a triangle with circumcentre O and incentre I. If O and I are the same point, prove that thetriangle must be equilateral.

[1 mark]

2. Let ABC be a triangle with circumcentre O and orthocentre H. Prove that ∠ABH = ∠CBO.[2 marks]

3. Let ABC be an acute triangle with altitudes AD, BE, CF.

(a) Prove that triangle AEF is similar to triangle ABC.

(b) Extend AD until it meets the circumcircle of triangle ABC at X and extend BE until it meets thecircumcircle of triangle ABC at Y. Prove that CX = CY.

[2 marks]

4. In the lectures, we tried to prove the following fact. If AB is a chord of a circle with centre O and Cis a point on the circle on the same side of AB as O, then ∠AOB = 2∠ACB. Unfortunately, our proofwas incomplete because we only considered the case when O lies inside triangle ABC. Prove that thestatement remains true in the case that O lies on one of the sides of triangle ABC and in the case that Olies outside triangle ABC.

[3 marks]

5. Extend the median AX of triangle ABC until it meets the circumcircle of triangle ABC at K. There is aunique circle which passes through A and B and is tangent to the line BC. If this circle meets AK at L,prove that triangle LBX and triangle KCX are congruent to each other. Hence, deduce that BKCL is aparallelogram.

[3 marks]

A

B CX

K

L

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6. Suppose that the circles C1 and C2 intersect at two points A and B and that C2 passes through the centreof C1, which we denote by O. Let X be a point on the arc AB which contains O and extend AX until itintersects C1 at Y. Prove that XY = XB.

[3 marks]

A

B

X

Y O

C1

C2

7. (a) In the lectures, we proved that the diameter of a circle subtends an angle of 90◦. Prove the converseof this statement — in other words, prove that if ABC is a triangle with ∠ACB = 90◦, then thecircle with diameter AB passes through C.

(b) Let ABCD be a cyclic quadrilateral such that the diagonals AC and BD are perpendicular andintersect at the point P. Let M be the midpoint of AB and let N be the point on CD such that PNis perpendicular to CD. Prove that the points M, P, N lie on a line.

[4 marks]

8. In the hyperbolic plane, there exist pentagons all of whose angles are equal to 90◦. Sketch one exampleof such a pentagon in the hyperbolic plane using the Poincare disk model. What is the area of thispentagon?

[2 marks]

9. Let D, E, F be points on sides AB, BC, CA of triangle ABC, respectively. If DE = BE and FE = CE,prove that the circumcentre of triangle ADF lies on the angle bisector of ∠DEF.

[BONUS]

Hints

1. Draw in the line segments AI, BI, CI or equivalently, AO, BO, CO. What is special about the lengths in the diagram? What isspecial about the angles in the diagram?

2. We already know how to work out all the angles in the diagram with triangle ABC and circumcentre O. We also already knowhow to work out all the angles in the diagram with triangle ABC and orthocentre H.

3. (a) Altitudes lead to right angles and right angles often lead to cyclic quadrilaterals. You should be able to use a cyclicquadrilateral or two to obtain some useful relations between angles.

(b) Equal chords in a circle subtend equal angles — but remember that equal angles are also subtended by equal chords.

4. Try to mimic the case that we covered in lectures as much as possible.

5. Since the problem involves a tangent, you should try to use the alternate segment theorem. Use this to prove that ∠LBX = ∠KCX.

6. How would you prove that triangle XYB is isosceles?

7. (a) Suppose that C lies outside the circle with diameter AB — can you find a contradiction?

(b) You should use the result from part (a), even if you are unable to prove it.

8. Do you remember how we found the sum of the angles in a polygon with n sides?

9. Do you really expect hints for a bonus question?

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6. ASSIGNMENTS, TESTS AND EXAMINATIONS Assignment 1 Solutions

1. Let ABC be a triangle with circumcentre O and incentre I. If O and I are the same point, prove that thetriangle must be equilateral.

[1 mark]

Proof. Let the angles of triangle ABC be 2a, 2b, 2c and recall that I lies on the angle bisectors. Thismeans that ∠BAI = ∠CAI = a, ∠CBI = ∠ABI = b, and ∠ACI = ∠BCI = c.

A

B C

O

A

B C

I

a a

bb cc

If O and I are the same point, then AI = BI = CI and this means that triangle ABI is isosceles. Hence,∠BAI = ∠ABI from which it follows that a = b. We can use the same argument to prove that b = c, soall angles of triangle ABC are equal. Therefore, triangle ABC is equilateral.

2. Let ABC be a triangle with circumcentre O and orthocentre H. Prove that ∠ABH = ∠CBO.[2 marks]

Proof. This problem is easy, because we already know that it’s possible to label every angle in thediagram involving triangle ABC and the circumcentre using only ∠CAB = a, ∠ABC = b and ∠BCA =c. We also know that it’s possible to label every angle in the diagram involving triangle ABC and theorthocentre using the same angles

A

B C

O

A

B C

H

E

First, note that BC is a chord of the circumcircle of triangle ABC. This means that the angle thatit subtends at O is twice the angle that it subtends at A. In other words, ∠BOC = 2∠BAC = 2a.Now we use the fact that triangle OBC is isosceles with ∠CBO = ∠BCO. If we sum up the anglesin triangle OBC, we obtain ∠CBO + ∠BCO + ∠BOC = 2∠CBO + 2a = 180◦ and this implies that∠CBO = 90◦ − a.

Next, note that if BE is an altitude of triangle ABC, then triangle ABE must be right-angled. Therefore,∠ABH = ∠ABE = 180◦ − ∠BAE − ∠AEB = 180◦ − a − 90◦ = 90◦ − a. In conclusion, we have∠ABH = ∠CBO = 90◦ − a.

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3. Let ABC be an acute triangle with altitudes AD, BE, CF.

(a) Prove that triangle AEF is similar to triangle ABC.

(b) Extend AD until it meets the circumcircle of triangle ABC at X and extend BE until it meets thecircumcircle of triangle ABC at Y. Prove that CX = CY.

[2 marks]

Proof. As we often do, let the angles of triangle ABC be ∠CAB = a, ∠ABC = b and ∠BCA = c.

(a) The first thing to observe is that the quadrilateral AEHF is cyclic since ∠AEH + ∠HFA = 180◦.In particular, the hockey theorem tells us that ∠AEF = ∠AHF. Since triangle AFH is right-angled, we know that ∠AHF + ∠FAH = 90◦. And since triangle ADB is right-angled, we knowthat ∠ABD + ∠DAB = 90◦. It’s clear that ∠FAH = ∠DAB, so these two equations imply that∠AHF = ∠ABD = b. So we’ve proven that ∠AEF = ∠AHF = b. The same argument can beused to show that ∠AFE = ∠AHE = c. And now we can use AAA to deduce that triangle AEF issimilar to triangle ABC.

A

B C

H

D

E

F

(b) When solving geometry problems, it’s a good habit to keep your diagrams as simple as possible.For example, the following diagram shows all that we need to solve this problem, without anyextraneous points or lines. We would like to show that CX = CY or, in other words, that thetriangle CXY is isosceles. So it makes sense to try to prove that ∠CXY = ∠CYX.

A

B C

E

H

X

Y

By the hockey theorem, we know that ∠CXY = ∠CBY. And since triangle BEC is right-angled,we know that ∠EBC + ∠ECB = 90◦ which can be written as ∠CBY = ∠EBC = 90◦ − c. So we’vemanaged to prove that ∠CXY = 90◦ − c.

It turns out that the very same argument can be used to show that ∠CYX = 90◦ − c as well.Therefore, ∠CXY = ∠CYX as desired.

149


4. In the lectures, we tried to prove the following fact. If AB is a chord of a circle with centre O and Cis a point on the circle on the same side of AB as O, then ∠AOB = 2∠ACB. Unfortunately, our proofwas incomplete because we only considered the case when O lies inside triangle ABC. Prove that thestatement remains true in the case that O lies on one of the sides of triangle ABC and in the case that Olies outside triangle ABC.

[3 marks]

Proof. There are two cases to consider here.

Case 1: O lies on one of the sides of triangle ABCWithout loss of generality, we can assume that O lies on the side AC. The beauty of considering thecentre is that we have the three equal radii OA = OB = OC. Equal lengths, for obvious reasons, oftenlead to isosceles triangles, and our diagram happens to have two of them. There’s the isosceles triangleOAB which means that we can label the equal angles ∠OAB = ∠OBA = x. There’s also the isoscelestriangle OBC, which means that we can label the equal angles ∠OBC = ∠OCB = y. Labelling equalangles like this is an extremely common and extremely useful trick.

A B

C

O

x xy

y

180◦ − 2x

Now it’s time for some angle chasing. In particular, since the angles in triangle OAB add up to180◦, we know that ∠AOB = 180◦ − 2x. Also, since AC is a diameter of the circle, we know that∠ABC = x + y = 90◦. This equation is, of course, the same thing as y = 90◦ − x. All we have to donow is recognise that

2∠ACB = 2y = 2(90◦ − x) = 180◦ − 2x = ∠AOB.

Case 2: O lies outside triangle ABCThe beauty of considering the centre is that we have the three equal radii OA = OB = OC. Equallengths, for obvious reasons, often lead to isosceles triangles, and our diagram happens to havethree of them. There’s the isosceles triangle OAB which means that we can label the equal angles∠OAB = ∠OBA = x. There’s also the isosceles triangle OBC, which means that we can label the equalangles ∠OBC = ∠OCB = y. And there’s also the isosceles triangle OCA, which means that we canlabel the equal angles ∠OCA = ∠OAC = z. Labelling equal angles like this is an extremely commonand extremely useful trick.

150


A B

CO

x− z x y

y− zz

z

180◦ − 2x

Now it’s time for some angle chasing. In particular, we can subtract angles to obtain the equations∠ACB = y− z and ∠BAC = x− z. Now let’s consider the sum of the angles in triangle ABC.

∠BAC + ∠ACB + ∠CBA = 180◦

We can replace all of these confusing angles with x’s, y’s and z’s in the following way.

(x− z) + (y− z) + (x + y) = 180◦

This equation is, of course, the same thing as 2(x + y− z) = 180◦. Let’s keep this equation in the backof our minds while we try and remember what it is exactly that we’re trying to do. We want to provethat ∠AOB = 2×∠ACB. Using the angle sum in triangle OAB, we can write ∠AOB = 180◦ − 2x andwe already know that ∠ACB = y− z. So what we’re actually aiming for is the following equation.

180◦ − 2x = 2(y− z)

But after rearranging, this is just the same thing as 2(x + y− z) = 180◦, which we already proved.

5. Extend the median AX of triangle ABC until it meets the circumcircle of triangle ABC at K. There is aunique circle which passes through A and B and is tangent to the line BC. If this circle meets AK at L,prove that triangle LBX and triangle KCX are congruent to each other. Hence, deduce that BKCL is aparallelogram.

[3 marks]

Proof. Since the problem involves a tangent, it makes sense to see if we can use the alternate segmenttheorem. The tangency occurs at B, so we should look for a chord of the circle which passes through Band this must be BL. The alternate segment theorem then tells us that ∠LAB = ∠LBX. Let’s call thisangle x.

Note that ∠LAB = ∠KAB since they coincide. But ∠KAB is an angle in the circle subtended by thechord BK. This chord also subtends an angle at C, so the hockey theorem tells us that ∠BCK = ∠BAK =∠LAB = x as well. In particular, we have deduced that ∠LBX = ∠BCK = ∠KCX.

151


A

B CX

K

L

To prove that triangle LBX is congruent to triangle KCX, we’re going to use the congruence test ASA.

We have just proved that ∠LBX = ∠KCX.

We know that BX = CX, since AX is a median in triangle ABC.

We have ∠LXB = ∠KXC, since they’re formed by the straight line segments AK and BC.

A parallelogram is defined to be a quadrilateral with two pairs of parallel sides. Since ∠LBX = ∠KCX,we know that the two line segments BL and KC must be parallel. Earlier, we proved that triangle LBXis congruent to triangle KCX. In particular, this means that the corresponding sides LX and KX areequal in length. Now we can use the congruence test SAS to prove that triangle LXC and triangle KXBare congruent to each other.

We have just observed that LX = KX.

We have ∠LXC = ∠KXB, since they are formed by the straight line segments AK and BC.

We know that CX = BX, since AX is a median in triangle ABC.

In particular, we have the equal corresponding angles ∠XLC = ∠XKC. However, this just means thatthe two line segments CL and KB are parallel to each other. It follows that BKCL is a parallelogram.

6. Suppose that the circles C1 and C2 intersect at two points A and B and that C2 passes through the centreof C1, which we denote by O. Let X be a point on the arc AB which contains O and extend AX until itintersects C1 at Y. Prove that XY = XB.

[3 marks]

Proof. Since we’d like to show that XY = XB,it suffices to show that ∠XYB = ∠XBY. So let∠XYB = x and note that since O is the centre ofthe circle C1,

∠AOB = 2∠AYB = 2∠XYB = 2x.

So we can use the hockey theorem to deduce that∠AXB = ∠AOB = 2x. The angle next door satis-fies

∠YXB = 180◦ −∠AXB = 180◦ − 2x.

A

B

X

Y O

C1

C2

152


Now consider the sum of the angles in triangle XYB. The equation ∠XYB + ∠XBY + ∠YXB = 180◦

implies that∠XBY = 180◦ −∠XYB−∠YXB = 180◦ − x− (180◦ − 2x) = x.

So we have managed to prove that ∠XYB = ∠XBY = x.

7. (a) In the lectures, we proved that the diameter of a circle subtends an angle of 90◦. Prove the converseof this statement — in other words, prove that if ABC is a triangle with ∠ACB = 90◦, then thecircle with diameter AB passes through C.

(b) Let ABCD be a cyclic quadrilateral such that the diagonals AC and BD are perpendicular andintersect at the point P. Let M be the midpoint of AB and let N be the point on CD such that PNis perpendicular to CD. Prove that the points M, P, N lie on a line.

[4 marks]

Proof. There are many ways to solve this problem and here is just one.

(a) Suppose that we draw the circle with AB as diameter and that it doesn’t pass through C. Then oneof the following two things must happen — either the circle meets the side AC at some point D asin the diagram below left, or you can extend the side AC until it meets the circle at some point Das in the diagram below right. Note that in both cases, we can use the result that ∠ADB = 90◦.

A B

C

D

O

A B

C

D

O

But in either case, we have a triangle BCD which has a right angle at C as well as a right angle atD. This violates the fact that the sum of the angles in a triangle is 180◦. From this contradiction,we can deduce that the circle with AB as diameter must actually pass through C. In particular, thecircumcircle of triangle ABC has AB as diameter.

(b) The result from part (a) tells us that AB is the diameter of triangle APB, so M is in fact thecircumcentre of triangle ABC. In particular, this means that MA = MP, because they’re both radiiof the circumcircle of triangle ABC. We can prove that M, P and N lie on a line by showing that∠MPA + ∠APD + ∠DPN = 180◦ or equivalently, ∠MPA + ∠DPN = 90◦.Let’s start by labelling ∠MPA = x. As discussed earlier, MA = MP so triangle AMP is isoscelesand we have the equal angles ∠MAP = ∠MPA = x. Now we can invoke the hockey theorem incyclic quadrilateral ABCD with BC as the goal to obtain ∠BDC = ∠BAC = ∠MAP = x. Usingthe angle sum in triangle PND, we obtain

∠DPN = 180◦ −∠PND−∠PDN = 180◦ − 90◦ −∠BDC = 90◦ − x.

