Math 161 - Notes

Neil Donaldson

Spring 2024

1 Geometry and the Axiomatic Method

1.1 The Early Origins of Geometry: Thales and Pythagoras

We begin with a condensed overview of geometric history. The word geometry comes from the an-

cient Greek geo (Earth), and metros (measure). Measurement (of distance, area, height, angle) had

obvious practical beneﬁts with regard to construction, taxation, commerce and navigation. Astron-

omy provided a related cultural driver of ancient geometry.

Ancient times (pre-500 BC) Egypt, Mesopotamia, China, India: basic rules for measuring lengths,

areas and volumes of simple shapes. Applications: surveying, tax collection, construction,

religious practice, astronomy, navigation. Typically worked examples without general for-

mulæ/abstraction.

Ancient Greece (from c. 600 BC) Philosophers such as Thales and Pythagoras began the process of

abstraction. General statements (theorems) formulated and proofs attempted. Concurrent de-

velopment of early scientiﬁc reasoning.

Euclid of Alexandria (c. 300 BC) Collected and expanded earlier work, especially that of the Pytha-

goreans. His compendium the Elements is one of the most important books in Western history

and remained a standard school textbook in to the 1900’s. The Elements is an early exemplar of

the axiomatic method at the heart of modern mathematics.

Later Greek Geometry Archimedes’ (c. 270–212 BC) work on area and volume included techniques

similar to those of modern calculus. Ptolemy (c. AD 100–170) writes the Almagest, a treatise on

astronomy which covers the foundations of trigonometry.

Post-Greek Geometry During the European Dark Ages, geometric understanding was developed

and enhanced by Indian and Islamic mathematicians who particularly developed trigonometry

and algebra.

Analytic Geometry In mid 1600’s France, Descartes and Fermat melded algebra with geometry with

the advent of co-ordinate systems (axes).

Modern Development Non-Euclidean geometries help provide the mathematical foundation for

Einstein’s relativity and the study of curvature. Following Klein (1872), modern geometry is

highly dependent on group theory.

Thales of Miletus (c. 624–546 BC) Thales was an olive trader from Miletus, a city-state on the west

coast of modern Turkey. Through trading and travelling, he absorbed mathematical ideas from

nearby cultures including Egypt and Mesopotamia. Here are ﬁve results partly attributable to Thales.

1. A circle is bisected by a diameter.

2. The base angles of an isosceles triangle are equal.

3. The pairs of angles formed by two intersecting lines are equal.

4. Two triangles are congruent if they have two angles and the included side equal.

5. An angle inscribed in a semicircle is a right angle.

The last is still known as Thales’ Theorem. Thales’ arguments were not rigorous by modern standards.

His real innovation was to state abstract, general principles: any circle is bisected by any of its diameters.

The Greek word θεωρεω (theoreo), from which we get theorem, has several meanings: ‘to look at,’

‘speculate,’ or ‘consider.’ Thales’ results were supposed to be clear just by looking at a picture.

The pictures show Thales’ Theorems 1, 2 and 5. Arguments for Theorems 1 and 2 could be as simple

as ‘fold.’ Theorem 5 follows from the observation that the radius of the circle splits the large triangle

into two isosceles triangles: Theorem 2 says that these have equal base angles (labelled), now check

that α + β is half the angles in a triangle, namely a right-angle.

Pythagoras of Samos (570—495 BC) Pythagoras grew up on Samos, an island in the Aegean Sea

not far from Miletus. He also travelled widely, eventually settling in Croton, southern Italy, around

530 BC where he founded a philosophical school devoted to the study of number, music and ge-

ometry. It has been claimed that the Pythagoreans ﬁrst classiﬁed the regular (Platonic) solids and

developed the musical relationship between the length of a vibrating string and its pitch. While it

is difﬁcult to verify such assertions, the Pythagorean obsession with number and the ‘music of the

universe’ certainly inspired later mathematicians and philosophers—particularly Euclid, Plato and

Aristotle—who believed they were reﬁning and clarifying this earlier work.

Of course, Pythagoras is best known for the result that bears his name.

Theorem 1.1 (Pythagoras). The square on the hypotenuse of a right triangle equals the sum of the

squares on the remaining sides.

Two important clariﬁcations are needed for modern readers.

1. By square, the Greeks meant an honest square! There is no algebra, no numerical lengths, and

the equation a

+ b

= c

won’t be seen for another 2000 years.

2. The word equals means equal area, though without a numerical concept of such. The Greeks

meant that the large square can be subdivided into pieces which may be rearranged to produce

the two squares on the remaining sides.

The result suddenly seems less easy! The pictures below provide a simple visualization.

A simple proof of Pythagoras’ Theorem

The problem with this ‘proof’ is that it relies on subtracting the areas of the four congruent triangles

from a very large square, whereas the Greeks idea of area was essentially additive. Book I of Euclid’s

Elements seems to have been structured precisely to correct this and provide a rigorous constructive

proof. Indeed it is possible (though very ugly!) to apply the 47 results leading up to and including his

proof of Pythagoras, in such as way as to explicitly subdivide the hypotenuse square and rearrange

it into the two smaller squares as required.

Much has been written about Pythagoras’ Theorem, including many, many proofs.

It is often

claimed that Pythagoras himself ﬁrst proved the result, but this is generally considered incorrect—

the ‘proof’ most often attributed to the Pythagoreans is based on contradictory ideas about numbers

which were debunked by the time of Aristotle. Moreover, other cultures, particularly ancient China,

are known to have used the result, at least in example form, several hundred years before Pythago-

ras. Regardless, an argument over attribution is fruitless without ﬁrst agreeing on what constitutes a

proof. This means that we need to spend some time considering Axiomatic Systems. . .

Exercises 1.1. 1. Let the side-lengths of the above triangles be a, b, c. Can you rephrase the proof

algebraically?

2. A theorem of Euclid states:

The square on the parts equals the sum of the squares on each part plus twice the

rectangle on the parts

By referencing the above picture, state Euclid’s result using modern algebra.

(Hint: let a and b be the ‘parts’. . . )

Including a proof by former US President James Garﬁeld. Would that current presidents were so learned. . .

In China, Pythagoras’ Theorem is known as the gou gu, which refers to the two non-hypotenuse sides of the triangle.

1.2 Axiomatic Systems

Arguably the most revolutionary aspect of the Euclid’s Elements was its axiomatic presentation.

Deﬁnition 1.2. An axiomatic system comprises four types of object.

1. Undeﬁned terms: Concepts accepted without deﬁnition/explanation.

2. Axioms: Logical statements regarding the undeﬁned terms which are accepted without proof.

3. Deﬁned terms: Concepts deﬁned in terms of 1 & 2.

4. Theorems: Logical statements deduced from 1–3.

Examples 1.3. Here are two systems viewed informally in this framework. In each case we provide

only examples of each type of object, not a full description of an axiomatic system.

Basic Geometry 1. Line and point.

2. There exists a line joining any two given points.

3. A triangle may be deﬁned in using three non-collinear points.

4. Thales’ and Pythagoras’ Theorems.

Chess 1. Pieces (as black/white objects) and the board.

2. Rules for how each piece moves.

3. Concepts such as check, stalemate or en-passant.

4. For example, Given a particular position, Black can win in 5 moves.

A proof is a logical argument demonstrating the truth of a theorem within an axiomatic system. In

practice, this is an ideal to which we aspire, and a proof is simply a convincing logical argument.

Deﬁnition 1.4. A model is a choice/deﬁnition of the undeﬁned terms such that all axioms are true.

Models are often abstract in that they depend on another axiomatic system. In a concrete model, the

undeﬁned terms are real-world objects (where contradictions are impossible(!)). The big idea is this:

Any theorem proved within an axiomatic system is true in any model of that system.

Mathematical discoveries often hinge on the realization that seemingly separate discussions can be

described in terms of models of a common axiomatic system.

Example 1.5. Monoid: If you’ve studied group theory, this should seem familiar.

1. A set G and a binary operation ∗.

2. (A1) Closure: ∀a, b ∈ G, a ∗b ∈ G

(A2) Associativity: ∀a, b, c ∈ G, a ∗ (b ∗c) = (a ∗b) ∗c

(A3) Identity: ∃e ∈ G such that ∀a ∈ G, a ∗ e = e ∗ a = a

3. Concepts such as square a

= a ∗ a, or commutativity a ∗b = b ∗ a.

4. For example, The identity is unique.

(G, ∗) = (Z, +) is an abstract model, where e = 0. If you really want a concrete model, consider a

single dot • on the page, equipped with the operation • ∗ • = •!

Deﬁnition 1.6. Certain properties are desirable in an axiomatic system.

Consistency The system is free of contradictions.

Independence An axiom is independent if it is not a theorem of the others. An axiomatic system is

independent if all its axioms are.

Completeness Every valid proposition within the theory is decidable: can be proved or disproved.

We unpack these ideas slightly. By necessity, our descriptions are vague. Many notions need to be

clariﬁed (e.g., what is meant by a valid proposition) before these ideas can be made rigorous.

Consistency May be demonstrated by exhibiting a concrete model. An abstract model demonstrates

relative consistency, dependent on the consistency of the underlying system. An inconsistent

system is essentially useless.

Independence To demonstrate the independence of an axiom, exhibit two models: one in which all

axioms are true, the other in which only the considered axiom is false.

Completeness This is very unlikely to hold for most useful axiomatic systems in mathematics, though

examples do exist. To show incompleteness, an undecidable

statement is required, which can

be viewed as a new independent axiom of an enlarged system.

Example (1.5, cont). The axiomatic system for a monoid is:

Consistent We have a (concrete) model.

Independent Consider three models:

• ( N, +) satisﬁes axioms A1 and A2 but not A3.

• ({e, a, b}, ∗) deﬁned by the following table satisﬁes A1 and A3 but not A2

∗ e a b

e e a b

a a e a

b b b a

e.g. a ∗ (b ∗b) = a ∗a = e = a = a ∗b = (a ∗b) ∗b

• ( Z \ {1}, +) satisﬁes axioms A2 and A3 but not A1.

Incomplete The proposition ‘A monoid contains at least two elements’ is undecidable just from the axioms.

For instance, ({0}, +) and (Z, +) are models with one/inﬁnitely many elements.

We could also ask if all elements have an inverse. That this is undecidable is the same as saying

that a new axiom is independent of A1, A2, A3.

(A4) Inverse: ∀g ∈ G, ∃g

−1

∈ G such that g ∗ g

−1

= g

−1

∗ g = e.

The new system deﬁned by the four axioms is also consistent and independent—this is the

structure of a group. Even this new system is incomplete; for instance, consider a new axiom of

commutativity. . .

A famous example of an undecidable statement from standard set theory is the Continuum Hypothesis, which states that

there is no uncountable set with cardinality strictly smaller than that of the real numbers.

Example 1.7 (Bus Routes). Here is a loosely deﬁned axiomatic system. Discuss the questions with

your classmates.

Undeﬁned Terms: Route, Stop

Axioms: (A1) Each route is a list of stops in some order. These are the stops visited by the route.

(A2) Each route visits at least four distinct stops.

(A3) No route visits the same stop twice, except the ﬁrst stop which is also the last stop.

(A4) There is a stop called downtown that is visited by each route.

(A5) Every stop other than downtown is visited by at most two routes.

1. Construct a model of this system with three routes. What is the fewest number of stops you

can use?

2. Your answer to 1 shows that this system is: complete, consistent, inconsistent, independent?

3. Is the following a model for the Bus Routes system? If not, determine which axioms are satisﬁed

by the model and which are not?

Stops: Downtown, Walmart, Albertsons, Main St., CVS, Trader Joe’s, Zoo

Route 1: Downtown, Walmart, Main St., CVS, Zoo, Downtown

Route 2: Main St., CVS, Zoo, Albertsons, Downtown, Main St.

Route 3: Walmart, Main St., Downtown, Albertsons, Main St., Walmart

4. Show that A3 is independent of the other axioms.

5. Demonstrate that ‘There are exactly three routes’ is not a theorem in this system by ﬁnding a

model in which it is not true.

We are only scratching the surface of axiomatics. If you really want to dive down the rabbit hole,

consider taking a class in formal logic or model theory. As an example of the ideas involved, we

ﬁnish with two results proved in 1931 by the German logician Kurt G

odel.

Theorem 1.8 (G¨odel’s incompleteness theorems).

1. Any consistent system containing the natural numbers is incomplete.

2. The consistency of such a system cannot be proved within the system itself.

odel’s ﬁrst theorem tells us that there is no ultimate consistent complete axiomatic system. Perhaps

this is reassuring—there will always be undecidable statements, so mathematics will never be ﬁn-

ished! However, the undecidable statements cooked up by G

odel are analogues of the famous liar

paradox (‘This sentence is false’), so the profundity of this is a matter of debate.

odel’s second theorem ﬂeshes out the difﬁculty in proving the consistency of an axiomatic system.

If a system is sufﬁciently complex to describe the natural numbers, its consistency can at best be

proved relative to some other axiomatic system. While an inconsistent system might be essentially

useless, good luck showing that what you have really is consistent!

Exercises 1.2. 1. Between two players are placed several piles of coins. On each turn a player takes

as many coins as they want from one pile, as long as they take at least one coin. The player who

takes the last coin wins.

If there are two piles where one pile has more coins than the other, prove that the ﬁrst player

can always win the game.

2. Consider a system where children in a classroom choose different ﬂavors of ice cream. Suppose

we have the following axioms:

(A1) There are exactly ﬁve ﬂavors of ice cream: vanilla, chocolate, strawberry, cookie

dough, and bubble gum.

(A2) Given any two distinct ﬂavors, there is exactly one child who likes these.

(A3) Every child likes exactly two ﬂavors of ice cream.

(a) How many children are in the classroom? Prove your assertion.

(b) Prove that any pair of children likes at most one common ﬂavor.

3. Consider an axiomatic system that consists of elements in a set S and a set P of pairings of

elements (a, b) that satisfy the following axioms:

(A1) If (a, b) is in P, then (b, a) is not in P.

(A2) If (a, b) is in P and (b, c) is in P, then (a, c) is in P.

(a) Let S = {1, 2, 3, 4} and P = {(1, 2), (2, 3), (1, 3)}. Is this a model for the axiomatic system?

Why/why not?

(b) Let S be the set of real numbers and let P consist of all pairs (x, y) where x < y. Is this a

model for the system? Explain.

of another independent axiom that could be added to the axioms A1 and A2 for which S

and P in part (a) is a model, but for which S and P from part (b) is not a model.

4. The undeﬁned terms of an axiomatic system are ‘brewery’ and ‘beer’. Here are some axioms.

(A1) Every brewery is a non-empty collection of at least two beers (every brewery

brews at least two beers!).

(A2) Any two distinct breweries have at most one beer in common.

(A3) Every beer belongs to exactly three breweries.

(A4) There exist exactly six breweries.

(a) Prove the following theorems.

i. There are exactly four beers.

ii. There are exactly two beers in each brewery.

iii. For each brewery, there is exactly one other brewery which has no beers in common.

(b) Prove that the axioms are independent.

(When negating A1, you should assume that a brewery is still a collection of beers, but that any

such could contain none or one beer)

2 Euclidean Geometry

2.1 Euclid’s Postulates and Book I of the Elements

Euclid’s Elements (c. 300 BC) formed a core part of European and Arabic curricula until the mid 20

century. Several examples are shown below.

Earliest Fragment c. AD 100 Full copy, Vatican, 9

C Pop-up edition, 1500s

Latin translation, 1572 Color edition, 1847 Textbook, 1903

Many of Euclid’s arguments can be found online, and you can read Byrne’s 1847 edition here: the

cover is Euclid’s proof of Pythagoras’. We present an overview of Book I.

Undeﬁned Terms E.g., point, line, etc.

Axioms/Postulates

A1 If two objects equal a third, then the objects are equal (= is transitive)

A2 If equals are added to equals, the results are equal (a = c & b = d =⇒ a + b = c + d)

A3 If equals are subtracted from equals, the results are equal

A4 Things that coincide are equal (in magnitude)

A5 The whole is greater than the part

P1 A pair of points may be joined to create a line

P2 A line may be extended

P3 Given a center and a radius, a circle may be drawn

P4 All right-angles are equal

P5 If a straight line crosses two others and the angles on one side sum to less than two right-

angles then the two lines (when extended) meet on that side.

In fact Euclid attempted to deﬁne these: ‘A point is that which has no part,’ and ‘A line has length but no breadth.’

In Euclid, an axiom is considered somewhat more general than a postulate. Here the postulates contain the geometry.

The ﬁrst three postulates describe the intuitive ruler and compass constructions. P4 allows Euclid to

compare angles at different locations. P5 is usually known as the parallel postulate.

Euclid’s system doesn’t quite ﬁt the modern standard. Some axioms are vague (what are ‘things’?)

and we’ll consider several more-serious shortcomings later. For now we clarify two issues and intro-

duce some notation.

Segments To Euclid, a line had ﬁnite extent—we call such a (line) segment. The segment joining points

A, B is denoted AB. In modern geometry, a line extends as far as is permitted.

Congruence Euclid uses equal where modern mathematicians say congruent. We’ll express, say, con-

gruent angles as ∠ABC

∼

∠DEF rather than ∠ABC = ∠DEF.

Basic Theorems `a la Euclid

Theorems were typically presented as a problem. Euclid ﬁrst provides a construction (P1–P3) before

proving that his construction solves the problem.

Theorem 2.1 (I. 1). Problem: to construct an equilateral triangle on a given segment.

The labelling I. 1 indicates Book I, Theorem 1.

Proof. Given a line segment AB:

By P3, construct circles centered at A and B with radius AB.

Call one of the intersection points C. By P1, construct AC and BC.

We claim that △ABC is equilateral.

Observe that AB and AC are radii of the circle centered at A, while

AB and BC are radii of the circle centered at B. By Axiom A1, the

three sides of △ABC are congruent.

Euclid proceeds to develop several well-known constructions and properties of triangles.

• (I. 4) Side-angle-side (SAS) congruence: if two triangles have two pairs of congruent sides and

the angles between these are congruent, then the remaining sides and angles are congruent in

pairs.











∼

∠ABC

∼

∠DEF

∼

=⇒











∼

∠BCA

∼

∠EFD

∠CAB

∼

∠FDE

• (I. 5) An isosceles triangle has congruent base angles.

• (I. 9) To bisect an angle.

• (I. 10) To ﬁnd the midpoint of a segment.

• (I. 15) If two lines/segments cut one another, opposite angles are congruent.

Have a look at some of Euclid’s arguments online. These are worth reading despite there being logical

issues with Euclid’s presentation. We’ll revisit these results in the Exercises and next two sections.

Parallel Lines: Construction & Existence

Deﬁnition 2.2. Lines are parallel if they do not intersect. Segments are parallel if no extensions of

them intersect.

In Euclid, a line is not parallel to itself. The next result is one of the most important in Euclidean

geometry, in that it describes how to create a parallel line through a given point.

Theorem 2.3 (I. 16 Exterior Angle Theorem). If one side of a

triangle is extended, then the exterior angle is larger than either

of the opposite interior angles.

In the picture, we have δ > α and δ > β.

Euclid did not quantify angles numerically: δ > α means that α is congruent to some angle inside δ.

Proof. Construct the bisector BM of AC (I. 10).

Extend BM to E such that BM

∼

ME (I. 2) and connect CE (P1).

The opposite angles at M are congruent (I. 15).

SAS (I. 4) applied to △AMB and △CME says ∠BAM

∼

∠EMC,

which is clearly smaller than the exterior angle at C.

Bisect BC and repeat the argument to see that β < δ.

The proof in fact constructs a parallel (CE) to AB through C, as the next result shows.

Theorem 2.4 (I. 27). If a line falls on two other lines such that the

alternate angles (α, β) are congruent, then the two lines are parallel.

The alternate angles in the exterior angle theorem are those at A and C: CE really is parallel to AB.

Proof. If the lines were not parallel, they would meet on one

side. WLOG suppose they meet on the right side at C.

The angle β at B, being exterior to △ABC, must be greater than

the angle α at A (I. 16): contradiction.

Euclid combines this with the vertical angles theorem (I. 15) to ﬁnish the ﬁrst half of Book I.

Corollary 2.5 (I. 28). If a line falling on two other lines makes con-

gruent angles, then the two lines are parallel.

Thus far, Euclid uses only postulates P1–P4. In any model in which these hold:

Given a line ℓ and a point C not on ℓ, there exists a parallel to ℓ through C

Parallel Lines: Uniqueness, Angle-sums & Playfair’s Postulate

Euclid ﬁnally invokes the parallel postulate to prove the converse of I. 27, showing that the congruent

alternate angle approach is the only way to have parallel lines.

Theorem 2.6 (I. 29). If a line falls on two parallel lines, then the alternate angles are congruent.

Proof. Given the picture, we must prove that α

∼

β.

Suppose not and WLOG that α > β.

But then β + γ < α + γ, which is a straight edge.

By the parallel postulate, the lines ℓ, m meet on the left side of

the picture, whence ℓ and m are not parallel.

ℓ

The most well-known result about triangles is now in our grasp, that the interior angles sum to a

straight edge. Euclid words this slightly differently.

Theorem 2.7 (I. 32). If one side of a triangle is extended, the exterior angle is congruent to the sum

of the opposite interior angles.

This is not a numerical sum, though for familiarity’s sake we’ll

often write 180° for a straight edge and 90° for a right-angle.

In the picture we’ve labelled angles as Greek letters for clarity.

The result amounts to showing that

α +

∼

α + β.

Proof. Construct CE parallel to BA as in I. 16, so that

∼

α.

BD falls on parallel lines AB and CE, whence

∼

β (Corollary of I. 29).

Axiom A2 shows that ∠ACD =

α +

∼

α + β.

The parallel postulate is stated in the negative (angles don’t sum to a straight edge, therefore lines are

not parallel). Though we cannot be sure, Euclid possibly chose this formulation in order to facilitate

proofs by contradiction. Unfortunately the effect is to obscure the meaning of the parallel postulate.

Here is a more modern interpretation.

Axiom 2.8 (Playfair’s Postulate). Given a line ℓ and a point C

not on ℓ , at most one parallel line m to ℓ passes through C.

ℓ

Our discussion up to now shows that the parallel postulate

implies Playfair.

• Let A, B ∈ ℓ and construct the triangle △ABC.

• The exterior angle theorem constructs E and thus a par-

allel m to ℓ by I. 27.

• I. 29 invokes the parallel postulate to prove that this is

the only such parallel.

ℓ

In fact the postulates are equivalent.

Theorem 2.9. In the presence of Euclid’s ﬁrst four postulates, Playfair’s postulate and the parallel

postulate (P5) are equivalent.

Proof. (P5 ⇒ Playfair) We proved this above.

(Playfair ⇒ P5) We prove the contrapositive. Assume postulates P1–P4 are true and that P5 is false.

Using quantiﬁers, and with reference to the picture in I. 29, we restate the parallel postulate:

P5: ∀ pairs of lines ℓ, m and ∀ crossing lines n, β + γ < 180° =⇒ ℓ , m not parallel.

