Euclidean plane and its relatives; a minimalist introduction
aa r X i v : . [ m a t h . HO ] J a n Euclidean plane and its relatives
A minimalist introductionAnton Petrunin his work is licensed under the Creative CommonsAttribution-ShareAlike 4.0 International License. To viewa copy of this license, visit http://creativecommons.org/licenses/by-sa/4.0/ ontents
Introduction 6
Prerequisite. Overview.
What is the axiomatic approach? What is a model? Metricspaces. Examples. Shortcut for distance. Isometries, motionsand lines. Half-lines and segments. Angles. Reals modulo2 · π . Continuity. Congruent triangles. Euclidean geometry
The axioms. Lines and half-lines. Zero angle. Straight angle.Vertical angles.
Sign of an angle. Intermediate value theorem. Same signlemmas. Half-planes. Triangle with the given sides.
Side-angle-side condition. Angle-side-angle condition. Isosce-les triangles. Side-side-side condition. On angle-side-side andside-angle-angle.
Right, acute and obtuse angles. Perpendicular bisector.Uniqueness of perpendicular lines. Reflection. Perpendicu-lar is shortest. Circles. Geometric constructions.3
CONTENTS
Similar triangles. Pythagorean theorem. Method of similartriangles.6.1 Ptolemy’s inequality . . . . . . . . . . . . . . . 46Ptolemy’s inequality.
Parallel lines. Transversal property. Angles of triangle. Par-allelograms. Method of coordinates.
Circumcircle and circumcenter. Altitudes and orthocenter.Medians and centroid. Angle bisectors. Equidistant property.Incenter.
Inversive geometry
Angle between a tangent line and a chord. Inscribed angle.Inscribed quadrilaterals. Method of additional circle. Arcs.
10 Inversion 70
Cross-ratio. Inversive plane and circlines. Method of inver-sion. Perpendicular circles. Angles after inversion.
Non-Euclidean geometry
11 Neutral plane 80
Two angles of triangle. Three angles of triangle. How to provethat something cannot be proved. Curvature.
12 Hyperbolic plane 89
Conformal disc model. Plan of the proof. Auxiliary state-ments. Axioms: I, II, III, IV, h-V. Hyperbolic trigonometry
13 Geometry of the h-plane 101
Angle of parallelism. Inradius of triangle. Circles, horocyclesand equidistants. Hyperbolic triangles. Conformal interpre-tation. Pythagorean theorem.
Additional topics
ONTENTS
14 Affine geometry 112
Affine transformations. Constructions. Matrix form. On in-versive transformations.
15 Projective geometry 119
Real projective plane. Euclidean space. Perspective projec-tion. Projective transformations. Moving points to infinity.Duality. Axioms.
16 Spherical geometry 128
Euclidean space. Pythagorean theorem. Inversion of thespace. Stereographic projection. Central projection.
17 Projective model 134
Special bijection on the h-plane. Projective model. Bolyai’sconstruction.
18 Complex coordinates 141
Complex numbers. Complex coordinates. Conjugation andabsolute value. Euler’s formula. Argument and polar co-ordinates. Method of complex coordinates. Fractional lin-ear transformations. Elementary transformations. Complexcross-ratio. Schwarz–Pick theorem.
19 Geometric constructions 152
Classical problems. Constructible numbers. Constructionswith a set square. More impossible constructions.
20 Area 160
Solid triangles. Polygonal sets. Definition of area. Vanishingarea and subdivisions. Area of solid rectangles, parallelogramsand triangles. Area method. Area in the neutral planes andspheres. Quadrable sets.
References
Hints 174Index 200Used resources 203 ntroduction
This book is meant to be rigorous, conservative, elementary and minimal-ist. At the same time it includes about the maximum what students canabsorb in one semester.Approximately one-third of the material used to be covered in highschool, but not any more.The present book is based on the courses given by the author at thePennsylvania State University as an introduction to the foundations ofgeometry. The lectures were oriented to sophomore and senior universitystudents. These students already had a calculus course. In particular,they are familiar with the real numbers and continuity. It makes it possi-ble to cover the material faster and in a more rigorous way than it couldbe done in high school.
Prerequisite
The students should be familiar with the following topics: ⋄ Elementary set theory: ∈ , ∪ , ∩ , \ , ⊂ , × . ⋄ Real numbers: intervals, inequalities, algebraic identities. ⋄ Limits, continuous functions and the intermediate value theorem. ⋄ Standard functions: absolute value, natural logarithm, exponentialfunction. Occasionally, trigonometric functions are used, but theseparts can be ignored. ⋄ Chapter 14 uses matrix algebra of 2 × ⋄ To read Chapter 16, it is better to have some previous experiencewith the scalar product , also known as dot product . ⋄ To read Chapter 18, it is better to have some previous experiencewith complex numbers. 6
ONTENTS Overview
We use the so called metric approach introduced by Birkhoff. It meansthat we define the Euclidean plane as a metric space which satisfies a listof properties ( axioms ). This way we minimize the tedious parts which areunavoidable in the more classical Hilbert’s approach. At the same timethe students have a chance to learn basic geometry of metric spaces.Here is a dependency graph of the chapters.
In (1) we give all the definitions necessary to formulate the axioms; itincludes metric space, lines, angle measure, continuous maps and congru-ent triangles.Further we do Euclidean geometry: (2) Axioms and immediate corol-laries; (3) Half-planes and continuity; (4) Congruent triangles; (5) Circles,motions, perpendicular lines; (6) Similar triangles and (7) Parallel lines— these are the first two chapters where we use Axiom V, an equivalentof Euclid’s parallel postulate. In (8) we give the most classical theoremof triangle geometry; this chapter is included mainly as an illustration.In the following two chapters we discuss geometry of circles on theEuclidean plane: (9) Inscribed angles; (10) Inversion. It will be used toconstruct the model of the hyperbolic plane.Further we discuss non-Euclidean geometry: (11) Neutral geometry— geometry without the parallel postulate; (12) Conformal disc model— this is a construction of the hyperbolic plane, an example of a neu-tral plane which is not Euclidean. In (13) we discuss geometry of theconstructed hyperbolic plane — this is the highest point in the book.In the reamining chapters we discuss some additional topics: (14)Affine geometry; (15) Projective geometry; (16) Spherical geometry; (17)Projective model of the hyperbolic plane; (18) Complex coordinates; (19)Geometric constructions; (20) Area. The proofs in these chapters are notcompletely rigorous.We encourage to use the visual assignments available at the author’s
CONTENTS website.
Disclaimer
It is impossible to find the original reference to most of the theoremsdiscussed here, so I do not even try to. Most of the proofs discussed inthe book already appeared in the Euclid’s Elements.
Recommended books ⋄ Kiselev’s textbook [10] — a classical book for school students; itshould help if you have trouble following this book. ⋄ Moise’s book, [15] — should be good for further study. ⋄ Greenberg’s book [9] — a historical tour in the axiomatic systemsof various geometries. ⋄ Prasolov’s book [16] is perfect to master your problem-solving skills. ⋄ Akopyan’s book [1] — a collection of problems formulated in figures. ⋄ Methodologically my lectures were very close to Sharygin’s textbook[18]. This is the greatest textbook in geometry for school students,I recommend it to anyone who can read Russian.
Acknowlegments
Let me thank Matthew Chao, Svetlana Katok, Alexander Lytchak, AlexeiNovikov and Lukeria Petrunina for useful suggestions and correcting themisprints. hapter 1
Preliminaries
What is the axiomatic approach?
In the axiomatic approach, one defines the plane as anything which sat-isfies a given list of properties. These properties are called axioms . Theaxiomatic system for the theory is like the rules for a game. Once theaxiom system is fixed, a statement is considered to be true if it followsfrom the axioms and nothing else is considered to be true.The formulations of the first axioms were not rigorous at all. Forexample, Euclid described a line as breadthless length and a straight line as a line which lies evenly with the points on itself . On the other hand,these formulations were sufficiently clear, so that one mathematician couldunderstand the other.The best way to understand an axiomatic system is to make one byyourself. Look around and choose a physical model of the Euclideanplane; imagine an infinite and perfect surface of a chalk board. Now tryto collect the key observations about this model. Assume for now thatwe have intuitive understanding of such notions as line and point .(i) We can measure distances between points.(ii) We can draw a unique line which passes thru two given points.(iii) We can measure angles.(iv) If we rotate or shift we will not see the difference.(v) If we change scale we will not see the difference.These observations are good to start with. Further we will develop thelanguage to reformulate them rigorously.90 CHAPTER 1. PRELIMINARIES
What is a model?
The Euclidean plane can be defined rigorously the following way:
Define a point in the Euclidean plane as a pair of real numbers ( x, y ) and define the distance between the two points ( x , y ) and ( x , y ) by thefollowing formula: p ( x − x ) + ( y − y ) . That is it! We gave a numerical model of Euclidean plane; it buildsthe Euclidean plane from the real numbers while the latter is assumed tobe known.Shortness is the main advantage of the model approach, but it is notintuitively clear why we define points and the distances this way.On the other hand, the observations made in the previous sectionare intuitively obvious — this is the main advantage of the axiomaticapproach.Another advantage lies in the fact that the axiomatic approach iseasily adjustable. For example, we may remove one axiom from the list,or exchange it to another axiom. We will do such modifications in Chapter11 and further.