Hence, we have ∠MPA + ∠DPN = x + (90◦ − x) = 90◦, as desired and M, P, N lie on a line.

153


A

B

C

DP

M

N

x

8. In the hyperbolic plane, there exist pentagons all of whose angles are equal to 90◦. Sketch one exampleof such a pentagon in the hyperbolic plane using the Poincare disk model. What is the area of thispentagon?

[2 marks]

Proof. The large circle below represents the Poincare disk model of the hyperbolic plane. The fivedotted curves represent five hyperbolic lines. As you can see from the diagram, they form a pentagonall of whose angles are right angles.

Now consider the following schematic diagram for the pentagon, where the hyperbolic lines arerepresented by normal straight lines. We’ve divided the pentagon into three triangles — labelled 1, 2and 3 — and labelled every single angle in the diagram.

154


a

b

c

d

e f

g

h

i1 3

2

Since each angle of the pentagon is a right angle, we have five equations that these labelled angles mustsatisfy. Here, we have labelled the angles using radians rather than degrees.

a + d + g =π

2, b =

π

2, c + e =

π

2, f + h =

π

2, i =

π

2

One of the advantages of using radians is that the area of a hyperbolic triangle is much easier todetermine. In fact, we know that the area of triangle 1 is π − a − b − c, the area of triangle 2 isπ − d− e− f , and the area of triangle 3 is π − g− h− i. Adding up the areas of these three triangles,we deduce that the area of the pentagon is

(π − a− b− c) + (π − d− e− f ) + (π − g− h− i)

= 3π − (a + b + c + d + e + f + g + h + i)

= 3π − (a + d + g)− (b)− (c + e)− ( f + h)− (i)

= 3π − π

2− π

2− π

2− π

2− π

2

=π

2.

9. Let D, E, F be points on sides AB, BC, CA of triangle ABC, respectively. If DE = BE and FE = CE,prove that the circumcentre of triangle ADF lies on the angle bisector of ∠DEF.

[BONUS]

Proof. Let’s start by labelling the angles of triangle ABC by a, b and c. If you draw a very accuratediagram — and I recommend that you do — then you’ll notice that the quadrilateral DEFP looksparticularly cyclic. So let’s try to prove this right now. The idea is that we can determine the oppositeangles ∠DEF and ∠FPD of the quadrilateral in terms of a, b, c. To calculate ∠DEF, we have

∠DEF = 180◦ −∠BED−∠CEF = 180◦ − (180◦ − 2b)− (180◦ − 2c) = 2b + 2c− 180◦.

To calculate ∠FPD, we note that it’s an angle subtended by the chord DF at the centre of the circumcircleof triangle ADF. Therefore, ∠FPD = 2∠FAD = 2a.

∠DEF + ∠FPD = (2b + 2c− 180◦) + 2a = 2(a + b + c)− 180◦ = 180◦.

Here, we’ve used the obvious fact that a + b + c = 180◦, since they are the angles of triangle ABC. Sowe’ve shown that ∠DEF + ∠FPD = 180◦ from which it follows that the quadrilateral DEFP is cyclic.

155


A

B C

D

E

FP

a

b

b

c

c

Remember that we’d like to show that ∠DEP = ∠FEP. But these angles are subtended by the chordsDP and FP, respectively, in the circle which passes through the vertices of the cyclic quadrilateralDEFP. These chords DP and FP are equal since they are both circumradii of triangle ADF by definition.Since equal chords subtend equal angles, it follows that ∠DEP = ∠FEP. Therefore, the circumcentre oftriangle ADF lies on the angle bisector of ∠DEF.

156


Release date: 1:30pm on Tuesday 11 May, 2010Due date: 11:00am on Tuesday 18 May, 2010


Problems

1. (a) Suppose that ABCD is a square with the vertices labelled counterclockwise. Let

GAC denote the glide reflection consisting of a reflection in the line AC followed by a transla-tion which takes A to C;TBD denote the translation which takes B to D;MAB denote the reflection in the line AB; andRB denote a counterclockwise rotation by 90◦ about B.

Identify the composition GAC ◦ TBD ◦MAB ◦ RB.

(b) If X denotes the composition GAC ◦ TBD ◦MAB ◦ RB, let n be the minimum number of reflectionswhose composition is equal to X. Determine, with proof, the value of n and carefully describe nreflections whose composition is equal to X.

[2 marks]

2. (a) Let M be a reflection and let R be a rotation by 90◦ counterclockwise. Is it possible for thecompositions M ◦ R and R ◦M to be the same?

(b) Let Mk be a reflection through the line k and let M` be a reflection through the line `. If someonetells you that M` ◦Mk = Mk ◦M`, what can you tell them about the lines k and `?

[3 marks]

3. (a) The following is a groupoku puzzle — a Cayley table for the group G, where some of the entriesare missing. Use the properties which you know about Cayley tables to fill in all of the missingentries. Give full reasoning only for the first entry of the table that you manage to fill in.

· a b c d

a * * * *b * c * *c b * d *d * * * *

(b) In the lectures, we have seen two groups with four elements — the cyclic group C4 and thedihedral group D2. Prove that G is isomorphic to one of these groups by writing down an explicitisomorphism. Furthermore, prove that what you have written down is in fact an isomorphism.How many isomorphisms are there?

(c) Prove that the cyclic group C4 and the dihedral group D2 are not isomorphic to each other, eventhough they are both abelian groups with four elements.

157


(d) Give a subset of the Euclidean plane whose symmetry group is isomorphic to G. Does there exista finite set of points in the Euclidean plane whose symmetry group is isomorphic to G? If so,determine the minimum number of points in such a set. If not, give a brief explanation of whysuch a set does not exist.

[4 marks]

4. (a) Write down the HVRG symbol for the frieze pattern below.

(b) Write down the HVRG symbol for the frieze pattern below.

LOΓOLOΓOLOΓOLOΓO

(c) For each of the seven types of frieze pattern, determine whether or not there exists a sequence ofcapital letters which can be written in a repeating pattern to form a frieze pattern of that type. Youshould assume that the capital letters look like the following.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

(d) Prove that if a symmetry group of a frieze pattern contains a glide reflection and a reflection in avertical mirror, then it must also contain a rotation by 180◦.

[3 marks]

5. (a) Identify the wallpaper pattern on the outside of Burnside Hall, using its RMG symbol.

(b) Identify the wallpaper pattern on the walls of room 1B24, using its RMG symbol.

(c) Write down the RMG symbol for the wallpaper pattern pictured below left.

(d) Write down the RMG symbol for the wallpaper pattern pictured below right. Draw the wallpaperpattern and mark clearly on your diagram a point which is the centre of a rotational symmetry oforder R, a point where M mirrors meet, and a point where G proper glide axes meet.

[4 marks]

158


6. Let ABC be a triangle with vertices labelled in a counterclockwise fashion. Three equilateral tri-angles BCX, CAY, ABZ are drawn on the sides of triangle ABC, outside of triangle ABC. Let thecentres of BCX, CAY, ABZ be K, L, M, respectively. Let RK be the rotation by 120◦ counterclockwiseabout K, let RL be the rotation by 120◦ counterclockwise about L, and let RM be the rotation by 120◦

counterclockwise about M.

(a) Prove that RM ◦ RK ◦ RL is a translation.

(b) Determine the isometry RM ◦ RK ◦ RL by considering the location of RM ◦ RK ◦ RL(A).

(c) Describe the location of RM ◦ RK ◦ RL(L). If P = RK(L), what can you say about the quadrilateralLKPM?

(d) Hence, what can you deduce about the triangle KLM?[3 marks]

7. Prove that Leonardo’s theorem does not hold in three dimensions. In other words, prove rigorouslythat there exists a subset of Euclidean space whose symmetry group is neither cyclic nor dihedral.

[1 mark]

8. Explain why it follows from S3 and C6 not being isomorphic that there is no shuffle of 3 cards that youcan repeat which goes through all of the possible orderings of the deck. Prove that there is no shuffleof n cards that you can repeat which goes through all of the possible orderings of the deck for n ≥ 3.Determine the largest number of possible orderings which you can go through if you repeat a shuffle ofn cards for n ≤ 10.

[BONUS]

9. The groups we’ve been dealing with have all been examples of cyclic groups, dihedral groups orsymmetric groups — however, there are many others around. Write down the Cayley table for thesmallest non-abelian group which is not a cyclic group, not a dihedral group and not a symmetricgroup.

[CHALLENGE]

159


1. (a) Suppose that ABCD is a square with the vertices labelled counterclockwise. Let

GAC denote the glide reflection consisting of a reflection in the line AC followed by a transla-tion which takes A to C;TBD denote the translation which takes B to D;MAB denote the reflection in the line AB; andRB denote a counterclockwise rotation by 90◦ about B.

Identify the composition GAC ◦ TBD ◦MAB ◦ RB.

(b) If X denotes the composition GAC ◦ TBD ◦MAB ◦ RB, let n be the minimum number of reflectionswhose composition is equal to X. Determine, with proof, the value of n and carefully describe nreflections whose composition is equal to X.

[2 marks]

Proof.

(a) With these sorts of problems, it’s useful to draw the rectangle on a grid of squares, like we’ve donebelow. We’ve also labelled some extra points in the diagram which we’ll soon need to use.

A B

CD

E

F

GHI

J

O

Note that GAC ◦ TBD ◦MAB ◦ RB must be a direct isometry, because it’s the composition of twodirect isometries and two opposite isometries. So we know that it’s either the identity, a translationor a rotation, and to determine which of these it is, we’ll calculate GAC ◦ TBD ◦MAB ◦ RB(P) for afew points P.

First, let’s consider what happens to the point A. It’s easy to check that RB(A) = E, MAB(E) = C,TBD(C) = F, and GAC(F) = G — in other words, GAC ◦ TBD ◦MAB ◦ RB(A) = G.

Now, let’s consider what happens to the point B. It’s easy to check that RB(B) = B, MAB(B) = B,TBD(B) = D, and GAC(D) = H — in other words, GAC ◦ TBD ◦MAB ◦ RB(B) = H.

Now, let’s consider what happens to the point C. It’s easy to check that RB(C) = A, MAB(A) = A,TBD(A) = I, and GAC(I) = J — in other words, GAC ◦ TBD ◦MAB ◦ RB(C) = J.

At this stage, we can be sure that the composition GAC ◦ TBD ◦MAB ◦ RB cannot be the identityor a translation. Since it takes the line segment AB to the line segment GH, we can try to guesswhere the centre of rotation is. It turns out that the centre of rotation is at the point O labelled onthe diagram. In fact, we can now guess that the composition is a 180◦ rotation about O. Since thisis certainly true for the vertices of triangle ABC, this confirms that the composition is indeed a180◦ rotation about O.

160


(b) In the lectures, we proved that every isometry is a composition of at most three reflections.However, the composition of an odd number of reflections is an opposite isometry and cannot bea rotation. Therefore, X can be written as the composition of 0 or 2 reflections. Since X is not equalto the identity, it follows that X must be a composition of 2 reflections.

Let Mh and Mv denote reflections through a horizontal line and a vertical which passes through O,respectively. The composition Mv ◦Mh is a rotation whose centre is the intersection point of thetwo mirrors and whose angle is twice the angle between the two mirrors. Hence, this is a rotationby 180◦ about O, so we know that X = Mv ◦Mh.

2. (a) Let M be a reflection and let R be a rotation by 90◦ counterclockwise. Is it possible for thecompositions M ◦ R and R ◦M to be the same?

(b) Let Mk be a reflection through the line k and let M` be a reflection through the line `. If someonetells you that M` ◦Mk = Mk ◦M`, what can you tell them about the lines k and `?

[3 marks]

Proof.

(a) Let M be a reflection through the mirror ` and let R be a rotation by 90◦ counterclockwise aboutthe centre O. Without loss of generality, we can rotate the diagram so that ` is a horizontal line.

Case 1: O lies on `

Take a point P on the line ` which is not the point O. Then R(P) is a point above `, so M ◦ R(P)must be a point below `. On the other hand, M(P) = P, so R ◦M(P) must be a point above `.Hence, M ◦ R(P) 6= R ◦M(P) and it’s impossible for the compositions M ◦ R and R ◦M to bethe same.Case 2: O does not lie on `

Without loss of generality, let O be above the line `. Then R(O) = O is a point above `, soM ◦ R(O) = M(O) must be a point below `. On the other hand, M(O) is a point below `,directly below O, so R ◦M(O) must be a point above `. Hence, M ◦ R(O) 6= R ◦M(O) andit’s impossible for the compositions M ◦ R and R ◦M to be the same.

(b) In the lectures, we proved that the composition M` ◦Mk is the identity, a translation or a rotation,depending on the following three cases.

Case 1: k and ` are the same lineIn this case, it’s obvious that M` ◦Mk = Mk ◦M`.Case 2: k and ` are parallelIn this case, M` ◦ Mk is a translation equal to twice the distance from k to `. On the otherhand, Mk ◦M` is a translation equal to twice the distance from ` to k. Therefore, these twotranslations are by the same distance, but in opposite directions. So it’s impossible for theequation M` ◦Mk = Mk ◦M` to hold.Case 3: k and ` meet at a point OIn this case, M` ◦ Mk is a rotation about O by twice the angle from k to `. On the otherhand, Mk ◦ M` is a rotation about O by twice the angle from ` to k. Therefore, these tworotations are by the same angle, but in opposite directions. So it’s possible for the equationM` ◦Mk = Mk ◦M` to hold if and only if these two rotations are by 180◦. This occurs preciselywhen the angle from k to ` or from ` to k is 90◦.

In summary, if someone tells you that M` ◦Mk = Mk ◦M`, then you you tell them that the lines kand ` are either the same line or perpendicular to each other.

161



· a b c d

a * * * *b * c * *c b * d *d * * * *

(b) In the lectures, we have seen two groups with four elements — the cyclic group C4 and thedihedral group D2. Prove that G is isomorphic to one of these groups by writing down an explicitisomorphism. Furthermore, prove that what you have written down is in fact an isomorphism.How many isomorphisms are there?

(c) Prove that the cyclic group C4 and the dihedral group D2 are not isomorphic to each other, eventhough they are both abelian groups with four elements.

(d) Give a subset of the Euclidean plane whose symmetry group is isomorphic to G. Does there exista finite set of points in the Euclidean plane whose symmetry group is isomorphic to G? If so,determine the minimum number of points in such a set. If not, give a brief explanation of whysuch a set does not exist.

[4 marks]

Proof.

(a) The first thing to notice from the Cayley table isthat the sudoku property implies that c · b can-not be equal to b, c or d. So we can deduce thatc · b = a. It’s now straightforward to fill in therest of the table using the sudoku property.

· a b c d

a c d b ab d c a bc b a d cd a b c d

(b) The group G is isomorphic to the group C4, which is the symmetry group of a decorated square. Ifwe write the elements of C4 as follows, then its Cayley table will look like the one below.

I: the identity isometryR1: the rotation by 90◦ counterclockwise

R2: the rotation by 180◦ counterclockwiseR3: the rotation by 270◦ counterclockwise

◦ I R1 R2 R3

I I R1 R2 R3

R1 R1 R2 R3 IR2 R2 R3 I R1

R3 R3 I R1 R2

162



F(a) = R1 F(b) = R3 F(c) = R2 F(d) = I.

This is certainly not the only isomorphism possible. To find one, you can use the fact that theidentity in G must correspond to the identity in C4. After you’ve done this, the problem canbe finished with a little trial and error. You should find that there are actually two possibleisomorphisms and the other one is given by the equations

F(a) = R3 F(b) = R1 F(c) = R2 F(d) = I.

(c) We can write the elements of D2 as follows.