Its negation (P5 false) is therefore:

∃ parallel lines ℓ, m and a crossing line n for

which β + γ < 180°

This is without loss of generality: if β + γ > 180°, consider

the angles on the other side of n.

By the the exterior angle theorem/I. 28, we may build a

parallel line

ℓ to ℓ through the intersection C of m and n

(in the picture,

∼

β). Crucially, this only requires postu-

lates P1–P4!

ℓ

Observe that

ℓ and m are distinct since

β + γ

∼

β + γ < 180°. We therefore have a line ℓ and a

point C not on ℓ, though which pass (at least) two parallels to ℓ : Playfair’s postulate is false.

Non-Euclidean Geometry

That Euclid waited so long before invoking the uniqueness of parallels suggests he was trying to

establish as much as he could about triangles and basic geometry in its absence. By contrast, every-

thing from I. 29 onwards relies on the parallel postulate, including the proof that the angle sum in a

triangle is 180°. For centuries, many mathematicians believed, though none could prove it, that such

a fundamental fact about triangles must be true independent of the parallel postulate.

Loosely speaking, a non-Euclidean geometry is a model for which a parallel through an off-line point

either doesn’t exist or is non-unique. It wasn’t until the 17–1800s and the development of hyperbolic

geometry (Chapter 4) that a model was found in which Euclid’s ﬁrst four postulates hold but for which

the parallel postulate is false.

We shall eventually see that every triangle in hyperbolic geometry has angle

sum less than 180°, though this will require a lot of work! For a more eas-

ily visualized non-Euclidean geometry consider the sphere. A rubber band

stretched between three points on its surface describes a spherical triangle: an

example with angle sum 270° is drawn. A similar game can be played on a

saddle-shaped surface: as in hyperbolic geometry, ‘triangles’ will have angle

sum less than 180°.

This shows that the parallel postulate is independent; in fact all Euclid’s postulates are independent. They are also

consistent (the ‘usual’ points and lines in the plane are a model), but incomplete: a sample undecidable is in Exercise 5.

Pythagoras’ Theorem

Following his discussion of parallels, Euclid shows that parallelograms with the same base and

height are equal (in area) (I. 33–41), before providing constructions of parallelograms and squares

(I. 42–46). Some of this is in Exercise 2. Immediately afterwards comes the capstone of Book I.

Theorem 2.10 (I. 47 Pythagoras’ Theorem). The square on the hypotenuse of a right triangle equals

(has the same area as) the sum of the squares on the other sides.

Proof. The given triangle △ABC is assumed to have a right-angle at A.

1. Construct squares on each side of △ABC (I. 46) and a

parallel AL to BD (I. 16).

2. AB

∼

FB and BD

∼

BC since sides of squares are con-

gruent. Moreover ∠ABD

∼

∠FBC since both contain

∠ABC and a right-angle.

3. Side-angle-side (I. 4) says that △ABD and △FBC are

congruent (identical up to rotation by 90°).

4. I. 41 compares areas of parallelograms and triangles

with the same base and height:

Area(□ABFG) = 2 Area(△FBC)

= 2 Area(△ABD)

= Area



BOLD



5. Similarly Area(□ACKH) = Area



OCEL



LD E

6. Sum the rectangles to obtain □BCED and complete the proof.

Euclid ﬁnishes Book I with the converse, which we state without proof. Euclid’s argument is very

sneaky—look it up!

Theorem 2.11 (I. 48). If the (areas of the) squares on two sides of a triangle equal the (area of the)

square on the third side, then the triangle has a right-angle opposite the third side.

The Elements contains thirteen books. Much of the remaining twelve discuss further geometric con-

structions, including in three dimensions. There is also a healthy dose of basic number theory includ-

ing what is now known as the Euclidean algorithm.

While undoubtedly a masterpiece of logical reasoning, Euclid’s presentation has several ﬂaws. Most

problematic is his reliance on pictorial reasoning: for instance, he ‘proves’ the SAS and SSS congru-

ence theorems (I. 4 & 8) by laying one triangle on top of another, a process not justiﬁed by his axioms

(look it up online or Byrne). In a modern sense, Euclid’s approach is part axiomatic system and part

model: his reasoning requires a visual/physical representation of lines, circles, etc. Because of these

issues, we now turn to a more modern description of Euclidean geometry, courtesy of David Hilbert.

Exercises 2.1. 1. (a) Prove the vertical angle theorem (I. 15): if two lines cut one another, opposite

angles are congruent.

(Hint: This is one place where you will need to use postulate 4 regarding right-angles)

(b) Use part (a) to complete the proof of the exterior angle theorem: i.e., explain why β < δ.

2. To help prove Pythagoras’, Euclid makes use of the following results. Prove them as best as

you can. Full rigor is tricky, but the pictures should help!

(a) (I. 11) At a given point on a line, to construct a perpendicular.

(b) (I. 46) To construct a square on a given segment.

(d) (I. 41) A parallelogram has twice the area of a triangle on the same base and with the same

height.

D EF

Theorem I. 11 Theorem I. 46 Theorem I. 35

3. Consider spherical geometry (page 12), where lines are paths of shortest distance (great circles).

(a) Which of Euclid’s postulates P1–P5 are satisﬁed by this geometry?

(b) (Hard) Where does the proof of the exterior angle theorem fail in spherical geometry?

4. (a) State the negation of Playfair’s postulate.

(b) Prove that Playfair’s postulate is equivalent to the following statement:

Whenever a line is perpendicular to one of two parallel lines, it must be perpen-

dicular to the other.

5. The line-circle continuity property states:

If point P lies inside and Q lies outside a circle α, then the segment PQ intersects α.

By considering the set of rational points in the plane Q

= {(x, y) : x, y ∈ Q}, and making a

sensible deﬁnition of line and circle, show that the line-circle continuity property is undecidable

within Euclid’s system.

6. The standard proof of the converse of Pythagoras’ theorem (I. 48) is, in fact, a corollary of the

original! Look it up and explain the argument as best you can.

2.2 Hilbert’s Axioms I: Incidence and Order

The long process of identifying and correcting the errors and omissions in Euclid’s Elements culmi-

nated in the 1899 publication of David Hilbert’s Grundlagen der Geometrie (Foundations of Geometry).

In the next two sections we consider some of the details of Hilbert’s approach, providing a modern

and logically superior description of Euclidean geometry.

Hilbert’s axioms for plane geometry

are listed on the next page. The undeﬁned terms consist of two

types of object (points and lines), and three relations (between ∗, on ∈ and congruence

∼

). For brevity

we’ll often use/abuse set notation, viewing a line as a set of points, though this is not necessary. At

various places, deﬁnitions and notations are required.

Deﬁnition 2.12. Throughout, A, B, C denote points and ℓ, m lines.

Line:

←→

AB denotes the line through distinct A, B. This exists and is unique by axioms I-1 and I-2.

Segment: AB := {A, B} ∪ {C : A ∗ C ∗ B} consists of distinct endpoints A, B and all interior points C

lying between them.

Ray:

−→

AB := AB ∪ {C : A ∗ B ∗C} is a ray with vertex A. In essence we extend AB beyond B.

Triangle: △ABC := AB ∪ BC ∪ CA where A, B, C are non-collinear. Triangles are congruent if their

sides and angles are congruent in pairs.

Sidedness: Distinct A, B, not on ℓ, lie on the same side of ℓ if AB ∩ ℓ = ∅. Otherwise A and B lie on

opposite sides of ℓ .

Angle: ∠BAC :=

−→

AB ∪

−→

AC has vertex A and sides

−→

AB and

−→

AC.

Parallelism: Lines ℓ and m intersect if there exists a point lying on both: ∃A ∈ ℓ ∩m. Lines are parallel

if they do not intersect. Segments/rays are parallel when the corresponding lines are parallel.

The pictures represent these notions in the usual model of Cartesian geometry.

Line

←→

AB Segment AB Ray

−→

AB Triangle △ABC

ℓ

Same side Opposite sides Angle ∠BAC Intersection A ∈ ℓ ∩ m

Like Euclid, Hilbert also covered 3D geometry—we only give the axioms for plane geometry. With regard to our

desired properties (Deﬁnition 1.6), his system is about as good as can be hoped. Essentially one only one model exists,

which is almost the same thing as completeness. In the absence of the continuity axiom, the axioms are consistent; in

line with G

odel’s theorems (1.8), consistency cannot be proved once continuity is included. As stated, the axioms are not

quite independent, though this can be remedied: O-3 does not require existence (follows from Pasch’s axiom), C-1 does

not require uniqueness (follows from uniqueness in C-4) and C-6 can be weakened slightly.

Hilbert’s Axioms for Plane Geometry

Undeﬁned terms

1. Points: use capital letters, A, B, C . . .

2. Lines: use lower case letters, ℓ , m, n, . . .

3. On: A ∈ ℓ is read ‘A lies on ℓ’

4. Between: A ∗B ∗C is read ‘B lies between

A and C’

5. Congruence:

∼

is a binary relation on

segments or angles

Axioms of Incidence

I-1 For any distinct A, B there exists a line ℓ

on which lie A, B.

I-2 There is at most one line through distinct

A, B (A and B both on the line).

Notation: line

←→

AB through A and B

I-3 On every line there exist at least two dis-

tinct points. There exist at least three

points not all on the same line.

Axioms of Order

O-1 If A ∗ B ∗ C, then A, B, C are distinct

points on the same line and C ∗ B ∗ A.

O-2 Given distinct A, B, there is at least one

point C such that A ∗ B ∗ C.

O-3 If A, B, C are distinct points on the same

line, exactly one lies between the others.

Deﬁnitions: segment AB and triangle △ABC

O-4 (Pasch’s Axiom) Let △ABC be a triangle

and ℓ a line not containing any of A, B, C.

If ℓ contains a point of the segment AB,

then it also contains a point of either AC

or BC.

Deﬁnitions: sides of line

←→

AB and ray

−→

Axioms of Congruence

C-1 (Segment transference) Let A, B be distinct

and r a ray based at A

′

. Then there exists a

unique point B

′

∈ r for which AB

∼

′

Moreover AB

∼

BA.

C-2 If AB

∼

EF and CD

∼

EF, then AB

∼

CD.

C-3 If A ∗ B ∗ C, A

′

∗ B

′

∗ C

′

, AB

∼

′

and

∼

′

, then AC

∼

′

Deﬁnitions: angle ∠ABC

C-4 (Angle transference) Given ∠BAC and

−−→

′

, there exists a unique ray

−−→

′

a given side of

←−→

′

for which ∠BAC

∼

∠B

′

C-5 If ∠ABC

∼

∠GHI and ∠DEF

∼

∠GHI,

then ∠ABC

∼

∠DEF. Moreover, ∠ABC

∼

∠CBA.

C-6 (Side-angle-side) Given triangles △ABC

and △A

′

, if AB

∼

′

, AC

∼

′

and ∠BAC

∼

∠B

′

, then the triangles

are congruent.

Axiom of Continuity

Suppose a line/segment is partitioned into non-

empty subsets Σ

, Σ

such that no point of Σ

lies

between two points of Σ

and vice versa.

Then there exists a unique point O satisfying

∗ O ∗ P

, if and only if O = P

, O = P

and

one of P

or P

lies in Σ

and the other in Σ

Playfair’s Axiom

Deﬁnition: parallel lines

Given a line ℓ and a point P /∈ ℓ, at most one line

through P is parallel to ℓ.

Its sides/angles are congruent in pairs. We extend congruence to other geometric objects similarly.

Axioms of Incidence: Finite Geometries

The axioms of incidence describe the relation on. An incidence geometry is any model satisfying axioms

I-1, I-2, I-3. Perhaps surprisingly, there exist incidence geometries with ﬁnitely many points!

Examples 2.13. By I-3, an incidence geometry requires at least three points.

A 3-point geometry exists, and is unique up to relabelling:

I-3 says the points A, B, C must be non-collinear. By I-1 and

I-2, each pair lies on a unique line, whence there are precisely

three lines

ℓ = {A, B}, m = {A, C}, n = {B, C}

Up to relabelling, there are two incidence geometries with four

points: one is drawn; how many lines has the other?

ℓ

3 points, 3 lines 4 points, 6 lines

The ﬁnal picture is a seven-point incidence geometry called the Fano

plane, which ﬁnds many applications particularly in combinatorics. Each

point lies on precisely three lines and each line contains precisely three

points—each dot is colored to indicate the lines to which it belongs.

Don’t be fooled by the black line looking ‘curved’ and seeming to cross

the blue line near the top, for the line really only contains three points!

We can even prove some simple theorems in incidence geometry.

Lemma 2.14. If distinct lines intersect, then they do so in exactly one point.

Proof. Suppose A, B are distinct points of intersection. By axiom I-2, there is at most one line through

A and B. Contradiction.

Lemma 2.15. Given any point, there exist at least two lines on which it lies.

The proof is an exercise. While incidence geometry is fun, our main goal is to understand Euclidean

geometry, so we move on to the next set of axioms.

Axioms of Order: Sides of a Line, Pasch’s Axiom & the Crossbar Theorem

The axioms of order describe the ternary relation between. Their inclusion in Hilbert’s axioms is due

in no small part to the work of Moritz Pasch, after whom Pasch’s axiom (O-4, c. 1882) is named. This

axiom is very powerful; in particular, it permits us to deﬁne the interiors of several geometric objects,

and to see that these are non-empty.

Lemma 2.16. Every segment contains an interior point.

We leave the proof to Exercise 5. By inducting on the Lemma, every segment contains inﬁnitely many

points, whence the above ﬁnite geometries are not valid models once the order axioms are included.

To get much further, it is necessary to establish that a line has precisely two sides (Deﬁnition 2.12). This

concept lies behind several of Euclid’s arguments, without being properly deﬁned in the Elements.

Theorem 2.17 (Plane Separation). A line ℓ separates all points not on ℓ into two half-planes: the two

sides of ℓ . To be explicit, suppose none of the points A, B, C lie on ℓ, then:

1. If A, B lie on the same side of ℓ and B, C lie on the same side, then A, C lie on the same side.

2. If A, B lie on opposite sides and B, C lie on opposite sides, then A, C lie on the same side.

3. If A, B lie on opposite sides and B, C lie on the same side, then A, C lie on opposite sides.

ℓ

Case 1 Case 2 Case 3

Proof. We prove the contrapositive of case 1. Suppose A, B, C are non-collinear. If AC intersects ℓ ,

then ℓ intersects one side of △ABC. By Pasch’s axiom, it also intersects either AB or BC.

The other cases are exercises, and we omit the tedious collinear possibilities.

Plane separation/sidedness allows us to properly deﬁne interiors of angles and triangles.

Deﬁnition 2.18. A point I is interior to angle ∠BAC if:

• I lies on the same side of

←→

AB as C, and,

• I lies on the same side of

←→

AC as B.

Otherwise said, I lies in the intersection of two half-planes.

A point I is interior to triangle △ABC if it is interior to all three of its angles ∠ABC, ∠BAC and

∠ACB. Otherwise said, I lies in the triple intersection of three of the half-planes deﬁned by the

triangle’s sides.

Interior points permit us to compare angles: if I is interior to ∠BAC, then ∠BAI < ∠BAC has obvious

meaning without resorting to numerical angle measure.

Corollary 2.19. Every angle has an interior point.

Proof. Given ∠BAC, consider any interior point I of the segment BC. This plainly lies on the same

side of

←→

AB as C and on the same side of

←→

AC as B.

In Exercise 8, we check that the interior of a triangle is non-empty.

Pasch’s axiom could be paraphrased: If a line enters a triangle, it must come out. We haven’t quite

established this crucial fact, however. What if the line passes through a vertex?

Theorem 2.20 (Crossbar Theorem). Suppose I is interior to ∠BAC.

Then

−→

AI intersects BC.

In particular, if a line passes through a vertex and an interior point of

a triangle, then it intersects the side opposite the vertex.

ℓ

Proof. Extend AB to a point D such that A lies between B and D (O-2). Since C is not on

←→

BD =

←→

AB we

have a triangle △BCD. Since

←→

AI intersects one edge of △BCD at A and does not cross any vertices

(think about why. . . ), Pasch says it intersects one of the other edges (BC or CD) at some point M.

The result follows from applying plane separation to the lines

←→

AB =

←→

BD and

←→

AC. First observe:

Since I, M lie on the same side of

←→

AB =

←→

BD as C, it follows that IM does not intersect

←→

AB.

Since A, I, M are collinear and A ∈

←→

AB, it follows that A /∈ I M.

If M ∈ BC, we are done. Our goal is to show that M ∈ CD is a contradiction.

correct arrangement

contradiction

Suppose, for contradiction, that M ∈ CD. Relative to

←→

AC:

• I and B lie on the same side since I is interior to ∠BAC;

• B and D lie on opposite sides, since B ∗ A ∗ D and

←→

AC =

←→

BD = AB;

• D and M lie on the same side since M ∈ CD and

←→

CD =

←→

AC.

By plane separation, I, M lie on opposite sides of

←→

AC. The collinearity of A, I, M then forces the

contradiction A ∈ IM.

Euclid repeatedly uses the crossbar theorem without justiﬁcation,

including in his construction of perpendiculars and angle/segment

bisectors (Theorems I. 9+10). We sketch the latter here.

Given ∠BAC, construct E such that AB

∼

AE. Construct D using

an equilateral triangle (I. 1). SSS (I. 8) shows that ∠BAC is bisected,

and SAS (I. 4) that

−→

AD bisects BE.

Quite apart from Euclid’s arguments for SAS and SSS being suspect

(we’ll deal with these in the next section), he gives no argument for

why D is interior to ∠BAC or why

−→

AD should intersect BE!

Even with Pasch’s axiom and the crossbar theorem, it requires some effort to repair Euclid’s proof. No

matter, we’ll provide an alternative construction of the bisector once we’ve considered congruence.

The pictures could be modiﬁed: e.g., I = M and A ∗ I ∗ M are also correct arrangements (M ∈ BC).

Exercises 2.2. 1. Label the vertices in the Fano plane 1 through 7 (any way you like). As we did in

Example 2.13 for the 3-point geometry, describe each line in terms of its points.

2. Prove Lemma 2.15.

3. Give a model for each of the 5-point incidence geometries. How many are there?

(Hint: remember that order doesn’t matter, so the only issue is how many points lie on each line)

4. Consider the proof of the crossbar theorem. Explain how we know that

←→

AI does not contain

any of the vertices of △BCD.

5. You are given distinct points A, B. Using the axioms of incidence and order and Lemma 2.14

(follows from I-2), show the existence of each of the points C, D, E, F in the picture in alphabetical

order. Hence conclude the existence of a point F lying between A and B (Lemma 2.16).

During your construction, address the following issues:

(a) Explain why D does not lie on

←→

AB.

(b) Explain why E does not lie on △ABD.

←→

CE exists).

(d) Explain why F lies on AB and not on BD.

6. We complete the proof of the plane separation theorem (2.17).

(a) Prove part 3 (it is almost a verbatim application of Pasch’s axiom).

(b) Suppose a line ℓ intersects all three sides of △ABC but no vertices.

This results in a very strange picture (we’ve labelled the intersec-

tions D, E, F and WLOG chosen D ∗ E ∗ F).

Apply Pasch’s axiom to △DBF and

←→

AC to obtain a contradiction.

Hence establish part 2 of the plane separation theorem.

ℓ

7. Suppose A, B, C are distinct points on a line ℓ.

(a) Explain why there exists a line m = ℓ such that B ∈ m.

(b) Prove that A ∗B ∗C ⇐⇒ A and C lie on opposite sides of m.

i. B is the only point common to the rays

−→

BA and

−→

BC.

ii. If D ∈ ℓ is any point other than B, prove that D lies in precisely one of

−→

BA or

−→

BC.

8. Prove that the interior of a triangle is non-empty.

(Hint: use Exercise 5 to construct a suitable I, then prove that it lies on the correct side of each edge)

2.3 Hilbert’s Axioms II: Congruence

Hilbert’s congruence axioms address two primary issues in Euclid.

1. Euclid’s use of equal is confusing. In Hilbert, segments/angles are now equal only when they

are precisely the same (this amounts to the reﬂexivity part of the next result).

2. Euclid’s frequent and unjustiﬁed use of pictorial reasoning. We previously discussed Euclid’s

erroneous approach to the SAS and SSS triangle congruence theorems. It was eventually real-

ized that one of the triangle congruences has to be an axiom: SAS is Hilbert’s C-6.

We start with a small piece of bookkeeping.

Lemma 2.21. Congruence of segments/angles is an equivalence relation.

Proof. (Reﬂexivity) Let AB be given. Apply C-1 to obtain A

′

such that AB

∼

′

. We sneakily use

this twice and apply C-2 to obtain

∼

′

and AB

∼

′

=⇒ AB

∼

(Symmetry) Assume AB

∼

CD. By reﬂexivity, CD

∼

CD. By C-2 we have CD

∼

AB.

(Transitivity) Suppose AB

∼

CD and CD

∼

EF. By symmetry, EF

∼

CD. Axiom C-2 now shows that

∼

EF.

Axioms C-4 and C-5 say essentially the same thing for angles (see Exercise 2).

Segment/Angle Transfer and Comparison

Hilbert’s axioms of segment and angle transference are crucial for comparing non-collinear segments

and angles with distinct vertices.

Deﬁnition 2.22. Let segments AB and CD be given.

By axiom C-1, let E be the unique point on

−→

CD such that CE

∼

AB:

we have transferred AB onto

−→

CD.

We write AB < CD if E lies between C and D, etc.

By O-3, any two segments are comparable: given AB & CD, precisely one of the following holds,

AB < CD, CD < AB, AB

∼

C-3 says that congruence respects the ‘addition’ of adjacent congruent segments. Unique angle trans-

fer, comparison and addition follow similarly from axiom C-4 and Deﬁnition 2.18 (interior points).

Neither Hilbert nor Euclid use or require an absolute notion of length/angle-measure: the compari-

son AB < CD does not indicate a relationship between numerical quantities (lengths). Introducing

numerical length requires the inclusion of the real numbers (and thus far more axioms)—for purity

reasons, we postpone this until Section 2.5.

The Triangle Congruence Theorems: SAS, ASA, SSS & SAA

Hilbert assumes side-angle-side (SAS) and proceeds to prove the remainder. Here is the ﬁrst of these;

we’ll cover SSS momentarily and SAA in Exercise 6.

Theorem 2.23 (Angle-Side-Angle/ASA, Euclid I. 26, case I). Suppose △ABC and △DEF satisfy

∠ABC

∼

∠DEF, AB

∼

DE, ∠BAC

∼

∠EDF

Then the triangles are congruent (∠ACB

∼

∠DFE, AC

∼

DF and BC

∼

EF).

Hilbert’s approach modiﬁes Euclid’s: instead of laying △ABC on top of △DEF, he creates a new

triangle △DEG

∼

△ABC and proves that G = F.

Proof. Segment transfer provides the unique point G ∈

−→

EF such that EG

∼

BC.