Metric spaces
The notion of metric space provides a rigorous way to say: “we can mea-sure distances between points” . That is, instead of (i) on page 9, we cansay “Euclidean plane is a metric space” . Let X be a nonempty set and d be a function whichreturns a real number d ( A, B ) for any pair A, B ∈ X . Then d is called metric on X if for any A, B, C ∈ X , the following conditions are satisfied:(a) Positiveness: d ( A, B ) > . (b) A = B if and only if d ( A, B ) = 0 . (c) Symmetry: d ( A, B ) = d ( B, A ) . (d) Triangle inequality: d ( A, C ) d ( A, B ) + d ( B, C ) . A metric space is a set with a metric on it. More formally, a metric spaceis a pair ( X , d ) where X is a set and d is a metric on X .The elements of X are called points of the metric space. Given twopoints A, B ∈ X , the value d ( A, B ) is called distance from A to B . Examples ⋄ Discrete metric.
Let X be an arbitrary set. For any A, B ∈ X , set d ( A, B ) = 0 if A = B and d ( A, B ) = 1 otherwise. The metric d iscalled discrete metric on X . ⋄ Real line.
Set of all real numbers ( R ) with metric defined as d ( A, B ) := | A − B | . ⋄ Metrics on the plane.
Let R denotes the set of all pairs ( x, y ) ofreal numbers. Assume A = ( x A , y A ) and B = ( x B , y B ). Considerthe following metrics on R : ◦ Euclidean metric, denoted by d and defined as d ( A, B ) = p ( x A − x B ) + ( y A − y B ) . ◦ Manhattan metric, denoted by d and defined as d ( A, B ) = | x A − x B | + | y A − y B | . ◦ Maximum metric, denoted by d ∞ and defined as d ∞ ( A, B ) = max {| x A − x B | , | y A − y B |} . Prove that the following functions are metrics on R :(a) d ; (b) d ; (c) d ∞ . Shortcut for distance
Most of the time, we study only one metric on the space. Therefore, wewill not need to name the metric function each time.Given a metric space X , the distance between points A and B will befurther denoted by AB or d X ( A, B );the latter is used only if we need to emphasize that A and B are pointsof the metric space X .For example, the triangle inequality can be written as AC AB + BC.
For the multiplication, we will always use “ · ”, so AB should not beconfused with A · B .2 CHAPTER 1. PRELIMINARIES
Isometries, motions and lines
In this section, we define lines in a metric space. Once it is done thesentence “We can draw a unique line which passes thru two given points.” becomes rigorous; see (ii) on page 9.Recall that a map f : X → Y is a bijection , if it gives an exact pairingof the elements of two sets. Equivalently, f : X → Y is a bijection, if ithas an inverse ; that is, a map g : Y → X such that g ( f ( A )) = A for any A ∈ X and f ( g ( B )) = B for any B ∈ Y . Let X and Y be two metric spaces and d X , d Y be theirmetrics. The map f : X → Y is called distance-preserving if d Y ( f ( A ) , f ( B )) = d X ( A, B ) for any A, B ∈ X .A bijective distance-preserving map is called an isometry .Two metric spaces are called isometric if there exists an isometry fromone to the other.The isometry from a metric space to itself is also called a motion ofthe space.
Show that any distance-preserving map is injective ; thatis, if f : X → Y is a distance-preserving map, then f ( A ) = f ( B ) for anypair of distinct points A, B ∈ X . Show that if f : R → R is a motion of the real line, theneither (a) f ( x ) = f (0) + x for any x ∈ R , or (b) f ( x ) = f (0) − x for any x ∈ R . Prove that ( R , d ) is isometric to ( R , d ∞ ) . Describe all the motions of the Manhattanplane, defined on page 11. If X is a metric space and Y is a subset of X , then a metric on Y can be obtained by restricting the metric from X . In other words, thedistance between two points of Y is defined to be the distance betweenthese points in X . This way any subset of a metric space can be alsoconsidered as a metric space. A subset ℓ of metric space is called a line , if it isisometric to the real line.
3A triple of points that lie on one line is called collinear . Note that if A , B and C are collinear, AC > AB and AC > BC , then AC = AB + BC .Some metric spaces have no lines, for example discrete metrics. Thepicture shows examples of lines on the Manhattan plane ( R , d ). Consider the graph y = | x | in R . In which of the fol-lowing spaces (a) ( R , d ) , (b) ( R , d ) , (c) ( R , d ∞ ) does it form a line?Why? How many points M are there on the line ( AB ) forwhich we have(a) AM = M B ?(b) AM = 2 · M B ? Half-lines and segments
Assume there is a line ℓ passing thru two distinct points P and Q . In thiscase we might denote ℓ as ( P Q ). There might be more than one line thru P and Q , but if we write ( P Q ) we assume that we made a choice of suchline.Let [
P Q ) denotes the half-line which starts at P and contains Q .Formally speaking, [ P Q ) is a subset of (
P Q ) which corresponds to [0 , ∞ )under an isometry f : ( P Q ) → R such that f ( P ) = 0 and f ( Q ) > Show that if X ∈ [ P Q ) , then QX = | P X − P Q | . The subset of line (
P Q ) between P and Q is called the segment be-tween P and Q and denoted by [ P Q ]. Formally, the segment can bedefined as the intersection of two half-lines: [
P Q ] = [
P Q ) ∩ [ QP ).4 CHAPTER 1. PRELIMINARIES
Angles
Our next goal is to introduce angles and angle measures ; after that, thestatement “we can measure angles” will become rigorous; see (iii) on page9.
O B Aα
An ordered pair of half-lines which startat the same point is called an angle . Theangle
AOB (also denoted by ∠ AOB ) is thepair of half-lines [ OA ) and [ OB ). In this casethe point O is called the vertex of the angle.Intuitively, the angle measure tells howmuch one has to rotate the first half-linecounterclockwise, so it gets the position ofthe second half-line of the angle. The full turn is assumed to be 2 · π ; itcorresponds to the angle measure in radians. The angle measure of ∠ AOB is denoted by ∡ AOB ; it is a real numberin the interval ( − π, π ].The notations ∠ AOB and ∡ AOB look similar; they also have closebut different meanings, which better not be confused. For example, theequality ∠ AOB = ∠ A ′ O ′ B ′ means that [ OA ) = [ O ′ A ′ ) and [ OB ) == [ O ′ B ′ ); in particular, O = O ′ . On the other hand the equality ∡ AOB == ∡ A ′ O ′ B ′ means only equality of two real numbers; in this case O maybe distinct from O ′ .Here is the first property of angle measure which will become a partof the axiom. Given a half-line [ OA ) and α ∈ ( − π, π ] there is a unique half-line [ OB ) such that ∡ AOB = α . Reals modulo 2 · π Consider three half-lines starting from the same point, [ OA ), [ OB ) and[ OC ). They make three angles AOB , BOC and
AOC , so the value ∡ AOC should coincide with the sum ∡ AOB + ∡ BOC up to full rotation. Thisproperty will be expressed by the formula ∡ AOB + ∡ BOC ≡ ∡ AOC, where “ ≡ ” is a new notation which we are about to introduce. The lastidentity will become a part of the axiom. For a while you may think that π is a positive real number that measures the sizeof a half turn in certain units. Its concrete value π ≈ .
14 will not be important for along time. α ≡ β or α ≡ β (mod 2 · π )if α = β + 2 · π · n for some integer n . In this case we say “ α is equal to β modulo · π ” . For example − π ≡ π ≡ · π and · π ≡ − · π. The introduced relation “ ≡ ” behaves as an equality sign, but · · · ≡ α − · π ≡ α ≡ α + 2 · π ≡ α + 4 · π ≡ . . . ;that is, if the angle measures differ by full turn, then they are consideredto be the same.With “ ≡ ”, we can do addition, subtraction and multiplication withinteger numbers without getting into trouble. That is, if α ≡ β and α ′ ≡ β ′ , then α + α ′ ≡ β + β ′ , α − α ′ ≡ β − β ′ and n · α ≡ n · β for any integer n . But “ ≡ ” does not in general respect multiplication withnon-integer numbers; for example π ≡ − π but · π
6≡ − · π. Show that · α ≡ if and only if α ≡ or α ≡ π . Continuity
The angle measure is also assumed to be continuous. Namely, the follow-ing property of angle measure will become a part of the axiom:
The function ∡ : ( A, O, B ) ∡ AOB is continuous at any triple of points ( A, O, B ) such that O = A and O = B and ∡ AOB = π . CHAPTER 1. PRELIMINARIES
To explain this property, we need to extend the notion of continuity tothe functions between metric spaces. The definition is a straightforwardgeneralization of the standard definition for the real-to-real functions.Further, let X and Y be two metric spaces, and d X , d Y be their metrics.A map f : X → Y is called continuous at point A ∈ X if for any ε > δ >
0, such that d X ( A, A ′ ) < δ ⇒ d Y ( f ( A ) , f ( A ′ )) < ε. (Informally it means that sufficiently small changes of A result in arbi-trarily small changes of f ( A ).)A map f : X → Y is called continuous if it is continuous at every point A ∈ X .One may define a continuous map of several variables the same way.Assume f ( A, B, C ) is a function which returns a point in the space Y fora triple of points ( A, B, C ) in the space X . The map f might be definedonly for some triples in X .Assume f ( A, B, C ) is defined. Then, we say that f is continuous atthe triple ( A, B, C ) if for any ε > δ > d Y ( f ( A, B, C ) , f ( A ′ , B ′ , C ′ )) < ε. if d X ( A, A ′ ) < δ , d X ( B, B ′ ) < δ and d X ( C, C ′ ) < δ . Let X be a metric space.(a) Let A ∈ X be a fixed point. Show that the function f ( B ) := d X ( A, B ) is continuous at any point B .(b) Show that d X ( A, B ) is continuous at any pair A, B ∈ X . Let X , Y and Z be a metric spaces. Assume that thefunctions f : X → Y and g : Y → Z are continuous at any point, and h = g ◦ f is their composition; that is, h ( A ) = g ( f ( A )) for any A ∈ X .Show that h : X → Z is continuous at any point.