I: the identity isometryR: rotation by 180◦ counterclockwise

Mh: reflection through a horizontal lineMv: reflection through a vertical line

One fact about the group C4 is that the element R1 composed with itself is not the identity. On theother hand, every element of D2 composed with itself is equal to the identity. Therefore, the cyclicgroup C4 and the dihedral group D2 cannot be isomorphic to each other.

(d) In the lectures, we used the decorated square as an example of a subset of the Euclidean planewhose symmetry group is C4. However, it’s possible to find a finite set — in fact, the eight verticesin the picture of the decorated square gives a subset of the Euclidean plane with symmetry groupC4. It should also be clear that there is no smaller set with this symmetry group.



LOΓOLOΓOLOΓOLOΓO

(c) For each of the seven types of frieze pattern, determine whether or not there exists a sequence ofcapital letters which can be written in a repeating pattern to form a frieze pattern of that type. Youshould assume that the capital letters look like the following.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

(d) Prove that if a symmetry group of a frieze pattern contains a glide reflection and a reflection in avertical mirror, then it must also contain a rotation by 180◦.

[3 marks]

Proof.

(a) HG

(b) G

(c) The task can be accomplished for the following five types of frieze pattern.

163


none F F F F F

V A A A A A

R N N N N N

HG E E E E E

HVRG H H H H H

In order to have a frieze group which contains a glide reflection but no reflection in a horizontalmirror, it would be necessary to include a letter whose reflection in a horizontal mirror produces adifferent letter. Since no such letter exists in the alphabet, the task cannot be accomplished for thefrieze pattern G nor the frieze pattern VRG.

(d) If we denote the glide reflection by G and the reflection in a vertical mirror by V, then the symmetrygroup of the frieze pattern must also contain G ◦V. This composition is a direct isometry, and weknow that the only direct symmetries which a frieze pattern can have are horizontal translationsor rotations by 180◦. Of course, we wish to prove that G ◦V is a rotation by 180◦.A picture of a right hand above the horizontal axis becomes a picture of a left hand above thehorizontal axis after applying V and then a picture of a right hand below the horizontal axis afterapplying G. This shows that the composition G ◦V is not a translation but a rotation by 180◦.Alternatively, we can express G as T ◦ H, where T is a horizontal translation and H is reflectionin a horizontal mirror. Then G ◦V = T ◦ H ◦V and, due to the way that reflections compose, weknow that H ◦V is a rotation by 180◦. The composition of a translation and a rotation by angle a isalways a rotation by angle a. It follows that G ◦V must be a rotation by 180◦.

5. (a) Identify the wallpaper pattern on the outside of Burnside Hall, using its RMG symbol.

(b) Identify the wallpaper pattern on the walls of room 1B24, using its RMG symbol.


(d) Write down the RMG symbol for the wallpaper pattern pictured below right. Draw the wallpaperpattern and mark clearly on your diagram a point which is the centre of a rotational symmetry oforder R, a point where M mirrors meet, and a point where G proper glide axes meet.

[4 marks]

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Proof.

(a) 220

(b) 222

(c) 200 — The maximum order of a rotation isat most 2 since white lizards only appear intwo distinct orientations. Furthermore, a rota-tion of order 2 exists whose centre is the pointwhere two lizards kiss. There are clearly noopposite symmetries since all lizards are fac-ing to their left and there are no lizards facingto their right.

(d) 423 — A point which is the centre of a rota-tional symmetry of order 4, a point which 2mirrors pass through, and a point where 3proper glide axes pass through are labelledin the diagram on the right in red, blue andgreen, respectively.

6. Let ABC be a triangle with vertices labelled in a counterclockwise fashion. Three equilateral tri-angles BCX, CAY, ABZ are drawn on the sides of triangle ABC, outside of triangle ABC. Let thecentres of BCX, CAY, ABZ be K, L, M, respectively. Let RK be the rotation by 120◦ counterclockwiseabout K, let RL be the rotation by 120◦ counterclockwise about L, and let RM be the rotation by 120◦

counterclockwise about M.

(a) Prove that RM ◦ RK ◦ RL is a translation.

(b) Determine the isometry RM ◦ RK ◦ RL by considering the location of RM ◦ RK ◦ RL(A).

(c) Describe the location of RM ◦ RK ◦ RL(L). If P = RK(L), what can you say about the quadrilateralLKPM?

(d) Hence, what can you deduce about the triangle KLM?[3 marks]

Proof. If someone gives you a geometry problem,then almost always your first step should be todraw a nice, large, accurate diagram, possiblyusing colours. For this particular problem, youshould end up with a diagram like the one on theright. The great thing about the diagram is that itsuggests very strongly what we’re trying to provein this problem — namely, that the triangle KLMis equilateral.

It turns out that this result is usually known asNapoleon’s theorem. Although it’s named afterNapoleon Bonaparte, we have reason to believethat it was almost certainly never proved by him.

A

B C

X

Y

Z

K

L

M

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(a) Let R1 be a rotation by angle a1 and R2 be a rotation by angle a2. In the lectures, we stated that thecomposition R2 ◦ R1 is

a rotation by angle a1 + a2 if a1 + a2 6= 360◦; ora translation if a1 + a2 = 360◦.

Using this proposition, we know that the composition RK ◦ RL is a rotation by 120◦ + 120◦ = 240◦.Therefore, RM ◦ RK ◦ RL = RM ◦ (RK ◦ RL) is the composition of a rotation by 240◦ followedby a rotation by 120◦. Since 240◦ + 120◦ = 360◦, it follows from the aforementioned fact thatRM ◦ RK ◦ RL is a translation.

(b) Here, we use the fact that since K is the centre of equilateral triangle BCX, we have ∠BKC = 120◦

and BK = CK. Similarly, we have ∠CLA = ∠AMB = 120◦, CL = AL and AM = BM. Hence, therotation RL takes A to C, the rotation RK takes C to B, and the rotation RM takes B to A. In otherwords,

RM ◦ RK ◦ RL(A) = RM ◦ RK(C) = RM(B) = A.

So we have deduced that RM ◦ RK ◦ RL is a translation which has the fixed point A. It follows thatRM ◦ RK ◦ RL is in fact the identity.

(c) Since RM ◦ RK ◦ RL is the identity, we know that RM ◦ RK ◦ RL(L) = L. However, RL(L) = Limplies that RM ◦ RK(L) = L. Another way to say this is that RM(P) = L where we let RK(L) = P.The fact that RM(P) = L means that ∠PML = 120◦ and PM = LM while the fact that RK(L) = Pmeans that ∠LKP = 120◦ and LK = PK. Therefore, the quadrilateral LKPM must look look likethe following.

L

KP

M120◦

120◦

In fact, it should be clear that triangles PML and LKP are similar by PAP. Furthermore, they sharethe common side LP so they are in fact congruent. Thus, the quadrilateral LKPM has four equalsides with two angles equal to 120◦ and the other two angles equal to 60◦.

(d) We now know that KLM must be an equilateral triangle because we have deduced that it isisosceles and contains a 60◦ angle. This completes the proof of Napoleon’s theorem.

7. Prove that Leonardo’s theorem does not hold in three dimensions. In other words, prove rigorouslythat there exists a subset of Euclidean space whose symmetry group is neither cyclic nor dihedral.

[1 mark]

Proof. Consider the triangular pyramid formed by gluing together the edges of four congruent equi-lateral triangles — this shape is called a regular tetrahedron. The regular tetrahedron has four verticeswhich we can label 1, 2, 3 and 4. Given any permutation of these four numbers, there is a uniqueisometry which permutes the vertices in that particular way. In other words, the symmetry group ofthe regular tetrahedron is the symmetric group S4, which has 4× 3× 2× 1 = 24 elements.

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The group S4 is certainly not isomorphic to the cyclic group C24, since it is non-abelian. Furthermore,it is certainly not isomorphic to the dihedral group D12. One way to see this is to note that the groupD12 contains the element g which represents rotation by 30◦. This element is such that g4 and g3 is notequal to the identity. On the other hand, you can check that for every element g in S4, either g4 equalsthe identity or g3 equals the identity.

So what we’ve shown is that symmetry groups of subsets of Euclidean space need not be cyclic nordihedral. This means that Leonardo’s theorem does not hold in three dimensions — at least not withoutmaking some changes.

8. Explain why it follows from S3 and C6 not being isomorphic that there is no shuffle of 3 cards that youcan repeat which goes through all of the possible orderings of the deck. Prove that there is no shuffleof n cards that you can repeat which goes through all of the possible orderings of the deck for n ≥ 3.Determine the largest number of possible orderings which you can go through if you repeat a shuffle ofn cards for n ≤ 10.

[BONUS]

Proof. If we number the cards in the deck 1, 2 and 3, then a shuffle of the deck corresponds preciselyto a permutation in S3. Suppose that it is possible to perform a shuffle s which goes through allsix of the possible orderings of the deck. Then the shuffles — in other words, the permutations —e, s, s2, s3, s4, s5 must be distinct. Here, we have used e to denote the identity permutation and sn todenote the composition

s ◦ s ◦ · · · ◦ s︸︷︷︸n times

.

But the cyclic group C6 consists of the elements I, R, R2, R3, R4, R5 where I denotes the identity isometryand R is a rotation by angle 60◦. This would give us an isomorphism f : S3 → C6 defined by

f (e) = I, f (s) = R, f (s2) = R2, f (s3) = R3, f (s4) = R4, f (s5) = R5.

However, we know that S3 and C6 cannot possibly be isomorphic because one is abelian while the otheris not. Therefore, there is no shuffle of three cards that you can repeat which goes through all of thepossible orderings of the deck.

Note that with a deck of n cards, there are n! = n× (n− 1)× (n− 2)× · · · × 3× 2× 1 possible orderingsof the deck. To prove that there is no shuffle of n cards that you can repeat which goes through all ofthese orderings for n ≥ 3, we need to prove that there is no isomorphism between Sn and Cn!. Butthis follows directly from the fact that the cyclic group Cn! is abelian while the symmetric group Sn isnon-abelian for n ≥ 3.

Let L(n) be the largest number of possible orderings which you can go through if you repeat a shuffleof n cards. The following table shows the values of L(n) for n ≤ 10.

n 1 2 3 4 5 6 7 8 9 10

L(n) 1 2 3 4 6 6 12 15 20 30

I’ll let you figure out on your own how to compute these numbers.

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9. The groups we’ve been dealing with have all been examples of cyclic groups, dihedral groups orsymmetric groups — however, there are many others around. Write down the Cayley table for thesmallest non-abelian group which is not a cyclic group, not a dihedral group and not a symmetricgroup.

[CHALLENGE]

Proof. It turns out that the smallest non-abelian group which is not a cyclic group, not a dihedral groupand not a symmetric group has eight elements and is known as the quaternion group Q8. Its Cayleytable is below, where we have labelled the elements 1, −1, i, −i, j, −j, k, −k.

◦ 1 −1 i −i j −j k −k

1 1 −1 i −i j −j k −k−1 −1 1 −i i −j j −k ki i −i −1 1 k −k −j j−i −i i 1 −1 −k k j −jj j −j −k k −1 1 i −i−j −j j k −k 1 −1 −i ik k −k j −j −i i −1 1−k −k k −j j i −i 1 −1

The quaternion group Q8 comes from the set of quaternions. Just as complex numbers can be used todeal with rotations in the plane, the quaternions can be used to deal with rotations in space. They werediscovered by an Irish mathematician named William Rowan Hamilton on 16 October 1843, while hewas going for a walk. At the time, he was so excited by the discovery that he carved the followingformulae into the bridge he was walking across.

i2 = j2 = k2 = ijk = −1

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Release date: 1:30pm on Wednesday 19 May, 2010Due date: 11:00am on Wednesday 26 May, 2010


Problems

1. (a) Show that at a party with five people, it is not necessarily true that there exist three people who allknow each other or three people who all don’t know each other.

(b) Note that if you take the dual of the tetrahedron, then you just get the tetrahedron again. Describeanother polyhedron whose dual is itself.

(c) Does there exist a map on the torus consisting of fifty triangles?

(d) For which values of n and d is there a graph — without loops or multiple edges — with n verticesof degree d?

(e) Draw a map on a genus two surface which consists of one polygon.

(f) Divide the letters of the alphabet into the smallest possible number of groups so that, within eachgroup, all letters are homeomorphic to each other. You should assume that the letters look like thefollowing and consist of infinitely thin line segments and curves.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z[3 marks]

2. Prove that there does not exist a graph — without loops or multiple edges — with eight vertices which

(a) has vertices of degrees 1, 1, 1, 1, 2, 2, 2, 8;

(b) has vertices of degrees 1, 1, 2, 3, 3, 4, 4, 5;

(c) has vertices of degrees 0, 1, 1, 2, 3, 4, 4, 7;

(d) has vertices of degrees 1, 1, 1, 2, 2, 3, 6, 6.

[2 marks]

3. (a) A domino consists of two squares, each of which is labelled with a number from 0 to 6. Verify thatthere are 28 different dominoes. Is it possible to arrange them all in a loop so that adjacent halvesof neighbouring dominoes are labelled by the same number?

(b) An n-domino consists of two squares, each of which is labelled with a number from 0 to n. Forwhich values of n is it possible to arrange them all in a loop so that adjacent halves of neighbouringn-dominoes are labelled by the same number?Hint: You should use the theorem which tells you when a graph is Eulerian.

[2 marks]

4. Consider a graph G — without multiple edges or loops — whose vertices all have degree greater thanor equal to six.

(a) If G has V vertices and E edges, use the handshaking lemma to prove that V ≤ E3 .

(b) Suppose that G is planar and can be drawn in the plane in such a way that there are F faces. Usethe handshaking lemma on the dual graph to prove that F ≤ 2E

3 .

169


(c) Explain how these two inequalities imply that there does not exist a planar graph — without loopsor multiple edges — whose vertices all have degree greater than or equal to six.

[3 marks]

5. Suppose that you have a polyhedron and you are told that each face is a quadrilateral or a hexagon andthat three faces meet at every vertex. Furthermore, every quadrilateral face shares an edge with fourhexagonal faces, while every hexagonal face shares an edge with three quadrilateral faces and threehexagonal faces.

(a) Deduce the number of quadrilateral faces and the number of hexagonal faces of the polyhedronand draw an example of such a polyhedron.

(b) Cut off each vertex of the polyhedron using a plane which passes through the midpoints of thethree edges which meet at that vertex. This process produces a convex polyhedron from a convexpolyhedron. For example, applying this process to a tetrahedron produces an octahedron. Howmany vertices, edges and faces does this new polyhedron have? How many triangular faces,quadrilateral faces, pentagonal faces, and hexagonal faces does this new polyhedron have?

[3 marks]

6. For each of the following edge words, determine whether or not the corresponding surface is orientableand calculate its Euler characteristic. Also, for each surface, identify it as the sphere, a connect sum oftori or a connect sum of projective planes.

(a) abc−1b−1da−1d−1c

(b) abacb−1ded−1e−1c

(c) a1a2a3 · · · ana−11 a−1

2 a−13 · · · a−1

n

Hint: Try this problem for several values of n until you find a pattern.[3 marks]

7. The shaded region, pictured below, is a shape which gives rise to a surface after gluing the edgestogether as indicated.

(a) Show that it can be described by the edge word abc−1d−1ecadb−1e−1.

(b) Determine whether or not the surface is orientable, calculate its Euler characteristic and identitythe surface using the classification of surfaces.

(c) What is the shortest possible edge word for this surface?

(d) If the surface is homeomorphic to T # X where T represents a torus and X represents a surface,then what must X be?