SAS applied to AB

∼

DE, ∠ABC

∼

∠DEG (= ∠DEF) , BC

∼

EG,

says ∠BAC

∼

∠EDG (

∼

∠EDF) (this last is by assumption).

Since F and G lie on the same side of

←→

DE, angle transfer (C-4) says

they lie on the same ray through D.

But then F and G both lie on two distinct lines (

←→

EF =

←→

EG and

←→

DF =

←→

DG). We conclude that F = G.

By SAS we conclude that △ABC

∼

△DEF.

D E

Geometry Without Circles

Circles are at the heart of Euclid’s constructions. Yet, for reason we’ll address in Section 2.4, Hilbert

essentially ignores them. We sketch a few of his alternative approaches to Euclid’s basic results.

Theorem 2.24 (Euclid I. 5). An isosceles triangle has congruent base angles.

Isosceles means equal legs: two sides of the triangle are congruent. The remaining side is the base.

Euclid’s argument relies on a famously complicated construction (look it up!). Hilbert does things

more speedily and sneakily, by relabelling the original triangle and applying SAS.

Proof. Suppose △ABC is isosceles where AB

∼

AC. Consider a ‘new’

triangle △A

′

= △ACB where the base points are switched:

′

:= A, B

′

:= C, C

′

:= B

Observe:

• ∠BAC

∼

∠CAB (axiom C-5) =⇒ ∠BAC

∼

∠B

′

• AB

∼

AC =⇒ AB

∼

′

and AC

∼

′

SAS says that ∠ABC

∼

∠A

′

∼

∠ACB.

A = A

′

B = C

′

C = B

′

Dropping a Perpendicular As with the majority of Book I, Euclid accomplishes this using circle

intersections.

Hilbert instead uses segment/angle transference and the concept of sidedness.

Suppose we are given a line ℓ and a point P not on ℓ. Our goal is to

construct a point M ∈ ℓ such that PM intersects ℓ in a right-angle.

Let A, B be distinct points on ℓ (axiom I-3) so that ℓ =

←→

AB.

By axioms C-4 and C-1, we may transfer AP to the other side of ℓ at

A, creating a new point Q.

Since P and Q lie on opposite sides of ℓ , the line intersects PQ at some

point M. There are two cases to consider.

• In the generic case M = A (pictured), SAS applied to △MAP

and △MAQ shows that ∠AMP

∼

∠AMQ. Since these angles

sum to a straight edge (PQ), they are both right-angles.

• In the extreme case M = A, there are no triangles and SAS

cannot be applied. Instead, observe that B does not lie on PQ

(which axioms/results make this clear?!) and apply the above

argument with B instead of A.

ℓ

A generalization of this construction facilitates a corrected argument for the SSS triangle congruence.

Theorem 2.25 (Side-Side-Side/SSS, Euclid I. 8). Suppose △ABC and △DEF have sides congruent

in pairs:

∼

DE, BC

∼

EF, AC

∼

Then the triangles are congruent (∠ABC

∼

∠DEF, ∠BCA

∼

∠EFD, ∠CAB

∼

∠FDE).

The strategy is similar to the proof of ASA. Hilbert creates a new triangle △DEG

∼

△ABC, though

this time with G on the opposite side of

←→

DE to F.

Proof. Transfer ∠BAC to D on the other side of

←→

DE from F to

obtain G (axioms C-4 and C-1).

SAS (AB

∼

DE, ∠BAC

∼

∠EDG, AC

∼

DG) shows that

∼

EF. Otherwise said, △DEG

∼

△ABC.

Join FG to produce isosceles triangles △FDG and △FEG

with base FG, both with congruent angles at F and G.

Sum angles at F and G and apply SAS (DF

∼

DG, ∠DFE

∼

∠DGE, EF

∼

EG) to see that △DEF

∼

△DEG.

We conclude that △ABC

∼

△DEG

∼

△DEF, as required.

To be completely formal, we should also carefully deal with

the situations where the sum is a subtraction or the triangle

is right-angled at A or B.

D E

Consider the picture for Thm. I. 11 in Exercise 2.2.2.

Exterior Angle Theorem (Thm. 2.3, I. 16) Euclid’s approach uses a bisector which he obtains from

circles. Hilbert does things a little differently.

Proof. Given △ABC, extend AB to D such that AC

∼

BD. For clarity, we label angles with Greek let-

ters as in the ﬁrst picture below. We show that γ < δ by proving that the alternatives are impossible.

1. (δ ≇ γ) Assume δ

∼

γ. By SAS, △ACB

∼

△DBC; in particular ϵ

∼

β. Since A and D lie on

opposite sides of

←→

BC, we see that ϵ + γ

∼

β + δ is a straight edge. But then A, D are distinct

points lying on two lines! Contradiction.

2. (δ < γ) Assume δ < γ. Transfer δ to C as shown to obtain η

∼

δ. By the crossbar theorem, we

obtain an intersection point E. But now δ is an exterior angle of △EBC congruent to an interior

angle η of the same triangle, contradicting part 1.

Step 1: δ

∼

γ is a contradiction Step 2: δ < γ is a contradiction

Take the vertical angle to δ at B and repeat the argument to see that α < δ.

The proof also shows that the sum of any two angles in a triangle is strictly less than a straight edge:

α + β < δ + β = 180°.

Is Euclid now ﬁxed? Almost! In the exercises we show how the following may be achieved:

• Construction of an isosceles triangle on a segment AB. With this one can construct segment

and angle bisectors (Euclid I. 9+10).

• SAA congruence (Euclid I. 26, case II), the last remaining triangle congruence theorem.

We’ve now recovered almost all of Book I prior to the application of the parallel postulate. Including

Playfair’s axiom completes the remainder, including Pythagoras’, all without circles!

Exercises 2.3. Except for question 8, answer everything without reference to the continuity axiom,

circles, or the uniqueness of parallels (e.g., Playfair’s axiom, (tri)angle sum = 180°).

1. Draw pictures to suggest why you don’t expect Angle-Angle-Angle (AAA) and Side-Side-

Angle (SSA) to be triangle congruence theorems.

2. Use Hilbert’s axioms C-4 and C-5 to prove that congruence of angles is an equivalence relation.

3. (a) Use ASA to prove that if the base angles are congruent then a triangle is isosceles.

(b) Find an alternative argument that relies the exterior angle theorem.

(Hint: this is essentially the same as the proof of Exercise 5 (a))

4. Given AB, axiom I-3 says ∃C ∈

←→

AB.

If △ABC is not isosceles, then WLOG assume ∠ABC < ∠BAC.

Transfer ∠ABC to A to produce D on the same side of

←→

AB as C with

∠ABC

∼

∠BAD, BC

∼

(a) Explain why rays

−→

AD and

−→

BC intersect (at some point M).

(b) Why is △MAB isosceles?

(d) Explain how to construct an angle bisector using the above discussion.

5. We prove Theorems I. 18, 19 and 20 on comparisons

of angles and sides in a triangle. For clarity, suppose

△ABC has sides and angles labelled as in the picture.

(a) (I. 18) Assume a < c. Prove that α < γ.

(Hint: let D on AB satisfy BD

∼

(b) (I. 19: converse to I. 18) α < γ =⇒ a < c.

(Hint: Prove the contrapositive)

(Hint: Let E lie on

−→

BC such that CE

∼

b and apply I. 19)

6. Prove the SAA congruence. If △ABC and △DEF satisfy

∼

DE, ∠ABC

∼

∠DEF and ∠BCA

∼

∠EFD

then the triangles are congruent: △ABC

∼

△DEF.

(Hint: Let G ∈

−→

BC be such that BG

∼

EF, apply SAS and the exterior

angle theorem)

7. Hilbert’s published SAS axiom is weaker than we’ve stated.

Given triangles △ABC and △A

′

, if AB

∼

′

, AC

∼

′

, and ∠BAC

∼

∠B

′

, then ∠ABC

∼

∠A

′

Use this to prove the full SAS congruence theorem (axiom C-6 as we’ve stated it).

(Hint: try a trick similar to the proof of ASA)

8. Construct the picture on the right, where BF is the perpendicular

bisector of AC, and

∼

AB, BE

∼

AD, BF

∼

Use Pythagoras’ Theorem to prove that △ACF is equilateral.

This construction requires Playfair’s axiom and thus unique parallels. It

does not require circle intersections (continuity) like Theorem 2.1 (I. 1).

2.4 Circles and Continuity

Deﬁnition 2.26. Let O and R be distinct points. The circle C with center O and

radius OR is the collection of points A such that OA

∼

OR.

A point P lies inside the circle C if P = O or OP < OR.

A point Q lies outside if OR < OQ.

Since all segments are comparable, any point lies inside, outside or on a given

circle.

A major weakness of Euclid is that many of his proofs rely on circle intersections, rather than lines.

To use circles in this manner requires the Axiom of Continuity. This much more technical than the

other axioms. It is likely for this reason that Hilbert barely mentions circles, instead wanting to build

as much geometry as possible using only the simplest axioms.

Here are the facts necessary before Euclid’s approach can be followed.

Theorem 2.27. Suppose C and D are circles.

1. (Elementary/line-circle Continuity Principle) If P lies inside and Q outside C, then PQ inter-

sects C in exactly one point.

2. (Circular Continuity Principle) If D contains a point inside and another outside C, then they

intersect in exactly two points. These lie on opposite sides of the line joining the circle centers.

The idea of the ﬁrst principle is to partition PQ into two pieces:

consists of the points lying on or inside C

consists of the points lying outside C

One shows that Σ

and Σ

satisfy the assumptions of the axiom. The

unique point O then exists and is shown to lie on C itself. Some of this are

in Exercise 6. The circular continuity principle is harder.

What is perhaps more interesting is to consider a geometry in which the axiom of continuity is false.

Example 2.28. The geometry Q

= {(x, y) ∈ R

: x, y ∈ Q} of points in the plane with rational

co-ordinates satisﬁes almost all of Hilbert’s axioms, however C-1 and continuity are false.

Axiom C-1 Given points A = (0, 0), B = (1, 0) and C = (1, 1), we see

that O = (

√

) is the unique point (in R

) on the ray r =

−→

such that AC

∼

AB. Clearly O is an irrational point and therefore

not in the geometry.

Continuity The circle centered at A = (0, 0) with radius 1 does not

intersect the segment AC. More properly, AC = Σ

∪Σ

may be

partitioned as shown and yet no point O in the geometry separates

, Σ

Equilateral triangles We can ﬁnally correct Euclid’s proof of the ﬁrst proposition of the Elements!

Theorem 2.29 (Euclid I.1). An equilateral triangle many be constructed on a given segment AB.

Proof. Following Euclid, take the circles α and β centered at A and B, with radii AB.

Axiom O-2: ∃D such that A ∗ B ∗ D.

Axiom C-1: let C ∈

−→

BD be such that BC

∼

AB.

Circular continuity principle: β contains A (inside α) and

C (outside α) so the circles intersect in two points P, Q.

Since P lies on both circles (and is therefore distinct from

A and B), we have AB

∼

BP whence △ABC is

equilateral.

If one allows Playfair’s axiom on unique parallels, Euclid’s result can be proved without using circles

or the continuity axiom (see Exercise 2.3.8). Nevertheless, we are ﬁnally able to say that every result

in Book I of Euclid is correct, even if his original axioms and arguments are insufﬁcient!

Basic Circle Geometry

We continue our survey of Euclidean geometry with a few results about circles, many of which are

found in Book III of the Elements. From this point on, we assume all Hilbert’s axioms including

Playfair and continuity. Indeed what follows often relies on their consequences, particularly angle-

sums in triangles and the circular continuity principle.

Deﬁnition 2.30. With reference to the picture:

• A chord AB is a segment joining two points on a circle.

• A diameter BC is a chord passing through the center O.

• An arc

)

AB is part of the circular edge between chord points

(major or minor by length).

• ∠AOB is a central angle and ∠APB an inscribed angle.

• △ABP is inscribed in its circumcircle.

Since these ideas shouldn’t be new, most of the details are left as exercises.

Theorem 2.31 (III. 20). The central angle is twice the inscribed angle: ∠AOB = 2∠APB.

For a sketch proof, join OP, breaking △ABP into three isosceles triangles and count angle sums.

Corollary 2.32. 1. (III. 21) If inscribed triangles share a side, the opposite angles are congruent.

2. (III. 22) An inscribed quadrilateral has opposite angles supplementary (summing to 180°).

3. (Thales’ Theorem III. 31) A triangle in a semi-circle is right-angled.

Theorem 2.33. Any triangle has a unique circumcircle.

This is similar to III. 1: construct the perpendicular bisectors of

two sides as in the picture.

Deﬁnition 2.34. A line is tangent to a circle if it intersects the circle exactly once.

Theorem 2.35 (III. 18, 19 (part)). A line is tangent to a circle if and only if it is perpendicular to the

radius at an intersection point.

Proof. (⇐) Suppose ℓ through T is perpendicular to the radius OT.

Let P be any another point on ℓ. But then (Exercise 2.3.5),

∠OPT < 90° = ∠OTP =⇒ OT < OP

thus P lies outside the circle. Every point on ℓ except T lies outside

the circle, so T is the unique intersection, and ℓ is therefore tangent.

The converse is an exercise.

ℓ

Theorem 2.36. Through a point outside a circle, exactly two lines are tangent to the circle.

Exercises 2.4. 1. Give formal proofs of all parts of Corollary 2.32.

2. Prove Theorem 2.33.

3. Complete the proof of Theorem 2.35 by showing the (⇒) direction.

(Hint: suppose T is an intersection and that the angle isn’t 90°, and drop a perpendicular to ℓ. . . )

4. Given a circle centered at O and a point P outside the circle, draw the circle centered at the

midpoint of OP passing through O and P. Explain why the intersections of these circles are the

points of tangency required in Theorem 2.36, and hence complete its proof.

5. (a) Prove Theorem 2.31 when O is an interior point to △ABP.

(b) Prove Theorem 2.31 when O is an exterior point to △ABP.

6. Suppose A ∗ C ∗ B and that O /∈

←→

AB. Use Exercise 2.3.5 to show that

OC < max



OA, OB



If A, B are interior to a circle centered at O, conclude that C is also.

(This is part of what’s needed to demonstrate the elementary continuity principle:

no point of Σ

lies between two points of Σ

. Can you prove the other condition?)

7. If a line contains a point inside a circle, show that it intersects the circle in two points.

(Hint: ﬁrst construct two points on the line lying outside the circle)

2.5 Similar Triangles, Length and Trigonometry

In the geometry of Euclid & Hilbert, there are no numerical measures of length or angle. Relative

measure is built in (Deﬁnition 2.22), and we’ve denoted right-angles and straight edges by 90° & 180°

purely for convenience. To avoid continued frustration it is time we introduced explicit numerical

measure, though to do so properly requires more axioms!

Axioms 2.37 (Length and Angle Measure).

L1 To each segment AB corresponds a unique length

, a positive real number

⇐⇒ AB

∼

⇐⇒ AB < CD (Deﬁnition 2.22)

L4 If A ∗ B ∗ C, then

A1 To each ∠ABC corresponds a unique degree measure m∠ABC, a real number between 0 and 180

A2 m∠ABC = m∠DEF ⇐⇒ ∠ABC

∼

∠DEF

A3 m∠ABC < m∠DEF ⇐⇒ ∠ABC < ∠DEF (Deﬁnition 2.18)

A4 If P is interior to ∠ABC, then m∠ABP + m∠PBC = m∠ABC

A5 Right-angles measure 90°

Don’t memorize these axioms, just observe how they ﬁt your intuition. Angle measure in Euclidean

geometry has two notable differences from what you might expect:

• (A1) All angles measure strictly between 0° and 180°. A straight edge isn’t an angle, though

such is commonly denoted 180°, and there are no reﬂex angles (> 180°).

• (A2) Angles are non-oriented, measuring the same in reverse (m∠ABC = m∠CBA).

The axioms for length and angle follow the same pattern except for us explicitly ﬁxing the scale of

angle measure (A5). To do the same for length requires only a choice of a reference segment of length

1. The following is a consequence of the continuity axiom.

Theorem 2.38 (Uniqueness of measure).

1. Given OP, there is a unique way to assign a length to every segment such that

= 1.

2. There is a unique way to assign a degree measure to every angle.

The segment OP in part 1 provides a length-scale

for a ruler. We measure the length of any segment

by moving this rule on top of the desired segment

(congruence!).

|QR| = 4.6

Area Measure If we also include Playfair’s axiom, then the discussion at the end of Book I of Euclid

becomes valid, and rectangles can be deﬁned (see e.g., Exercise 2.1.2).

Deﬁnition 2.39. The area (measure) of a rectangle is the product of its base and height (measures).

Given a length measure, a square with side length 1 necessarily has area 1. Relative to a base segment,

the height of a triangle is the length of the perpendicular dropped from the vertex.

Since every rectangle is a parallelogram and a triangle half a parallelogram, Eu-

clid’s discussion (Thm. I. 35) amounts to the familiar area formulæ:

area(parallelogram) = bh, area(triangle) =

While these expressions are nice to have, they are not necessary. Indeed every-

thing that follows depends only on area ratios: e.g.,

Lemma 2.40. If triangles have congruent bases, then their areas are in the same ratio as their heights.

The same holds with the roles of heights and bases reversed.

Similarity and the AAA Theorem Similar triangles are the concern of Book VI of the Elements.

Deﬁnition 2.41. Triangles are similar, written △ABC ∼ △XYZ, if

their sides are in the same length ratio

Euclid discusses these using non-numerical ratios of segments (e.g., AB : XY = BC : YZ). This is

unnecessarily confusing for modern readers, indeed some of the most difﬁcult parts of the Elements

are where he describes what this should mean, particularly for irrational ratios (Books V & X).

Our primary result comes in two versions, where the second (which we’ll prove) is a special case of

the ﬁrst.

Theorem 2.42 (Angle-Angle-Angle/AAA, Euclid VI. 2–5).

1. Triangles are similar if and only if their angles come in mutually con-

gruent pairs.

2. Suppose a line intersects two sides of a triangle. The smaller triangle

so created is similar to the original if and only if the line is parallel to

the third side of the triangle.

ℓ

The picture should convince you that 1 ⇒ 2 follows from the uniqueness of parallels (Playfair’s

axiom, Corollary 2.5 & Theorem 2.6). This reliance is crucial! We should not expect AAA similarity

in non-Euclidean geometry, and indeed shall see later that it is false in hyperbolic geometry (Chapter

4), where AAA is a theorem for congruent triangles! The converse (2 ⇒ 1) is left to Exercise 10.

Proof (AAA similarity, part 2). Suppose ℓ intersects △ABC at

points D, E as shown. Drop perpendiculars to create distances

h, k, d

, d

as indicated. We prove:

ℓ is parallel to BC ⇐⇒ △ABC ∼ △ADE

( ⇒) Suppose ℓ is parallel to BC. Playfair

tells us that d

= d

By Lemma 2.40, triangles with the same height have areas

proportional to their bases:

ℓ

area(BDE)

area(ADE)

(△BDE, △ADE have same height h)

area(CDE)

area(ADE)

(△CDE, △ADE have same height k)

Since △BDE and △CDE share base DE, we see that

ℓ is parallel to BC ⇐⇒ d

= d

⇐⇒ area(BDE) = area(CDE)

⇐⇒

(∗)

Add 1 =

to both sides to obtain one part of the similarity ratio

It remains to see that this ratio equals

. Again using common heights (h, k) of triangles,

area(ABE)

area(BDE)

area(BCE)

area(ABE)

(†)

=⇒

(∗)

(†)

area(BCE)

area(BDE)

where the last equality follows since △BCE and △BDE have common height d

= d

( ⇐) Suppose △ABC ∼ △ADE. By Playfair, let m be the

unique parallel to BC through D. This intersects AC at a

point G. We must prove that G = E (consequently m = ℓ).

By the (⇒) direction above,

△ABC ∼ △ADG

ℓ

However, ∼ is plainly transitive (it is an equivalence relation), whence △ADE ∼ △ADG. The

similarity ratio is 1 =

, whence

= 1 =⇒

=⇒ E = G

= d

⇐⇒ ℓ parallel to BC is Playfair! Compare Exercise 2.1.2 (Thm I. 46) on the construction of a square. . .

Applications of Similarity: Trigonometric Functions, Cevians and the Butterﬂy Theorem

We ﬁnish with several applications of similarity which hopefully give an idea of what can be done

without co-ordinates. None of these ideas were known to Euclid.

Deﬁnition 2.43. Given an acute angle ∠ABC (m∠ABC < 90°), drop a perpendicular from A to

−→

at D so that ∠ADB is a right-angle. Deﬁne

sin ∠ABC :=

cos ∠ABC :=

Early trigonometry dates to a few hundred years after Euclid, though the approach was different.

Theorem 2.44. Angles have the same sine (cosine) if and only if they are congruent.

Proof. Assume ∠ABC

∼

∠A

′

as in the picture, and drop perpendiculars to D, D

′

Since △ABD and △A

′

have two pairs of mutually

congruent angles, the third pair is congruent also. AAA

applies, the triangles are similar and

′

In particular, sin ∠ABC = sin ∠A

′

and cos ∠ABC = cos ∠A

′

The converse is an exercise.

After Giovanni Ceva (1647–1734), a cevian is a segment joining a vertex to the opposite side of a

triangle. Here is a beautiful result from the height of Euclidean geometry—good luck trying to prove

it using co-ordinates!

Theorem 2.45 (Ceva’s Theorem). Given △ABC and cevians AX, BY, CZ,

= 1 ⇐⇒ the cevians meet at a common point P

Proof. (⇐) This is simply a repeated application of Lemma 2.40.

area(ABX)

area(AXC)

area(PBX)

area(PXC)

=⇒

(∗)

area(ABX) −area(PBX)

area(AXC) − area(PXC)

area(ABP)

area(APC)

Repeat for the other ratios and multiply to get 1.

A simple justiﬁcation of (∗) and the converse are an exercise.

The ancient forerunners of sine and cosine were deﬁned using chords of circles rather than triangles. The word

trigonometry (literally triangle measure) wasn’t coined until 1595.

Theorem 2.46 (Butterﬂy Theorem). We are given the following data

as in the picture:

• PQ is a chord of a circle with midpoint M.

• AC and BD are chords meeting at M.

• X, Y are the chord-intersections as shown.

Then M is the midpoint of XY.

This beautiful result dates to 1803-5 and has several proofs. We present an argument relying on

similar triangles.

Proof. For convenience we introduce several numerical

lengths:

• z =

, x =

and y =

• Drop perpendiculars from X, Y to chords AD, BC,

and label the lengths x

, x

, y

as shown.

The four colored pairs of angles are congruent: vertical an-

gles at M, and inscribed angles at A, B, C, D.

We compare sides of several similar triangles:

•

and

=⇒

•

and

The result follows by putting these observations together and applying Exercise 1 twice:

( z − x)(z + x)

( z + y)(z − y)

− x

−y

=⇒ x

( z

−y

) = y

( z

− x

)

=⇒ x

= y

=⇒ x = y

as required.

Exercises 2.5. 1. Let AD and PQ be chords of a circle which intersect at X. Use similar triangles to

prove that

The denominators are equal by applying the Exercise to the chords BC and PQ.