Show that any distance-preserving map is continuousat any point.
Congruent triangles
Our next goal is to give a rigorous meaning for (iv) on page 9. To dothis, we introduce the notion of congruent triangles so instead of “if werotate or shift we will not see the difference” we say that for triangles,7the side-angle-side congruence holds; that is, two triangles are congruentif they have two pairs of equal sides and the same angle measure betweenthese sides.An ordered triple of distinct points in a metric space X , say A, B, C ,is called a triangle
ABC (briefly △ ABC ). Note that the triangles
ABC and
ACB are considered as different.Two triangles A ′ B ′ C ′ and ABC are called congruent (written as △ A ′ B ′ C ′ ∼ = △ ABC ) if there is a motion f : X → X such that A ′ = f ( A ) , B ′ = f ( B ) and C ′ = f ( C ) . Let X be a metric space, and f, g : X → X be two motions. Note thatthe inverse f − : X → X , as well as the composition f ◦ g : X → X arealso motions.It follows that “ ∼ =” is an equivalence relation ; that is, any trianglecongruent to itself, and the following two conditions hold: ⋄ If △ A ′ B ′ C ′ ∼ = △ ABC , then △ ABC ∼ = △ A ′ B ′ C ′ . ⋄ If △ A ′′ B ′′ C ′′ ∼ = △ A ′ B ′ C ′ and △ A ′ B ′ C ′ ∼ = △ ABC , then △ A ′′ B ′′ C ′′ ∼ = △ ABC.
Note that if △ A ′ B ′ C ′ ∼ = △ ABC , then AB = A ′ B ′ , BC = B ′ C ′ and CA = C ′ A ′ .For a discrete metric, as well as some other metrics, the conversealso holds. The following example shows that it does not hold in theManhattan plane: Example.
Consider three points A = (0 , B = (1 ,
0) and C = ( − , R , d ). Note that d ( A, B ) = d ( A, C ) = d ( B, C ) = 2 . A BC
On one hand, △ ABC ∼ = △ ACB.
Indeed, it follows since the map ( x, y ) ( − x, y ) is a motion of ( R , d ), whichsends A A , B C and C B .On the other hand, △ ABC ≇ △ BCA.
Indeed, arguing by contradiction, assume that △ ABC ∼ = △ BCA ; that is,there is a motion f of ( R , d ) which sends A B , B C and C A .8 CHAPTER 1. PRELIMINARIES
We say that M is a midpoint of A and B if d ( A, M ) = d ( B, M ) = · d ( A, B ) . Note that a point M is a midpoint of A and B if and only if f ( M ) is amidpoint of B and C .The set of midpoints for A and B is infinite, it contains all points ( t, t )for t ∈ [0 ,
1] (it is the dark gray segment on the picture above). On theother hand, the midpoint for B and C is unique (it is the black point onthe picture). Thus, the map f cannot be bijective — a contradiction. hapter 2 Axioms
A system of axioms appears already in Euclid’s “Elements” — the mostsuccessful and influential textbook ever written.The systematic study of geometries as axiomatic systems was triggeredby the discovery of non-Euclidean geometry. The branch of mathematics,emerging this way, is called “Foundations of geometry”.The most popular system of axiom was proposed in 1899 by DavidHilbert. This is also the first rigorous system by modern standards. Itcontains twenty axioms in five groups, six “primitive notions” and three“primitive terms”; these are not defined in terms of previously definedconcepts.Later a number of different systems were proposed. It is worth men-tioning the system of Alexandr Alexandrov [2] which is very intuitive andelementary, the system of Friedrich Bachmann [3] which is based on theconcept of symmetry, and the system of Alfred Tarski [19], which wasdesigned for analysis using mathematical logic.We will use another system, which is very close to the one proposedby George Birkhoff in [6]. This system is based on the key observations (i)–(v) listed on page 9. The axioms use the notions of metric space,lines, angles, triangles, equalities modulo 2 · π ( ≡ ), the continuity of mapsbetween metric spaces, and the congruence of triangles ( ∼ =). All thisdiscussed in the preliminaries.Our system is build upon metric spaces. In particular, we use the realnumbers as a building block. By that reason our approach is not purelyaxiomatic — we build the theory upon something else; it resembles amodel-based introduction to Euclidean geometry discussed on page 10.We used this approach to minimize the tedious parts which are unavoid-able in purely axiomatic foundations.190 CHAPTER 2. AXIOMS
The axioms
I. The
Euclidean plane is a metric space with at least two points.II. There is one and only one line, that contains any two givendistinct points P and Q in the Euclidean plane.III. Any angle AOB in the Euclidean plane defines a real numberin the interval ( − π, π ]. This number is called angle measureof ∠ AOB and denoted by ∡ AOB . It satisfies the followingconditions:(a) Given a half-line [ OA ) and α ∈ ( − π, π ], there is a uniquehalf-line [ OB ), such that ∡ AOB = α .(b) For any points A , B and C , distinct from O we have ∡ AOB + ∡ BOC ≡ ∡ AOC. (c) The function ∡ : ( A, O, B ) ∡ AOB is continuous at any triple of points (
A, O, B ), such that O = A and O = B and ∡ AOB = π .IV. In the Euclidean plane, we have △ ABC ∼ = △ A ′ B ′ C ′ if and onlyif A ′ B ′ = AB, A ′ C ′ = AC, and ∡ C ′ A ′ B ′ = ± ∡ CAB.
V. If for two triangles
ABC , AB ′ C ′ in the Euclidean plane and for k > B ′ ∈ [ AB ) , C ′ ∈ [ AC ) ,AB ′ = k · AB, AC ′ = k · AC, then B ′ C ′ = k · BC, ∡ ABC = ∡ AB ′ C ′ , ∡ ACB = ∡ AC ′ B ′ . From now on, we can use no information about the Euclidean plane whichdoes not follow from the five axioms above.
Show that there are (a) an infinite set of points, (b) aninfinite set of lines on the plane. Lines and half-lines X Any two distinct lines intersect at most at onepoint.Proof.
Assume that two lines ℓ and m intersect at two distinct points P and Q . Applying Axiom II, we get that ℓ = m . Suppose A ′ ∈ [ OA ) and A ′ = O . Show that [ OA ) = [ OA ′ ) . X Given r > and a half-line [ OA ) there is a unique A ′ ∈ [ OA ) such that OA ′ = r .Proof. According to definition of half-line, there is an isometry f : [ OA ) → [0 , ∞ ) , such that f ( O ) = 0. By the definition of isometry, OA ′ = f ( A ′ ) for any A ′ ∈ [ OA ). Thus, OA ′ = r if and only if f ( A ′ ) = r .Since isometry has to be bijective, the statement follows. Zero angle X ∡ AOA = 0 for any A = O .Proof. According to Axiom IIIb, ∡ AOA + ∡ AOA ≡ ∡ AOA.
Subtract ∡ AOA from both sides, we get that ∡ AOA ≡ − π < ∡ AOA π ; therefore ∡ AOA = 0.
Assume ∡ AOB = 0 . Show that [ OA ) = [ OB ) . X For any A and B distinct from O , we have ∡ AOB ≡ − ∡ BOA.
Proof.
According to Axiom IIIb, ∡ AOB + ∡ BOA ≡ ∡ AOA
By Proposition 2.5, ∡ AOA = 0. Hence the result. X A statement marked with “ X ” if Axiom V was not used in its proof. Ignore thismark for a while; it will be important in Chapter 11, see page 80. CHAPTER 2. AXIOMS
Straight angle If ∡ AOB = π , we say that ∠ AOB is a straight angle . Note that byProposition 2.7, if ∠ AOB is a straight, then so is ∠ BOA .We say that point O lies between points A and B , if O = A , O = B and O ∈ [ AB ]. X The angle
AOB is straight if and only if O lies between A and B . B O A
Proof.
By Proposition 2.4, we may assumethat OA = OB = 1. “If ” part. Assume O lies between A and B . Set α = ∡ AOB .Applying Axiom IIIa, we get a half-line [ OA ′ ) such that α = ∡ BOA ′ .By Proposition 2.4, we can assume that OA ′ = 1. According to Axiom IV, △ AOB ∼ = △ BOA ′ . Let f denotes the corresponding motion of the plane; that is, f is a motionsuch that f ( A ) = B , f ( O ) = O and f ( B ) = A ′ . B O AA ′ αα Then( A ′ B ) = f ( AB ) ∋ f ( O ) = O. Therefore, both lines ( AB ) and ( A ′ B ) contain B and O . By Axiom II, ( AB ) = ( A ′ B ).By the definition of the line, ( AB ) contains exactly two points A and B on distance 1 from O . Since OA ′ = 1 and A ′ = B , we get that A = A ′ .By Axiom IIIb and Proposition 2.5, we get that2 · α = ∡ AOB + ∡ BOA ′ == ∡ AOB + ∡ BOA ≡≡ ∡ AOA == 0Therefore, by Exercise 1.12, α is either 0 or π .Since [ OA ) = [ OB ), we have that α = 0, see Exercise 2.6. Therefore, α = π . “Only if ” part. Suppose that ∡ AOB = π . Consider the line ( OA ) andchoose a point B ′ on ( OA ) so that O lies between A and B ′ .From above, we have that ∡ AOB ′ = π . Applying Axiom IIIa, we getthat [ OB ) = [ OB ′ ). In particular, O lies between A and B .A triangle ABC is called degenerate if A , B and C lie on one line.The following corollary is just a reformulation of Theorem 2.8.3 X A triangle is degenerate if and only if one of its anglesis equal to π or . Moreover in a degenerate triangle the angle measuresare , and π . Show that three distinct points A , O and B lie on oneline if and only if · ∡ AOB ≡ . Let A , B and C be three points distinct from O . Showthat B , O and C lie on one line if and only if · ∡ AOB ≡ · ∡ AOC.