[2 marks]

a c

d

b

b a

c

d

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8. Given an edge word W, let σ(W) denote the surface corresponding to W. Prove the following facts,where a and b represent letters, W, X, Y, Z represent words, and P and T represent the projective planeand the torus, respectively.

(a) σ(aXa−1Y) ∼= σ(b−1XbY)

(b) σ(XY) ∼= σ(YX)

(c) σ(X) ∼= σ(X−1)

(d) σ(aa−1X) ∼= σ(X)

(e) σ(aXaY) ∼= P # σ(XY−1)

(f) σ(bWxXb−1Yx−1Z) ∼= T # σ(WZYX)

(g) T #P ∼= P #P #P[2 marks]

9. Use the results from question 8 to prove the classification of surfaces — in other words, prove thatevery surface is either the sphere, a connect sum of tori or a connect sum of projective planes.

[BONUS]

10. Consider the pair (S, p) where S is a surface and p is a point on the surface. To the pair (S, p) weassociate a set G consisting of the paths in S which start at p and end at p. However, we consider twopaths a and b to be the same if you can slide the path a along the surface S, while keeping its start andend points at p, until you get the path b. If a and b are two loops on S which start and end at p, then wecan compose them to form the loop a · b which traverses a and then traverses b.

(a) Prove that the set G is a group under this composition.

(b) Prove that the group G does not depend on the choice of p — in other words, if you form thegroup G1 using the pair (S, p) and the group G2 using the pair (S, q), then G1

∼= G2.

(c) Let a and b be two paths in a torus which start and end at some point p such that a and b intersectonly once. Show that a · b = b · a.

(d) Describe the group G when the surface is a torus, a projective plane and a genus two surface.[CHALLENGE]

171


1. (a) Show that at a party with five people, it is not necessarily true that there exist three people who allknow each other or three people who all don’t know each other.

(b) Note that if you take the dual of the tetrahedron, then you just get the tetrahedron again. Describeanother polyhedron whose dual is itself.

(c) Does there exist a map on the torus consisting of fifty triangles?

(d) For which values of n and d is there a graph — without loops or multiple edges — with n verticesof degree d?

(e) Draw a map on a genus two surface which consists of one polygon.

(f) Divide the letters of the alphabet into the smallest possible number of groups so that, within eachgroup, all letters are homeomorphic to each other. You should assume that the letters look like thefollowing and consist of infinitely thin line segments and curves.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z[3 marks]

Proof.

(a) The diagram below left shows a party of five people where there do not exist three people who allknow each other nor three people who all don’t know each other. The people are represented byvertices, two people are connected by a blue edge if they know each other, and two people areconnected by a red edge if they don’t know each other. This example shows that at a party withfive people, it is not necessarily true that there exist three people who all know each other or threepeople who all don’t know each other.

(b) Consider a pyramid with a square base, as shown in the diagram above right. If you take the dualof this polyhedron, then you will obtain a pyramid with a square base.

(c) The following diagram shows a map on the torus consisting of 50 triangles. After the edges of therectangle are glued together in the manner indicated by the arrows, you obtain a map on the toruswhich has 25 vertices, 75 edges and 50 faces.

a

a

b b

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Another way to construct a map on the torus consisting of 50 triangles is to start with a map on thetorus consisting of just 2 triangles. This is easy to do, since we can start with a rectangle, divideit into two triangles using one of its diagonals, and then glue the edges together to form a torus.Now we simply note that any map consisting of n triangles can be turned into a map consisting ofn + 2 triangles. This is because we can divide any triangular face into three triangular faces usingone vertex and three edges.

(d) The handshaking lemma implies that a graph with n vertices of degree d cannot exist if nd is anodd number. We claim that a graph with n vertices of degree d does exist if nd is an even number.

If d is even, then place the n vertices around a circle so that they are evenly spaced. Now connecteach vertex with an edge to the d

2 vertices to the left of it and to the d2 vertices to the right of it.

This will result in a graph with n vertices of degree d.

If d is odd, then it must be the case that n is even. Again we place the n vertices around a circleso that they are evenly spaced. Note that, from the previous argument, we can construct a graphwith n vertices of degree d− 1. Now simply add in the n

2 edges which join each vertex to the pointdiametrically opposite on the circle. This will result in a graph with n vertices of degree d.

In conclusion, a graph with n vertices of degree d exists if and only if nd is an even number.

(e) A map on a surface which consists of one polygon is the same thing as a polygon model for thatsurface. We know that a genus two surface can be represented by a polygon model with eightsides and edge word aba−1b−1cdc−1d−1. After gluing the sides of this polygon, one obtains a mapon a genus two surface which consists of one polygon, as shown in the diagram below.

(f) The letters of the alphabet can be divided into the following groups so that, within each group,all letters are homeomorphic to each other. Furthermore, this is the smallest possible number ofgroups for which this is possible.

A R

B

C G I J L M N S U V W Z

D O

E F T Y

H K

P

Q

X

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2. Prove that there does not exist a graph — without loops or multiple edges — with eight vertices which

(a) has vertices of degrees 1, 1, 1, 1, 2, 2, 2, 8;

(b) has vertices of degrees 1, 1, 2, 3, 3, 4, 4, 5;

(c) has vertices of degrees 0, 1, 1, 2, 3, 4, 4, 7;

(d) has vertices of degrees 1, 1, 1, 2, 2, 3, 6, 6.

[2 marks]

Proof.

(a) A vertex can only be connected at most once to each of the remaining 7 vertices. Therefore, themaximum degree of a vertex in a graph with eight vertices is 7 and there does not exist a graphwith vertices of degrees 1, 1, 1, 1, 2, 2, 2, 8.

(b) By the handshaking lemma, the sum of the degrees of the vertices in a graph must be an evennumber. Since 1 + 1 + 2 + 3 + 3 + 4 + 4 + 5 = 23, there does not exist a graph with vertices ofdegrees 1, 1, 2, 3, 3, 4, 4, 5.

(c) If there exists a vertex of degree 0, then no vertex can be connected to it by an edge. In particular,no vertex in the graph can be connected to all other vertices by an edge. Therefore, there cannotsimultaneously be a vertex of degree 0 and a vertex of degree 7 in a graph with eight vertices andthere does not exist a graph with vertices of degrees 0, 1, 1, 2, 3, 4, 4, 7.

(d) Suppose that there exists a graph with vertices of degrees 1, 1, 1, 2, 2, 3, 6, 6. Let the two vertices ofdegree 6 be A and B and let the remaining vertices be C1, C2, C3, C4, C5, C6. Then A and B mustboth be connected by an edge to at least five of the vertices in the set {C1, C2, C3, C4, C5, C6}. Thismeans that at least four vertices in the set {C1, C2, C3, C4, C5, C6} are connected to both A andB. Therefore, it is impossible for there to be three vertices of degree 1 in the graph. So we canconclude that there does not exist a graph with vertices of degrees 0, 1, 1, 2, 3, 4, 4, 7.

3. (a) A domino consists of two squares, each of which is labelled with a number from 0 to 6. Verify thatthere are 28 different dominoes. Is it possible to arrange them all in a loop so that adjacent halvesof neighbouring dominoes are labelled by the same number?

(b) An n-domino consists of two squares, each of which is labelled with a number from 0 to n. Forwhich values of n is it possible to arrange them all in a loop so that adjacent halves of neighbouringn-dominoes are labelled by the same number?Hint: You should use the theorem which tells you when a graph is Eulerian.

[2 marks]

Proof.

(a) Here, the best thing to do is just draw out all 28 different dominoes and try to arrange them all ina loop so that adjacent halves of neighbouring dominoes are labelled by the same number. Oneway to do this is to arrange the dominoes as follows, where (ab) stands for the domino with onesquare labelled a and one square labelled b.

(00) (01) (11) (12) (22) (23) (33) (34) (44) (45) (55) (56) (66) (60)(02) (24) (46) (61) (13) (35) (50) (03) (36) (62) (25) (51) (14) (40)

Let’s try and solve this problem in a more graph theoretic manner, which will help us with themore general problem concerning n-dominoes. In the world of graph theory, each edge connectstwo vertices, while in the world of dominoes, each domino arises from connecting two numberedsquares. This suggests that we should form the graph with seven vertices labelled 0, 1, 2, 3, 4, 5

174


and 6 with an edge between every pair of vertices as well as a loop from each vertex to itself. Nowevery domino corresponds to an edge in this graph. Furthermore, the problem of arranging themall in a loop so that adjacent halves of neighbouring dominoes are labelled by the same number isequivalent to the following question.

Is it possible to walk around the graph, traversing every edge exactly once, and finish at the samevertex that you started at?

Of course, you can just find such a path by tracing around the edges of the graph, but a muchslicker way is to invoke Euler’s theorem, which tells us that a connected graph — possibly withmultiple edges and loops — is Eulerian if and only if it has zero vertices with odd degree or twovertices with odd degree. Recall that a graph is said to be Eulerian if it’s possible to walk aroundthe graph, traversing every edge exactly once.

5

60

1

2

3

4

The graph we’re dealing with is certainly connected and every vertex has even degree. Therefore,Euler’s theorem tells us that it’s possible to walk around the graph, traversing every edge exactlyonce. The fact that we finish at the same vertex that we started at follows from the fact that thegraph has zero vertices of odd degree. For if we didn’t finish at the same vertex that we started at,then the start and end vertices would necessarily have odd degree.

(b) Our earlier reasoning suggests that we should form the graph with n + 1 vertices labelled0, 1, 2, . . . , n with an edge between every pair of vertices as well as a loop from each vertexto itself. Now every n-domino corresponds to an edge in this graph. Furthermore, the problem ofarranging them all in a loop so that adjacent halves of neighbouring dominoes are labelled by thesame number is equivalent to the following question.

Is it possible to walk around the graph, traversing every edge exactly once, and finish at the samevertex that you started at?

Again, we would like to invoke Euler’s theorem, which tells us that a connected graph — possiblywith multiple edges and loops — is Eulerian if and only if it has zero vertices with odd degree ortwo vertices with odd degree. If n is even, then the graph we’re dealing with is certainly connectedand every vertex has even degree — degree n + 2, in fact. Therefore, Euler’s theorem tells us thatit’s possible to walk around the graph, traversing every edge exactly once. The fact that we finishat the same vertex that we started at follows from the fact that the graph has zero vertices of odddegree. For if we didn’t finish at the same vertex that we started at, then the start and end verticeswould necessarily have odd degree.

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If n is odd, then the graph we’re dealing with is certainly connected but every vertex has odddegree — degree n + 2, in fact. Therefore, Euler’s theorem tells us that it’s not possible to walkaround the graph, traversing every edge exactly once. In summary, it is possible to arrange all ofthe n-dominoes in a loop so that adjacent halves of neighbouring n-dominoes are labelled by thesame number if and only if n is even.

4. Consider a graph G — without multiple edges or loops — whose vertices all have degree greater thanor equal to six.

(a) If G has V vertices and E edges, use the handshaking lemma to prove that V ≤ E3 .

(b) Suppose that G is planar and can be drawn in the plane in such a way that there are F faces. Usethe handshaking lemma on the dual graph to prove that F ≤ 2E

3 .

(c) Explain how these two inequalities imply that there does not exist a planar graph — without loopsor multiple edges — whose vertices all have degree greater than or equal to six.

[3 marks]

Proof.

(a) Since every vertex has degree greater than or equal to six, the sum of the degrees must be greaterthan or equal to 6V. However, the handshaking lemma states that the sum of the degrees isactually equal to 2E. Therefore, 2E ≥ 6V which rearranges to give V ≤ E

3 .

(b) If G is planar and can be drawn in the plane in such a way that there are F faces, then note thateach face must have at least three edges around it. This is due to our assumption that G has nomultiple edges or loops. The handshaking lemma on the dual graph states that the sum of thenumbers of edges around each face is equal to 2E. By our previous observation, the sum of thenumbers of edges around each face is at least 3F. Therefore, 2E ≥ 3F which rearranges to giveF ≤ 2E

3 .

(c) Suppose that there does exist a planar graph G — without multiple edges or loops — whosevertices all have degree greater than or equal to six. Then Euler’s formula states that V− E + F = 2.However, we can substitute the inequalities V ≤ E

3 and F ≤ 2E3 to obtain E

3 − E + 2E3 ≥ 2 which

rearranges to give 0 ≥ 2. This is clearly a contradiction, so we can deduce that there does not exista planar graph G — without multiple edges or loops — whose vertices all have degree greaterthan or equal to six.

5. Suppose that you have a polyhedron and you are told that each face is a quadrilateral or a hexagon andthat three faces meet at every vertex. Furthermore, every quadrilateral face shares an edge with fourhexagonal faces, while every hexagonal face shares an edge with three quadrilateral faces and threehexagonal faces.

(a) Deduce the number of quadrilateral faces and the number of hexagonal faces of the polyhedronand draw an example of such a polyhedron.

(b) Cut off each vertex of the polyhedron using a plane which passes through the midpoints of thethree edges which meet at that vertex. This process produces a convex polyhedron from a convexpolyhedron. For example, applying this process to a tetrahedron produces an octahedron. Howmany vertices, edges and faces does this new polyhedron have? How many triangular faces,quadrilateral faces, pentagonal faces, and hexagonal faces does this new polyhedron have?

[3 marks]

176


Proof.

(a) Let V be the number of vertices, E the number of edges, Q the number of quadrilateral faces, andH the number of hexagonal faces of the polyhedron. We will deduce these four values by writingdown four equations that they satisfy and solving them.

The first equation comes from applying the handshaking lemma to the polyhedron. Since threefaces meet at every vertex, every vertex of the polyhedron has degree three. Therefore, the sum ofthe degrees is simply 3V and we obtain the equation 3V = 2E.

The second equation comes from applying the handshaking lemma to the dual of the polyhedron.Since there are Q quadrilaterals and H hexagons, the sum of the numbers of edges around eachface is 4Q + 6H and we obtain the equation 4Q + 6H = 2E or equivalently, E = 2Q + 3H.

The third equation comes from a clever counting argument. The trick here is to count the numberof times a quadrilateral face shares an edge with a hexagonal face. This happens four times foreach quadrilateral face — in other words, the answer is 4Q. Arguing in a different way, we cansay that this happens three times for each hexagonal face — in other words, the answer is 3H. Ofcourse, these are just two answers to the same question, so we must have 4Q = 3H.

The fourth equation is simply Euler’s formula. Since 4Q = 3H, we may substitute H = 43 Q

into the equation E = 2Q + 3H — this gives us E = 6Q. And using the equation 3V = 2E, weobtain V = 4Q. Since every face is a quadrilateral or a hexagon, we also know that F = Q + Hor equivalently, F = 7

3 Q. And finally, we can substitute these expressions for V, E, F into Euler’sformula to obtain

V − E + F = 2 ⇒ 4Q− 6Q +73

Q = 2 ⇒ Q = 6.

The fact that H = 43 Q then leads to H = 8. This information can be used to draw the polyhedron

and you should obtain something like the diagram below.

(b) Note that the polyhedron drawn above has 24 vertices. After applying the process, each ofthe 14 faces of the original polyhedron becomes a little smaller, but retains its number of sides.However, 24 triangular faces are created, one for each vertex of the original polyhedron. Therefore,the resulting polyhedron has 38 faces in total — 24 triangular faces, 6 quadrilateral faces and 8hexagonal faces.

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By applying the handshaking lemma to the dual of the resulting polyhedron, we find that itsnumber of edges must satisfy 2E = 24× 3 + 6× 4 + 8× 6 or equivalently, E = 72. Therefore, theresulting polyhedron has 72 edges in total. Another way to obtain this result is to note that theresulting polyhedron has three edges for every vertex of the original polyhedron.