2. Let △ABC have a right-angle at C. Drop a perpendicular from C to

←→

AB at D.

(a) Prove that D lies between A and B.

(b) Prove that you have three similar triangles

△ACB ∼ △ADC ∼ △CDB

(Use the picture, where a, b, c, x, y are lengths)

c = x + y

3. Prove a simpliﬁed version of the SAS similarity theorem:

⇐⇒ △ABC ∼ △AGH

(Hint: construct BJ parallel to GH and appeal to AAA)

4. By excluding the other possibilities, prove the converse of length axiom L4:

If A, B, C are distinct and

, then B lies between A and C.

5. Use Pythagoras’ to prove that sin 45° = cos 45° =

√

, that sin 60° =

√

and cos 60° =

6. Prove the converse of Theorem 2.44: if sin ∠ABC = sin ∠A

′

, then ∠ABC

∼

∠A

′

(Hint: create right-triangles and prove they are similar. Label the side lengths o, a, h, etc.)

7. We complete the proof of Ceva’s theorem.

(a) If p, q, r, s are non-zero real numbers, verify that α =

=⇒ α =

p−r

q−s

(b) Assume X, Y, Z satisfy Ceva’s formula.

Deﬁne P as the intersection of BY and CZ and let

−→

meet BC at X

′

Prove the (⇒) direction of Ceva’s theorem by using

the (⇐) direction to show that X

′

= X.

′

8. (a) A median of a triangle is a segment from a vertex to the midpoint of the opposite side. Use

Ceva’s Theorem to prove that the medians of a triangle meet at a point (the centroid).

(b) (Hard) Medians split a triangle into six sub-triangles. Prove that all have the same area.

9. Prove that similarity of triangles is an equivalence relation.

(Don’t use AAA since its proof requires this fact!)

10. (Hard) Explain how to prove (2 ⇒ 1) in the AAA Theorem (2.42).

3 Analytic Geometry

Geometry in the style of Euclid and Hilbert is synthetic: axiomatic, without co-ordinates or explicit

formulæ for length, area, volume, etc. By contrast, the practice of elementary geometry nowadays

is typically analytic: reliant on co-ordinates & algebra, vectors. The critical invention was the axis,

developed by Ren

e Descartes and Pierre de Fermat in the early 1600s: a ﬁxed reference ruler against

which objects can be measured using co-ordinates.

3.1 The Cartesian Co-ordinate System

Since Cartesian geometry (Descartes’ geometry) should be familiar, we merely sketch the core ideas.

• Perpendicular axes meet at the origin O.

• The co-ordinates of a point are measured by projecting onto the axes;

since these are real numbers we denote the set of these



(x, y) : x, y ∈ R



E.g., P has co-ordinates (1, 2), we usually just write P = (1, 2).

• Algebra is introduced via addition and scalar multiplication

−2

−1

−2 −1 1 2 3

P + Q = (p

, p

) + (q

, q

) = (p

+ q

, p

+ q

) λP = (λp

, λp

)

• The length of a segment uses Pythagoras’ Theorem

d(P, Q) =

( q

− p

)

+ (q

− p

)

In the picture

√

+ 2

√

5. As in Section 2.5, segments are congruent if and only if

they have the same length.

• Curves are deﬁned using equations. E.g. x

+ y

= 1 describes a circle.

Analytic geometry was conceived as a computational toolkit built on top of Euclid. At ﬁrst, math-

ematicians felt the need to justify analytic arguments synthetically lest no-one believe their work.

Synthetic geometry is not without its beneﬁts, but its study has increasingly become a fringe activity;

co-ordinates are just too useful to ignore.

We may therefore assume anything from Euclid and mix strategies as appropriate. To see this at

work, consider a simple result.

Lemma 3.1. Non-collinear points O = (0, 0), A = (x, y), B = (v, w) and

C := (x + v, y + w), form a parallelogram OACB.

Proof. Opposite sides have the same length (

+ y

, etc.)

and are thus congruent. SAS shows △OAC

∼

△CBO. Euclid’s discussion

of alternate angles (pages 10–11) forces opposite sides to be parallel.

This attitude persisted for some time. For instance, when Issac Newton published his groundbreaking Principia in

1687, his presentation was largely synthetic, even though he had used co-ordinates in his derivations.

Lemma 3.2. The points X

on the line

←→

PQ are in 1–1 correspondence with the real numbers via

= P + t(Q − P) = ( 1 −t)P + tQ

Moreover, d(P, X

) =

so that t measures the (signed) distance along the line.

The proof is an exercise. As an example of how easy it can be to work in analytic geometry, we

repeatedly apply the Lemma to re-establish a famous result.

Theorem 3.3. The medians of a triangle meet at a point 2/3 of the way along each median.

Proof. Given △ABC, label the midpoints of each side as shown. By the Lemma, these are

M =

(B + C), N =

(A + B), P =

(A + C)

The point

of the way along median AM is then

A +

(M − A) = A +

(B + C −2A) =

(A + B + C)

By symmetry (check directly if you like!), this is also the point

of the

way along the other two medians.

The three points are therefore identical: the medians meet at the centroid G =

(A + B + C).

Compare this to Exercise 2.5.8 where we used Ceva’s Theorem!

Exercises 3.1. 1. By completing the square, identify the curve described by the equation

+ y

−4x + 2y = 10

2. (a) Perform a pure co-ordinate proof of Theorem 3.3. For simplicity, arrange the triangle so

that A = (0, 0) is the origin, and B points along the positive x-axis.

(b) Descartes and Fermat did not have a ﬁxed perpendicular second axis! Their approach was

equivalent to choosing a second axis at an angle which made the problem as simple as

possible.

Given △ABC, let A be the origin and choose axes which point along the edges AB and

BC. What are the co-ordinates of B and C with respect to these axes? Now give an even

simpler proof of the centroid theorem.

3. Prove Lemma 3.2.

4. A parabola is a curve whose points are equidistant from a ﬁxed

point F, the focus, and a ﬁxed line d (the directrix). Choose axes as

shown in the picture so that F = (0, a) and d has equation y = −a.

Find the equation of the parabola.

3.2 Angles and Trigonometry

Angles are deﬁned differently to Section 2.5, though the approach should feel familiar.

Deﬁnition 3.4. Suppose A, B, C are distinct points in the plane. Take

any circular arc centered at A and deﬁne the radian measure

∡BAC :=

arc-length

radius

∈ [0, 2π)

where arc-length is measured counter-clockwise from

−→

AB to

−→

AC.

2π −θ

Since arc-length scales with radius, the deﬁnition is independent of the radius of the circular arc. It

is important to appreciate the difference between angle measures in our two geometries.

Euclidean geometry All angles < 180°. Reversed legs ⇝ congruent angles and same degree measure:

∠CAB

∼

∠BAC ⇐⇒ m∠CAB = m∠BAC

Analytic geometry Reﬂex angles exist (≥ π). Reversed legs ⇝ different radian measure:

∡CAB = 2π −θ = 2π −∡BAC = ∡BAC (unless a straight edge)

In the picture, ∡CAB is not the radian measure (θ) of ∠CAB! However,

Angles congruent ⇐⇒ radian measures equal and < π

As such, it is common to label angles in a triangle by their radian measure;

standard convention is shown: e.g., (A, a, α) for (point,length,angle).

Deﬁnition 3.5 (Trigonometric Functions). Let O be the origin and I = (1, 0).

Let P = (x, y) lie on a circle of radius r and θ = ∡IOP. We deﬁne:

cos θ :=

sin θ :=

tan θ :=

(x = 0)

AAA similarity (Thm. 2.42) says these are well-deﬁned, independent of r.

Example 3.6. Basic trig identities should be obvious from the picture: e.g.,

cos

θ + sin

θ = 1 (Pythagoras!) and sin θ = cos(

−θ)

What well-known facts regarding sine and cosine do the following illustrate?

π −θ

cos θ

sin θ

√

Solving Triangles A triangle is described by six values: three side lengths and three angle mea-

sures. Euclid’s triangle congruence theorems (SAS, ASA, SSS, SAA) say that three of these in suitable

combination is enough to recover the rest. In analytic geometry, these calculations typically use the

sine and cosine rules.

Theorem 3.7. Label the sides/angles of △ABC following the standard convention (page 37):

Sine Rule If d is the diameter of the circumcircle (Defn. 2.30), then

sin α

sin β

sin γ

Cosine Rule c

= a

+ b

−2ab cos γ

Proof. We prove the sine rule and leave the cosine rule as an exercise.

Everything relies on Corollary 2.32. Draw the circumcircle of △ABC.

Construct △BCD with diameter BD; this is right-angled at C by Thales’

Theorem. There are two cases:

1. If A lies on the same side of

←→

BC as D, then A and D share the same

arc, whence ∡BDC = α and

a = d sin ∡BDC = d sin α

2. If A lies on the opposite side, then the quadrilateral ABDC lies on

a circle. Opposite angles at A, D are supplementary, whence

sin α = sin(π −α) = sin ∡BDC =

The two other angle-side combinations follow by permutation.

π −α

Examples 3.8. 1. The SSS congruence corresponds to solving a triangle using the cosine rule. For

instance, the given triangle has angles

α =

+ 7

−3

2 ·6 ·7

= cos

−1

≈ 25° β =

+ 7

−6

2 ·3 ·7

= cos

−1

≈ 58°

γ =

+ 6

−7

2 ·3 ·6

= cos

−1

≈ 96°

Once you have α, you could alternatively switch to the sine rule to ﬁnd β, before

computing γ = π −α − β.

2. To solve a triangle with data corresponding to the ASA congruence,

ﬁnd the remaining angle γ = π −

−

5π

and apply the sine rule

sin

= sin

5π

= cos

=⇒ a =

√

2 cos

≈ 0.732

b =

√

2 cos

≈ 0.897

Multiple-angle formulæ The picture provides a very simple

proof of the expressions

sin(α + β) = sin α cos β + cos α sin β

cos(α + β) = cos α cos β −sin α sin β

at least when α + β <

. A little algebraic manipulation pro-

duces the double-angle and difference formulæ, and veriﬁes

that these hold for all possible angle inputs.

cos β

sin β

sin α cos β

cos α sin β

sin 2α = 2 sin α cos α sin(α − β) = sin α cos β −cos α sin β

cos 2α = cos

α −sin

α = 2 cos

α −1 cos(α − β) = cos α cos β + sin α sin β

Exercises 3.2. 1. A triangle has angle of

2π

radians between sides of lengths 2 and

√

3 −1. Find the

length of the remaining side, and the remaining angles.

2. Describe how to solve a triangle given data in line with the SAA congruence theorem.

3. Two measurements for the height of a mountain are taken at sea level 5000 ft apart in a line

pointing away from the mountain. The angles of elevation to the mountain top from the hori-

zontal are 15° and 13° respectively. What is the height of the mountain?

4. Use a multiple angle formula to ﬁnd an exact value for cos

and thus exact values for the side

lengths of the triangle in Example 3.8.2.

5. The area of a triangle is

(base)·(height). By using each side of the triangle alternately as the

‘base,’ ﬁnd an alternative proof of the sine rule without the relationship to the circumcircle.

6. You are given SSA data for a triangle: sides with lengths a = 1 and b =

√

3 and angle α =

Show that there are two triangles satisfying this data. Can you generalize to general SSA data?

7. (a) By dropping a perpendicular from B to

←→

AC at D, construct a

proof of the cosine rule.

(Hint: apply Pythagoras’ to the two right-triangles)

(b) Is your argument valid if D is not interior to AC?

8. The dot product of A = (a

, a

) and B = (b

, b

) is A · B := a

+ a

. Apply the cosine rule to

△OAB to prove that

A · B =

cos ∡AOB

9. Derive the multiple-angle formula for sin(α − β).

(Remember that 0 ≤ α, β, α − β < 2π so you can’t simply switch the sign of β!)

10. Given the arrangement pictured, ﬁnd x, the radian-measure α and

the exact value of cos α.

(Hint: ﬁrst show that you have similar isosceles triangles)

x x

1 − x

3.3 Isometries

At the heart of elementary geometry is congruence, the idea that geometric ﬁgures can be essentially

the same without necessarily being equal. In analytic geometry, congruence may be described alge-

braically using functions. This follows from the idea that two segments have the same length if and

only if they are congruent.

Deﬁnition 3.9. A function f : R

→ R

is a (Euclidean) isometry if it preserves lengths:

∀P, Q ∈ R

, d



f (P), f (Q)



Two ﬁgures (segments, angles, triangles, etc.) are congruent precisely when there is an isometry

f : R

→ R

mapping one to the other.

Example 3.10. We check that the map f (x, y) =



3x + 4y, 4x − 3y



+ (3, 1) is an isometry. If

P = (x, y) and Q = (v, w), then



f (P), f (Q)





3v + 4w −3x −4y





4v −3w −4x + 3y



+ 4



( v − x)

+ (w −y)



Isometric segments are certainly congruent. We should make sure the same holds for angles.

Lemma 3.11. Isometries preserve (non-oriented) angles: if f : R

→ R

is an isometry, then

∠PQR

∼

∠ f (P) f (Q) f (R)

Proof. Since f is an isometry, the sides of △PQR and △f (P) f (Q) f (R) are mutually congruent in

pairs. The SSS triangle congruence theorem says that the angles are also mutually congruent.

Example (3.10, cont). Warning: Isometries can reverse orientation!

In the picture,

∡ABC =

but ∡ f (A) f (B) f (C) =

3π

= 2π −∡ABC

Our next task is to conﬁrm our intuition that isometries are rotations, reﬂections and translations.

Given an isometry f , deﬁne g(X) = f (X) − f (O), where O is the origin. Then g is an isometry

g(P) − g(Q) = f (P) − f (Q) =⇒ d



g(P), g(Q)



= d



f (P), f (Q)



which moreover ﬁxes the origin: g(O) = O. We conclude that every isometry f is the composition of

an origin-preserving isometry g followed by a translation “+C:”

f (X) = g(X) + C

In ancient Greek, iso-metros is literally same measure (length/distance).

It thus sufﬁces to describe the origin-preserving isometries g. For these, we make two observations.

1. Suppose

= 1 and let X

= rQ for some r ∈ R. Then

• g(X

) is a distance

from the origin O = g(O).

• g(X

) is a distance

1 −r

from g(Q).

g(X

) therefore lies on the intersection of two circles, which in-

tersect at a single point: we conclude that

g(rQ) = rg(Q)

The picture shows the case 0 < r < 1, where the uniqueness of

intersection follows from 1 =

1 −r

|r|

|1 − r |

g(Q)

g(X

)

2. g(1, 0) lies on the unit circle and therefore has the form

g(1, 0) = S



cos θ, sin θ



for some θ ∈ [0, 2π). By preservation of length and angle

(Lemma 3.11), any other point S

= (cos ϕ, sin ϕ) on the unit

circle must therefore be mapped to one of two points

g( S

) = S

θ±ϕ



cos(θ ± ϕ), sin(θ ±ϕ)



The angle ϕ is transferred to one side of the ray

−−→

0 1

θ+ϕ

θ−ϕ

Putting these together by writing X = rS

= (r cos ϕ, r sin ϕ) in polar co-ordinates, we conclude that

g has one of two forms:

X = rS

g(X) = rS

θ+ϕ

−ϕ

X = rS

g(X) = rS

θ−ϕ

Rotation counter-clockwise by θ Reﬂection across the line making

angle

with positive x-axis

Theorem 3.12. Every isometry of R

has the form

f (X) = g(X) + C

where g is either a rotation about the origin, or a reﬂection across a line through the origin.

Calculating with isometries

This beneﬁts from column-vector notation and matrix multiplication. Writing x =

(

)



r cos ϕ

r sin ϕ



for

the position vector of X

= (x, y) = rS

and applying the multiple-angle formulæ, rotation becomes

g( x) = r



cos(θ + ϕ)

sin(θ + ϕ)



= r



cos θ cos ϕ −sin θ sin ϕ

sin θ cos ϕ + cos θ sin ϕ





cos θ −sin θ

sin θ cos θ



For reﬂections, the sign of the second column is reversed:



cos θ sin θ

sin θ −cos θ



. Every isometry therefore has

the form f (x) = Ax + c where A is an orthogonal matrix.

Examples 3.13. 1. We revisit Example 3.10 in matrix format:

f (x) =



3x + 4y

4x −3y









3 4

4 −3









Since

sin θ

cos θ

4/5

3/5

, we see that its effect is to reﬂect across the line through the origin making

angle

tan

−1

≈ 26.6° with the positive x-axis, before translating by (3, 1).

2. △

has vertices (0, 0), (1, 0), (2, −1) and is congruent to △

, two of whose vertices are (1, 2) and

(1, 3). Find all isometries transforming △

to △

and the location(s) of the third vertex of △

Let f = Ax + c be the isometry. Since d



(1, 2), (1, 3)



= 1 these points must be the images

under f of (0, 0) and ( 1, 0). There are four distinct isometries:

Cases 1, 2: If f (0, 0) = (1, 2) and f (1, 0) = (1, 3), then c = f









and





+ c =





=⇒ A









=⇒ A =



0 a

1 a



for some a

, a

. Since A is orthogonal, the options are A =



0 ∓1

1 0



and

we obtain two possible isometries:

• f

( x) =



0 −1

1 0



x +





rotates by 90°, then translates by





• f

( x) =



0 1

1 0



x +





reﬂects across y = x, then translates by





The third point of △

is f

(2, −1) = (2, 4) or f

(2, −1) = (0, 4).

Cases 3, 4: f (0, 0) = (1, 3) and f ( 1, 0) = (1, 2) results in two further

isometries f

and f

. The details are an exercise.

All four possible triangles △

are drawn in the picture.

−1

1 2

△

In 1872, Felix Klein suggested that the geometry of a set is the study of its invariants: properties

preserved by its group of structure-preserving transformations. In Euclidean geometry, this is the

group of Euclidean isometries (Exercise 9). Klein’s approach provided a method for analyzing and

comparing the non-Euclidean geometries beginning to appear in the late 1800s. By the mid 1900s, the

resulting theory of Lie groups had largely classiﬁed classical geometries. Klein’s algebraic approach

remains dominant in modern mathematics and physics research.

An orthogonal matrix satisﬁes A

A = I. All such have the form



cos θ ∓sin θ

sin θ ±cos θ





a ∓b

b ±a



where a

+ b

= 1.

Exercises 3.3. 1. Let f : R

→ R

be the isometry, “reﬂect across the line through the origin making

angle

with the positive x-axis.” Find a 2 ×2 matrix A such that f (x) = Ax.

2. Describe the geometric effect of the isometry f (x) =



√

−

√

3 1



x +



−2



3. Find the remaining isometries f

, f

and the third points of △

in Exercise 3.13.2.

4. Find the reﬂection of the point (4, 1) across the line making angle

tan

−1

≈ 33.7° with the

positive x-axis.

(Hint: if tan θ =

, what are cos θ and sin θ?)

5. An origin-preserving isometry f (v) = Av moves the point (7, 4) to (−1, 8).

(a) If f is a rotation, ﬁnd the matrix A. Through what angle does it rotate?

(b) If f is a reﬂection, ﬁnd the matrix A. Across which line does it reﬂect?

6. Let ABCD be the rectangle with vertices A = (0, 0), B = (4, 0), C = (4, 3), D = (0, 3). Suppose

an isometry f : R

→ R

maps ABCD to a new rectangle PQRS where

P = f (A) := (2, 4) and R = f (C) := (2, 9)

Find all possible isometries f and the remaining points Q = f (B) and S = f (D).

7. (a) If A =



cos θ −sin θ

sin θ cos θ



and p is constant, explain why f (x) = A(x −p) + p = Ax + (I − A)p

rotates by θ around the point with position vector p.

(b) Suppose f (x) = Ax + c rotates the plane around the point P = (−2, 1) by an angle θ =

tan

−1

. Find A and c.

i. If θ + ϕ = 2π, show that f ◦ g is a rotation: by what angle and about which point?

ii. What happens instead if θ + ϕ = 2π?

8. Make an argument involving circle intersections (see page 41) to prove that for any isometry f ,



(1 − t)P + tQ



= (1 −t) f (P) + t f (Q)

9. Throughout this question, we use the notation f

A,c

: x 7→ Ax + c.

(a) Prove that isometries obey the composition law f

A,c

◦ f

B,d

= f

AB,c+Ad

(b) Find the inverse function of the isometry f

A,c

. Otherwise said, if f

A,c

◦ f

C,d

= f

I,0

, where I

is the identity matrix, how do B, d depend on A, c?

A,c

◦ f

I,d

◦ f

−1

A,c

is a translation.

Part (a) can be written using augmented matrices: (A |c)(B |d) := (AB |c + Ad).

If you know group theory, parts (a) and (b) are the closure and inverse properties of the group of Euclidean

isometries E. Part (c) says the translations T form a normal subgroup. We may therefore write E as a

semi-direct product of T and the orthogonal group of origin-preserving isometries

E = T ⋊ O

(R )

3.4 The Complex Plane

Complex numbers date to 16

century Italy. Their application to geometry really begins with Leon-

hard Euler (1707–1783) who identiﬁed the set of complex numbers C with the plane (what is now

known as the Argand diagram).

Deﬁnition 3.14. Let i be an abstract symbol satisfying the property

= −1.

Given real numbers x, y, the complex number z = x + iy is simply the

point (x, y) in the standard Cartesian plane.

Given z = x + iy, its:

• Complex conjugate z = x −iy is its reﬂection across the real axis.

• Modulus

√

zz =

+ y

is its distance from the origin.

• Argument arg(z) is the angle (measured counter-clockwise) be-

tween the positive real axis and the ray

−→

0z.

1 2 3 4

z = 2 −3i

z = 2 + 3i

|z | =

√

−i

−2i

−3i

arg(z) = tan

−1

Addition, scalar multiplication (by real numbers) and complex multiplication follow the usual algebraic

rules while using i

= −1 to simplify.

Example 3.15. A simple example of multiplication of complex numbers:

(2 + 3i)(4 + 5i) = 2 ·4 + 2 ·5i + 3i ·4 + 3i ·5i (multiply out)

= 8 + 10i + 12i −15 (use i

= −1 to simplify)

= −7 + 22i

The algebra screams geometry! Deﬁnition 3.14 already length, angle and reﬂection in the real axis.

Two other aspects of basic geometry are immediate:

• Addition by z translates all points by z.

• Scalar multiplication scales distances from the origin (similarity).

The algebraic property distinguishing the complex numbers from the standard Cartesian plane is

complex multiplication. To start visualizing this, consider multiplication by i,

iz = i(x + iy) = −y + ix

This is the result of rotating z counter-clockwise

radians about the origin. To obtain all rotations

and reﬂections, we need an alternative description of a complex number.

Lemma 3.16. 1. (Euler’s Formula) For any θ ∈ R, e

iθ

= cos θ + i sin θ.

2. (Exponential laws) e

iθ

iϕ

= e

i(θ+ϕ)

and (e

iθ

)

= e

inθ

for any n ∈ Z.