Show that there is a nondegenerate triangle.
Vertical angles
A pair of angles
AOB and A ′ OB ′ is called vertical if the point O liesbetween A and A ′ and between B and B ′ at the same time. X The vertical angles have equal measures.
A A ′ OB B ′ Proof.
Assume that the angles
AOB and A ′ OB ′ arevertical. Note that ∠ AOA ′ and ∠ BOB ′ are straight.Therefore, ∡ AOA ′ = ∡ BOB ′ = π .It follows that0 = ∡ AOA ′ − ∡ BOB ′ ≡≡ ∡ AOB + ∡ BOA ′ − ∡ BOA ′ − ∡ A ′ OB ′ ≡≡ ∡ AOB − ∡ A ′ OB ′ . Since − π < ∡ AOB π and − π < ∡ A ′ OB ′ π , weget that ≡ ∡ AOB = ∡ A ′ OB ′ . Assume O is the midpoint for both segments [ AB ] and [ CD ] . Prove that AC = BD . hapter 3 Half-planes
This chapter contains long proofs of intuitively evident statements. It isokay to skip it, but make sure you know definitions of positive/negativeangles and that your intuition agrees with 3.7, 3.9, 3.10, 3.12 and 3.17.
Sign of an angle
The positive and negative angles can be visualized as counterclockwise and clockwise directions; formally, they are defined the following way: ⋄ The angle
AOB is called positive if 0 < ∡ AOB < π ; ⋄ The angle
AOB is called negative if ∡ AOB <
Show that ∠ AOB is positive if and only if ∠ BOA isnegative.
Let ∠ AOB be straight. Then ∠ AOX is positive if andonly if ∠ BOX is negative.Proof.
Set α = ∡ AOX and β = ∡ BOX . Since ∠ AOB is straight, ➊ α − β ≡ π. It follows that α = π ⇔ β = 0 and α = 0 ⇔ β = π . In these two casesthe sign of ∠ AOX and ∠ BOX are undefined.In the remaining cases we have that | α | < π and | β | < π . If α and β have the same sign, then | α − β | < π ; the latter contradicts ➊ . Hence thestatement follows. 245 Assume that the angles
AOB and
BOC are positive.Show that ∡ AOB + ∡ BOC + ∡ COA = 2 · π. if ∠ COA is positive, and ∡ AOB + ∡ BOC + ∡ COA = 0 . if ∠ COA is negative.
Intermediate value theorem
Let f : [ a, b ] → R be a continuousfunction. Assume f ( a ) and f ( b ) have opposite signs, then f ( t ) = 0 forsome t ∈ [ a, b ] . f ( b ) f ( a ) t ba The intermediate value theorem is assumedto be known; it should be covered in any cal-culus course. We will use only the followingcorollary: X Assume that for any t ∈∈ [0 , we have three points in the plane O t , A t and B t , such that(a) Each function t O t , t A t and t B t is continuous.(b) For for any t ∈ [0 , , the points O t , A t and B t do not lie on oneline.Then ∠ A O B and ∠ A O B have the same sign.Proof. Consider the function f ( t ) = ∡ A t O t B t .Since the points O t , A t and B t do not lie on one line, Theorem 2.8implies that f ( t ) = ∡ A t O t B t = 0 nor π for any t ∈ [0 , f is a continuous function.By the intermediate value theorem, f (0) and f (1) have the same sign;hence the result follows. Same sign lemmas X Assume Q ′ ∈ [ P Q ) and Q ′ = P . Then for any X / ∈ / ∈ ( P Q ) the angles P QX and
P Q ′ X have the same sign. CHAPTER 3. HALF-PLANES PQ ′ Q X
Proof.
By Proposition 2.4, for any t ∈ [0 ,
1] thereis a unique point Q t ∈ [ P Q ) such that
P Q t = (1 − t ) · P Q + t · P Q ′ . Note that the map t Q t is continuous, Q = Q, Q = Q ′ and for any t ∈ [0 , P = Q t .Applying Corollary 3.5, for P t = P , Q t and X t = X , we get that ∠ P QX has the same sign as ∠ P Q ′ X . X In arbitrary nondegenerate tri-angle
ABC , the angles
ABC , BCA and
CAB have the same sign.
ZAB C
Proof.
Choose a point Z ∈ ( AB ) so that A liesbetween B and Z .According to Lemma 3.6, the angles ZBC and
ZAC have the same sign.Note that ∡ ABC = ∡ ZBC and ∡ ZAC + ∡ CAB ≡ π. Therefore, ∠ CAB has the same sign as ∠ ZAC which in turn has the same sign as ∡ ABC = ∡ ZBC .Repeating the same argument for ∠ BCA and ∠ CAB , we get the re-sult. X Assume [ XY ] does not intersect ( P Q ) , then the angles P QX and
P QY have the same sign.
PQ X Y
The proof is nearly identical to the one above.
Proof.
According to Proposition 2.4, for any t ∈∈ [0 ,
1] there is a point X t ∈ [ XY ], such that XX t = t · XY.
Note that the map t X t is continuous. More-over, X = X , X = Y and X t / ∈ ( QP ) for any t ∈ [0 , P t = P , Q t = Q and X t , we get that ∠ P QX has the same sign as ∠ P QY .7 Half-planes
Assume
X, Y / ∈ ( P Q ) . Then the angles P QX and
P QY have the same sign if and only if [ XY ] does not intersect ( P Q ) . QP XYZ
Proof.
The if-part follows from Lemma 3.8.Assume [ XY ] intersects ( P Q ); let Z denotes thepoint of intersection. Without loss of generality, wecan assume Z = P .Note that Z lies between X and Y . ByLemma 3.2, ∠ P ZX and ∠ P ZY have opposite signs.This proves the statement if Z = Q .If Z = Q , then ∠ ZQX and ∠ QZX have oppo-site signs by 3.7. The same way we get that ∠ ZQY and ∠ QZY haveopposite signs.If Q lies between Z and P , then by Lemma 3.2 two pairs of angles ∠ P QX , ∠ ZQX and ∠ P QY , ∠ ZQY have opposite signs. It follows that ∠ P QX and ∠ P QY have opposite signs as required.In the remaining case [ QZ ) = [ QP ) and therefore ∠ P QX = ∠ ZQX and ∠ P QY = ∠ ZQY . Hence again ∠ P QX and ∠ P QY have oppositesigns as required. X The complement of a line ( P Q ) in the plane canbe presented in a unique way as a union of two disjoint subsets called half-planes such that(a) Two points X, Y / ∈ ( P Q ) lie in the same half-plane if and only ifthe angles P QX and
P QY have the same sign.(b) Two points
X, Y / ∈ ( P Q ) lie in the same half-plane if and only if [ XY ] does not intersect ( P Q ) . O ABA ′ B ′ We say that X and Y lie on one side of ( P Q ) ifthey lie in one of the half-planes of (
P Q ) and we saythat P and Q lie on the opposite sides of ℓ if they liein the different half-planes of ℓ . Assume that the angles
AOB and A ′ OB ′ are vertical. Show that the line ( AB ) does notintersect the segment [ A ′ B ′ ] . Consider the triangle
ABC . The segments [ AB ], [ BC ] and [ CA ] arecalled sides of the triangle .The following theorem follows from Corollary 3.10:8 CHAPTER 3. HALF-PLANES
A B Cℓ X Assume line ℓ does not passthru any vertex a triangle. Then it intersects either twoor zero sides of the triangle.Proof. Assume that line ℓ intersects side [ AB ] of thetriangle ABC and does not pass thru A , B and C .By Corollary 3.10, the vertexes A and B lie on opposite sides of ℓ .The vertex C may lie on the same side with A and on opposite sidewith B or the other way around. By Corollary 3.10, in the first case, ℓ intersects side [ BC ] and does not intersect [ AC ]; in the second case, ℓ intersects side [ AC ] and does not intersect [ BC ]. Hence the statementfollows. Show that two points
X, Y / ∈ ( P Q ) lie on the same sideof ( P Q ) if and only if the angles P XQ and
P Y Q have the same sign.
PQ X Y BA A ′ B ′ C Let △ ABC be a nondegenerate triangle, A ′ ∈ [ BC ] and B ′ ∈ [ AC ] . Show that the segments [ AA ′ ] and [ BB ′ ] intersect. Assume that the points X and Y lie on opposite sidesof the line ( P Q ) . Show that the half-line [ P X ) does not intersect [ QY ) . Note that the following quantity ˜ ∡ ABC = ( π if ∡ ABC = π − ∡ ABC if ∡ ABC < π can serve as the angle measure; that is, the axioms hold if one exchanges ∡ to ˜ ∡ everywhere.Show that ∡ and ˜ ∡ are the only possible angle measures on the plane.Show that without Axiom IIIc, this is no longer true. Triangle with the given sides
Consider the triangle
ABC . Set a = BC, b = CA, c = AB.