Now we can apply Euler’s formula to determine that the number of vertices of the resultingpolyhedron must satisfy V − E + F = 2 or equivalently, V = 36. Another way to obtain this resultis to note that the resulting polyhedron has one vertex for each edge of the original polyhedron.

6. For each of the following edge words, determine whether or not the corresponding surface is orientableand calculate its Euler characteristic. Also, for each surface, identify it as the sphere, a connect sum oftori or a connect sum of projective planes.

(a) abc−1b−1da−1d−1c

(b) abacb−1ded−1e−1c

(c) a1a2a3 · · · ana−11 a−1

2 a−13 · · · a−1

n

Hint: Try this problem for several values of n until you find a pattern.[3 marks]

Proof. The classification of surfaces implies that we can recognise a surface from its orientability and itsEuler characteristic. For this question, we will use this fact to identify surfaces only from their edgewords.

(a) The polygon model for the surface is shown below. It turns out that all of the vertices get gluedtogether.

Since we have V = 1, E = 4 and F = 1, we can calculate that the Euler characteristic is χ =V − E + F = 1− 4 + 1 = −2. The surface is orientable since every letter in the edge word appearswith its inverse. It follows that the surface is the connect sum of two tori — in other words, agenus 2 surface.

a

b

c

b

d

a

d

c

(b) The polygon model for the surface is shown on the right, with the numbers on the verticesindicating which vertices get glued together.

Since we have V = 2, E = 5 and F = 1, we can calculate that the Euler characteristic is χ =V − E + F = 2− 5 + 1 = −2. The surface is non-orientable since the letter a in the edge word doesnot appear with its inverse. It follows that the surface is the connect sum of four projective planes.

178


a

b

a

c

b d

e

d

e

c

2

2

1

2

1

2

1

2

2

2

(c) The polygon model for the surface in this case is hard to draw, but the surface is orientable sinceevery letter appears with its inverse. Furthermore, we know that E = n and F = 1, so it onlyremains to determine the value of V. Note that the following vertices get glued together, wheretailk denotes the vertex which is at the tail of the edge labelled ak and tipk denotes the vertex whichis at the tip of the edge labelled ak.

tail1 → tip2 → tail3 → tip4 → · · · → tailn−1 → tipn

Similarly, the following vertices get glued together as well.

tip1 → tail2 → tip3 → tail4 → · · · → tipn−1 → tailn

ana1

a2

a3

an a1

a2

a3

Now if you’ve played around with the problem enough, you will have realised that there are twocases to consider. When n is odd, tipn gets glued to tail1 and tailn gets glued to tip1, so there aretwo vertices, corresponding to the two sequences above.

On the other hand, when n is even, tipn gets glued to tip1 and tailn gets glued to tail1, so the twovertices corresponding to the two sequences above get glued together to give only one vertex.

So when n is odd, the Euler characteristic is 2− n + 1 = 3− n and the surface is a connect sumof n−1

2 tori. And when n is even, the Euler characteristic is 1− n + 1 = 2− n and the surface is aconnect sum of n

2 tori.

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7. The shaded region, pictured below, is a shape which gives rise to a surface after gluing the edgestogether as indicated.

(a) Show that it can be described by the edge word abc−1d−1ecadb−1e−1.

(b) Determine whether or not the surface is orientable, calculate its Euler characteristic and identitythe surface using the classification of surfaces.

(c) What is the shortest possible edge word for this surface?

(d) If the surface is homeomorphic to T # X where T represents a torus and X represents a surface,then what must X be?

[2 marks]

a c

d

b

b a

c

d

Proof.

(a) The idea here is that we can turn this shape with gluing instructions into a polygon with gluinginstructions which represents the same surface. We already know how to deal with polygonmodels, which will allow us to write down an edge word for the surface and identify exactly whatit is. The way to construct a polygon model is to simply cut the shape open to create a polygonand remember how the cuts glue together by inserting some gluing instructions. You can do allsorts of cuts, but the one we’ll choose is to cut from the top-left corner of the outer square to thetop-left corner of the inner square.

a c

d

b

b a

c

d

a c

d

b

b a

c

d

e

e

So an edge word for this surface can be obtained by reading around the boundary of the polygon.

180


If we do this starting in the top-left corner and moving around the polygon counterclockwise, weobtain abc−1d−1ecadb−1e−1.

(b) The polygon model for the surface is shown below. It turns out that all of the vertices get gluedtogether. Since we have V = 1, E = 5 and F = 1, we can calculate the Euler characteristicχ = V − E + F = 1− 5 + 1 = −3. The surface is non-orientable since the letter a in the edge worddoes not appear with its inverse. It follows that the surface is the connect sum of five projectiveplanes.

b

ea

b

c

d

e c

a

d

(c) This edge word contains ten letters and is, in fact, the equal shortest possible edge word for theconnect sum of five projective planes. Another way to express the connect sum of five projectiveplanes by an edge word with ten letters is aabbccddee.

(d) We know that T # X is the connect sum of five projective planes and hence, is non-orientable. SinceT is orientable, it follows that X must be non-orientable. Furthermore, we can use the relationχ(T # X) = χ(T) + χ(X)− 2 to deduce that χ(X) = −1. We may now use the classification ofsurfaces to deduce that X is the connect sum of three projective planes.

8. Given an edge word W, let σ(W) denote the surface corresponding to W. Prove the following facts,where a and b represent letters, W, X, Y, Z represent words, and P and T represent the projective planeand the torus, respectively.

(a) σ(aXa−1Y) ∼= σ(b−1XbY)

(b) σ(XY) ∼= σ(YX)

(c) σ(X) ∼= σ(X−1)

(d) σ(aa−1X) ∼= σ(X)

(e) σ(aXaY) ∼= P # σ(XY−1)

(f) σ(bWxXb−1Yx−1Z) ∼= T # σ(WZYX)

(g) T #P ∼= P #P #P[2 marks]

Proof.

(a) Clearly, the following two polygon models represent the same surface, since the gluing is per-formed in precisely the same way. Hence, we have σ(aXa−1Y) ∼= σ(b−1XbY).

181


a

a

XY

b

b

XY

(b) One polygon model can correspond to different edge words, depending on where we choose tostart reading the edges from. Hence, σ(XY) ∼= σ(YX).

X

Y

(c) One polygon model can also correspond to different edge words, depending on which directionwe choose to start reading the edges from. Hence, σ(X) ∼= σ(X−1).

X X

(d) As shown in the following diagram, if a appears next to a−1 in the edge word, then the twocorresponding edges in the polygon model can be glued together and removed from the edgeword. Hence, σ(aa−1X) ∼= σ(X).

a a

X

a

X

(e) The polygon model corresponding to the edge word aXaY is shown below left. Consider cuttingalong the diagonal labelled b and pasting the two edges labelled a together. The result is thepolygon model corresponding to the edge word bbXY−1 shown below right. Hence, σ(aXaY) ∼=σ(bbXY−1) ∼= σ(bb) # σ(XY−1) ∼= P # σ(XY−1).

182


a

a

XYb

b

X

bY

(f) The polygon model corresponding to the edge word aWbXa−1Yb−1Z is shown below left. Con-sider cutting along the diagonal labelled c and pasting the two edges labelled a together. Theresult is the polygon model corresponding to the edge word cbXWc−1ZYb shown below right.

X

a

Y

b

Z

a

W

b

c

W

c

Z

Y

b

c

b

X

Now consider cutting along the diagonal labelled d and pasting the two edges labelled b together.The result is the polygon model corresponding to the edge word cdZYXWc−1d−1 shown belowright.

W

c

Z

Y

b

c

b

Xd

Y

X

W

c

d

c

d

Z

Hence, σ(aWbXa−1Yb−1Z) ∼= σ(cdZYXWc−1d−1) ∼= σ(c−1d−1cdZYXW) ∼= T # σ(WZYX).

(g) The polygon model corresponding to the edge word aba−1b−1cc is shown below left. Considercutting along the diagonal labelled d and pasting the two edges labelled c together. This gives the

183


polygon model corresponding to the edge word a−1da−1b−1db−1 as shown below centre. Nowconsider cutting along the diagonal labelled e and pasting the two edges labelled a together. Thisgives the polygon model corresponding to the edge word d−1b−1db−1ee as shown below right.Finally, consider cutting along the diagonal labelled f and pasting the two edges labelled b together.This gives the polygon model corresponding to the edge word d−1d−1 f f ee or equivalently, xxyyzz.

b

c

c

a

b

a

d

b

d

b

a

d

ae

b

d

b

e

e

d

f

So we’ve managed to prove that σ(aba−1b−1cc) ∼= σ(xxyyzz), but this is equivalent to the equationT #P ∼= P #P #P.

9. Use the results from question 8 to prove the classification of surfaces — in other words, prove thatevery surface is either the sphere, a connect sum of tori or a connect sum of projective planes.

[BONUS]

Proof. The proof of the classification of surfaces consists of the sketch proof provided in the lecturenotes as well as the results from question 8.

10. Consider the pair (S, p) where S is a surface and p is a point on the surface. To the pair (S, p) weassociate a set G consisting of the paths in S which start at p and end at p. However, we consider twopaths a and b to be the same if you can slide the path a along the surface S, while keeping its start andend points at p, until you get the path b. If a and b are two loops on S which start and end at p, then wecan compose them to form the loop a · b which traverses a and then traverses b.

(a) Prove that the set G is a group under this composition.

(b) Prove that the group G does not depend on the choice of p — in other words, if you form thegroup G1 using the pair (S, p) and the group G2 using the pair (S, q), then G1

∼= G2.

(c) Let a and b be two paths in a torus which start and end at some point p such that a and b intersectonly once. Show that a · b = b · a.

(d) Describe the group G when the surface is a torus, a projective plane and a genus two surface.[CHALLENGE]

I won’t provide a proof to this question, but leave it as a challenge for you. The group G that I’ve definedis called the fundamental group of the surface S and is usually written as π1(S, p). The fundamentalgroup is a very useful tool and you can read about it in just about any introductory text on topology.

184

6. ASSIGNMENTS, TESTS AND EXAMINATIONS Sample Midterm Test

FAMILY NAME

GIVEN NAME(S)

STUDENT NUMBER

DATE any date

START TIME any time

FINISH TIME one hour after the start time

EXAMINER Norman Do

Instructions

1. Fill in your name and student number clearly in the space provided above.

2. Do not remove any pages from this booklet — all of your writing, even rough work, must be handed in.

3. Calculators, books and notes are not permitted.

4. You are encouraged to spend the first few minutes reading through the problems. The problems are notordered in terms of difficulty and the number of marks available does not indicate the difficulty of theproblem.

5. Please write your rough work and solution in the space provided on the page where the question isprinted. If you use up all of that space, then you may write on the back of the page or on one of thepages provided at the back of the booklet. If you use up all of that space as well, then please raise yourhand and ask for another sheet of paper.

6. The test booklet should consist of this cover, pages 1 through 6 which contain the questions, and pages7 through 10 which are blank. Please inform me if your booklet is defective.

185


1. For each of the following statements, indicate whether it is true or false. If you indicate that a statementis true, then you must explain why it is true and if you indicate that a statement is false, then you mustexplain why it is false.

(a) The centroid of a triangle always lies inside the triangle. (Recall that the centroid of a triangle iswhere the medians meet.)

(b) Hyperbolic geometry obeys all of Euclid’s axioms.

(c) The composition of a reflection followed by a rotation can only be a reflection.

(d) There exists a graph with seven vertices of degrees 0, 1, 1, 2, 2, 3, 3.

(e) It is possible to build a house with exactly six rooms such that each room has exactly five doorsand there are exactly three doors which lead outside.

[2 + 2 + 2 + 2 + 2 = 10 marks]

186


2. (a) Prove that the diameter of a circle subtends an angle of 90◦. In other words, if AB is the diameterof a circle and C is a point on the circle, then ∠ACB = 90◦. (You may only use basic facts aboutisosceles triangles, the sum of the angles in a triangle, and so on.)

(b) Start with a cyclic quadrilateral ABCD.

Let W be the point on BD such that AW is perpendicular to BD.Let X be the point on AC such that BX is perpendicular to AC.Let Y be the point on BD such that CY is perpendicular to BD.Let Z be the point on AC such that DZ is perpendicular to AC.

Prove that the quadrilateral WXYZ is cyclic.[3 + 3 = 6 marks]

187


3. (a) Fill in each entry of the middle column with the word direct or opposite and each entry of the rightcolumn with the word yes or no.

ISOMETRY DIRECT OR OPPOSITE FIXED POINTS

translation

rotation

reflection

glide reflection

(b) Suppose that ABCD is a rectangle with the vertices labelled counterclockwise and such thatBC = 2AB. Let

GCD denote the glide reflection consisting of a reflection in the line CD followed by a transla-tion which takes C to D;TDB denote the translation which takes D to B;RB denote a counterclockwise rotation by 90◦ about B; andMAB denote the reflection in the line AB.

Identify the composition MAB ◦ RB ◦ TDB ◦ GCD.[2 + 4 = 6 marks]

188



· a b c d e f g h

a g * e f h * d *b h f * c * b * ac * a f * d c h gd f * b g c d a ee c g * h * * b df a b c d e f * hg d e * * b g f *h * d g b a h c *

(b) In the lectures, we have seen two groups with eight elements — the cyclic group C8 and thedihedral group D4. Prove that G is isomorphic to one of these groups by writing down an explicitisomorphism. (You do not have to prove that it is an isomorphism.)

(c) Prove that the cyclic group C8 and the dihedral group D4 are not isomorphic.[2 + 2 + 2 = 6 marks]

189




HXHXHXHXHXHXHXHX

(c) Write down the RMG symbol for the wallpaper pattern pictured below left. Mark clearly on thediagram a point which is the centre of a rotational symmetry of order R, a point where M mirrorsmeet, and a point where G proper glide axes meet.

(d) Write down the RMG symbol for the wallpaper pattern pictured below right.

(e) Prove that if a symmetry group of a frieze pattern contains a reflection in a vertical mirror and arotation by 180◦, then it must also contain a glide reflection.

[1 + 1 + 2 + 1 + 2= 7 marks]

190


6. Suppose that you have a polyhedron with V vertices, E edges and F faces. You are told that each face isa triangle or a quadrilateral and that four faces meet at every vertex. Furthermore, every triangular faceshares an edge with three quadrilateral faces, while every quadrilateral face shares an edge with fourtriangular faces.

(a) Use the handshaking lemma applied to the polyhedron to prove that 4V = 2E.

(b) Let T be the number of triangular faces and Q be the number of quadrilateral faces of thepolyhedron. Use the handshaking lemma applied to the dual of the polyhedron to prove thatE = 3

2 T + 2Q.

(c) Prove the equation 3T = 4Q.

(d) Use these equations — along with anything else you know about the polyhedron — to deduce thevalues of T and Q.

[1 + 1 + 1 + 2 = 5 marks]

191

6. ASSIGNMENTS, TESTS AND EXAMINATIONS Sample Midterm Test Solutions

FAMILY NAME DO

GIVEN NAME(S) NORMAN

STUDENT NUMBER NOT APPLICABLE

DATE Sunday 16 May, 2010

START TIME 6:00pm

FINISH TIME 7:00pm

EXAMINER Norman Do

Instructions







192



(a) The centroid of a triangle always lies inside the triangle. (Recall that the centroid of a triangle iswhere the medians meet.)

(b) Hyperbolic geometry obeys all of Euclid’s axioms.