Evaluating at θ = π yields the famous Euler identity e

iπ

= −1. Part 1 can be taken as a deﬁnition. To

see that it is a reasonable deﬁnition requires either power series or elementary differential equations,

topics best described elsewhere. Part 2 is an exercise.

In the language of linear algebra, C is a vector space over R with basis {1, i}.

Deﬁnition 3.17. Let z = x + iy be a non-zero complex number.

Writing x = r cos θ and y = r sin θ, we obtain the polar form

z = re

iθ

= r(cos θ + i sin θ)

where r =

is the modulus and θ = arg(z) the argument of z.

0 1 2

1 +

√

3i = 2e

iπ

Now consider the effect of multiplying a complex number z = re

iϕ

by e

iθ

= cos θ + i sin θ: according

to the Lemma

iθ

z = re

iθ

iϕ

= re

i(θ+ϕ)

which has the same modulus (r) as z but a new argument.

Theorem 3.18. The complex number e

iθ

z is the result of rotating z counter-clockwise about the origin

through an angle θ.

Example 3.19. To rotate z = 1 + 2i counter-clockwise by

3π

radians, we multiply by

3πi

= cos

3π

+ i sin

3π

√

( −1 + i)

to obtain

3πi

z =

√

( −1 + i)(1 + 2i) = −

√

(3 + i)

−2 −1 1

3πi

3π

−i

You could try to keep things in polar form, though it doesn’t result in a nice answer:

z =

√

i tan

−1

=⇒ e

3πi

z =

√

3πi

+i tan

−1

Reﬂections may be described by combining rotations with complex con-

jugation. To reﬂect across the line making angle θ with the positive real

axis, we rotate the plane so that the reﬂection appears to be vertical:

1. Rotate the plane clockwise by θ, that is z 7→ e

−iθ

2. Reﬂect across the real axis by complex conjugation.

3. Rotate counter-clockwise by θ.

Combining these steps gives the formula.

Theorem 3.20. To reﬂect z across the line making angle θ with the positive real axis, we compute

z 7→ e

iθ

( e

−iθ

z) = e

2iθ

Example 3.21. Reﬂect z = −2 + 3i across the line through the origin and w =

√

3 + i.

First compute θ = arg(w) = tan

−1

√

. The desired point is therefore

iπ

( −2 −3i) =

√

( −2 −3i) =

√

−1

−



√

3 +



To describe general rotations and reﬂections about arbitrary points/lines, we combine our approach

with translations (compare Exercise 3.3.7).

Corollary 3.22. 1. To rotate z by θ about a point w, compute z 7→ e

iθ

( z −w) + w.

2. To reﬂect z across the line with slope θ through a point w, compute z 7→ e

2iθ

( z −w) + w.

Example 3.23. The combination of translation by −i, rotation by

around the origin, then translation

by 1, may be expressed

z 7→ e



z −i



+ 1 = i + e



z −i



+ 1 − i

Alternatively, this is rotation by

around i followed by translation by 1 −i.

We have now described all the Euclidean isometries of the previous section in the language of com-

plex numbers. Here is the full dictionary.

Isometry/Transformation Complex numbers Matrices/vectors

Addition/Translation z + w = (x + iy) + (u + iv) z + w =









Scaling

λz = (λx) + i(λy) λz =



λx

λy



Rotation CCW by

z 7→ iz z 7→



0 −1

1 0



Rotation CCW by θ z 7→ e

iθ

z z 7→



cos θ −sin θ

sin θ cos θ



Vertical reﬂection z 7→ z z 7→



1 0

0 −1



Reﬂection across line with slope

z 7→ e

iθ

z z 7→



cos θ sin θ

sin θ −cos θ



It is perhaps surprising to modern readers, but complex numbers came before vectors and matrix-

geometry! During the 1800s mathematicians tried unsuccessfully to replicate the complex number

approach in higher dimensions. This ultimately led (via Hamilton’s quaternions) to the adoption of

vectors and linear algebra/matrix calculations.

One reason for the desire to keep the complex number description is that it may be used to describe

further (non-isometric) transformations of the plane: for instance z 7→ z

−1

is reﬂection in a circle! We’ll

discuss some of this at the end of Chapter 4.

Scaling isn’t an isometry, but it is worth including nonetheless!

Exercises 3.4. 1. Use complex numbers to compute the result of the following transformations: you

can answer in either standard or polar form.

(a) Rotate 3 −5i counter-clockwise around the origin by

3π

radians.

(b) Reﬂect 2 −i across the line joining 1 + i

√

3 and the origin.

radians with the positive

real axis.

2. Find the reﬂection of the point (2, 3) across the line making angle

3π

with the positive x-axis.

Give your answer using both complex numbers and matrices/vectors.

3. Repeat the previous question for the point (3, 4) and the angle

5π

= 75°.

4. Describe the geometric effect of the map z 7→

√

( −1 −i)



z −3 + 4i



(Hint: compare Example 3.23)

5. (Hard) Consider the line ℓ through the origin and



2 +

√

2 −

√



. Compute the result

of reﬂecting −2 + 3i across ℓ.

6. By letting n = 3 in Lemma 3.16, prove that

cos 3θ = 4 cos

θ −3 cos θ

Find a corresponding trigonometric identity for sin 3θ.

7. Prove part 2 of Lemma 3.16.

(Hint: use the multiple-angle formulae (page 39) to expand e

i(θ+ϕ)

)

3.5 Birkhoff’s Axiomatic System for Analytic Geometry (non-examinable)

Analytic geometry was originally conceived as an addition to Euclidean geometry. In 1932, courtesy

of George David Birkhoff, it was axiomatized in its own right.

Background Assume the usual properties/axioms of the real numbers as a complete ordered ﬁeld.

Birkhoff’s system is typical of modern axiomatic systems in that it is built on top of pre-existing

systems (set theory, complete ordered ﬁelds, etc.).

Undeﬁned terms Two objects: Point, line. Two function: distance d, angle measure ∡. If the set of

points is S, then,

d : S ×S → R

, ∡ : S ×S × S → [ 0, 2π)

Axioms Euclidean Given two distinct points, there exists a unique line containing them.

Ruler Points on a line ℓ are in bijective correspondence with the real numbers in such a way that if

, t

correspond to A, B ∈ ℓ, then

−t

= d(A, B).

Protractor The rays emanating from a point O are in bijective correspondence with the set [0, 2π) so

that if α, β correspond to rays

−→

OA,

−→

OB, then ∡AOB ≡ β − α (mod 2π). This correspondence is

continuous in A, B.

SAS similarity

If triangles have a pair of angles with equal measure, and the sides adjacent to said

angles are in the same ratio, then the remaining angles have equal measure and the ﬁnal sides

are in the same ratio.

Deﬁnitions As with Hilbert, some of these are required before later axioms make sense. In partic-

ular, the deﬁnition of ray is required before the protractor axiom.

Betweenness B lies between A and C if d(A, B) + d(B, C) = d(A, C)

Segment AB consists of the points A, B and all those between

Ray

−→

AB consists of the segment AB and all points C such that B lies between A and C.

Basic shapes Triangles, circles, etc.

Analytic Geometry as a Model

The axioms should feel familiar. Being shorter than Hilbert’s list, and being built on familiar notions

such as the real line, it is somewhat easier for us to understand what the axioms are saying and to

visualize them. There is something to prove however; indeed the major point of Birkhoff’s system!

Theorem 3.24. Cartesian analytic geometry is a model of Birkhoff’s axioms.

Recall what this requires: we must provide a deﬁnition of each of the undeﬁned terms and prove that

these satisfy each of Birkhoff’s axioms. Here are suitable deﬁnitions for Cartesian analytic geometry:

As with Hilbert, Birkhoff makes SAS an axiom: Birkhoff’s version is stronger, for it also applies to similar triangles

Point An ordered pair (x, y) of real numbers.

Distance d(A, B) =

− B

)

+ (A

− B

)

Line All points satisfying a linear equation ax + by + c = 0.

Angle Deﬁne column vectors as differences (v = P −O and w = Q −O) and consider the matrix

J =



0 −1

1 0



. Now deﬁne angle via

cos ∡POQ =

v · w

where ∡POQ ∈

(

[0, π] ⇐⇒ w · Jv ≥ 0

( π, 2π) ⇐⇒ w · Jv < 0

(∗)

In essence, J is ‘rotate counter-clockwise by

.’ Cosine may be deﬁned using power series, so

no pre-existing geometric meaning is required.

Proof. (Euclidean axiom) If (x

, y

) and (x

, y

) satisfy ax + by + c = 0 then

a(x

− x

) + b(y

−y

) = 0

whence a = y

−y

, b = x

− x

up to scaling. It follows that the line has equation

( y

−y

)x + (x

− x

) y + x

− x

= 0

unique up to multiplication of all three of a, b, c by a non-zero constant.

The remaining axioms are exercises.

Exercises 3.5. 1. Prove that the ruler axiom is satisﬁed:

(a) First show that if P = Q lie on ℓ, then any point A on the line has the form

A = P +

d(P, Q)

(Q −P) where t

∈ R

(b) Use this formula to verify that d(A, B)

= (t

−t

)

2. Let i =





. Given any non-zero point B, deﬁne b = B −O and let β = cos

−1

i·b

in accordance

with (∗). This is a continuous function of b.

(a) If

B is any other point on the same ray

−→

OB, explain why we get the same value β.

(β is thus a continuous function of B)

(b) If B = (x, y), what is are values of cos β and sin β?

prove that the protractor axiom is satisﬁed.

3. Use the cosine rule (Theorem 3.7) to prove that the SAS similarity axiom is satisﬁed.

4 Hyperbolic Geometry

4.1 History: Saccheri, Lambert and Absolute Geometry

For 2000 years after Euclid, many mathematicians believed that his parallel postulate could not be

an independent axiom. Rigorous work on this problem was undertaken by Giovanni Saccheri (1667–

1733) & Johann Lambert (1728–1777); both attempted to force contradictions by assuming the nega-

tion of the parallel postulate. While this approach ultimately failed, their insights supplied the foun-

dation of a new non-Euclidean geometry. Before considering their work, we deﬁne some terms and

recall our earlier discussion of parallels (pages 10–13).

Deﬁnition 4.1. Absolute or neutral geometry is the axiomatic system comprising all of Hilbert’s

axioms except Playfair. Euclidean geometry is therefore a special case of neutral geometry.

A non-Euclidean geometry is (typically) a model satisfying most of Hilbert’s axioms but for which

parallels might not exist or are non-unique:

There exists a line ℓ and a point P ∈ ℓ through which there are no parallels or at least two.

For instance, spherical geometry is non-Euclidean since there are no parallel lines—Hilbert’s axioms

I-2 and O-3 are false, as is the exterior angle theorem.

Results in absolute geometry The conclusions of Euclid’s ﬁrst 28 theorems are valid.

• Basic constructions: bisectors, perpendiculars, etc.

• Triangle congruence theorems: SAS, ASA, SAA, SSS.

• Exterior angle theorem and its consequences:

ℓ

– Side/angle comparison and triangle inequality (Exercise 2.3.5).

– Existence of a parallel m to a line ℓ through a point P ∈ ℓ via congruent angles

∼

β =⇒ ℓ ∥ m

Arguments making use of unique parallels The following results were proved using Playfair’s

axiom or the parallel postulate, whence the arguments are false in absolute geometry:

• A line crossing parallel lines makes congruent angles: in the picture, ℓ ∥ m =⇒ α

∼

β. This is

the uniqueness claim in Playfair: the parallel m to ℓ through P is unique.

• Angles in a triangle sum to 180°.

• Constructions of squares/rectangles.

• Pythagoras’ Theorem.

While our arguments for the above are false in absolute geometry, we cannot instantly claim that the

results are false, for there might be alternative proofs! To show that these results truly require unique

parallels, we must exhibit a model in which they are false—such will be described in the next section.

The existence of this model explains why Saccheri and Lambert failed in their endeavors; the parallel

postulate (Playfair) is indeed independent of Euclid’s (Hilbert’s) other axioms.

The Saccheri–Legendre Theorem

We work in absolute geometry, starting with an extension of

the exterior angle theorem based on Euclid’s proof.

Suppose △ABC has angle sum Σ

△

and construct M and E fol-

lowing Euclid to the arrangement pictured. Observe:

1. ∡ACB + ∡CAB = ∡ACB + ∡ACE < 180° is the exterior angle theorem. More generally, the

exterior angle theorem says that the sum of any two angles in a triangle is strictly less than 180°.

2. △ABC and △EBC have the same angle sum

△

•

Just look at the picture—remember that we do not know whether Σ

△

= 180°!

3. △EBC has at least one angle (∡EBC or ∡BEC) measuring ≤

∡ABC.

Iterate this construction: if ∡EBC ≤

∡ABC, start by bisecting CE; otherwise bisect BC . . . The result

is an inﬁnite sequence of triangles △

= △EBC, △

, △

, . . . with two crucial properties:

(a) All triangles have same angle sum Σ

△

= Σ

△

= Σ

△

= ···.

(b) △

has at least one angle measuring α

≤

∡ABC.

Now suppose Σ

△

= 180° + ϵ is strictly greater than 180°. Since lim

= 0, we may choose n large

enough to guarantee α

< ϵ. But then the sum of the other two angles in △

would be greater than

180°, contradicting the exterior angle theorem (observation 1)! We have proved a famous result.

Theorem 4.2 (Saccheri–Legendre). In absolute geometry, triangles have angle sum Σ

△

≤ 180°.

Saccheri’s failed hope was to prove equality without invoking the parallel postulate.

Saccheri and Lambert Quadrilaterals

Two families of quadrilaterals in absolute geometry are named in honor of these pioneers.

Deﬁnition 4.3. A Saccheri quadrilateral ABCD satisﬁes

∼

BC and ∡DAB = ∡CBA = 90°

AB is the base and CD the summit.

The interior angles at C and D are the summit angles.

A Lambert quadrilateral has three right-angles; for instance AMND

in the picture.

We draw these with curved sides to indicate that the summit angles need not be right-angles, though

we haven’t yet exhibited a model which shows they could be anything else. Regardless of how they

are drawn, AD, BC and CD are all segments!

The apparent symmetry of a Saccheri quadrilateral is not an illusion.

Lemma 4.4. 1. If the base and summit of a Saccheri quadrilateral

are bisected, we obtain congruent Lambert quadrilaterals.

2. The summit angles of a Saccheri quadrilateral are congruent.

3. In Euclidean geometry, Saccheri and Lambert quadrilaterals

are rectangles (four right-angles).

Parts 1 and 2 are exercises. We could interpret part 3 as saying that Saccheri and Lambert quadrilat-

erals are as close as we can get to rectangles in absolute geometry.

Proof of 3. By part 1 we need only prove this for a Saccheri quadrilateral. Following the exterior angle

theorem,

←→

AB is a crossing line making congruent right-angles, whence AD ∥ BC.

However

←→

CD also crosses the same parallel lines. By the parallel postulate, the summit angles sum

to a straight edge. Since these are congruent, they are both right-angles.

We now show that drawing acute summit angles is justiﬁed by the Saccheri–Legendre Theorem.

Theorem 4.5. The summit angles of a Saccheri quadrilateral measure ≤ 90°.

Proof. Suppose ABCD is a Saccheri quadrilateral with base AB.

Extend CB to E (opposite side of AB to C) such that BE

∼

DA.

Let M be the midpoint of AB.

SAS implies ∠DAM

∼

△EBM; the vertical angles at M are con-

gruent, whence M lies on DE.

By Saccheri–Legendre, the (congruent) summit angles at C and D sum to

∡ADC + ∡DCB = ∡ADM + ∡EDC + ∡DCE = ∡CED + ∡EDC + ∡DCE ≤ 180°

Exercises 4.1. Work in absolute geometry; you cannot use Playfair’s Axiom or the parallel postulate!

1. Use the ﬁrst picture to prove parts 1 and 2 of Lemma 4.4.

2. Use the ﬁrst picture to give an alternative proof of Theorem 4.5.

3. Suppose □ABCD has four right-angles (second picture). Show

that AC splits □ABCD into two congruent triangles, and con-

clude that the opposite sides are congruent.

Why is this question easier in Euclidean geometry?

4. Suppose two Saccheri quadrilaterals have congruent bases

(AB) and perpendicular sides (AD, BC). Prove that they are

congruent.

5. (Hard!) Suppose Saccheri quadrilaterals have congruent sum-

mits and perpendicular sides. Prove that the quadrilaterals are

congruent.

4.2 Models of Hyperbolic Geometry

In the 1820-30s, J

anos Bolyai, Carl Friedrich Gauss and Nikolai Lobachevsky independently took the

next step, each describing versions of non-Euclidean geometry.

Rather than attempting to establish

the parallel postulate as a theorem within Euclidean geometry, a new geometry was deﬁned based on

the ﬁrst four of Euclid’s postulates plus an alternative to the parallel postulate:

Axiom 4.6 (Bolyai–Lobachevsky/Hyperbolic Postulate). Given a line ℓ and a point P ∈ ℓ, there

exist at least two parallel lines to ℓ through P.

The resulting axiomatic system

is called hyperbolic geometry. Consistency was proved in the late

1800s by Beltrami, Klein and Poincar

e, each of whom created models by deﬁning point, line, etc., in

novel ways. One of the simplest is named for Poincar

e, though was ﬁrst proposed by Beltrami.

Deﬁnition 4.7. The Poincar´e disk is the interior of the unit circle



(x, y) ∈ R

: x

+ y

< 1





z ∈ C :

< 1



A hyperbolic line is a diameter or a circular arc meeting the unit circle at

right-angles.

In the picture we have a hyperbolic line ℓ and a point P: also drawn

are several parallel hyperbolic lines to ℓ passing through P.

Points on the boundary circle are termed omega-points: these are not in

the Poincar

e disk and are essentially ‘points at inﬁnity.’

ℓ

Since it depends only on the incidence axioms, there exists a unique hyperbolic line joining any two

points in the Poincar

e disk.

Hyperbolic lines may straightforwardly be described using equations in analytic geometry.

Lemma 4.8. Every hyperbolic line in the Poincar

e disk model is

one of the following:

• A diameter passing through (c, d) = (0, 0) with Euclidean

equation dx = cy.

• The arc of a (Euclidean) circle with equation

+ y

−2ax −2by + 1 = 0 where a

+ b

> 1

and (Euclidean) center and radius

C = (a, b) and r =

+ b

−1

C = (a, b)

Bolyai indeed is the source of the term ‘absolute geometry.’

We assume all of Hilbert’s axioms, replacing Playfair axiom with the hyperbolic postulate.

Example 4.9. We compute the hyperbolic line through P = (0,

) and Q = (

) in the Poincar

disk: this is the picture shown in Lemma 4.8.

Substitute into x

+ y

−2ax −2by + 1 = 0 to obtain a system of equations for a, b:

(

−b + 1 = 0

− a −

b + 1 = 0

=⇒ (a, b) =





The required hyperbolic line

←→

PQ therefore has equation

+ y

−

x −

y + 1 = 0 or



x −





y −



545

648

The undeﬁned terms point, line, on and between now make sense. To complete the model, we need to

deﬁne congruence of hyperbolic segments and angles.

Deﬁnition (4.7 continued). The hyperbolic distance between points P, Q in the Poincar

e disk is

d(P, Q) := cosh

−1

1 +

(1 −

)(1 −

)

where

is the Euclidean distance and

are the Euclidean

distances of P, Q from the origin.

Hyperbolic segments are congruent if they have the same length.

The angle between hyperbolic rays is that between their tangent lines:

angles are congruent if they have the same measure.

Lemma 4.10. The hyperbolic distance of P from the origin is

d(O, P) = cosh

−1

1 +

1 −

= ln

1 +

1 −

Example 4.11. We calculate the sides and angles in the isosceles right-triangle

with vertices O = (0, 0), P = (

, 0) and Q = (0,

d(O, P) = d(O, Q) = ln

1 +

1 −

= ln 3 = cosh

−1

≈ 1.099

d(P, Q) = cosh

−1

1 +

2 ·

(1 −

)

= cosh

−1

≈ 1.681

It seems reasonable for hyperbolic functions to play some role in hyperbolic geometry! As a primer:

cosh x =

+ e

−x

, sinh x =

−e

−x

, tanh x =

sinh x

cosh x

−e

−x

+ e

−x

and cosh

−1

x = ln(x +

−1)

To ﬁnd the interior angle θ, implicitly differentiate the equation for the hyperbolic line

←→

PQ:

+ y

−

x −

y + 1 = 0 =⇒



4x −5

5 −4y



= −

=⇒ θ = tan

−1

≈ 30.96°

By symmetry, we have the same angle at Q. With a right-angle at O, we conclude that the angle sum

is approximately Σ

△

= 151.93°!

As a sanity check, we compare data for △OPQ and the Euclidean triangle with the same vertices

Property Hyperbolic Triangle Euclidean Triangle

Edge lengths 1.099 : 1.099 : 1.681 0.5 : 0.5 : 0.707

Relative edge ratios 1 : 1 : 1.530 1 : 1 : 1.414

Angles 30.06°, 30.96°, 90° 45°, 45°, 90°

The hyperbolic triangle has longer sides and a relatively longer hypotenuse. Moreover, its side lengths

do not satisfy the Pythagorean relation a

+ b

= c

(though cosh a cosh b = cosh c . . .).

The next result is an exercise; it says that distance increases

smoothly as one moves along a hyperbolic line.

Lemma 4.12. Fix P and a hyperbolic line through P. Then the

distance function Q 7→ d(P, Q) maps the set of points on one side

of P differentiably and bijectively onto the interval (0, ∞).

The Lemma means that hyperbolic circles are well-deﬁned and

look like one expects: the circle of hyperbolic radius δ centered at

P is the set of points Q such that d(P, Q) = δ.

In the picture are several hyperbolic circles and their centers; one has several of its radii drawn.

Observe how the centers are closer (in a Euclidean sense) to the boundary circle than one might

expect: this is since hyperbolic distances measure greater the further one is from the origin.

In fact (Exercise 4.2.5) hyperbolic circles in the Poincar

e disk model are also Euclidean circles! Their

hyperbolic radii moreover intersect the circles at right-angles, as we’d expect.

Theorem 4.13. The Poincar

e disk is a model of hyperbolic geometry.

Sketch Proof. A rigorous proof would require us to check the hyperbolic postulate and all Hilbert’s

axioms except Playfair. Instead we verify Euclid’s postulates 1–4 and the hyperbolic postulate 5.

1. Lemma 4.8 says we can join any given points in the Poincar

e disk by a unique segment.

2. A hyperbolic segment joins two points inside the (open) Poincar

e disk. The distance formula in-

creases (Lemma 4.12) unboundedly as P moves towards the boundary circle, so we can always

make a hyperbolic line longer.

3. Hyperbolic circles are deﬁned above.

4. All right-angles are equal since the notion of angle is unchanged from Euclidean geometry.

5. The ﬁrst picture on page 53 shows multiple parallels!

Other Models of Hyperbolic Space: non-examinable

There are several other models of hyperbolic space. Here are three of the most common.