Without loss of generality, we may assume that a b c. Then all three triangle inequalities for △ ABC hold if and only if c a + b. The following theorem states that this is the only restriction on a , b and c . X Assume that < a b c a + b . Then there is atriangle with sides a , b and c ; that is, there is △ ABC such that a = BC , b = CA and c = AB . A CBs ( β, r ) r rβ The proof is given at the end of the section.Assume r > π > β >
0. Consider the triangle
ABC such that AB = BC = r and ∡ ABC = β . Theexistence of such a triangle follows from Axiom IIIaand Proposition 2.4.Note that according to Axiom IV, the values β and r define the triangle ABC up to the congruence. Inparticular, the distance AC depends only on β and r .Set s ( β, r ) := AC. X Given r > and ε > , there is δ > such that if < β < δ , then s ( r, β ) < ε. Proof.
Fix two points A and B such that AB = r . AB C D ZYXr r Choose a point X such that ∡ ABX is positive.Let Y ∈ [ AX ) be the point such that AY = ε ; itexists by Proposition 2.4.Note that X and Y lie on the same side of( AB ); therefore, ∠ ABY is positive. Set δ == ∡ ABY .Assume 0 < β < δ , ∡ ABC = β and BC = r .Applying Axiom IIIa, we can choose a half-line [ BZ ) such that ∡ ABZ = · β . Note that A CHAPTER 3. HALF-PLANES and Y lie on opposite sides of ( BZ ). Therefore, ( BZ ) intersects [ AY ]; let D denotes the point of intersection.Since D ∈ ( BZ ), we get that ∡ ABD = β or β − π . The latter isimpossible since D and Y lie on the same side of ( AB ). Therefore, ∡ ABD = ∡ DBC = β . By Axiom IV, △ ABD ∼ = △ CBD . In particular, AC AD + DC == 2 · AD · AY == ε and hence the result. X Fix a real number r > and two distinct points A and B . Then for any real number β ∈ [0 , π ] , there is a unique point C β such that BC β = r and ∡ ABC β = β . Moreover, the map β C β is acontinuous map from [0 , π ] to the plane.Proof. The existence and uniqueness of C β follows from Axiom IIIa andProposition 2.4.Note that if β = β , then C β C β = s ( r, | β − β | ) . Therefore, Proposition 3.18 implies that the map β C β is continu-ous. Proof of Theorem 3.17.
Fix the points A and B such that AB = c . Given β ∈ [0 , π ], let C β denotes the point in the plane such that BC β = a and ∡ ABC = β .According to Corollary 3.19, the map β C β is continuous. There-fore, the function b ( β ) = AC β is continuous (formally, it follows fromExercise 1.13 and Exercise 1.14).Note that b (0) = c − a and b ( π ) = c + a . Since c − a b c + a , by theintermediate value theorem (3.4) there is β ∈ [0 , π ] such that b ( β ) = b .Hence the result.Given a positive real number r and a point O , the set Γ of all pointson distance r from O is called a circle with radius r and center O . Show that if two circles intersect if and only if | r − r | d r + r , where d denotes the distances between their centers and r , r — theirradiuses. hapter 4 Congruent triangles
Side-angle-side condition
Our next goal is to give conditions which guarantee congruence of twotriangles.One of such conditions is given in Axiom IV; it states that if two pairsof sides of two triangles are equal, and the included angles are equal upto sign, then the triangles are congruent. This axiom is also called side-angle-side congruence condition , or briefly,
SAS congruence condition . Angle-side-angle condition X Assume that AB = A ′ B ′ , ∡ ABC = ± ∡ A ′ B ′ C ′ , ∡ CAB = ± ∡ C ′ A ′ B ′ and △ A ′ B ′ C ′ is nondegenerate. Then △ ABC ∼ = △ A ′ B ′ C ′ . A ′ B ′ C ′ C ′′ Note that for degenerate triangles the statementdoes not hold. For example, consider one triangle withsides 1, 4, 5 and the other with sides 2, 3, 5.
Proof.
According to Theorem 3.7, either ➊ ∡ ABC = ∡ A ′ B ′ C ′ , ∡ CAB = ∡ C ′ A ′ B ′ CHAPTER 4. CONGRUENT TRIANGLES or ➋ ∡ ABC = − ∡ A ′ B ′ C ′ , ∡ CAB = − ∡ C ′ A ′ B ′ . Further we assume that ➊ holds; the case ➋ is analogous.Let C ′′ be the point on the half-line [ A ′ C ′ ) such that A ′ C ′′ = AC .By Axiom IV, △ A ′ B ′ C ′′ ∼ = △ ABC . Applying Axiom IV again, weget that ∡ A ′ B ′ C ′′ = ∡ ABC = ∡ A ′ B ′ C ′ . By Axiom IIIa, [ B ′ C ′ ) = [ BC ′′ ). Hence C ′′ lies on ( B ′ C ′ ) as well ason ( A ′ C ′ ).Since △ A ′ B ′ C ′ is not degenerate, ( A ′ C ′ ) is distinct from ( B ′ C ′ ). Ap-plying Axiom II, we get that C ′′ = C ′ .Therefore, △ A ′ B ′ C ′ = △ A ′ B ′ C ′′ ∼ = △ ABC . Isosceles triangles
A triangle with two equal sides is called isosceles ; the remaining side iscalled the base . X Assume △ ABC is an isosceles triangle with the base [ AB ] . Then ∡ ABC ≡ − ∡ BAC.
Moreover, the converse holds if △ ABC is nondegenerate.
A BC
The following proof is due to Pappus of Alexan-dria.
Proof.
Note that CA = CB, CB = CA, ∡ ACB ≡ − ∡ BCA.
Therefore, by Axiom IV, △ CAB ∼ = △ CBA.
Applying the theorem on the signs of angles of triangles (3.7) and Ax-iom IV again, we get that ∡ BAC ≡ − ∡ ABC.
To prove the converse, we assume that ∡ CAB ≡ − ∡ CBA . By ASAcondition 4.1, △ CAB ∼ = △ CBA . Therefore, CA = CB .A triangle with three equal sides is called equilateral . Let △ ABC be an equilateral triangle. Show that ∡ ABC = ∡ BCA = ∡ CAB. Side-side-side condition X △ ABC ∼ = △ A ′ B ′ C ′ if A ′ B ′ = AB, B ′ C ′ = BC and C ′ A ′ = CA.
Note that this condition is valid for degenerate triangles as well.
Proof.
Choose C ′′ so that A ′ C ′′ = A ′ C ′ and ∡ B ′ A ′ C ′′ = ∡ BAC . Ac-cording to Axiom IV, △ A ′ B ′ C ′′ ∼ = △ ABC.
It will suffice to prove that ➌ △ A ′ B ′ C ′ ∼ = △ A ′ B ′ C ′′ . The condition ➌ trivially holds if C ′′ = C ′ . Thus, it remains to considerthe case C ′′ = C ′ . A ′ B ′ C ′ C ′′ Clearly, the corresponding sides of △ A ′ B ′ C ′ and △ A ′ B ′ C ′′ are equal. Hencethe triangles △ C ′ A ′ C ′′ and △ C ′ B ′ C ′′ areisosceles. By Theorem 4.2, we have ∡ A ′ C ′′ C ′ ≡ − ∡ A ′ C ′ C ′′ , ∡ C ′ C ′′ B ′ ≡ − ∡ C ′′ C ′ B ′ . Adding them, we get that ∡ A ′ C ′′ B ′ ≡ − ∡ A ′ C ′ B ′ . Applying Axiom IV again, we get ➌ . BA A ′ B ′ C Let M be the mid-point of the side [ AB ] of △ ABC and M ′ be themidpoint of the side [ A ′ B ′ ] of △ A ′ B ′ C ′ . Assume C ′ A ′ = CA , C ′ B ′ = CB and C ′ M ′ = CM .Prove that △ A ′ B ′ C ′ ∼ = △ ABC.
Let △ ABC be an isosceles triangle with the base [ AB ] .Suppose that the points A ′ ∈ [ BC ] and B ′ ∈ [ AC ] are such that CA ′ == CB ′ . Show that CHAPTER 4. CONGRUENT TRIANGLES (a) △ AA ′ C ∼ = △ BB ′ C ;(b) △ ABB ′ ∼ = △ BAA ′ . Show that if AB + BC = AC then B ∈ [ AC ] . Let △ ABC be a nondegenerate triangle and let f be amotion of the plane such that f ( A ) = A, f ( B ) = B and f ( C ) = C. Show that f is the identity; that is, f ( X ) = X for any point X on theplane. On angle-side-side and side-angle-angle
In each of the conditions SAS, ASA, and SSS we specify three correspond-ing parts of the triangles. Let us discuss other triples of correspondingparts.The fist triple is called side-side-angle , or briefly SSA; it specifies twosides and a non-included angle. This condition is not sufficient for con-gruence; that is, there are two nondegenerate triangles
ABC and A ′ B ′ C ′ such that AB = A ′ B ′ , BC = B ′ C ′ , ∡ BAC ≡ ± ∡ B ′ A ′ C ′ , but △ ABC = △ A ′ B ′ C ′ and moreover AC = A ′ C ′ . A BC A ′ B ′ C ′ We will not use this negativestatement in the sequel and there-fore there is no need to prove it for-mally. An example can be guessedfrom the diagram.The second triple is side-angle-angle , or briefly SAA; it specifies one side and two angles one of which isopposite to the side. This provides a congruence condition; that is, if oneof the triangles
ABC and A ′ B ′ C ′ is nondegenerate then AB = A ′ B ′ , ∡ ABC ≡ ± ∡ A ′ B ′ C ′ , ∡ BCA ≡ ± ∡ B ′ C ′ A ′ implies △ ABC ∼ = △ A ′ B ′ C ′ .The SAA condition will not be used directly in the sequel. One proofof this condition can be obtained from ASA and the theorem on sum ofangles of triangle proved below (see 7.10). For a more direct proof, seeExercise 11.6.Another triple is called angle-angle-angle , or briefly AAA; by Axiom V,it is not sufficient for congruence. However AAA turns out to be a con-gruence condition in the hyperbolic plane; see 13.8. hapter 5 Perpendicular lines
Right, acute and obtuse angles ⋄ If | ∡ AOB | = π , we say that ∠ AOB is right ; ⋄ If | ∡ AOB | < π , we say that ∠ AOB is acute ; ⋄ If | ∡ AOB | > π , we say that ∠ AOB is obtuse .On the diagrams, the right angles will be markedwith a little square, as shown.If ∠ AOB is right, we say also that [ OA ) is perpendicular to [ OB ); itwill be written as [ OA ) ⊥ [ OB ).From Theorem 2.8, it follows that two lines ( OA ) and ( OB ) are ap-propriately called perpendicular , if [ OA ) ⊥ [ OB ). In this case we alsowrite ( OA ) ⊥ ( OB ). Assume point O lies between A and B and X = O . Showthat ∠ XOA is acute if and only if ∠ XOB is obtuse.