(c) The composition of a reflection followed by a rotation can only be a reflection.

(d) There exists a graph with seven vertices of degrees 0, 1, 1, 2, 2, 3, 3.

(e) It is possible to build a house with exactly six rooms such that each room has exactly five doorsand there are exactly three doors which lead outside.

[2 + 2 + 2 + 2 + 2 = 10 marks]

(a) TRUE. A median is a line segment which joins a vertex of a triangle to the midpoint of the oppositeside. Therefore, the three medians of a triangle lie inside the triangle and the point where theymeet — namely, the centroid — must also lie inside the triangle.

(b) FALSE. We know that in Euclidean geometry — in other words, the geometry which follows fromEuclid’s axioms — the parallel postulate holds. The parallel postulate states that, given a line anda point not on the line, there exists a unique line through the given point, parallel to the given line.

On the other hand, in hyperbolic geometry, given a line and a point not on the line, there existinfinitely many lines through the given point, parallel to the given line. This means that hyperbolicgeometry does not obey all of Euclid’s axioms.

(c) FALSE. The composition of a reflection by a rotation can be a glide reflection. For example,consider a square ABCD with the vertices labelled counterclockwise. Let M be the reflection in theline AC and let R be the rotation by 90◦ clockwise about D. You can check that the compositionR ◦M is a glide reflection along the axis XY where X is the midpoint of BC and Y is the midpointof AD.

(d) TRUE. You can find an example of such a graph pictured below.

0

1

1

2

2 3

3

(e) FALSE. Consider the graph which has one vertex corresponding to each room in the house aswell as one vertex for the outside — this makes seven vertices in total. Let the edges in the graphcorrespond to the doors in the house.

Since each room has exactly five doors, each vertex corresponding to a room must have degreeexactly five. Since there are exactly three doors which lead outside, the vertex corresponding tothe outside must have degree exactly three. Therefore, the sum of the degrees in the graph mustbe 6× 5 + 1× 3 = 33. The handshaking lemma asserts that the sum of the degrees in any graph is

193


equal to twice the number of edges in the graph. So we’ve deduced that the graph must have 16 12

edges, which is clearly a contradiction. Hence, it is not possible to build a house with exactly sixrooms such that each room has exactly five doors and there are exactly three doors which leadoutside.

194


2. (a) Prove that the diameter of a circle subtends an angle of 90◦. In other words, if AB is the diameterof a circle and C is a point on the circle, then ∠ACB = 90◦. (You may only use basic facts aboutisosceles triangles, the sum of the angles in a triangle, and so on.)

(b) Start with a cyclic quadrilateral ABCD.

Let W be the point on BD such that AW is perpendicular to BD.Let X be the point on AC such that BX is perpendicular to AC.Let Y be the point on BD such that CY is perpendicular to BD.Let Z be the point on AC such that DZ is perpendicular to AC.

Prove that the quadrilateral WXYZ is cyclic.[3 + 3 = 6 marks]

(a) Let O be the centre of the circle. The beauty of considering the centre is that we have the threeequal radii OA = OB = OC. Equal lengths, for obvious reasons, often lead to isosceles triangles,and our diagram happens to have two of them. There is the isosceles triangle OAC which meansthat we can label the equal angles ∠OAC = ∠OCA = x. There’s also the isosceles triangle OBC,which means that we can label the equal angles ∠OBC = ∠OCB = y. Labelling equal angles likethis is an extremely common and extremely useful trick.

A B

C

O

x y

x y

Now it’s time for some angle chasing. In particular, let’s consider the sum of the angles in triangleABC.

∠BAC + ∠ACB + ∠CBA = 180◦

We can replace all of these confusing angles with x’s and y’s in the following way.

x + (x + y) + y = 180◦

This equation is, of course, the same thing as 2(x + y) = 180◦ or, equivalently, x + y = 90◦. All wehave to do now is recognise that ∠ACB = x + y so we have proven that ∠ACB = 90◦.

(b) Hopefully, you remember that right angles often lead to cyclic quadrilaterals. In this particulardiagram, we happen to have four right angles and these lead to two cyclic quadrilaterals. One ofthem is quadrilateral ABWX, which is cyclic because ∠AWB = ∠AXB = 90◦. The other one isquadrilateral CDYZ, which is cyclic because ∠CYD = ∠CZD = 90◦. Our goal is to show that thequadrilateral WXYZ is also cyclic and we essentially know only two ways to do this. Either we

195


show that it has opposite angles which add to 180◦ or we show that it obeys the hockey theorem.For this particular problem, the latter is the right way to go.

A

B C

D

W

XY

Zx

So let’s start by labelling ∠XWY = x. The natural thing to do, of course, is label the anglenext door ∠XWB = 180◦ − x. Now we can use the fact that the quadrilateral ABWX is cyclicto label the opposite angle ∠XAB = x. Since the large quadrilateral ABCD is cyclic, we canuse the hockey theorem to label ∠CDB = ∠CAB = ∠XAB = x. In particular, this means that∠CDY = x. Now we can use the fact that the quadrilateral CDYZ is cyclic to label the oppositeangle ∠YZC = 180◦ − x. And again the natural thing to do is label the angle next door ∠XZY = x.But just hold on a second — we now know that ∠XWY = ∠XZY so, by the converse of the hockeytheorem, the quadrilateral WXYZ is indeed cyclic.

196




translation

rotation

reflection

glide reflection

(b) Suppose that ABCD is a rectangle with the vertices labelled counterclockwise and such thatBC = 2AB. Let

GCD denote the glide reflection consisting of a reflection in the line CD followed by a transla-tion which takes C to D;TDB denote the translation which takes D to B;RB denote a counterclockwise rotation by 90◦ about B; andMAB denote the reflection in the line AB.

Identify the composition MAB ◦ RB ◦ TDB ◦ GCD.[2 + 4 = 6 marks]

(a) The completed table is as follows.


translation DIRECT NO

rotation DIRECT YES

reflection OPPOSITE YES

glide reflection OPPOSITE NO

(b) With these sorts of problems, it’s useful to draw the rectangle on a grid of squares. It’s also usefulto keep in mind the table above and the fact that it applies to transformations which are not theidentity. In other words, the identity is a special case of an isometry which doesn’t fit into thetable.

A

B C

D

EF G

H

I

J

First, we note that MAB ◦ RB ◦ TDB ◦ GCD must be a direct isometry, because it’s the compositionof two direct isometries and two opposite isometries. So we know that it is either the identity,

197


a translation or a rotation, and to determine which of these it is, we’ll calculate MAB ◦ RB ◦TDB ◦ GCD(P) for a few points P. First, let’s consider what happens to the point C. It’s easyto check that GCD(C) = D, TDB(D) = B, RB(B) = B, and MAB(B) = B — in other words,MAB ◦ RB ◦ TDB ◦ GCD(C) = B.

Now, let’s consider what happens to the point A. It’s easy to check that GCD(A) = E as shownin the diagram, TDB(E) = D, RB(D) = F as shown in the diagram, and MAB(F) = G — in otherwords, MAB ◦ RB ◦ TDB ◦ GCD(A) = G.

At this stage, we can be sure that the composition MBC ◦ GAB ◦ RD ◦ TAC cannot be the identityor a translation. Since it takes the line segment AC to the line segment GB, we can try to guesswhere the centre of rotation is. It turns out that the centre of rotation is at the point H labelled inthe diagram. To prove this, all we need to do is show that H is a fixed point of the composition.It’s easy to check that GCD(H) = I as shown in the diagram, TDB(I) = H, RB(H) = J as shown inthe diagram, and MAB(J) = H — in other words, MAB ◦ RB ◦ TDB ◦ GCD(H) = H. Therefore, His the centre of the rotation and, since it takes A to G, it must be a rotation by ∠AHG = 90◦ in theclockwise direction or equivalently, 270◦ in the counterclockwise direction.

198



· a b c d e f g h

a g * e f h * d *b h f * c * b * ac * a f * d c h gd f * b g c d a ee c g * h * * b df a b c d e f * hg d e * * b g f *h * d g b a h c *

(b) In the lectures, we have seen two groups with eight elements — the cyclic group C8 and thedihedral group D4. Prove that G is isomorphic to one of these groups by writing down an explicitisomorphism. (You do not have to prove that it is an isomorphism.)

(c) Prove that the cyclic group C8 and the dihedral group D4 are not isomorphic.[2 + 2 + 2 = 6 marks]

(a) The missing entry h · a is already in the same row as a, b, c, d, g, h and already in the same columnas a, c, d, f , g, h. Since it must be one of the letters a, b, c, d, e, f , g, h, we can use the sudoku propertyto deduce that h · a = e. Using the same sort of strategy, we can complete the Cayley table for G asfollows.

· a b c d e f g h

a g c e f h a d bb h f d c g b e ac b a f e d c h gd f h b g c d a ee c g a h f e b df a b c d e f g hg d e h a b g f ch e d g b a h c f

(b) The group G is isomorphic to the group D4, the symmetry group of the square. If we draw thesquare ABCD with the vertices labelled counterclockwise and A in the top left corner, then wecan write the elements of D4 as follows.

I : the identity isometryR1 : rotation by 90◦ counterclockwiseR2 : rotation by 180◦ counterclockwiseR3 : rotation by 270◦ counterclockwise

Mh : reflection in a horizontal mirrorMv : reflection in a vertical mirrorMAC : reflection in the line ACMBD : reflection in the line BD

199


We may now write out the Cayley table for D4.










An explicit isomorphism F : G → D4 is given by the equations

F(a) = R1 F(b) = Mh F(c) = MBD F(d) = R3

F(e) = Mv F( f ) = I F(g) = R2 F(h) = MAC.This is certainly not the only possible isomorphism. To find one, you can use the fact that theidentity in G must correspond to the identity in D4. Furthermore, in G there are six elementswhich square to give the identity and in D4 there are six elements which square to give the identity— these six elements in G must correspond to these six elements in D4. In particular, this also tellsus that the elements a and d in G must correspond to the elements R1 and R3 in D4. After youhave obtained some facts like this, the problem can be finished with a little trial and error.

(c) All cyclic groups are abelian so, in particular, C8 is abelian. On the other hand, the dihedral groupD4 is not abelian since MBD ◦Mv 6= Mv ◦MBD. Therefore, the cyclic group C8 and the dihedralgroup D4 are not isomorphic to each other.

200




HXHXHXHXHXHXHXHX



(e) Prove that if a symmetry group of a frieze pattern contains a reflection in a vertical mirror and arotation by 180◦, then it must also contain a glide reflection.

[1 + 1 + 2 + 1 + 2= 7 marks]

(a) VRG

(b) HVRG

(c) 211 — A point which is the centre of a rota-tional symmetry of order 2, a point which 1 mir-ror passes through, and a point where 1 properglide axis passes through are labelled in the dia-gram on the right in red, green and blue, respec-tively.

(d) 101

201


(e) If we denote the reflection in a vertical mirror by V and the rotation by 180◦ by R, then thesymmetry group of the frieze pattern must also contain R ◦V. This composition is an oppositeisometry and it is easy to see that it is either reflection in a horizontal mirror or a glide reflectionalong a horizontal axis. (Remember that the centre of rotation for R does not have to lie on themirror for V.) Since the symmetry group of a frieze pattern contains a horizontal translation bydefinition, in either case, the symmetry group of the frieze pattern must contain a glide reflectionalong a horizontal axis.

So how do you know that the composition R ◦V is a reflection in a horizontal mirror or a glidereflection along a horizontal axis? One way to see this is to consider what happens to the pictureof a left footprint walking right, above the centre of rotation of R. After applying the reflectionV, this becomes a right footprint walking left, above the centre of rotation of R. And then afterapplying R, this becomes a right footprint walking right below the centre of rotation of R. Theonly way to start with a left footprint walking right and end up with a right footprint walkingright is via a reflection in a horizontal mirror or a glide reflection along a horizontal axis.

202


6. Suppose that you have a polyhedron with V vertices, E edges and F faces. You are told that each face isa triangle or a quadrilateral and that four faces meet at every vertex. Furthermore, every triangular faceshares an edge with three quadrilateral faces, while every quadrilateral face shares an edge with fourtriangular faces.


(b) Let T be the number of triangular faces and Q be the number of quadrilateral faces of thepolyhedron. Use the handshaking lemma applied to the dual of the polyhedron to prove thatE = 3

2 T + 2Q.

(c) Prove the equation 3T = 4Q.

(d) Use these equations — along with anything else you know about the polyhedron — to deduce thevalues of T and Q.

[1 + 1 + 1 + 2 = 5 marks]

(a) Let’s apply the handshaking lemma to the polyhedron. Since four faces meet at every vertex, everyvertex of the polyhedron has degree four. Therefore, the sum of the degrees is simply 4V and weobtain the equation 4V = 2E.

(b) Now let’s apply the handshaking lemma to the dual of the polyhedron. Since there are T trianglesand Q quadrilaterals, the sum of the numbers of edges around each face is 3T + 4Q and we obtainthe equation 3T + 4Q = 2E or equivalently, E = 3

2 T + 2Q.

(c) The trick here is to count the number of times a triangular face shares an edge with a quadrilateralface. This happens three times for each triangular face — in other words, the answer is 3T. Arguingin a different way, we can say that this happens four times for each quadrilateral face — in otherwords, the answer is 4Q. Of course, these are just two answers to the same question, so we musthave 3T = 4Q.

(d) Since 3T = 4Q, we may substitute T = 43 Q into the equation from part (b) — this gives us E = 4Q.

And using the equation from part (a), we obtain V = 2Q.

Since every face is a triangle or a quadrilateral, we also know that F = T + Q or equivalently,F = 7

3 Q. And finally, we can substitute these expressions for V, E, F into Euler’s formula to obtain

V − E + F = 2 ⇒ 2Q− 4Q +73

Q = 2 ⇒ Q = 6.

The fact that T = 43 Q then leads to T = 8.

203

6. ASSIGNMENTS, TESTS AND EXAMINATIONS Midterm Test

FAMILY NAME

GIVEN NAME(S)

STUDENT NUMBER

DATE Thursday 20 May, 2010

START TIME 11:05 am

FINISH TIME 12:20 pm

EXAMINER Norman Do

Instructions







204



(a) Let ABC be an acute triangle with circumcentre O and orthocentre H. If O and H are the samepoint, then the triangle must be equilateral. (Recall that the circumcentre of a triangle is where theperpendicular bisectors meet and the orthocentre of a triangle is where the altitudes meet.)

(b) In spherical geometry, given a line `, there exist infinitely many lines which are parallel to `.

(c) Every glide reflection which is not a reflection can be expressed as the composition of threereflections but cannot be expressed as the composition of fewer than three reflections.

(d) There are exactly three graphs — without loops or multiple edges — with three vertices, wherewe do not distinguish between two graphs if they are isomorphic.

(e) It is possible at a party with 100 people to have one person who knows no other people and oneperson who knows 99 other people. (You should assume that acquaintance is mutual so that ifperson A knows person B, then person B knows person A.)

[2 + 2 + 2 + 2 + 2 = 10 marks]

205


2. (a) Use the hockey theorem to prove that if ABCD is a cyclic quadrilateral, then

∠ABC + ∠CDA = 180◦ and ∠BCD + ∠DAB = 180◦.

(b) Let D, E, F be points on the sides BC, CA, AB of triangle ABC, respectively. Prove that thecircumcircles of triangles AEF, BFD and CDE intersect at a point.

Hint: Let the circumcircle of triangles BFD and CDE meet at the points D and P and prove thatthe quadrilateral AEPF is cyclic.