Klein Disk Model This is similar to the Poincar

e disk, though lines are chords of the unit circle

(‘Euclidean’ straight lines!) and the distance function is different:

(P, Q) =



PΘ

QΩ

PΩ

QΘ



where Ω, Θ are where the chord

←→

PQ meets the boundary circle.

The cost is that the notion of angle is different. The picture shows

perpendicularity: Given a hyperbolic line ﬁnd the tangents to where

it meets the boundary circle. Any chord whose extension passes

through the intersection of these tangents is perpendicular to the orig-

inal line. Measuring other angles is difﬁcult!

Ω

Gauss’ famous theorem egregium says that this problem is unavoidable; there is no model in which

lines and angles both have the same meaning as in Euclidean geometry.

Poincar´e Half-plane Model Widely used in complex analysis, the

points comprise the upper half-plane (y > 0) in R

, while hyperbolic

lines are verticals or semicircles centered on the x-axis

x = constant or (x −a)

+ y

= r

and angles are the same as in Euclidean space. The expression for

hyperbolic distance remains horriﬁc! The picture shows several hy-

perbolic lines and a hyperbolic triangle.

Hyperboloid Model Points comprise the upper sheet (z ≥ 1) of the hyperboloid x

+ y

= z

−1.

A hyperbolic line is the intersection of the hyperboloid with a plane through the origin. Isometries

(congruence) can be described using matrix-multiplication and hyperbolic distance is relatively easy:

given P = (x, y, z) and Q = (a, b, c), hyperbolic distance is

d(P, Q) = cosh

−1

( cz −ax −by)

Difﬁculties include working in three dimensions and the fact

that angles are awkward.

The relationship to the Poincar

e disk is via projection. Place

the disk in the x, y-plane centered at the origin and draw a

line through the disk and the point (0, 0, −1). The intersec-

tion of this line with the hyperboloid gives the correspon-

dence.

Exercises 4.2. Answer all questions within the Poincar

e disk model.

1. (a) Find the equation of the hyperbolic line joining P = (

, 0) and Q = (0,

(b) Find the side lengths of the hyperbolic triangle △OPQ where O = (0, 0) is the origin.

the sides opposite O, P, Q respectively, check that the Pythagorean theorem p

+ q

= o

is false. Now compute cosh p cosh q: what do you observe?

2. Let P =





and Q =



, −



(a) Compute the hyperbolic distances d(O, P), d(O, Q) and d(P, Q), where O is the origin.

(b) Compute the angle ∡POQ.

←→

PQ has equation

−

x + y

+ 1 = 0

(d) Calculate

and hence show that a tangent vector to ℓ at P is

√

15i + 7j. Use this to

compute ∡OPQ.

3. We extend Example 4.11. Let c ∈ (0, 1) and label O = (0, 0), P = (c, 0) and Q = (0, c).

(a) Compute the hyperbolic side lengths of △OPQ.

(b) Find the equation of the hyperbolic line joining P = (c, 0) and Q = (0, c).

−1

1−c

1+c

What happens as c → 0

and as c → 1

−

4. Let 0 < r < 1 and ﬁnd the hyperbolic side lengths and interior angles of the equilateral triangle

with vertices (r, 0), (−

√

) and (−

, −

√

).What do you observe as r → 0

and r → 1

−

5. (a) Use the cosh distance formula to prove that the hyperbolic circle of hyperbolic radius

ρ = ln 3 and center C = (

, 0) in the Poincar

e disk has Euclidean equation



x −



+ y

(b) Prove that every hyperbolic circle in the Poincar

e disk is in fact a Euclidean circle.

6. We sketch a proof of Lemma 4.12.

(a) Prove that f (x) = cosh

−1

x = ln(x +

√

−1) is strictly increasing on the interval (1, ∞).

(b) By part (a), it is enough to show that

1−

increases as Q moves away from P along a

hyperbolic line. Appealing to symmetry, let P = (0, c) lie on the hyperbolic line with

equation x

+ y

−2by + 1 = 0. Prove that

1 −

( b −c)y + bc −1

1 −by

and hence show that this is an increasing function of y when c < y <

4.3 Parallels, Perpendiculars & Angle-Sums

From now on, all examples will be illustrated within the Poincar

disk model. Recall (page 50) that we may use anything from ab-

solute geometry; as a sanity check, think through how the picture

illustrates the following result.

Lemma 4.14. Through a point P not on a line ℓ there exists a

unique perpendicular to ℓ .

We now consider a major departure from Euclidean geometry.

ℓ

Theorem 4.15 (Fundamental Theorem of Parallels). Given P ∈ ℓ, drop

the perpendicular PQ. Then there exist precisely two parallel lines m, n to ℓ

through P with the following properties:

1. A ray based at P intersects ℓ if and only if it lies between m and n in

the same fashion as

−→

PQ.

2. m and n make congruent acute angles µ with

−→

PQ.

ℓ

Deﬁnition 4.16. The lines m, n are the limiting, or asymptotic, parallels to ℓ through P. Every other

parallel is an ultraparallel. The angle of parallelism at P relative to ℓ is the acute angle µ.

More generally, parallel lines ℓ, m are limiting if they ‘meet’ at an omega-point.

The proof depends crucially on ideas from analysis, particularly continuity & suprema. As you read

through, consider how everything except the last line is valid in Euclidean geometry!

Proof. Points R ∈ ℓ are in continuous bijective correspondence with the real numbers (Lemma 4.12).

It follows that we have a continuous increasing function

f : R → (−90°, 90°) where f (r) = ∡QPR

By Saccheri–Legendre, ±90° ∈ range f . Since dom f = R is an interval, the intermediate value

theorem forces range f to be a subinterval I ⊆ (−90°, 90°).

Given R ∈ ℓ, transfer QR to the other side of Q to obtain S ∈ ℓ. By SAS,

∡QPS = −∡QPR whence I = range f is symmetric: θ ∈ I ⇐⇒ −θ ∈ I.

Deﬁne µ := sup I ∈ (0°, 90°] to be the least upper bound; by symmetry,

inf I = −µ. Let m and n be the lines making angles ±µ respectively.

Plainly every ray making angle θ ∈ (−µ, µ) intersects ℓ.

Suppose m intersected ℓ at M. Let

M ∈ ℓ lie on the other side of M from

Q. Since f is increasing, we see that ∡QP

M > µ, which contradicts

µ = sup I. It follows that m is parallel to ℓ. Similarly n ∥ ℓ and we have

part 1.

ℓ

Finally m = n ⇐⇒ µ = 90°. In such a case there would exist only one parallel to ℓ through P,

contradicting the hyperbolic postulate.

The picture suggests a bijective relationship between µ and the perpendicular distance. Here it is; we

postpone a simpliﬁed argument to Exercise 4.3.3; the full result follows from a discussion of omega-

triangles in the next section.

Corollary 4.17. The perpendicular distance δ = d(P, Q) and the angle of parallelism are related via

cosh δ = csc µ or equivalently tan

= e

−δ

Examples 4.18. 1. Let ℓ be the hyperbolic line x

+ y

−4x + 1 = 0.

Intersect with x

+ y

= 1 to ﬁnd Ω =



√



and Θ =



, −

√



By symmetry, the perpendicular from P = (0, 0) to ℓ has equation

y = 0 and results in Q = (2 −

√

3, 0).

The limiting parallels through P have equations y = ±

√

3x, from

which the angle of parallelism is µ = tan

−1

√

3 = 60°.

In accordance with Corollary 4.17, we easily verify that

Ω

δ = d(P, Q) = ln

1 + (2 −

√

1 −(2 −

√

= ln

√

3 ↭ e

−δ

√

= tan

2. We ﬁnd the limiting parallels and the angle of parallelism when

P =



−



and x

+ y

+ 2x + 4y + 1 = 0

First ﬁnd the omega-points by intersecting with x

+ y

= 1:

Ω = (−1, 0), Θ =



, −



Plainly

←→

PΘ is the diameter y = −

x with slope −

Ω

For

←→

PΩ, substitute into the usual expression x

+ y

−2ax −2by + 1 = 0 and implicitly differ-

entiate:

+ y

+ 2x −

y + 1 = 0 =⇒



16( 1 + x)

13 −16y



16 ·

13 −

The angle of parallelism is half that between the tangent vectors



−33

−56



and



−4



µ =

cos

−1



−33

−56





−4







−33

−56







−4





cos

−1

≈ 33.69°

Corollary 4.17 can now be used to ﬁnd the perpendicular distance d(P, Q) = ln

√

Without the development of later machinery, it is very tricky to compute Q. If you want a serious

challenge, see if you can convince yourself that Q =



93(−29+2

√

117)

1865

26(−29+2

√

117)

1865



Angles in Triangles, Rectangles and the AAA Congruence

We ﬁnish this section three important differences between hyperbolic and Euclidean geometry.

Theorem 4.19. In hyperbolic geometry:

1. There are no rectangles (quadrilaterals with four right-angles). In particular, the summit angles

of a Saccheri quadrilateral are acute.

2. The angles in a triangle sum to strictly less than 180°.

3. (AAA congruence) If the angles of △ABC and △DEF are congruent in pairs, then the triangles

are congruent (△ABC

∼

△DEF).

Note that AAA is a congruence theorem in hyperbolic geometry, not a similarity theorem (compare

with Theorem 2.42). Also revisit the observations on page 50. These results largely show that Euclid’s

arguments making use of the parallel postulate actually require it!

Proof. Given a rectangle □ABCD, reﬂect across CD (Exercise 4.1.4) and repeat to obtain an inﬁnite

family of congruent rectangles. Let P ∈ CD and drop perpendiculars to R ∈ AB and C

as shown.

□PRBC is a rectangle: if not, then one of □ARPD or □PRBC would have angle sum exceeding 360°,

contradicting Saccheri–Legendre (Theorem 4.2). Similarly □DPC

is a rectangle.

By Exercise 4.1.3,

−→

BP splits □PRBC into a pair of congruent triangles. In particular,

−→

BP crosses CD at

the same angle as it leaves B. We ﬁnd ourselves in the original conﬁguration: the ray

−→

BP emanates

from the upper-right vertex of □DPC

at the same angle as it does for □ABCD!

Iterate the process to obtain the picture, each time dropping the perpendicular from P

to CD to

produce the equidistant sequence Q

, Q

, . . . Since CD is ﬁnite, this eventually

passes D: some

lies on the opposite side of

←→

AD. It follows that P

∈

−→

BP does also, whence

−→

BP intersects

←→

AD.

Since P ∈ CD was generic, we see that any ray based at B on the same side as AD intersects

←→

AD. The

angle of parallelism of B with respect to

←→

AD is therefore 90°, contradicting the hyperbolic postulate.

Parts 2 and 3 are addressed in Exercises 4 and 5.

This is the Archimedean property from analysis: a > b =⇒ ∃n ∈ N such that nb > a.

Exercises 4.3. 1. Prove the following in hyperbolic geometry (use Theorem 4.19).

(a) Two hyperbolic lines cannot have more than one common perpendicular.

(b) Saccheri quadrilaterals with congruent summits and summit angles are congruent.

2. Let ℓ be the line x

+ y

−4x + 2y + 1 = 0 and drop a perpendicular from O to Q ∈ ℓ.

(a) Explain why Q has co-ordinates (

√

t, −

√

t) for some t ∈ (0, 1).

(b) Show that the hyperbolic distance δ = d(O, Q) of ℓ from the origin is ln

√

explicitly and check that cosh δ = csc µ.

3. Suppose P = (0, 0) is the origin, let 0 < r < 1 and Q = (r, 0). Also let ℓ be the hyperbolic line

passing through Q at right-angles to PQ.

(a) Find the equation of ℓ and prove that its limiting parallels through P have equations

±2ry = (1 −r

(Hint: what does symmetry tell you about the location of the Euclidean center of ℓ?)

(b) Let µ be the angle of parallelism of P relative to ℓ and δ = d(P, Q) the hyperbolic distance.

Prove that cosh δ = csc µ.

(Hint: csc

µ = 1 + cot

µ = 1 +

tan

= . . .)

4. We work in absolute geometry.

(a) Suppose A, B and P are non-collinear and drop the per-

pendicular from P to Q ∈

←→

AB.

If P lies between the perpendiculars ℓ, m to

←→

AB through A

and B, prove that Q is interior to AB.

(Hint: show that the other cases are impossible)

(b) Suppose there exists a triangle with angle sum 180°. Show

that there exists a right-triangle with angle sum 180° and

therefore a rectangle.

ℓ

(Since rectangles are impossible in hyperbolic geometry, this proves part 2 of Theorem 4.19)

5. We prove the AAA congruence theorem (Theorem 4.19, part 3).

Suppose △ABC and △DEF are non-congruent but have angles congruent in pairs. WLOG as-

sume DE < AB. By uniqueness of angle/segment transfer, there exist unique points G ∈ AB

and H ∈

−→

AC such that (SAS) △DEF

∼

△AGH.

The picture shows the three possible arrangements.

(a) H is interior to AC.

(b) H = C.

In each case, explain why we have a contradiction.

C = H?

4.4 Omega-triangles

Recall that limiting parallels (Deﬁnition 4.16) ‘meet’ at an omega-point.

Deﬁnition 4.20. An omega-triangle or ideal-triangle is a ‘triangle’ one or

more of whose vertices is an omega-point. At least two of the sides of

an omega-triangle form a pair of limiting parallels.

The three types of omega-triangle depend on how many omega-points

they have. In the picture, △PQΩ has one omega-point, △PΩΘ has

two and △ΩΘΞ three!

Ω

Amazingly, many of the standard results of absolute geometry also apply to omega-triangles! The

ﬁrst can be thought of as the AAA congruence theorem where one ‘angle’ is zero.

Theorem 4.21 (Angle-Angle Congruence for Omega-triangles). Suppose △ABΩ and △PQΘ are

omega-triangles, each with a single omega-point. If the angles are congruent in pairs

∠ABΩ

∼

∠PQΘ ∠BAΩ

∼

∠QPΘ

then the ﬁnite sides of each triangle are also congruent: AB

∼

PQ.

It doesn’t really make sense to speak of the ‘inﬁnite’ sides, or the ‘angles’ at omega-points, being

congruent. However, if one deﬁnes congruence in terms of isometries (Section 4.6), then this idea is

more reasonable.

Proof. Let T ∈

−→

AB be such that AT

∼

PQ. If T = B we are done.

Otherwise, WLOG and for contradiction, assume AB < AT. The hy-

pothesis asserts that the angles at B and T are congruent.

Let M be the midpoint of BT and drop the perpendicular to N ∈

←→

BΩ.

Let L ∈

←→

ΩT be on the opposite side of

←→

BT to N such that TL

∼

BN.

We now have Side-Angle-Side data, whence △NBM

∼

△LTM. The

vertical angles at M are congruent, whence M lies on LN and we have

a right-angle(!) at L.

The angle of parallelism of L relative to

←→

BΩ is now ∠NLΩ = 90°, which

contradicts Theorem 4.15.

There are two other possible orientations:

• N could lie on the opposite side of B from Ω. In this case SAS is

applied to the same triangles but with respect to the congruent

magenta angles.

Ω

• In the special case that N = B, the cyan angles are right-angles and the same contradiction

appears: the angle of parallelism of T with respect to

←→

BΩ is 90°.

Theorem 4.22 (Exterior Angle Theorem for Omega-Triangles). Suppose △PQΩ has a single

omega-point and let P ∗ Q ∗ R. Then ∠RQΩ > ∠QPΩ.

Proof. We show that the two other cases are impossible.

(∠RQΩ

∼

∠QPΩ) This is the contradictory arrangement described in

the previous proof: P = T, Q = B, R = A, and the magenta an-

gles cannot be congruent.

(∠RQΩ < ∠QPΩ) Transfer the latter to Q to produce a ray

−→

QX interior to

∠PQΩ with ∠RQX

∼

∠QPΩ.

Since

←→

QΩ is a limiting parallel to

←→

PΩ, the Fundamental Theorem

says that

−→

QX intersects

←→

PΩ at some point Y.

We now have △PQY contradicting the standard exterior angle the-

orem (∠RQY

∼

∠QPY).

Ω

The ﬁnal congruence theorem is an exercise based on the previous picture.

Corollary 4.23 (Side-Angle Congruence for Omega-triangles). Suppose △PQΩ and △ABΘ have

a single omega-point. If ∠QPΩ

∼

∠BAΘ and PQ

∼

AB then ∠PQΩ

∼

∠ABΘ.

A triangle with one omega-point only has three pieces of data: two ﬁnite angles and one ﬁnite edge.

The AA and SA congruence theorems say that two of these determine the third.

Other observations

Pasch’s Axiom: Versions of this are theorems for omega-triangles.

• If a line crosses a side of an omega-triangle and does not pass through any vertex (including

Ω), then it must pass through exactly one of the other sides.

• (Omega Crossbar Thm) If a line passes through an interior point and exactly one vertex (includ-

ing Ω) of an omega-triangle, then it passes through the opposite side. This is partly embedded

in the proof of Theorem 4.22.

Perpendicular Distance and the Angle of Parallelism: Applied to right-angled omega-triangles, the AA

and SA theorems prove that the angle of parallelism is a bijective function of the perpendicular dis-

tance. Moreover, by transferring the right-angle to the positive x-axis and the other vertex to the

origin, we obtain the arrangement in Exercise 4.3.3, thus completing the proof of Corollary 4.17.

Exercises 4.4. 1. Let △PQΩ be an omega-triangle. Prove that ∡PQΩ + ∡QPΩ < 180°

2. Let ℓ and m be limiting parallels. Explain why they cannot have a common perpendicular.

3. Prove the Side-Angle congruence theorem for omega-triangles with one omega-point.

4. What would an ‘omega-triangle’ look like in Euclidean geometry? Comment on the three re-

sults in this section: are they still true?

4.5 Area and Angle-defect

In this section we consider one of the triumphs of Johann Lambert: the relationship between the

sum of the angles in a triangle and its area. We start with a loose axiomatization of area as a relative

measure. Until explicitly stated otherwise, we work in absolute geometry.

Axiom I Two geometric ﬁgures have the same area if and only if they may be sub-divided into

ﬁnitely many pairs of mutually congruent triangles.

Axiom II The area of a triangle is positive.

Axiom III The area of a union of disjoint ﬁgures is the sum of the areas of the ﬁgures.

Deﬁnition 4.24 (Angle defect). Let Σ

△

be the sum of the angles in a triangle. Measured in radians,

the angle-defect of △ is π −Σ

△

Since triangles in absolute geometry have Σ

△

≤ π (Theorem 4.2), it follows that

0 ≤ π −Σ

△

≤ π

In Euclidean geometry the defect is always zero, while in hyperbolic geometry the defect is strictly

positive (Theorem 4.19). A ‘triangle’ with three omega-points would have defect π.

Lemma 4.25. Angle-defect is additive: If a triangle is split into two sub-

triangles, then the defect of the whole is the sum of the defects of the parts.

This is immediate from the picture:

[

π −(α + γ + ϵ)

]

[

π −(β + δ + ζ)

]

= π − (α + β + γ + δ)

since ϵ + ζ = π. Notice that angle-sum is not additive!

Theorem 4.26 (Area determines angle-sum in absolute geometry). If triangles have the same area,

then their angle-sums are identical.

Of course this trivial in Euclidean geometry where all triangles have the same angle-sum!

Proof. The lemma provides the induction step: if △

and △

have the same area, then their interiors

are disjoint unions of a ﬁnite collection of mutually congruent triangles:

△

[

k=1

△

1,k

and △

[

k=1

△

2,k

where △

1,k

∼

△

2,k

Each pair △

1,k

, △

2,k

has the same angle-defect, whence the angle-defects of △

and △

are equal:

defect(△

) =

∑

k=1

defect(△

1,k

) =

∑

k=1

defect(△

2,k

) = defect(△

)

To allow inﬁnitely many inﬁnitesimal sub-triangles would require ideas from calculus and complicate our discussion.

Angle-sum determines area in hyperbolic geometry

The converse in hyperbolic geometry relies on a beautiful and reversible construction relating tri-

angles and Saccheri quadrilaterals. The construction itself is valid in absolute geometry, though the

ultimate conclusion that angle-sum determines area is not. If the initial discussion seems difﬁcult,

pretend you are in Euclidean geometry and think about rectangles.

Lemma 4.27. 1. Given △ABC, choose a side BC. Bisect the remaining sides at E, F and drop per-

pendiculars from A, B, C to

←→

EF. Then HICB is a Saccheri quadrilateral with base HI.

2. Conversely, given a Saccheri quadrilateral HICB with summit BC, let A be any point such that

←→

HI bisects AB at E. Then the intersection F =

←→

HI ∩ AC is the midpoint of AC.

Both constructions yield the same picture and the following

conclusions:

• The triangle and quadrilateral have equal area.

• The sum of the summit angles of the quadrilateral

equals the angle sum of the triangle.

We chose BC to be the longest side of △ABC—this isn’t nec-

essary, though it helpfully forces E, F to lie between H, I.

H I

Proof. 1. Two applications of the SAA congruence (follow the arrows!) tell us that

△BEH

∼

△AEG and △CFI

∼

△AFG

We conclude that BH

∼

CI whence HICB is a Sac-

cheri quadrilateral. The area and angle-sum correspon-

dence is immediate from the picture.

2. Suppose the midpoint of AC were at J = F. By part 1, we

may create a new Saccheri quadrilateral with summit BC

using the midpoints E, J.

The perpendicular bisector of BC bisects the bases of both

Saccheri quadrilaterals (Lemma 4.4), creating △EUV

with two right-angles: contradiction.

H I

We now prove a special case of the main result.

Lemma 4.28. Suppose hyperbolic triangles △ABC and △PQR have congruent sides BC

∼

QR and

the same angle-sum. Then the triangles have the same area.

Proof. Construct the quadrilaterals corresponding to △ABC and △PQR with summits BC

∼

QR.

These have congruent summits and summit angles: by Exercise 4.3.1b they are congruent.

The ﬁnal observation is what makes this special to hyperbolic geometry. In the Euclidean case, Saccheri

quadrilaterals are rectangles, and congruent summits do not force congruence of the remaining sides.

Theorem 4.29. In hyperbolic geometry, if △ABC and △PQR have the same angle-sum then they

have the same area.

Proof. If the triangles have a pair of congruent edges, the previous result says we are done. Other-

wise, we use Lemma 4.27 to create a new triangle △LBC which matching the same Saccheri quadri-

lateral as △ABC.

WLOG suppose

and construct the Saccheri quadrilateral with summit BC. Select K on

←→

EF such that

and extend such that K is the midpoint of BL.

• By Lemma 4.27,

Area(△LBC) = Area(HICB) = Area(△ABC)

• By Theorem 4.26, △LBC has the same angle-sum as

△ABC and thus △PQR.

• △LBC and △PQR share a congruent side (LB

∼

PQ)

and have the same angle-sum. Lemma 4.28 says their

areas are equal.