Perpendicular bisector
Assume M is the midpoint of the segment [ AB ]; that is, M ∈ ( AB ) and AM = M B .The line ℓ which passes thru M and perpendicular to ( AB ), is calledthe perpendicular bisector to the segment [ AB ]. X Given distinct points A and B , all points equidistantfrom A and B and no others lie on the perpendicular bisector to [ AB ] . CHAPTER 5. PERPENDICULAR LINES
A BMP
Proof.
Let M be the midpoint of [ AB ].Assume P A = P B and P = M . Accord-ing to SSS (4.4), △ AM P ∼ = △ BM P . Hence ∡ AM P = ± ∡ BM P.
Since A = B , we have “ − ” in the above for-mula. Further, π = ∡ AM B ≡≡ ∡ AM P + ∡ P M B ≡≡ · ∡ AM P.
That is, ∡ AM P = ± π . Therefore, P lies onthe perpendicular bisector.To prove the converse, suppose P is any point on the perpendicularbisector to [ AB ] and P = M . Then ∡ AM P = ± π , ∡ BM P = ± π and AM = BM . By SAS, △ AM P ∼ = △ BM P ; in particular, AP = BP . Let ℓ be the perpendicular bisector to the segment [ AB ] and X be an arbitrary point on the plane.Show that AX < BX if and only if X and A lie on the same sidefrom ℓ . Let △ ABC be nondegenerate. Show that
AC > BC ifand only if | ∡ ABC | > | ∡ CAB | . Uniqueness of a perpendicular X There is one and only one line which passes thru agiven point P and is perpendicular to a given line ℓ . A Bℓ PP ′ According to the above theorem, there is aunique point Q ∈ ℓ such that ( QP ) ⊥ ℓ . Thispoint Q is called the foot point of P on ℓ . Proof. If P ∈ ℓ , then both, existence anduniqueness, follow from Axiom III. Existence for P ℓ . Let A and B be two dis-tinct points of ℓ . Choose P ′ so that AP ′ == AP and ∡ P ′ AB ≡ − ∡ P AB . Accordingto Axiom IV, △ AP ′ B ∼ = △ AP B . Therefore, AP = AP ′ and BP = BP ′ .7According to Theorem 5.2, A and B lie on the perpendicular bisectorto [ P P ′ ]. In particular, ( P P ′ ) ⊥ ( AB ) = ℓ . Uniqueness for P ℓ . From above we can choose a point P ′ in such away that ℓ forms the perpendicular bisector to [ P P ′ ].Assume m ⊥ ℓ and m ∋ P . Then m is a perpendicular bisector tosome segment [ QQ ′ ] of ℓ ; in particular, P Q = P Q ′ . Q Q ′ PP ′ ℓ m Since ℓ is the perpendicular bisector to[ P P ′ ], we get that P Q = P ′ Q and P Q ′ = P ′ Q ′ .Therefore, P ′ Q = P Q = P Q ′ = P ′ Q ′ . By Theorem 5.2, P ′ lies on the perpendicularbisector to [ QQ ′ ], which is m . By Axiom II, m = ( P P ′ ). Reflection
Assume the point P and the line ( AB ) are given. To find the reflection P ′ of P in ( AB ), one drops a perpendicular from P onto ( AB ), and continuesit to the same distance on the other side.According to Theorem 5.5, P ′ is uniquely determined by P .Note that P = P ′ if and only if P ∈ ( AB ). X Assume P ′ is a reflection of the point P in theline ( AB ) . Then AP ′ = AP and if A = P , then ∡ BAP ′ ≡ − ∡ BAP .Proof.
Note that if P ∈ ( AB ), then P = P ′ . By Corollary 2.9, ∡ BAP = 0or π . Hence the statement follows. A BPP ′ If P / ∈ ( AB ), then P ′ = P . By the con-struction of P ′ , the line ( AB ) is perpendic-ular bisector of [ P P ′ ]. Therefore, accord-ing to Theorem 5.2, AP ′ = AP and BP ′ == BP . In particular, △ ABP ′ ∼ = △ ABP .Therefore, ∡ BAP ′ = ± ∡ BAP .Since P ′ = P and AP ′ = AP , we getthat ∡ BAP ′ = ∡ BAP . That is, we are leftwith the case ∡ BAP ′ = − ∡ BAP. CHAPTER 5. PERPENDICULAR LINES X The reflection in a line is a motion of the plane. More-over, if △ P ′ Q ′ R ′ is the reflection of △ P QR , then ∡ Q ′ P ′ R ′ ≡ − ∡ QP R.
Proof.
From the construction, it follows that the composition of two re-flections in the same line is the identity map. In particular, any reflectionis a bijection.Assume P ′ , Q ′ and R ′ denote the reflections of the points P , Q and R in ( AB ). Let us show that ➊ P ′ Q ′ = P Q and ∡ AP ′ Q ′ ≡ − ∡ AP Q.
Without loss of generality, we may assume that the points P and Q are distinct from A and B . By Proposition 5.6, ∡ BAP ′ ≡ − ∡ BAP, ∡ BAQ ′ ≡ − ∡ BAQ,AP ′ = AP, AQ ′ = AQ.
It follows that ∡ P ′ AQ ′ ≡ − ∡ P AQ . Therefore △ P ′ AQ ′ ∼ = △ P AQ and ➊ follows.Repeating the same argument for P and R , we get that ∡ AP ′ R ′ ≡ − ∡ AP R.
Subtracting the second identity in ➊ , we get that ∡ Q ′ P ′ R ′ ≡ − ∡ QP R.
Show that any motion of the plane can be presented as acomposition of at most three reflections.
Applying the exercise above and Corollary 5.7, we can divide the mo-tions of the plane in two types, direct and indirect motions . The motion f is direct if ∡ Q ′ P ′ R ′ = ∡ QP R for any △ P QR and P ′ = f ( P ), Q ′ = f ( Q ) and R ′ = f ( R ); if instead wehave ∡ Q ′ P ′ R ′ ≡ − ∡ QP R for any △ P QR , then the motion f is called indirect. Let X and Y be the reflections of P in the lines ( AB ) and ( BC ) correspondingly. Show that ∡ XBY ≡ · ∡ ABC. Perpendicular is shortest X Assume Q is the foot point of P on the line ℓ . Thenthe inequality P X > P Q holds for any point X on ℓ distinct from Q . If P , Q and ℓ are as above, then P Q is called the distance from P to ℓ . Xℓ PQP ′ Proof. If P ∈ ℓ , then the result follows since P Q = 0.Further we assume that
P / ∈ ℓ .Let P ′ be the reflection of P in the line ℓ . Note that Q is the midpoint of [ P P ′ ] and ℓ is the perpendicular bisectorof [ P P ′ ]. Therefore P X = P ′ X and P Q = P ′ Q = · P P ′ Note that ℓ meets [ P P ′ ] only at the point Q . Therefore, X / ∈ [ P P ′ ]; by the triangle inequality and Exercise 4.7, P X + P ′ X > P P ′ and hence the result: P X > P Q . Assume ∠ ABC is right or obtuse. Show that
AC > AB.
Circles
Recall that a circle with radius r and center O is the set of all pointson distance r from O . We say that a point P lies inside of the circle if OP < r ; if
OP > r , we say that P lies outside of the circle. Let Γ be a circle and P / ∈ Γ . Assume a line ℓ is passingthru the point P and intersects Γ at two distinct points, X and Y . Showthat P is inside Γ if and only if P lies between X and Y . A segment between two points on a circle is called a chord of the circle.A chord passing thru the center of the circle is called its diameter . Assume two distinct circles Γ and Γ ′ have a commonchord [ AB ] . Show that the line between centers of Γ and Γ ′ forms aperpendicular bisector to [ AB ] . CHAPTER 5. PERPENDICULAR LINES X A line and a circle can have at most two points ofintersection.Proof.
Assume A , B and C are distinct points which lie on a line ℓ anda circle Γ with the center O .Then OA = OB = OC ; in particular, O lies on the perpendicularbisectors m and n to [ AB ] and [ BC ] correspondingly. Note that the A B Cℓ m n midpoints of [ AB ] and [ BC ] are distinct. Therefore, m and n are distinct.The latter contradicts the uniqueness of the perpendicular (Theorem 5.5). Show that two distinct circles can have at most twopoints of intersection.