[3 + 3 = 6 marks]

206




translation

rotation

reflection

glide reflection

(b) Suppose that ABCD is a square with the vertices labelled counterclockwise. Let

TAC denote the translation which takes A to C;GBC denote the glide reflection consisting of a reflection in the line BC followed by a translationwhich takes B to C;RD denote a counterclockwise rotation by 90◦ about D; andMAD denote the reflection in the line AD.

Identify the composition TAC ◦ GBC ◦ RD ◦MAD.[2 + 3 = 5 marks]

207



· a b c d e f


(b) In the lectures, we have seen two groups with four elements — the cyclic group C6 and thedihedral group D3. Prove that G is isomorphic to one of these groups by writing down an explicitisomorphism. (You do not have to prove that it is an isomorphism.)

(c) Draw a subset of the Euclidean plane whose symmetry group is isomorphic to G. Does there exista finite set of points in the Euclidean plane whose symmetry group is isomorphic to G? If so, drawan example which uses the minimum number of points. If not, briefly explain why such a set doesnot exist.

[2 + 2 + 2 = 6 marks]

208



(b) Find a sequence of two or more distinct capital letters from the set {N, O, R, M} which can bewritten in a repeating pattern to form a frieze pattern whose HVRG symbol is R.



(e) Prove that if the symmetry group of a frieze pattern contains a rotation by 180◦ and a glidereflection, then it must also contain a reflection in a vertical mirror.

[1 + 1 + 2 + 1 + 2 = 7 marks]

209


6. Suppose that you have a polyhedron with V vertices, E edges and F faces. You are told that each faceis a triangle or a pentagon and that four faces meet at every vertex. Furthermore, every triangularface shares an edge with three pentagonal faces, while every pentagonal face shares an edge with fivetriangular faces.


(b) Let T be the number of triangular faces and P be the number of pentagonal faces of the polyhedron.Use the handshaking lemma applied to the dual of the polyhedron to prove that E = 3

2 T + 52 P.

(c) Prove the equation 3T = 5P.

(d) Use these equations — along with anything else you know about the polyhedron — to deduce thevalues of T and P.

(e) Draw an example of a planar graph corresponding to such a polyhedron. Draw an example ofsuch a polyhedron.

[1 + 1 + 1 + 2 + 1 = 6 marks]

210

6. ASSIGNMENTS, TESTS AND EXAMINATIONS Midterm Test Solutions

FAMILY NAME DO

GIVEN NAME(S) NORMAN

STUDENT NUMBER NOT APPLICABLE

DATE Thursday 20 May, 2010

START TIME 11:05 am

FINISH TIME 12:20 pm

EXAMINER Norman Do

Instructions







211



(a) Let ABC be an acute triangle with circumcentre O and orthocentre H. If O and H are the samepoint, then the triangle must be equilateral. (Recall that the circumcentre of a triangle is where theperpendicular bisectors meet and the orthocentre of a triangle is where the altitudes meet.)

(b) In spherical geometry, given a line `, there exist infinitely many lines which are parallel to `.

(c) Every glide reflection which is not a reflection can be expressed as the composition of threereflections but cannot be expressed as the composition of fewer than three reflections.

(d) There are exactly three graphs — without loops or multiple edges — with three vertices, wherewe do not distinguish between two graphs if they are isomorphic.

(e) It is possible at a party with 100 people to have one person who knows no other people and oneperson who knows 99 other people. (You should assume that acquaintance is mutual so that ifperson A knows person B, then person B knows person A.)

[2 + 2 + 2 + 2 + 2 = 10 marks]

(a) TRUE. Let the angles of triangle ABC be a, b, c. By considering the angle sum in triangle BCE andtriangle BCF, we can deduce that ∠HBC = 90◦ − c and ∠HCB = 90◦ − b.

A

B C

O

A

B CD

EF

H

If O and H are the same point, then AH = BH = CH and this means that triangle BCH is isosceles.Hence, ∠HBC = ∠HCB from which it follows that 90◦ − c = 90◦ − b and b = c. We can use thesame argument to prove that a = b, so all angles of triangle ABC are equal. Therefore, triangleABC is equilateral.

(b) FALSE. Remember that a line in spherical geometry is represented by a great circle — in otherwords, a circle on the sphere whose centre coincides with the centre of the sphere. Any two distinctgreat circles on the sphere intersect exactly twice. So in spherical geometry, given a line `, thereexist no lines which are parallel to `.

(c) TRUE. Recall that every isometry is the composition of at most three reflections. Since a glidereflection is an example of an opposite isometry, it must be the case that it’s the composition ofexactly one or three reflections. A glide reflection which is not a reflection clearly cannot be thecomposition of exactly one reflection. Therefore, it can be expressed as the composition of threereflections but cannot be expressed as the composition of fewer than three reflections.

212


(d) FALSE. There are actually four graphs — without loops or multiple edges — with three vertices,as shown in the diagram below.

(e) FALSE. If there is a loner at the party who knows no other people, then no one at the party canknow the loner. In particular, no one at the party can know everybody else at the party. So it’simpossible for someone to know 99 other people.

213


2. (a) Use the hockey theorem to prove that if ABCD is a cyclic quadrilateral, then

∠ABC + ∠CDA = 180◦ and ∠BCD + ∠DAB = 180◦.

(b) Let D, E, F be points on the sides BC, CA, AB of triangle ABC, respectively. Prove that thecircumcircles of triangles AEF, BFD and CDE intersect at a point.

Hint: Let the circumcircle of triangles BFD and CDE meet at the points D and P and prove thatthe quadrilateral AEPF is cyclic.

[3 + 3 = 6 marks]

(a) If we draw in the diagonals AC and BD, the hockey theorem tells us that there are equal anglesgalore. For example, we can label ∠ACB = ∠ADB = w, ∠BDC = ∠BAC = x, ∠CAD =∠CBD = y and ∠DBA = ∠DCA = z. You can go crazy labelling equal angles like this wheneverthere is a cyclic quadrilateral somewhere in your geometry diagram.

A B

CD

x zy

wz

xw

y

Now we’re going to add up all of the angles in triangle ABC and the answer should be 180◦.

∠CAB + ∠ABC + ∠BCa = 180◦

We can replace all of these confusing angles with w’s, x’s, y’s and z’s in the following way.

x + (y + z) + w = 180◦

This equation is, of course, the same thing as w + x + y + z = 180◦. All we have to do now isrecognise that ∠ABC = y + z and ∠CDA = w + x, so that

∠ABC + ∠CDA = (y + z) + (w + x) = 180◦.

You could also have chosen to recognise that ∠BCD = w + z and ∠DAB = x + y, so that

∠BCD + ∠DAB = (w + z) + (x + y) = 180◦.

(b) Let’s do as the hint suggests and let the circumcircle of triangles BFD and CDE meet at the pointsD and P. Our goal is to prove that the quadrilateral AEPF is cyclic.

214


B

C AE

FD

P

B

C AE

FD

P

x

Let’s remove the two circles from the diagram — since they are obfuscating what should be quitesimple — and just remember that the quadrilaterals BFPD and CDPE are cyclic. Our goal is toshow that the quadrilateral AEPF is also cyclic and we essentially know only two ways to do this.Either we show that it has opposite angles which add to 180◦ or we show that it obeys the hockeytheorem. For this particular problem, the former is the right way to go.

So let’s start by labelling ∠PEA = x. The natural thing to do, of course, is label the angle nextdoor ∠PEC = 180◦ − x. Now we can use the fact that the quadrilateral CDPE is cyclic to labelthe opposite angle ∠PDC = x. And again the natural thing to do is label the angle next door∠PDB = 180◦ − x. Now we can use the fact that the quadrilateral BFPD is cyclic to label theopposite angle ∠PFB = x. And once again the natural thing to do is label the angle next door∠PFA = 180◦ − x. But just hold on a second — we now know that

∠PEA + ∠PFA = x + (180◦ − x) = 180◦.

Therefore, the quadrilateral AEPF is indeed cyclic.

215




translation

rotation

reflection

glide reflection


TAC denote the translation which takes A to C;GBC denote the glide reflection consisting of a reflection in the line BC followed by a translationwhich takes B to C;RD denote a counterclockwise rotation by 90◦ about D; andMAD denote the reflection in the line AD.

Identify the composition TAC ◦ GBC ◦ RD ◦MAD.[2 + 3 = 5 marks]

(a) The completed table is as follows.


translation DIRECT NO

rotation DIRECT YES

reflection OPPOSITE YES

glide reflection OPPOSITE NO

(b) With these sorts of problems, it’s useful to draw the rectangle on a grid of squares. It’s also usefulto keep in mind the table above and the fact that it applies to transformations which are not theidentity. In other words, the identity is a special case of an isometry which doesn’t fit into thetable.

A B

CD

E

F

G

H

I J

K

O

216


First, we note that TAC ◦ GBC ◦ RD ◦MAD must be a direct isometry, because it’s the compositionof two direct isometries and two opposite isometries. So we know that it is either the identity, atranslation or a rotation, and to determine which of these it is, we’ll calculate TAC ◦GBC ◦RD ◦MAD

for a few points P. First, let’s consider what happens to the point A. It’s easy to check thatMAD(A) = A, RD(A) = C, GBC(C) = E, and TAC(E) = F — in other words, TAC ◦ GBC ◦ RD ◦MAD(A) = F.

Now, let’s consider what happens to the point B. It’s easy to check that MAD(B) = G, RD(G) = B,GBC(B) = C, and TAC(C) = H — in other words, TAC ◦ GBC ◦ RD ◦MAD(B) = H.

Now, let’s consider what happens to the point C. It’s easy to check that MAD(C) = I, RD(I) = A,GBC(A) = J, and TAC(J) = K — in other words, TAC ◦ GBC ◦ RD ◦MAD(C) = K.

At this stage, we can be sure that the composition TAC ◦ GBC ◦ RD ◦MAD cannot be the identity ora translation. So it’s a rotation and we can try to guess where its centre is. Note that the centre ofrotation must lie on the perpendicular bisector of AF as well as the perpendicular bisector of CH,so it appears that the centre of rotation is at the point O labelled in the diagram.

Note that ∠AOF = ∠BOH = ∠COK = 90◦ and we have the equations AO = FO, BO = HO,CO = KO. This means that the composition TAC ◦ GBC ◦ RD ◦MAD agrees with a rotation by 90◦

clockwise about O for the vertices of triangle ABC. It follows that the composition TAC ◦ GBC ◦RD ◦MAD is equal to a rotation by 90◦ clockwise about O.

217



· a b c d e f


(b) In the lectures, we have seen two groups with four elements — the cyclic group C6 and thedihedral group D3. Prove that G is isomorphic to one of these groups by writing down an explicitisomorphism. (You do not have to prove that it is an isomorphism.)

(c) Draw a subset of the Euclidean plane whose symmetry group is isomorphic to G. Does there exista finite set of points in the Euclidean plane whose symmetry group is isomorphic to G? If so, drawan example which uses the minimum number of points. If not, briefly explain why such a set doesnot exist.

[2 + 2 + 2 = 6 marks]

(a) If x is the identity of G, then it is the only element of the group which satisfies e · x = e. However,we can read from the Cayley table that e · e = e so it follows that e is the identity of G. We maynow label a · e = a and, in fact, we can label all entries in the row and column labelled by e. Theremainder of the Cayley table for G can be filled out using the sudoku property and the result isas follows.

· a b c d e f

a e d f b a cb d f e c b ac f e d a c bd b c a f d ee a b c d e ff c a b e f d

(b) The group G is isomorphic to the group C6, which consists of the rotational symmetries of a regularhexagon. Therefore, we can label the elements of C6 as follows.

I : the identity isometryR1 : rotation by 60◦ counterclockwiseR2 : rotation by 120◦ counterclockwise

R3 : rotation by 180◦ counterclockwiseR4 : rotation by 240◦ counterclockwiseR5 : rotation by 300◦ counterclockwise

We may now write out the Cayley table for C6.

218


◦ I R1 R2 R3 R4 R5

I I R1 R2 R3 R4 R5

R1 R1 R2 R3 R4 R5 IR2 R2 R3 R4 R5 I R1

R3 R3 R4 R5 I R1 R2

R4 R4 R5 I R1 R2 R3

R5 R5 I R1 R2 R3 R4


F(a) = R3 F(b) = R1 F(c) = R5

F(d) = R4 F(e) = I F( f ) = R2.This is certainly not the only possible isomorphism. To find one, you can use the fact that theidentity in G must correspond to the identity in C6. Furthermore, in G the only non-identityelement which squares to give the identity is a while in C6 the only non-identity element whichsquares to give the identity is R3. Hence, we can deduce that any isomorphism must take a to R3.After you’ve obtained some facts like this, the problem can be finished with a little trial and error.

(c) Usually, we describe C6 as the symmetry group of a decorated regular hexagon as shown belowleft. The decorated regular hexagon is an infinite set of points in the plane, but we can considerjust the end points of the six segments which make up the regular hexagon as shown below right.This gives a set of twelve points whose symmetry group is also C6. In fact, this is the smallest setof points in the Euclidean plane whose symmetry group is C6.

219



(b) Find a sequence of two or more distinct capital letters from the set {N, O, R, M} which can bewritten in a repeating pattern to form a frieze pattern whose HVRG symbol is R.




[1 + 1 + 2 + 1 + 2 = 7 marks]

(a) HVRG(b) · · · NONONONONO · · ·(c) 111

A point which is the centre of a rotational sym-metry of order 1, a point which 1 mirror passesthrough, and a point where 1 proper glide axispasses through are labelled in the diagram onthe right in red, green and blue, respectively.

(d) 101

(e) If we denote the rotation by 180◦ by R and the glide reflection in a horizontal axis by G, then thesymmetry group of the frieze pattern must also contain R ◦ G. This composition is an oppositeisometry and it’s easy to see that it must be a reflection in a vertical mirror. One way to see thisis to consider what happens to the picture of a left footprint walking right, above the horizontalaxis. After applying the glide reflection G, this becomes a right footprint walking right, below thehorizontal axis. And then after applying R, this becomes a right footprint walking left above thehorizontal axis. The only way to start with a left footprint walking right and end up with a rightfootprint walking left is via a reflection in a vertical mirror.

220


6. Suppose that you have a polyhedron with V vertices, E edges and F faces. You are told that each faceis a triangle or a pentagon and that four faces meet at every vertex. Furthermore, every triangularface shares an edge with three pentagonal faces, while every pentagonal face shares an edge with fivetriangular faces.


(b) Let T be the number of triangular faces and P be the number of pentagonal faces of the polyhedron.Use the handshaking lemma applied to the dual of the polyhedron to prove that E = 3

2 T + 52 P.

(c) Prove the equation 3T = 5P.

(d) Use these equations — along with anything else you know about the polyhedron — to deduce thevalues of T and P.

(e) Draw an example of a planar graph corresponding to such a polyhedron. Draw an example ofsuch a polyhedron.

[1 + 1 + 1 + 2 + 1 = 6 marks]

(a) Since four faces meet at every vertex, every vertex of the polyhedron has degree four. Therefore,the sum of the degrees is simply 4V and we obtain the equation 4V = 2E.

(b) Since there are T triangular faces and P pentagonal faces, the sum of the numbers of edges aroundeach face is 3T + 5P and we obtain the equation 3T + 5P = 2E or equivalently,

E =32

T +52

P.

(c) The trick here is to count the number of times a triangular face shares an edge with a pentagonalface. This happens three times for each triangular face — in other words, the answer is 3T. Arguingin a different way, we can say that this happens five times for each pentagonal face — in otherwords, the answer is 5P. Of course, these are just two answers to the same question, so we musthave 3T = 5P.