Since both area and angle-defect are additive, we immediately conclude:

Corollary 4.30. The angle-defect of a hyperbolic triangle is an additive function of its area. By

normalizing the deﬁnition of area,

we may conclude that

π −Σ

△

= Area △

Note ﬁnally how the AAA congruence (Theorem 4.19, part 3) is related to the corollary:

△ABC

∼

△DEF

AAA

⇐⇒ angles congruent in pairs

⇓ ⇓

equal area

Cor 4.30

⇐⇒ same angle-defect

We have really only proved that π −Σ

△

is proportional to Area △. However, it can be seen that these quantities are

equal if we use the area measure arising naturally from the hyperbolic distance function (see page 79).

Corollary 4.30 is a special case of the famous Gauss–Bonnet theorem from differential geometry: for any triangle on a

surface with Gauss curvature K, we have

△

−π =

△

K dA

We’ve now met all three special constant-curvature examples of this:

Euclidean space is ﬂat (K = 0) so the angle-defect is always zero.

Hyperbolic space has constant negative curvature K = −1, whence

△

dA = −(Σ

△

−π) is the angle-defect.

Spherical geometry A sphere of radius 1 has constant positive curvature K = 1 and

△

dA is the angle-excess Σ

△

−π.

Example (4.11, cont). The isosceles right-triangle with vertices O, P = (

, 0) and Q = (0,

) has

angle-sum and area

+ 2 tan

−1

≈ 151.93° =⇒ area = π −



+ 2 tan

−1



−2 tan

−1

≈ 0.490

A Euclidean triangle with the same vertices has area

= 0.125.

Generalizing this (Exercise 4.2.3), the triangle with vertices O, P =

( c, 0) and Q = (0, c) has area

π −



+ 2 tan

−1

1 −c

1 + c



−2 tan

−1

1 −c

1 + c

As expected, lim

c→0

area(c) = 0. In the other limit, the triangle becomes

an omega-triangle with two omega-points and lim

c→1

−

area(c) =

: an

inﬁnite ‘triangle’ with ﬁnite ‘area’!

The limit c → 1

−

Our discussion in fact provides an explicit method for cutting a triangle into sub-triangles and rear-

ranging its pieces to create a triangle with equal area.

△

Suppose △

and △

have equal area and construct the quadrilaterals S

and S

. Let L, K be chosen

so that BL

∼

QR and K is the midpoint of BL. We now have:

• △

, △

, S

have the same area.

• The summit angles of S

, S

are congruent (half the angle-sum of each triangle).

• S

, S

are congruent since they have congruent summits and summit angles.

We can now follow the steps in Lemma 4.27 to transform △

to △

△

→ S

→ △

→ S

∼

→ △

where each arrow represents cutting off two triangles and moving them. Indeed this works even for

triangles in Euclidean geometry!

Exercises 4.5. 1. Use Corollary 4.30 to ﬁnd the area of the hyperbolic triangle with given vertices.

Your answers to exercises from Section 4.2 should supply the angles!

(a) O = (0, 0), P = (

) and Q = (

, −

(b) O = (0, 0), P = (

, 0), Q = (0,



−

√



, R =



−

, −

√



where 0 < r < 1.

2. In the proof of Theorem 4.29, explain why we can ﬁnd K such that

3. Show that there is no ﬁnite triangle in hyperbolic geometry that achieves the maximum area

bound π.

(Hard!) For a challenge, try to prove that omega-triangles also satisfy the angle-defect formula:

Area = π − Σ

△

, so that only triangles with three omega-points have maximum area.

4. Let Ω

, . . . , Ω

be n distinct omega-points arranged counter-clockwise around the boundary

circle of the Poincar

e disk. A region is bounded by the n hyperbolic lines

←−→

Ω

←−→

Ω

, . . . ,

←−−→

Ω

What is the area of the region? Hence argue that the ‘area’ of hyperbolic space is inﬁnite.

5. An omega-triangle has vertices O = (0, 0), Ω = (1, 0) and P = (0, h) where h > 0.

(a) Prove that the hyperbolic segment PΩ is an arc of a circle with equation

(x −1)

+ (y −k)

= k

for some k > 0.

(b) Prove that the area of △OPΩ is given by

A(h) = sin

−1

1 + h

6. Suppose two Saccheri quadrilaterals in hyperbolic geometry have the same area and congruent

summits. Prove that the quadrilaterals are congruent.

4.6 Isometries and Calculation

There are (at least!) two major issues in our approach to hyperbolic geometry.

Calculations are difﬁcult In analytic (Euclidean) geometry we typically choose the origin and orient

axes to ease calculation. We’d like to do the same in hyperbolic geometry.

We assumed too much We deﬁned distance, angle and line separately, but these concepts are not inde-

pendent! In Euclidean geometry, the distance function, or metric, deﬁnes angle measure via the

dot product,

and (with some calculus) the arc-length of any curve. One then proves that the

paths of shortest length (geodesics) are straight lines: the metric deﬁnes the notion of line!

Isometries provide a related remedy for these issues. To describe these it is helpful to use an alterna-

tive deﬁnition of the Poincar

e disk and its distance function.

Deﬁnition 4.31. The Poincar´e disk is the set D := {z ∈ C :

< 1}

equipped with the distance function

d( z, w) :=



z − Ω

w − Θ

z − Θ

w − Ω



where Ω, Θ are the omega-points for the hyperbolic line through z, w

(deﬁned as circular arcs intersecting the boundary perpendicularly).

Ω

We’ll see shortly (Corollary 4.38) that this is the same as the original cosh formula (page 54); it is

already easy to check that d(z, 0) = ln

1−

as in Lemma 4.10 (if w = 0, then Ω, Θ = ±

For candidate isometries we need functions f : D → D for which d



f (z), f (w)



= d(z, w). These

follow from some standard results of complex analysis that we state without proof.

Theorem 4.32 (M¨obius/fractional-linear transformations). If a, b, c, d ∈ C and ad − bc = 0, then the

function f (z) =

az+b

cz+d

has the following properties:

1. (Invertibility) f : C ∪ {∞} → C ∪ {∞} is bijective, with inverse f

−1

( z) =

dz−b

−cz+a

2. (Conformality) If curves intersect, then their images under f intersect at the same angle.

3. (Line/circle preservation) Every line/circle

is mapped by f to another line/circle.

4. (Cross-ratio preservation) Given distinct z

, z

, we have



f (z

) − f (z

)



f (z

) − f (z

)





f (z

) − f (z

)



f (z

) − f (z

)



( z

−z

)(z

−z

)

( z

−z

)(z

−z

)

Writing

for the length of a line segment, we see that for any u, v,

u ·v =



u + v

−



so that the metric deﬁnes the dot product. Now deﬁne angle measure via u ·v =

cos θ.

In C ∪ {∞} a line is just a circle containing ∞!

The isometries of the Poincar

e disk are a subset of the M

obius transformations.

Theorem 4.33. The orientation-preserving

isometries of the Poincar

e disk have the form

f (z) = e

iθ

α − z

αz −1

where

< 1 and θ ∈ [0, 2π) (∗)

All isometries can be found by composing f with complex conjugation (reﬂection in the real axis).

Referring to the properties in Theorem 4.32:

1. The isometries are precisely the set of M

obius transformations which map D bijectively to itself;

omega-points are also mapped to omega-points.

2. Isometries preserve angles.

3. The class of hyperbolic lines is preserved: any circle or line intersecting the unit circle at right-

angles is mapped to another such (angle-preservation is used here).

4. If Ω, Θ are the omega-points on

←→

zw, then (by 2 and 3), f (Ω) and f ( Θ) are the omega-points for

the hyperbolic line through f (z), f (w). Preservation of the cross-ratio says that f is an isometry:

d( f (z), f (w)) =



f (z) − f (Ω)

f (w) − f (Θ)

f (z) − f (Θ)

f (w) − f (Ω)



z − Ω

w − Θ

z − Θ

w − Ω



= d(z, w)

How does this help us compute? The isometry (∗) moves α to the origin, where calculating distances

and angles is easy!

Example 4.34. Let P =

and Q =

√

i. Move P to the origin using

an isometry

with α = P:

f (z) =

α − z

αz −1

1 −2z

z −2

=⇒ f (P) = O

f (Q) =

1 −

−

√

−2 +

√

= −

1 + 2

√

−4 +

√

Let us compare distances:

O = f (P)

f (Q)



f (P), f (Q)



= ln

1 +

f (Q)

1 −

f (Q)

= ln

1 +

√

1 −

√

= ln

√

2 + 1

√

2 −1

= ln(3 + 2

√

2) (Deﬁnition 4.31)

d(P, Q) = cosh

−1

1 +

(1 −

)(1 −

)

= cosh

−1

1 +

(1 −

)(1 −

)

= cosh

−1

3 = ln(3 +

−1) = ln(3 + 2

√

2) = d



f (P), f (Q)



If we trust the original cosh-formula (page 54), then the points really are the same distance apart!

Indeed the hyperbolic segment PQ has been transformed by f to a segment f (P) f (Q) of the y-axis.

If C is to the left of

−→

AB, then f (C) is to the left of

−−−−−−→

f (A) f (B). This is the usual ‘right-hand rule.’

We could also include a rotation (e

iθ

= −i) to move f (Q) to the positive x-axis, but there is no real beneﬁt.

Recall (e.g., Example 4.11) how we previously computed angles. Isometries make this much easier.

Example 4.35. Given A = −

, B = −

and C =

(3 −i), we ﬁnd d(A, B), d(A, C) and ∡BAC.

Start by moving A to the origin and consider f (B), f (C):

f (z) =

−

−z

z −1

2z + i

2 −iz

=⇒ f (B) =

−

+ i

2 −

f ( C) =

(3 −i) + i

2 −

(3 −i)

2( 3 −i) + 5i

10 −i(3 − i)

2 + i

3 −i

(2 + i)(3 + i)

1 + i

By mapping A to the origin, two sides of the triangle are now Eu-

clidean straight lines and the computations are easy:

d(A, B) = d



O, f (B)



= ln

1 +

1 −

= ln 2

d(A, C) = d



O, f (C)



= ln

1 +

√

1 −

√

= 2 ln(

√

2 + 1)

∡BAC = ∡ f (B) f (A) f (C) = arg

−arg

1 + i

−

To compute the ﬁnal side and angles, isometries moving B and then

C to the origin could be used.

O = f (A)

f (B)

f (C)

Interpretation of Isometries (non-examinable)

As in Euclidean geometry, isometries can be interpreted as rotations, reﬂections and translations.

Here is the dictionary in hyperbolic space.

Translations Move α to the origin via T

−α

( z) =

α−z

αz−1

The picture shows repeated applications of T

−α

to seven initial points.

Compose these to translate α to β:

◦ T

−α

( z) =

( αβ −1)z + α − β

( α − β)z + αβ −1

Rotations R

( z) = e

iθ

z rotates counter-clockwise around the origin. To ro-

tate around α, one computes the composition

◦ R

◦ T

−α

The picture shows repeated rotation by 30° =

around α.

Reﬂections P

( z) = e

2iθ

z reﬂects across the line making angle θ with the real

axis. Composition permits more general reﬂections, e.g.,

◦ P

◦ T

−α

Hyperbolic Trigonometry

By employing isometries in the abstract, we can develop formulas

allowing us to solve triangles

without the explicit requirement to compute with isometries at all!

Given a right-triangle, we may suppose an isometry has already moved the right-angle to the origin

and the other sides to the positive axes as in the picture. The non-hypotenuse side-lengths are

a = ln

1 + p

1 − p

= cosh

−1

1 + p

1 − p

, b = cosh

−1

1 + q

1 −q

To measure the hypotenuse, translate p to the origin via an isometry

f (z) =

p − z

pz −1

=⇒ f (iq) =

p −iq

ipq −1

=⇒

f (iq)

+ q

+ 1

f (p)

f (iq)

We therefore see that

cosh c =

1 +

f (iq)

1 −

f (iq)

1 + p

+ p

+ q

1 + p

− p

−q

1 + p

1 − p

1 + q

1 −q

= cosh a cosh b

Moreover, applying the hyperbolic identity sinh

b = cosh

b −1, we obtain

sinh b =

1 −q

=⇒ tanh b =

sinh b

cosh b

1 + q

Writing f (iq) in real and imaginary parts allows us to ﬁnd the slope

f (iq) =

p −iq

ipq −1

−p(1 + q

) + iq(1 − p

)

+ 1

=⇒ tan B =

q(1 − p

)

p(1 + q

)

tanh b

sinh a

Applying trigonometric identities such as csc

B = 1 + cot

B, we eventually conclude:

Theorem 4.36. In a hyperbolic right-triangle with adjacent a, opposite b, and hypotenuse c,

sin B =

sinh b

sinh c

cos B =

tanh a

tanh c

tan B =

tanh b

sinh a

cosh c = cosh a cosh b

This last is Pythagoras’ Theorem for hyperbolic right-triangles.

Example 4.37. A right-triangle has non-hypotenuse sides a = cosh

−1

3 ≈ 1.76, b = cosh

−1

5 ≈ 2.29.

cosh c = cosh a cosh b = 15 =⇒ c = cosh

−1

15 ≈ 3.40

sin A =

sinh a

sinh c

cosh

a −1

cosh

c −1

−1

√

=⇒ A ≈ 10.9°

sin B =

sinh b

sinh c

cosh

b −1

cosh

c −1

−1

=⇒ B ≈ 19.1°

You could use the other trig expressions to calculate: e.g., tan A =

tanh a

sinh b

sinh a

cosh a sinh b

√

. . .

Treat the formulas of hyperbolic trigonometry as open-book—they are not worth memorizing!

The goal of trigonometry is to ‘solve’ triangles: given minimal numerical data, to compute the re-

maining sides and angles. As in Euclidean geometry, you can attack general problems by dropping

perpendiculars and using the results of Theorem 4.36, though it is helpful to generalize this by de-

veloping the sine and cosine rules.

Corollary 4.38 (Sine/Cosine rules and the Cosh-distance formula). Label a general triangle with

angle-measures A, B, C opposite sides with (hyperbolic) lengths a, b, c.

Sine Rule Drop a perpendicular from C and observe that sin A =

sinh h

sinh b

and

sin B =

sinh h

sinh a

. Eliminate sinh h to obtain the ﬁrst equality in

sinh a

sin A

sinh b

sin B

sinh c

sin C

Drop a different altitude for the other equality.

Cosine Rule I Repeat the argument of Theorem 4.36 for a triangle with vertices 0, p and qe

to obtain

cosh c = cosh a cosh b −sinh a sinh b cos C

Expressing the right-hand side in terms of p, q (cosh a =

1+p

1−p

, etc.) and applying the Euclidean

cosine rule (cos C = ···) yields the original cosh-formula for distance (page 54).

Cosine Rule II Hyperbolic geometry admits a second version:

cos C = sin A sin B cosh c −cos A cos B

A proof is in Exercise 14.

In hyperbolic geometry, the triangle congruence theorems (SAS, ASA, SSS, SAA and AAA) pro-

vide suitable minimal data. The second version of the cosine rule has no analogue in Euclidean

geometry—it is particularly helpful for solving triangles given ASA or AAA data.

Examples 4.39. 1. (SAS) An isosceles triangle has angle C =

and sides a = b = cosh

−1

2 ≈ 1.32.

We have sinh a = sinh b =

cosh

a −1 =

√

3. By the cosine rule,

cosh c = 2 ·2 −

√

3 ·

=⇒ c = cosh

−1

≈ 1.57

Apply the sine rule for the ﬁnal angle:

sin B = sin A =

sin C sinh a

sinh c

√

21/4

√

=⇒ A = B ≈ 40.9°

The area of the triangle is therefore π −

−2 sin

−1

≈ 0.67

cosh

−1

cosh

−1

Unlike in Euclidean geometry, knowing two angles doesn’t automatically give you the third! For SAS and SSS start

with the cosine rule. SAA data is best solved by dropping a perpendicular and using Theorem 4.36.

2. (Equilateral AAA) An equilateral triangle has interior angles 30°. We compute its side-length

using the second version of the cosine rule:

cosh c =

cos A cos B + cos C

sin A sin B

√

= 3

√

=⇒ a = b = c = cosh

−1

√

3) ≈ 2.33

30°

3. (Right-angled AAA) A triangle has angles

and

: ﬁnd its sides.

Rather than using the second version of the cosine rule, we instead

indicate part of its proof by employing the tan-formula twice,

√

= tan

tanh a

sinh b

sinh a

cosh a sinh b

1 = tan

tanh b

sinh a

sinh b

sinh a cosh b

Multiply together and use hyperbolic Pythagoras,

√

cosh a cosh b

cosh c

=⇒ c = cosh

−1

√

3 = ln(

√

3 +

√

2) ≈ 1.15

Since sinh c =

cosh

c −1 =

√

2, the sine-rule yields the other sides:

sinh b = sin

sinh c

sin π

= 1 =⇒ b = sinh

−1

1 = cosh

−1

√

2 ≈ 0.88

=⇒ cosh a =

cosh c

cosh b

=⇒ a = cosh

−1

≈ 0.66

4. (ASA) Solve a triangle with angles

and a distance cosh

−1

3 between them.

Apply the second version of the cosine rule:

cos C = sin A sin B cosh c −cos A cos B

√

·3 −

√

=⇒ C =

The triangle is isosceles, whence a = cosh

−1

3 ≈ 1.76 also.

cosh

−1

The remaining side can be found using multiple methods: here is the cosine rule

cosh b = cosh a cosh c −sinh a sinh c cos B = 9 −



−1



= 5

=⇒ b = cosh

−1

5 = ln(5 +

√

24) ≈ 2.29

Hyperbolic Tilings (just for fun!)

Example 4.39.3 can be used to make a regular tiling of hyperbolic space.

Take eight congruent copies of the triangle and arrange

them around the origin as in the picture. Now reﬂect

the quadrilateral over each of its edges and repeat the

process in all directions. We obtain a regular tiling of

hyperbolic space comprising four-sided ﬁgures with six

meeting at every vertex!

In hyperbolic space, many different regular tilings are

possible. Suppose such is to be made using regular m-

sided polygons, n of which are to meet at each vertex:

each polygon comprises 2m copies of the fundamental

right-triangle, whose angles are therefore

and

Since the angles sum to less than π radians, we see that

there exists a regular tiling of hyperbolic space when-

ever m, n satisfy

< π ⇐⇒ (m −2)(n −2) > 4

The ﬁrst example is m = 4 and n = 6, where the fun-

damental triangle is clear. In the second example four

pentagons meet at each vertex and the interiors of the

polygons have been colored. This was produced using

the tools found here and here: have a play!

The multitude of possible tilings in hyperbolic geome-

try is in contrast to Euclidean geometry, where a regular

tiling requires equality

( m −2)(n −2) = 4

The three solutions (m, n) = (3, 6), (4, 4), (6, 3) cor-

respond to the only tilings of Euclidean geometry by

regular polygons (equilateral triangles, squares and

hexagons). However, all can be scaled to arbitrary side-

lengths. In hyperbolic geometry, there are inﬁnitely

many distinct tilings, but each has a unique side-length.

For related fun, look up M.C. Escher’s Circle Limit art-

works, some of which are based on hyperbolic tilings.

If you want an excuse to play video games while pre-

tending to study geometry, have a look at Hyper Rogue,

which relies on (sometimes irregular) tilings.

The fundamental triangle

( m, n) = (4, 6)

( m, n) = (5, 4)

Exercises 4.6. 1. Use Deﬁnition 4.31 to prove that d(z, 0) = ln

1−

2. (a) Use an isometry to ﬁnd angle ∡ABC when A = 0, B =

, and C =

1+i

(b) Now compute ∡ACB, and thus ﬁnd the angle sum and area of the triangle.

3. Associate a M

obius transformation f (z) =

az+b

cz+d

with the matrix



a b

c d



in the obvious way. If g

is another M

obius transformation, prove that the composition f ◦ g is associated to the product

of the matrices associated to f , g. Verify

that f

−1

( z) =

dz−b

a−cz

4. (a) A triangle has vertices A =

, B =

and C, where C lies in the upper half-plane (positive

imaginary part) such that ∡BAC = 45° and b = d(A, C) = cosh

−1

Compute a = d(B, C) using the hyperbolic cosine rule.

(b) The isometry f (z) =

−z

z−1

1−3z

z−3

moves A to the origin. What is f (B) and therefore f (C)?

(Hint: remember that f is orientation preserving)

the cosh distance formula to recover your answer to part (a).

5. Suppose f (z) =

α−z

αz−1

for some constant α ∈ C with

= 1. If

= 1, prove that

f (z)

= 1.

Argue that the functions f in Theorem 4.33 really do map the interior of the unit disk to itself.

6. Provided α = β, show that the isometry T

◦ T

−α

which translates α to β (page 71) is the trans-

lation T

−γ

where γ =

β−α

αβ−1

followed by a rotation around the origin.

7. Use the power series cosh x = 1 +

+ ··· to expand the hyperbolic Pythagorean

theorem cosh c = cosh a cosh b to order 4 (a

, a

, etc.). What do you observe?

8. A hyperbolic right-triangle has non-hypotenuse sides a = cosh

−1

2 and b = cosh

−1

3. Find the

hypotenuse, the angles and the area of the triangle.

9. Given ASA data c = cosh

−1

(

√

2 +

√

3) , A =

, B =

, ﬁnd the remaining data for the triangle.

10. An equilateral hyperbolic triangle has side-length a and angle A. Prove that cosh a =

cos A

1−cos A

If A = 45°, what is the side-length?

11. Find the interior angles and side-lengths for the quadrilateral and pentagonal tiles on page 75.

12. Use the hyperbolic cosine rule to prove that the cosh distance formula is valid.

13. Suppose you are given isosceles ASA data: angles A = B and side c between them. Prove that

c ≤ cosh

−1

(2 csc

A −1). What happens when this is equality?

14. (a) Prove the second cosine rule when C =

(see the trick in Example 4.39.3).

(b) (Hard!) Prove the full version by dropping a perpendicular from B = B

+ B

and observ-

ing that

cos A

sin B

cos C

sin B

cos C

sin(B−B

)

. . .

Since multiplying a, b, c, d by a non-zero scalar doesn’t change f , we see that the group of M

obius transformations is

isomorphic to the projective special linear group PSL

(R). The isometries of hyperbolic space form a proper subgroup.

The Poincar´e Disk for Differential Geometers (non-examinable)

Most of this last optional section should be accessible to anyone who’s taken basic vector-calculus.

All we really need is the Poincar

e disk model with its distance function d(z, w) and a description of

the isometries (Theorems 4.32, 4.33).

Consider the inﬁnitesimally separated points z and z + dz. Map z to

the origin via an isometry

f : ξ 7→

z − ξ

zξ − 1

Then z + dz is mapped to

P := f (z + dz) =

−dz

z( z + dz) −1

1 −

z + dz

z + dw

where we deleted z dz since it is inﬁnitesimal compared to 1 −

Since isometries preserve length and angle, this construction has several consequences.