In consequence of the above lemma, a line ℓ and a circle Γ mighthave 2, 1 or 0 points of intersections. In the first case the line is called secant line , in the second case it is tangent line ; if P is the only point ofintersection of ℓ and Γ, we say that ℓ is tangent to Γ at P .Similarly, according Exercise 5.15, two circles might have 2, 1 or 0points of intersections. If P is the only point of intersection of circles Γand Γ ′ , we say that Γ is tangent to Γ at P . X Let ℓ be a line and Γ be a circle with the center O .Assume P is a common point of ℓ and Γ . Then ℓ is tangent to Γ at P ifand only if ( P O ) ⊥ ℓ .Proof. Let Q be the foot point of O on ℓ .Assume P = Q . Let P ′ denotes the reflection of P in ( OQ ).Note that P ′ ∈ ℓ and ( OQ ) is the perpendicular bisector of [ P P ′ ].Therefore, OP = OP ′ . Hence P, P ′ ∈ Γ ∩ ℓ ; that is, ℓ is secant to Γ.If P = Q , then according to Lemma 5.10, OP < OX for any point X ∈ ℓ distinct from P . Hence P is the only point in the intersection Γ ∩ ℓ ;that is, ℓ is tangent to Γ at P . Let Γ and Γ ′ be two distinct circles with centers at O and O ′ correspondingly. Assume Γ meets Γ ′ at the point P . Show that Γ is tangent to Γ ′ if and only if O , O ′ and P lie on one line. Let Γ and Γ ′ be two distinct circles with centers at O and O ′ and radii r and r ′ . Show that Γ is tangent to Γ ′ if and only if OO ′ = r + r ′ or OO ′ = | r − r ′ | . Assume three circles intersect at twopoints as shown on the diagram. Prove that the centers ofthese circles lie on one line.
Geometric constructions
The ruler-and-compass constructions in the plane is the construction ofpoints, lines, and circles using only an idealized ruler and compass. Theseconstruction problems provide a valuable source of exercises in geometry,which we will use further in the book. In addition, Chapter 19 is devotedcompletely to the subject.The idealized ruler can be used only to draw a line thru the given twopoints. The idealized compass can be used only to draw a circle with agiven center and radius. That is, given three points A , B and O we candraw the set of all points on distance AB from O . We may also mark newpoints in the plane as well as on the constructed lines, circles and theirintersections (assuming that such points exist).We can also look at the different set of construction tools. For example,we may only use the ruler or we may invent a new tool, say a tool whichproduces a midpoint for any given two points.As an example, let us consider the following problem: Construct the midpoint of the givensegment [ AB ] . A BPQ M
Construction.
1. Construct the circle with center at A whichis passing thru B .2. Construct the circle with center at B whichis passing thru A .3. Mark both points of intersection of these cir-cles, label them with P and Q .4. Draw the line ( P Q ).5. Mark the point of intersection of (
P Q ) and[ AB ]; this is the midpoint.Typically, you need to prove that the construction produces what wasexpected. Here is a proof for the example above.2 CHAPTER 5. PERPENDICULAR LINES
Proof.
According to Theorem 5.2, (
P Q ) is the perpendicular bisectorto [ AB ]. Therefore, M = ( AB ) ∩ ( P Q ) is the midpoint of [ AB ]. Make a ruler-and-compass construction of a line thrua given point which is perpendicular to a given line.
Make a ruler-and-compass construction of the center ofa given circle.
Make a ruler-and-compass construction of the lines tan-gent to a given circle which pass thru a given point.
Given two circles Γ and Γ and a segment [ AB ] makea ruler-and-compass construction of a circle with the radius AB , which istangent to each circle Γ and Γ . hapter 6 Similar triangles
Similar triangles
Two triangles A ′ B ′ C ′ and ABC are called similar (briefly △ A ′ B ′ C ′ ∼∼ △ ABC ) if (1) their sides are proportional; that is, ➊ A ′ B ′ = k · AB, B ′ C ′ = k · BC and C ′ A ′ = k · CA for some k >
0, and (2) the corresponding angles are equal up to sign: ➋ ∡ A ′ B ′ C ′ = ± ∡ ABC, ∡ B ′ C ′ A ′ = ± ∡ BCA, ∡ C ′ A ′ B ′ = ± ∡ CAB.
Remarks. ⋄ According to 3.7, in the above three equalities, the signs can beassumed to be the same. ⋄ If △ A ′ B ′ C ′ ∼ △ ABC with k = 1 in ➊ , then △ A ′ B ′ C ′ ∼ = △ ABC . ⋄ Note that “ ∼ ” is an equivalence relation . That is,(i) △ ABC ∼ △
ABC for any △ ABC .(ii) If △ A ′ B ′ C ′ ∼ △ ABC , then △ ABC ∼ △ A ′ B ′ C ′ . (iii) If △ A ′′ B ′′ C ′′ ∼ △ A ′ B ′ C ′ and △ A ′ B ′ C ′ ∼ △ ABC , then △ A ′′ B ′′ C ′′ ∼ △ ABC.
CHAPTER 6. SIMILAR TRIANGLES
Using the new notation “ ∼ ”, we can reformulate Axiom V: If for the two triangles △ ABC , △ AB ′ C ′ , and k > we have B ′ ∈ [ AB ) , C ′ ∈ [ AC ) , AB ′ = k · AB and AC ′ = k · AC , then △ ABC ∼ △ AB ′ C ′ . In other words, the Axiom V provides a condition which guaranteesthat two triangles are similar. Let us formulate three more such similarityconditions . Two triangles △ ABC and △ A ′ B ′ C ′ aresimilar if one of the following conditions holds:(SAS) For some constant k > we have AB = k · A ′ B ′ , AC = k · A ′ C ′ and ∡ BAC = ± ∡ B ′ A ′ C ′ . (AA) The triangle A ′ B ′ C ′ is nondegenerate and ∡ ABC = ± ∡ A ′ B ′ C ′ , ∡ BAC = ± ∡ B ′ A ′ C ′ . (SSS) For some constant k > we have AB = k · A ′ B ′ , AC = k · A ′ C ′ , CB = k · C ′ B ′ . Each of these conditions is proved by applying Axiom V with the SAS,ASA and SSS congruence conditions correspondingly (see Axiom IV andthe conditions 4.1, 4.4).
Proof.
Set k = ABA ′ B ′ . Choose points B ′′ ∈ [ A ′ B ′ ) and C ′′ ∈ [ A ′ C ′ ), sothat A ′ B ′′ = k · A ′ B ′ and A ′ C ′′ = k · A ′ C ′ . By Axiom V, △ A ′ B ′ C ′ ∼∼ △ A ′ B ′′ C ′′ .Applying the SAS, ASA or SSS congruence condition, depending onthe case, we get that △ A ′ B ′′ C ′′ ∼ = △ ABC . Hence the result.A bijection X ↔ X ′ from a plane to itself is called angle preservingtransformation if ∡ ABC = ∡ A ′ B ′ C ′ for any triangle ABC and its image △ A ′ B ′ C ′ . Show that any angle-preserving transformation of theplane multiplies all the distance by a fixed constant. Pythagorean theorem
A triangle is called right if one of its angles is right. The side oppositethe right angle is called the hypotenuse . The sides adjacent to the rightangle are called legs . Assume △ ABC is a right triangle with the right angleat C . Then AC + BC = AB . Proof.
Let D be the foot point of C on ( AB ). A BCD
According to Lemma 5.10,
AD < AC < AB and
BD < BC < AB.
Therefore, D lies between A and B ; in particular, ➌ AD + BD = AB.
Note that by the AA similarity condition, we have △ ADC ∼ △
ACB ∼ △
CDB.
In particular, ➍ ADAC = ACAB and
BDBC = BCBA .
Let us rewrite the two identities in ➍ : AC = AB · AD and BC = AB · BD.
Summing up these two identities and applying ➌ , we get that AC + BC = AB · ( AD + BD ) = AB . The idea in the proof above appears in the Elements, see [8, X.33], butthe proof given there [8, I.47] is different; it uses area method discussedin Chapter 20.
Assume A , B , C and D are as in the proof above. Showthat CD = AD · BD.
The following exercise is the converse to the Pythagorean theorem.
Assume that
ABC is a triangle such that AC + BC = AB . Prove that the angle at C is right. CHAPTER 6. SIMILAR TRIANGLES
Method of similar triangles
A BZCX Y
The proof of the Pythagorean theorem given above isa good example using the method of similar triangles .To apply this method, one has to search for pairs ofsimilar triangles and then use the proportionality ofcorresponding sides and/or equalities of correspondingangles. Finding such pairs might be tricky at first.
Let
ABC be a nondegenerate triangleand the points X , Y and Z as on the diagram. Assume ∡ CAY ≡ ∡ XBC .Find four pairs of similar triangles with these six points as the vertexesand prove their similarity. A quadrilateral is defined as an ordered quadruple of distinct points in theplane. These 4 points are called vertexes of quadrilateral. The quadrilat-eral ABCD will be also denoted by (cid:3)
ABCD .Given a quadrilateral
ABCD , the four segments [ AB ], [ BC ], [ CD ]and [ DA ] are called sides of (cid:3) ABCD ; the remaining two segments [ AC ]and [ BD ] are called diagonals of (cid:3) ABCD . In any quadrilateral, the product of diago-nals can not exceed the sum of the products of its opposite sides; that is, AC · BD AB · CD + BC · DA for any (cid:3) ABCD . We will present a classical proof of this inequality using the methodof similar triangles with an additional construction. This proof is givenas an illustration — it will not be used further in the sequel.