(d) Since 3T = 5P, we may substitute T = 53 P into the equation from part (b) — this gives us E = 5P.

Now we can use the equation from part (a) to deduce that V = 12 E = 5

2 P. Since every face is atriangle or a pentagon, we also know that F = T + P or equivalently, F = 8

3 P. And finally, we cansubstitute these expressions for V, E, F into Euler’s formula to obtain

V − E + F = 2 ⇒ 52

P− 5P +83

P = 2 ⇒ P = 12.

The fact that T = 53 P then leads to T = 20.

(e) Once we’ve worked out the values of T and P, it’s a simple matter to deduce that V = 30, E = 60and F = 32. You can try to draw the corresponding planar graph and the easiest way to do thisis in a symmetric fashion, as shown on the next page on the left. You can then try to draw thepolyhedron as a three-dimensional figure and you should get something like the diagram on thenext page on the right.

221


222

6. ASSIGNMENTS, TESTS AND EXAMINATIONS Sample Final Examination

FAMILY NAME

GIVEN NAME(S)

STUDENT NUMBER

DATE any date

START TIME any time

FINISH TIME three hours after the start time

EXAMINER Norman Do

Instructions





5. Please write your rough work and solution in the space provided on the page where the question isprinted. If you use up all of that space, then you may write on the back of the page or on one of thepages provided at the back of the booklet. If you use up all of that space as well, then please raise yourhand and ask the invigilator for another sheet of paper.

6. The examination booklet should consist of this cover, pages 1 through 10 which contain the questions,and pages 11 through 15 which are blank. Please inform the invigilator if your booklet is defective.

223



(a) The incentre of a triangle always lies inside the triangle. (Recall that the incentre of a triangle iswhere the angle bisectors meet.)

(b) Every isometry can be expressed as the composition of four or fewer reflections.

(c) If the circumcentre of triangle ABC and the centroid of triangle ABC are the same point, then trian-gle ABC must be equilateral. (Recall that the circumcentre of a triangle is where the perpendicularbisectors meet and the centroid of a triangle is where the medians meet.)

(d) In the hyperbolic plane, there exists a triangle with angles 1◦, 10◦ and 100◦.(e) Let e be the identity in a group G. If g and h are two elements in G such that g · h = e, then it is

also true that h · g = e.[2 + 2 + 2 + 2 = 10 marks]

224


2. Let the incentre of triangle ABC be I and let its incircle touch the sides AB and AC at R and Q,respectively. Let the line segment AI meet the incircle at J.

(a) Prove that the quadrilateral AQIR is cyclic.

(b) Prove that 2∠JRQ = ∠ARQ.

(c) Deduce that J is the incentre of triangle ARQ.[2 + 6 + 2 = 10 marks]

225




translation

rotation

reflection

glide reflection


GCB denote the glide reflection consisting of a reflection in the line CB followed by a translationwhich takes C to B; andMCD denote the reflection in the line CD;RC denote a counterclockwise rotation by 90◦ about C; andGAC denote the glide reflection consisting of a reflection in the line AC followed by a transla-tion which takes A to C.

Identify the composition GCB ◦MCD ◦ RC ◦ GAC.

(c) If X denotes the composition GCB ◦MCD ◦ RC ◦ GAC, let n be the minimum number of reflectionswhose composition is equal to X. Determine the value of n and carefully describe n reflectionswhose composition is equal to X.

[2 + 4 + 4 = 10 marks]

226



◦ a b c d e f

a * * * c b db a * * * * *c * * b * f *d f d * * * ae * * d * * cf * * a e * *

(b) In the lectures, we have seen two groups with six elements — C6 and D3 — which are notisomorphic to each other. Prove that G is isomorphic to one of these groups by writing down theexplicit isomorphism. How many isomorphisms are there?

(c) Give a subset of the Euclidean plane whose symmetry group is isomorphic to G. Does there exista finite set of points in the Euclidean plane whose symmetry group is isomorphic to G? If so,determine the minimum number of points in such a set. If not, give a brief explanation of whysuch a set does not exist.

(d) Why are the cyclic group C6 and the dihedral group D3 not isomorphic to each other?[3 + 3 + 2 + 2 = 10 marks]

227




VΛNVΛNVΛNVΛNVΛNVΛNV


(d) Write down the RMG symbol for the wallpaper pattern pictured below right. Mark clearly on thediagram a point which is the centre of a rotational symmetry of order R, a point where M mirrorsmeet, and a point where G proper glide axes meet.


[1 + 1 + 2 + 4 + 2 = 10 marks]

228


6. Suppose that you have a polyhedron with V vertices, E edges and F faces. You are told that each faceis a pentagon or a hexagon and that three faces meet at every vertex. Furthermore, every pentagonalface shares an edge with five hexagonal faces, while every hexagonal face shares an edge with threepentagonal faces and three hexagonal faces.

(a) Let P be the number of pentagonal faces and H be the number of hexagonal faces of the polyhedron.Prove that the following equations hold.

V =53

P + 2H E =52

P + 3H

(b) Prove the equation 5P = 3H.

(c) Use these equations — along with anything else you know about the polyhedron — to deduce thevalues of P and H.

(d) Draw a planar graph corresponding to such a polyhedron and draw an example of such apolyhedron.

(e) Cut off each vertex of the polyhedron using a plane which passes through the midpoints of thethree edges which meet at that vertex. This process produces a convex polyhedron from a convexpolyhedron. For example, applying this process to a tetrahedron produces an octahedron. Howmany vertices, edges and faces does this new polyhedron have? How many triangular faces,quadrilateral faces, pentagonal faces, and hexagonal faces does this new polyhedron have?

[3 + 1 + 2 + 2 + 2 = 10 marks]

229


7. (a) Fill in each entry of the middle column with the word orientable or non-orientable and each entry ofthe right column with a number or mathematical expression for the Euler characteristic.


sphere

connect sum of g tori

connect sum of n projective planes

(b) Determine whether or not the surface corresponding to the edge word aceb−1deabd−1c−1 is ori-entable and calculate its Euler characteristic.

(c) Identify the surface using the table above.

(d) Write down the shortest possible edge word for the surface.

(e) If the surface is homeomorphic to T # X where T represents a torus and X represents some surface,then what is X?

[4 + 2 + 2 + 2 = 10 marks]

230


8. (a) I would like to tile a 10× 10 square with a× b rectangles, where a and b are positive integerswhich may or may not be equal. Determine all values of a and b for which this is possible.

(b) Use a colouring argument to prove that it is not possible to tile a 10× 10 rectangle with 4× 1rectangles.

(c) Hence, deduce that it is not possible to tile a 10× 10 rectangle with 4× 5 rectangles.

(d) It is possible to tile a 10× 10 rectangle with thirty-three 3× 1 rectangles and one 1× 1 square.Determine all possible locations for the 1× 1 square and prove that these are the only ones possible.

[3 + 3 + 1 + 3 = 10 marks]

231


9. Let GXY denote the glide reflection which is the composition of a reflection in the line XY followed by atranslation which takes X to Y.

(a) If ABCD is a rectangle, show that the composition GDA ◦ GCD ◦ GBC ◦ GAB is equal to the identity.In other words, if you glide around a rectangle, then the result is the identity.

(b) If you glide around an arbitrary quadrilateral, show that the result is not necessarily the identity.

(c) Which quadrilaterals can you glide around so that the result is the identity?[6 + 2 + 2 = 10 marks]

232


10. (a) Determine all surfaces which can be obtained by gluing the sides of a quadrilateral in pairs.

(b) Determine all surfaces which can be obtained by gluing the sides of a hexagon in pairs.[6 + 4 = 10 marks]

233

6. ASSIGNMENTS, TESTS AND EXAMINATIONS Final Examination

FAMILY NAME

GIVEN NAME(S)

STUDENT NUMBER

DATE Wednesday 2 June, 2010

START TIME 2:00 pm

FINISH TIME 5:00 pm

EXAMINER Norman Do

ASSOCIATE EXAMINER Jacques Hurtubise

Instructions





5. Please write your rough work and solution in the space provided on the page where the question isprinted. If you use up all of that space, then you may write on the back of the page or on one of thepages provided at the back of the booklet. If you use up all of that space as well, then please raise yourhand and ask the invigilator for another sheet of paper.

6. The examination booklet should consist of this cover, pages 1 through 9 which contain the questions,and pages 10 through 15 which are blank. Please inform the invigilator if your booklet is defective.

234



(a) The orthocentre of a triangle always lies inside the triangle. (Recall that the orthocentre of atriangle is where the altitudes meet.)

(b) In the hyperbolic plane, given a line ` and a point P not on `, there exist infinitely many linesthrough P which are parallel to `.

(c) If the circumcentre of triangle ABC and the incentre of triangle ABC are the same point, then trian-gle ABC must be equilateral. (Recall that the circumcentre of a triangle is where the perpendicularbisectors meet and the incentre of a triangle is where the angle bisectors meet.)

(d) Suppose that the earth is a perfect sphere. It is possible to start at a town A, walk in a straight lineto the town B, turn left 135◦, walk in a straight line to the town C, turn left 135◦, walk in a straightline to the town A, turn left 135◦ and be facing the town B again.

[3 + 3 + 3 + 3 = 12 marks]

235


2. Let AD, BE, CF be the altitudes of an acute triangle ABC. Extend AD until it meets the circumcircle oftriangle ABC at X, as shown in the diagram below.

A

B CD

E

F

H

X

(a) If H is the orthocentre of triangle ABC, prove that ∠BHD = ∠BCA.

(b) Prove that triangle BHD is congruent to triangle BXD.

(c) Prove that triangle BHC is congruent to triangle BXC.

(d) Use the previous part to explain why the circumradius of triangle HBC is equal to the circumradiusof triangle ABC.

(e) Find two other triangles in the diagram which have the same circumradius as triangle ABC.[3 + 3 + 2 + 2 + 2 = 12 marks]

236


3. (a) Fill in each entry of the table below with one or more of the letters T, R, M, G where T denotes atranslation (which is not the identity), R denotes a rotation (which is not the identity), M denotesa reflection and G denotes a glide reflection. You should ignore the case when the composition oftwo isometries is the identity.

For example, the entry which is already filled in corresponds to the fact that a translation followedby a reflection can be a reflection or a glide reflection.

◦ translation rotation reflection glide reflection

translation

rotation

reflection R G

glide reflection


MBD denote the reflection in the line BD;RC denote a counterclockwise rotation by 90◦ about C;GBC denote the glide reflection consisting of a reflection in the line BC followed by a translationwhich takes B to C; andTAB denote the translation which takes A to B.

Identify the composition MBD ◦ RC ◦ GBC ◦ TAB.

(c) If X denotes the composition MBD ◦ RC ◦ GBC ◦ TAB, let n be the minimum number of reflectionswhose composition is equal to X. Determine the value of n and carefully describe n reflectionswhose composition is equal to X.

(d) Prove whether or not it is possible to write X as the composition of two glide reflections which arenot reflections.

[4 + 4 + 2 + 2 = 12 marks]

237



◦ a b c d e f

a * * * c b db a * * * * *c * * b * f *d f d * * * ae * * d * * cf * * a e * *

(b) In the lectures, we have seen two groups with six elements — C6 and D3 — which are notisomorphic to each other. Prove that G is isomorphic to one of these groups by writing down anexplicit isomorphism. How many isomorphisms are there?

(c) Give a subset of the Euclidean plane whose symmetry group is isomorphic to G. Does there exista finite set of points in the Euclidean plane whose symmetry group is isomorphic to G? If so,determine the minimum number of points in such a set. If not, give a brief explanation of whysuch a set does not exist.

(d) Why are the cyclic group C6 and the dihedral group D3 not isomorphic to each other?[3 + 3 + 2 + 2 = 10 marks]

238



(b) Find a sequence of one or more capital letters from the set {H, V, R, G} which can be written in arepeating pattern to form a frieze pattern whose HVRG symbol is the sequence itself.



(e) Prove that if the symmetry group of a frieze pattern contains a glide reflection and a reflection in avertical mirror, then it must also contain a rotation by 180◦.

[2 + 2 + 3 + 2 + 3 = 12 marks]

239


6. Suppose that you have a polyhedron with V vertices, E edges and F faces. You are told that each face isa triangle or a quadrilateral and that three faces meet at every vertex. Furthermore, every triangularface shares an edge with three quadrilateral faces, while every quadrilateral face shares an edge withtwo triangular faces and two quadrilateral faces.

(a) Let T be the number of triangular faces and Q be the number of quadrilateral faces of the polyhe-dron. Prove that the following equations hold.

3V = 2E E =32

T + 2Q

(b) Prove the equation 3T = 2Q.

(c) Use these equations — along with anything else you know about the polyhedron — to deduce thevalues of T and Q.

(d) Draw a planar graph corresponding to such a polyhedron and draw an example of such apolyhedron.

(e) Suppose that we mark two points on each edge of the polyhedron, one-third and two-thirds of theway along the edge. Cut off each vertex of the polyhedron using a plane which passes through themarked points of the three edges which meet at that vertex and which are closer to that vertex.This process — often called truncation — produces a convex polyhedron from a convex polyhedron.How many vertices, edges and faces does this new polyhedron have? How many triangular facesand hexagonal faces does this new polyhedron have?

[2 + 1 + 4 + 2 + 3 = 12 marks]

240


7. (a) Fill in each entry of the middle column with the word orientable or non-orientable and each entry ofthe right column with a number or mathematical expression for the Euler characteristic.


sphere

connect sum of g tori

connect sum of n projective planes

(b) The diagram below shows four squares which give rise to a surface after gluing the edges togetheras indicated. Show that an edge word for this surface is ac−1e−1dab−1cdbe.

b

c

d

g

f

g

b

e

h

a

f

a

e

c

h

d

(c) Determine whether or not the surface corresponding to the edge word ac−1e−1dab−1cdbe is ori-entable and calculate its Euler characteristic.

(d) Identify the surface using the table above.

(e) Write down the shortest possible edge word for the surface.

(f) If the surface is homeomorphic to T # X where T represents a torus and X represents some surface,then what is X?

[3 + 2 + 4 + 2 + 2 + 2 = 15 marks]

241


8. (a) I would like to tile a 4× 7 rectangle with a× b rectangles, where a and b are positive integerswhich may or may not be equal. Write down all values of a and b for which this is possible.

(b) Prove that it is not possible to tile a 4× 7 rectangle with 3× 1 rectangles.

(c) Prove that it is not possible to tile a 4× 7 rectangle with 2× 2 squares.

(d) It is possible to tile a 4× 7 rectangle with nine 3× 1 rectangles and one 1× 1 square. Determineall possible locations for the 1× 1 square and prove that these are the only ones possible.

(e) What is the maximum area of a 4× 7 rectangle that can be tiled with isosceles right-angled triangleswhose hypotenuses have length 2?

[4 + 2 + 2 + 5 + 2 = 15 marks]

242


9. You may only receive marks for one of the following four questions. Furthermore, you will notreceive any marks unless you have seriously attempted every other question on this examination.

Let OABC and OPQR be two squares with vertices labelled counterclockwise and which have acommon vertex O. Let M and N be the centres of the two squares and let K and L be the midpoints ofAR and CP. Prove that the quadrilateral KMLN is a square.

OR

Prove that in any finite symmetry group, either every isometry is direct or there is an equal number ofdirect and opposite isometries.

OR

Show that at any party with nine people, there exist four who all know each other or three who alldon’t know each other.

OR

Prove the Bolyai–Gerwien theorem, which asserts that two polygons are scissors congruent if and onlyif they have the same area.

[BONUS]

243

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