Inﬁnitesimal distance, arc-length, and geodesics

The hyperbolic distance from z to z + dz is

d( z, z + dz) = d(O, P) = ln

1 +

1 −

= ln(1 +

) −ln(1 −

) = 2

1 −

z( t)

(∗)

where the approximation ln(1 ±

) = ±

is used since

is inﬁnitesimal. If z(t) parametrizes a

curve in the disk, then the inﬁnitesimal distance formula allows us to compute its arc-length

′

( t)

1 −

z( t)

Example 4.40 (Circles and ‘hyperbolic π’). Suppose that a circle has hyperbolic radius δ. By moving

its center to the origin via an isometry, we may parametrize in the usual manner:

z( t) = r



cos θ

sin θ



, θ ∈ [0, 2π) where δ = ln

1 + r

1 −r

equivalently r =

−1

+ 1

Its circumference (hyperbolic arc-length) is then

2π

1 −r

dθ =

4πr

1 −r

= 2π sinh δ = 2π



δ +

+ ···



> 2πδ

where we used the Maclaurin series to compare.

A hyperbolic circle has a larger circumference : diameter ratio than for a Euclidean circle (π). More-

over, this ratio is not constant: one might say that the hyperbolic version of π is a function (

π sinh δ

Our arc-length integral approach also allows us to show that hyperbolic lines are really what we

want them to be: lines of shortest distance between points.

Theorem 4.41. The geodesics—paths of minimal length between two points—in the Poincar

e disk

are precisely the hyperbolic lines.

Following the comments on page 69, the distance function really does deﬁne the concept of hyper-

bolic line.

Proof. First suppose b lies on the positive x-axis. Parametrize a curve from 0 to b via

z( t) = x(t) + iy(t) where 0 ≤ t ≤ 1, z(0) = 0, z(1) = b

Its arc-length satisﬁes

′

( t)

1 −

z( t)

dt =

′2

+ y

′2

1 − x

−y

dt ≥

′

1 − x

dt ≥

′

( t)

1 − x(t)

dt =

2 dx

1 − x

= ln

1 + b

1 −b

= d(0, b)

where we have equality if and only if y(t) ≡ 0 and x(t) is increasing. The length-minimizing path is

therefore along the x-axis.

More generally, given points A, B, apply an isometry f such that f (A) = 0 and f (B) = b lies on the

positive x-axis. The geodesic from A to B is therefore the image of the segment 0b under the inverse

isometry f

−1

. By the properties of M

obius transforms, this is an arc of a Euclidean circle through A, B

intersecting the unit circle at right-angles, our original deﬁnition of a hyperbolic line.

Area Computation

If dx and idy are inﬁnitesimal horizontal and vertical changes in z = x + iy, then the area of the

inﬁnitesimal rectangle spanned by z → z + dx and z → z + idy is the area element

dA =

2 dx

1 −

2 dy

1 −

4 dx dy

(1 − x

−y

)

The area of a region R in the Poincar

e disk is therefore given by the double integral

4 dxdy

(1 − x

−y

)

4r dr dθ

(1 −r

)

sinh δ dδ dθ

where the last expression is written in polar co-ordinates using the hyperbolic distance δ. In this way

the measure of area also depends on the distance function.

Example (4.40, cont). The area of a hyperbolic circle with hyperbolic radius δ is

2π

sinh δ dδ dθ = 2π(cosh δ −1) = π



+ ···



> πδ

Again, this is larger than you’d expect in Euclidean geometry.

Angle Measure and the First Fundamental Form

If we repeat the distance translation (∗, page 77) for a second inﬁnitesimal segment z → z + dw, it can

be checked that the angle between the original segments is precisely that between the inﬁnitesimal

vectors dz and dw. This is precisely the conformality observation in Theorem 4.32 and moreover

shows how the distance function determines the angle measure.

If you’ve studied differential geometry, then a more formal way to think about this is to use the

ﬁrst fundamental form or metric: essentially the dot product of inﬁnitesimal tangent vectors. For the

Poincar

e disk model, (∗) says that this is

I =

4( dx

+ dy

)

(1 − x

−y

)

4( dr

+ r

dθ

)

(1 −r

)

Since this is a scalar multiple of the standard Euclidean metric dx

+ dy

, angle measures are identi-

cal.

It also gels with the fact that arc-length is the integral



′

( t), z

′

( t)



Using this language, two of the major theorems of introductory differential geometry quickly put a

couple of remaining issues to bed.

Gauss’ Theorem Egregium The ﬁrst fundamental form determines the Gaussian curvature K. In this

case K = −1 is constant and negative, as you should easily be able to verify if you’ve studied

differential geometry.

Gauss–Bonnet Theorem The angle-sum Σ

△

of a geodesic triangle in a space with Gaussian curva-

ture K satisﬁes

△

−π =

△

This establishes our earlier assertion that Area △ = π − Σ

△

(Corollary 4.30).

Recall that the angle ψ between vectors u, v satisﬁes u ·v =

cos ψ. For inﬁnitesimal vectors we use I = λ

(dx

) instead of the dot product, where λ =

1−r

. The resulting angle is the same as if we use the Euclidean metric

+ dy

, since factors of λ

cancel on both sides.

5 Fractal Geometry

5.1 Natural Geometry, Self-similarity and Fractal Dimension

Classical geometry typically considers objects (lines, curves, spheres, etc.) which seem ﬂatter and

less interesting as one zooms in: a differentiable curve at small scales looks like a line segment!

By contrast, real-world objects tend to exhibit greater detail at smaller scales. A seemingly spherical

orange is dimpled on closer inspection. Is its surface area that of a sphere, or is it greater due to the

dimples? What if we zoom in further? Under a microscope, the dimples are seen to have minute

cracks and ﬁssures. With modern technology, we can see almost to the molecular level; what does

surface area even mean at such a scale?

The Length of a Coastline In 1967 Benoit Mandelbrot asked a related question in a now-famous

paper, How Long Is the Coast of Britain? Statistical Self-Similarity and Fractional Dimension. His essential

point was that the question has no simple answer:

Should one measure by walking along the mean

high tide line? But where is this? Do we ‘walk’ round every pebble? Round every grain of sand?

Every molecule? As one shrinks the scale, the measured length becomes absurdly large. We sketch

Mandelbrot’s approach.

• Given a ruler of length R, measure how many N are required to trace round the coastline when

laid end-to-end.

• Plotting log N against log(1/R) for several sizes of ruler seems to give a straight line!

log N ≈ log k + D log(1/R) = log(kR

−D

) =⇒ N ≈ kR

−D

The number D is Mandelbrot’s fractal dimension of the coastline.

Mandelbrot’s fractal dimension is purely empirical, though it does seem to capture something about

the ‘bumpiness’ of a coastline: the bumpier, the greater its fractal dimension. For mainland Britain

with its smooth east and rugged west coasts, D ≈ 1.25. Given its many fjords, Norway has a far

rougher coastline and a higher fractal dimension D ≈ 1.52.

Example 5.1. As a sanity check, consider a smooth circular ‘coastline.’

Approximate the circumference using N rulers of length R: clearly

R = 2 sin

As N → ∞, the small angle approximation for sine applies,

R ≈

2π

=⇒ N ≈ 2πR

−1

where the approximation improves as N → ∞. The fractal dimension

of a circle is therefore 1.

2π

The ofﬁcial answer from the Ordnance Survey (the UK government mapping ofﬁce) is, ‘It depends.’ The all-knowing

CIA states 7723 miles, though offers no evidence as to why.

For more detail see the Fractal Foundation’s website. Mandelbrot coined the word fractal, though he didn’t invent the

concept from nothing. Rather he applied earlier ideas of Hausdorff, Minkowski and others, and observed how the natural

world contains many examples of fractal structures.

Our goal is to describe self-similar objects and thus create a new notion of dimension related to

Mandelbrot’s. To begin consideration of self-similarity, we ﬁrst consider some of the standard objects

of pre-fractal geometry.

Segment A segment can be viewed as N copies of itself scaled by a factor

r =

Square A square comprises N copies of itself scaled by a factor r =

√

Cube A cube comprises N copies of itself scaled by a factor r =

√

In each case observe that N =





where D is the usual dimension of the

object (1, 2 or 3). Inspired by this, we make a loose deﬁnition.

Deﬁnition 5.2. A geometric ﬁgure is self-similar if it may be subdivided into N similar copies of

itself, each scaled by a magniﬁcation factor r < 1. The fractal dimension of such a ﬁgure is

D := log

1/r

N =

log N

log( 1/r)

= −

log N

log r

Example 5.3. The botanical pictures below offer some evidence for non-integer fractal dimension

and that self-similarity is a natural phenomenon. The ‘tree’ comprises N = 3 copies of itself, each

scaled by a factor of r = 0.4. Its fractal dimension is D = −

log 3

log 0.4

≈ 1.199.

The fern has N = 7 and r = 0.3 for a fractal dimension D = −

log 7

log 0.3

≈ 1.616.

Tree fractal D ≈ 1.199 Fern fractal D ≈ 1.616

The pictures illustrate the interpretation of fractal dimension. Both objects seem to occupy more

space than mere lines, but neither has positive area. Moreover, the fern seems to occupy more space

than the tree. The ‘trunk’ and ‘branches’ in the ﬁrst picture aren’t part of the fractal, and are drawn

only to give the picture a skeleton.

Example 5.4 (Cantor’s Middle-third Set). This famous example dates from the late 1800s.

Starting with the unit interval C

= [0, 1], deﬁne a se-

quence of sets (C

) where C

n+1

is obtained by deleting the

open ‘middle-third’ of each interval in C

; for instance





∪



, 1



Cantor’s set is essentially the limit of this sequence:

C :=

∞

n=0

Cantor’s set has several strange properties. . .

Zero length If the length of a set is the sum of the lengths of its disjoint sub-intervals, then

length(C

) =





since we delete

of the remaining set at each step. It follows that

∀n ∈ N

, length(C) ≤





=⇒ length(C) = 0

Otherwise said, C contains no subintervals.

Uncountable There exists a bijection between C and the original interval [0, 1]!

Self-similarity Abusing notation somewhat,

n+1

∪





where we mean that C

n+1

consists of two copies of C

, each shrunk by a factor of

and one

shifted

to the right. The upshot is that the Cantor set itself satisﬁes

C =

C ∪



C +



Being similar to two disjoint subsets of itself, its fractal dimension is D =

log 2

log 3

≈ 0.631. The

above image links to an animation showing how the full set may be doubled to produce itself.

The Cantor set has many generalizations. Look up the Sierpi

nski triangle (D =

log 3

log 2

≈ 1.585) and

carpet (Examples 5.8, D =

log 8

log 3

≈ 1.893), and the Menger sponge (D =

log 20

log 3

≈ 2.727).

Henry Smith discovered this set in 1874 while investigating integrability, in which context the ‘length’ of a set was later

formalized using measure theory. Cantor’s description in 1883 was more focused on topological properties. Self-similarity

was less of a concern at the time.

Example 5.5 (The Koch Curve and Snowﬂake). The Koch curve is another generalization of the

Cantor set, produced as the limit of a sequence of curves.

• Let K

be a segment of length 1.

• Replace the middle third of K

with the other two sides of

an equilateral triangle to create K

• Replacing the middle third of each segment in K

as before

to create K

• Repeat this process ad inﬁnitum.

The curve is drawn, along with the Koch snowﬂake obtained by

arranging three copies around an equilateral triangle.

The relation to the Cantor set should be obvious in the construc-

tion. Indeed if K

= [0, 1], then the intersection of this with the

Koch curve is the Cantor set!

The Koch curve is self-similar in that it comprises N = 4 copies

of itself shrunk by a factor of r =

. Its fractal dimension is

therefore

log 4

log 3

≈ 1.2619.

We may also consider the curve’s length. Let s

be the number

of segments in K

, each having length t

. Also let ℓ

= t

the length of the curve K

. We easily see that

= 4

, t

=⇒ ℓ





→ ∞

from which the Koch curve is inﬁnitely long!

Koch Curve

Koch Snowﬂake

Self-similarity

Exercises 5.1. 1. By removing a constant middle fraction of each interval, construct a fractal analo-

gous to the Cantor set but with dimension

2. Prove that the area inside the n

iteration of the construction of the Koch snowﬂake is



1 +



1 −







√

The area inside the complete snowﬂake is therefore

that of the original triangle.

3. Suppose r(t), t ∈ [0, 1] is a regular (smooth) curve in the plane.

(a) Use the arc-length formula L =

′

( t)

dt together with Riemann sums and the linear

approximation r(t + ϵ) ≈ r(t) + ϵr

′

( t) with ϵ =

to argue that

L ≈

N−1

∑

k=0





k + 1



−r







(∗)

(b) Suppose the curve is parametrized such that each segment on the right side of (∗) has the

same length R. Prove that L ≈ NR.

Any regular curve thus has fractal dimension 1 in the sense stated by Mandelbrot (pg. 80).

5.2 Contraction Mappings & Iterated Function Systems

Thus far we have only dealt with fractals where the whole consists of pieces scaled by the same

factor. In general we can mix up scaling factors. To do this it is helpful to borrow some language

from topology.

Deﬁnition 5.6. A contraction mapping is a function S on a subset of R

such that ∃c ∈ [0, 1) with

S(x) − S(y)

≤ c

x −y

A contraction mapping therefore moves points closer together. It should be clear that every con-

traction mapping is continuous. The main idea of this section is that fractals may be generated by

repeatedly applying contraction mappings to an initial shape. We have already seen a example:

Example (5.4, mk. II). Consider the following functions S

, S

: R → R

(x) =

These are certainly contraction mappings

∀x, y ∈ R,

(x) − S

( y)

(x) − S

( y)

x −y

with scale factor c =

. More importantly, these functions deﬁne the Cantor set: at each stage of its

construction, we have

n+1

:= S

) ∪ S

)

As the limit of this process, the self-similarity of the Cantor set can be expressed in the same manner:

C = S

( C) ∪ S

( C).

Surprisingly, it barely seems to matter what initial set C

we choose. For example, we could start

with the singleton set C

= {0}, from which

= {0,

}, C

= {0,

}, C

= {0,

}, . . .

We draw the ﬁrst few iterations below. In the second picture, we start with a very different initial set

= [0.2, 0.5] ∪ [0.6, 0.7]. Iterating this also appears to produce the Cantor set!

Iterated Function Systems

It certainly seems as if the Cantor set might be generated by the contraction maps S

, S

indepen-

dently of the initial data C

. The following result shows in what sense this is the case, though it relies

on some heavy lifting from topology. If you’ve done some analysis, then several of the concepts will

be familiar. We summarize the discussion without proof.

• A subset of R

is compact if it is closed (contains its boundary points) and bounded (all points lie

within some ball centered at the origin).

• The set of all compact subsets of R

is a metric space H. This means that the distance d(X, Y)

between two compact sets X, Y ∈ H may sensibly be deﬁned, though it is a little tricky. . .

• Since H is a metric space, we can discuss convergent sequences (K

) of compact sets

lim

n→∞

= K ⇐⇒ lim

n→∞

d(K

, K) = 0

It also makes sense to speak of Cauchy sequences in H. Moreover, H is complete in that every

Cauchy sequence (K

) ⊆ H converges to some K ∈ H.

• The Banach Fixed Point Theorem now applies.

If S : H → H is a contraction mapping on a complete metric space H, then S has a

unique ﬁxed point (some F ∈ H such that S(F) = F). Moreover, if F

∈ H is any

initial value, then the sequence deﬁned iteratively by F

k+1

:= S(F

) converges to F.

This powerful result has applications throughout mathematics.

Theorem 5.7. Let S

, . . . , S

be contraction mappings on R

with ratios c

, . . . , c

. Deﬁne

S : H → H by S(D) =

[

i=1

(D)

1. S is a contraction mapping on H, with contraction ratio c = max{c

2. S has a unique ﬁxed set F ∈ H given by F = lim

k→∞

(E) for any non-empty E ∈ H.

Part 1 is not difﬁcult to prove if you’re willing to work with the deﬁnition of the Hausdorff metric

(try it if you’re comfortable with analysis!). Part 2 is Banach’s theorem.

The upshot is this: if we take any non-empty compact set E and repeatedly apply contraction map-

pings, the process will converge to a limit which is independent of E! We call the limit set F for fractal.

Such fractals are often called attractors: being limit-sets, they ‘attract’ data towards themselves.

This is the Hausdorff metric. Given Y ∈ H, and x ∈ R

, deﬁne d

(x) = inf

y∈Y

x −y

to be the distance from x to the

‘nearest’ point of Y. Deﬁne d

(y) similarly. The Hausdorff distance between X and Y is then

d(X, Y) := max

(

sup

x∈X

(x), sup

y∈Y

(y)

)

Roughly speaking, ﬁnd x ∈ X which is as far away d

(x) as possible from anything in Y, and ﬁnd y ∈ Y similarly; d(X, Y)

is the larger of these distances.

Examples 5.8. 1. (Cantor set Ex. 5.4) Theorem 5.7 shows that we may let C

be any closed bounded

subset of R. Repeatedly applying the contraction mappings S

and S

will always result in the

same set C.

A nice application is that one can easily ﬁnd all sorts of interesting points in the Cantor set. For

instance, suppose x, y ∈ R are a pair such that y = S

(x) and x = S

( y): otherwise said

y =

x and x =

( y + 2)

Since E = {x, y} is a compact set satisfying E ⊆ S(E), it follows that E ⊆ lim S

(E) = C, from

which x, y both lie in the Cantor set! However, we can easily solve to see that (x, y) = (

This seems paradoxical:

does not lie at the end of any deleted interval (denominators of the

form 3

) but yet the Cantor set contains no intervals. How does

end up in there?!

2. (Koch curve, Ex. 5.5) Deﬁne four mappings S

: R

→ R

each with scale factor c =

Mapping Effect

(x, y) =





Scale

(x, y) =



x −

√

y +

√

x +



Scale

, rotate 60°, translate

(x, y) =



x +

√

y +

√

x −

y +

√



Scale

, rotate −60°, translate

(x, y) =





Scale

, translate

Applied to the Koch curve, the image of each map corresponds by color. The picture links to a

series of animated constructions of the curve starting with different initial sets E.

3. (Sierpi

nski carpet) Eight contraction mappings produce this fractal,

each reducing the whole by a (length-scale) factor of

As with the Koch curve, the image links to several alternative con-

structions using different initial starting sets.

4. (A Fractal Fern) This is built from three contraction mappings:

: Scale by

, rotate 5° clockwise, and translate by (0,

)

: Scale by

, rotate 60° counter-clockwise, and translate by (0,

)

: Scale by

, rotate 60° clockwise, and translate by (0,

)

Fractal Dimension Revisited

Since Theorem 5.7 permits several different contraction factors, we need a new

approach to computing fractal dimension. We ask how many disks of a given

radius ϵ are required to cover a set. In the picture, the unit square requires four

disks of radius ε = 0.4. For smaller ε, we will plainly need more disks. . .

Deﬁnition 5.9. Let A be a compact subset of R

1. If ε > 0, the closed ε-ball centered at x ∈ A consists of the points at most a distance ε from x:

(x) = {y ∈ R

: d(x, y) ≤ ε}

2. The minimal ε-covering number for A is

N(A, ε) = min

(

M : ∃x

, . . . , x

∈ A with A ⊆

[

n=1

)

3. Given a compact set A ⊆ R

, its fractal dimension is the limit

D = lim

ε→0

log N(A, ε)

log( 1/ε)

We don’t claim to prove that D must exist, though a simple example should at least convince you

that the deﬁnition is reasonable!

Example 5.10. Let A = [0, 1] be the interval of length 1. It is not hard to see that

ε ≥

⇐⇒ N(ε) = 1, and

≤ ε <

⇐⇒ N(ε) = 2

etc. More generally, N and ϵ are related via

≤ ϵ <

2(N −1)

The dimension of the line (1) may therefore be recovered via the squeeze theorem

D = lim

ϵ→0

log N

log( 1/ε)

= 1

Thankfully an easier-to-use modiﬁcation is available using boxes.

Theorem 5.11 (Box-counting). Let A be compact and cover R

by boxes of side length

. Let N

(A)

be the number of boxes intersecting A. Then

D = lim

n→∞

log N

(A)

log 2

We ﬁnish with a formula satisﬁed by the dimension of an iterated function system (Theorem 5.7).

Theorem 5.12. Let {S

}

n=1

be an iterated function system with attractor (limiting fractal) F and

where each contraction S

has scale factor c

∈ (0, 1). At each stage of the construction, suppose

portions of the fractal generated by each contraction map meet only at boundary points. Then the

fractal dimension is the unique D satisfying

∑

n=1

= 1

Examples 5.13. 1. If all scale-factors are identical c

= r, we recover Deﬁnition 5.2,

= 1 =⇒ D =

−log M

log r

log M

log( 1/r)

2. The fractal fern (Examples 5.8) is generated by three contraction maps with scale factors

Its dimension is the solution to the equation













= 1 =⇒ D ≈ 1.3267

3. Numerical approximation is usually required to solve for D, though sometimes an exact solu-

tion is possible. For instance, if c

= c

and c

= c

, then





+ 3





= 1

Writing α =





yields the quadratic equation

2α + 3α

= 1 =⇒ α =

=⇒ D = log

3 ≈ 1.584

Other methods of creating fractals

The contraction mapping approach is one of many ways

to create fractals. Two other famous examples are the logis-

tic map (related to numerical approximations to non-linear

differential equations) and the Mandelbrot set (pictured).

The Mandelbrot set arises from a construction in the com-

plex plane. For a given c ∈ C, we iterate the function

( z) = z

+ c

If f ( f ( f (··· f (c) ···))) remains bounded, no matter how

many times f is applied, then c lies in the Mandelbrot set.

Much better pictures and some trippy videos can be found

online. . .

−1

−i

Exercises 5.2. 1. Let S

(x) =

x and S

(x) =

x +

be the contraction mappings deﬁning the

Cantor set and suppose x, y, z ∈ R satisfy

y = S

(x), z = S

( y), x = S

( z)

Show that x, y, z lie in the Cantor set, and ﬁnd their values.

2. The construction of a Cantor-type set starts by removing the open intervals (0.1, 0.2) and

(0.6, 0.8) from the unit interval.

(a) Sketch the ﬁrst three iterations of this fractal.

(b) This construction may be described using three contraction mappings; what are they?

package to estimate its value.

3. A variation on the Koch curve is constructed using the following contraction mappings. Each

is built by ﬁrst scaling the whole picture by a factor c, rotating the picture through an angle

counter-clockwise, and then translating the picture by adding a constant. The resulting fractal

is drawn.

map scale rotate translate (add (x, y) )

0 0

90° (

, 0)

0 (

)

−90° (

)

0 (

, 0)

(a) Suppose you start with the straight line segment from (0, 0) to (1, 0). Draw the ﬁrst two

iterations of the fractal’s construction.

(b) The dimension of the fractal is the unique solution D to the equation





















= 1

By observing that





, convert this to a quadratic equation in the variable α :=





Hence compute the dimension of the fractal.

log 4

log 3

of the Koch curve.

Explain what this means.

4. Verify the details of Example 5.10, including the computation of the limit.

5. In Theorem 5.12, prove that D exists and is unique.

(Hint: You’ll need the intermediate value theorem from calculus)