AB CDX
Proof.
Consider the half-line [ AX ) such that ∡ BAX = ∡ CAD . In this case ∡ XAD = ∡ BAC since adding ∡ BAX or ∡ CAD to the correspond-ing sides produces ∡ BAD . Note that we canchoose X so that AX = ABAC · AD.
In this case we have
AXAD = ABAC , AXBA = ADAC . .1. PTOLEMY’S INEQUALITY △ BAX ∼ △
CAD, △ XAD ∼ △
BAC.
Therefore
BXCD = ABAC , XDBC = ADAC , or, equivalently AC · BX = AB · CD, AC · XD = BC · AD.
Adding these two equalities we get AC · ( BX + XD ) = AB · CD + BC · AD.
It remains to apply the triangle inequality, BD BX + XD .Using the proof above together with Theorem 9.10, one can show thatthe equality holds only if the vertexes A , B , C and D appear on a line ora circle in the same cyclic order; see also 10.12 for another proof of theequality case. Exercise 18.2 below suggests another proof of Ptolemy’sinequality using complex coordinates. hapter 7 Parallel lines
Parallel lines
In consequence of Axiom II, any two distinct lines ℓ and m have either one point in common or none. In the first casethey are intersecting (briefly ℓ ∦ m ); in the second case, ℓ and m are said to be parallel (briefly ℓ k m ); in addition,a line is always regarded as parallel to itself.To emphasize that two lines on a diagram are parallel we will markthem with arrows of the same type. X Let ℓ , m and n be three lines. Assume that n ⊥ m and m ⊥ ℓ . Then ℓ k n .Proof. Assume the contrary; that is, ℓ ∦ n . Then there is a point, say Z ,of intersection of ℓ and n . Then by Theorem 5.5, ℓ = n .Since any line is parallel to itself, we have that ℓ k n — a contradiction. For any point P and any line ℓ there is a unique line m which passes thru P and is parallel to ℓ . The above theorem has two parts, existence and uniqueness. In theproof of uniqueness we will use the method of similar triangles.
Proof; existence.
Apply Theorem 5.5 two times, first to construct the line m thru P which is perpendicular to ℓ , and second to construct the line n thru P which is perpendicular to m . Then apply Proposition 7.1. Uniqueness. If P ∈ ℓ , then m = ℓ by the definition of parallel lines.Further we assume P / ∈ ℓ . 489Let us construct the lines n ∋ P and m ∋ P as in the proof of existence,so m k ℓ .Assume there is yet another line s ∋ P which is distinct from m andparallel to ℓ . Choose a point Q ∈ s which lies with ℓ on the same sidefrom m . Let R be the foot point of Q on n . PRD Q Zℓm s n Let D be the point of intersection of n and ℓ . According to Proposi-tion 7.1 ( QR ) k m . Therefore, Q , R and ℓ lie on the same side of m . Inparticular, R ∈ [ P D ).Choose Z ∈ [ P Q ) such that
P ZP Q = P DP R .
By SAS similarity condition (or equivalently by Axiom V) we have that △ RP Q ∼ △
DP Z ; therefore ( ZD ) ⊥ ( P D ). It follows that Z lies on ℓ and s — a contradiction. Assume ℓ , m and n are lines such that ℓ k m and m k n .Then ℓ k n .Proof. Assume the contrary; that is, ℓ ∦ n . Then there is a point P ∈ ℓ ∩ n .By Theorem 7.2, n = ℓ — a contradiction.Note that from the definition, we have that ℓ k m if and only if m k ℓ .Therefore, according to the above corollary, “ k ” is an equivalence relation .That is, for any lines ℓ , m and n the following conditions hold:(i) ℓ k ℓ ;(ii) if ℓ k m , then m k ℓ ;(iii) if ℓ k m and m k n , then ℓ k n . Let k , ℓ , m and n be lines such that k ⊥ ℓ , ℓ ⊥ m and m ⊥ n . Show that k ∦ n . Make a ruler-and-compass construction of a line thru agiven point which is parallel to a given line. CHAPTER 7. PARALLEL LINES
AB CD
Transversal property
If the line t intersects each line ℓ and m at one point,then we say that t is a transversal to ℓ and m . Forexample, on the diagram, line ( CB ) is a transversalto ( AB ) and ( CD ). ( AB ) k ( CD ) if and only if ➊ · ( ∡ ABC + ∡ BCD ) ≡ . Equivalently ∡ ABC + ∡ BCD ≡ or ∡ ABC + ∡ BCD ≡ π. Moreover, if ( AB ) = ( CD ) , then in the first case A and D lie on theopposite sides of ( BC ) , in the second case A and D lie on the same sidesof ( BC ) . The following lemma will be used several times in the sequel. X Let
ABC be a nondegenerate triangle. Then there is apoint
X / ∈ ( AC ) such that ∡ BAX = ∡ ABC and, moreover, X lies with B on the same side of ( AC ) . ABC XM
Proof.
Let M be the midpoint of [ AB ]. Choose X ∈∈ ( CM ) distinct from C so that XM = CM .Note that the angles CM B and
XM A are vertical;in particular, ➋ ∡ CM B = ∡ XM A.
By construction, AM = BM and CM = XM . Therefore, △ CM B ∼ = △ XM A ;in particular ∡ ABC ≡ ± ∡ BAX . Since angles of △ AM X have the samesign (3.7), by ➋ we can not have a minus sign in the last equality; therefore ∡ ABC ≡ ∡ BAX.
By construction, the line segments [ BM ] and [ M X ] do not cross theline ( AC ). Therefore the points X and B lie on the same side of ( AC )(see 3.10); in particular, X / ∈ ( AC ).1 Proof of transversal property; “only-if ” part.
We need to show that if( AB ) ∦ ( CD ), then ➊ does not hold.Since ( AB ) ∦ ( CD ), there is a unique point Z ∈ ( AB ) ∩ ( CD ). Since Z lies on ( AB ) and on ( CD ), Exercise 2.11 implies that2 · ∡ ABC ≡ · ∡ ZBC, · ∡ BCD ≡ · ∡ BCZ.
Note that the triangle
ZBC is nondegenerate. By Lemma 7.7, thereis a point
X / ∈ ( ZB ) such that ∡ CBX = ∡ BCZ . Applying Exercise 2.10and the equalities above, we get that
ABCDX Z · ∡ ZBX ≡≡ · ( ∡ ZBC + ∡ CBX ) ≡ · ( ∡ ZBC + ∡ BCZ ) ≡ · ( ∡ ABC + ∡ BCD );that is, ➊ does not hold. “If ”-part. Given points A , B and C there is unique line ( CD ) such that ➊ holds. Indeed, suppose there are two such lines ( CD ) and ( CD ′ ), then2 · ( ∡ ABC + ∡ BCD ) ≡ · ( ∡ ABC + ∡ BCD ′ ) ≡ . Therefore 2 · ∡ BCD ≡ · ∡ BCD ′ . It follows that 2 · ∡ DCD ′ ≡
0; that is, D ′ ∈ ( CD ), or equivalently the line( CD ) coinsides with the line ( CD ′ ).From the “only-if” part we know that if ➊ holds then ( CD ) k ( AB ).By Theorem 7.2 there is unique line thru C that is parallel to ( AB ).Therefore it must be the same line ( CD ) such that ➊ hold. Let △ ABC be a nondegenerate triangle. Assume B ′ and C ′ are points on sides [ AB ] and [ AC ] such that ( B ′ C ′ ) k ( BC ) . Showthat △ ABC ∼ △ AB ′ C ′ . Trisect a given segment with a ruler and a compass.
Angles of triangles
In any △ ABC , we have ∡ ABC + ∡ BCA + ∡ CAB ≡ π. Proof.
First note that if △ ABC is degenerate, then the equality followsfrom Corollary 2.9. Further we assume that △ ABC is nondegenerate.2
CHAPTER 7. PARALLEL LINES
ABC X
By Lemma 7.7 there is a point X that lies with B on the same side of ( AC ) such that ∡ BAX = ∡ ABC .In particular 2 · ( ∡ XAB + ∡ ABC ) ≡ . By transversal property, we get that ( AX ) k ( CB ).Now, let us apply the transversal property (7.6) to the parallel lines( BC ) and ( AX ) and their transversal ( CA ). Since B and X lie on thesame side of ( CA ), we have ➌ ∡ BCA + ∡ CAX ≡ π. Since ∡ BAX = ∡ ABC , we have ∡ CAX ≡ ∡ CAB + ∡ ABC
The latter identity and ➌ imply the theorem. AB CD
Let △ ABC be a nonde-generate triangle. Assume there is a point D ∈ [ BC ] such that ∡ BAD ≡ ∡ DAC, BA = AD = DC.
Find the angles of △ ABC . Show that | ∡ ABC | + | ∡ BCA | + | ∡ CAB | = π for any △ ABC . A B CD
Let △ ABC be an isosceles non-degenerate triangle with the base [ AC ] . Considerthe point D on the extension of the side [ AB ] suchthat AB = BD . Show that ∠ ACD is right.
Let △ ABC be an isosceles nondegenerate triangle withbase [ AC ] . Assume that a circle is passing thru A , centered at a point on [ AB ] and tangent to ( BC ) at the point X . Show that ∡ CAX = ± π . Show that for any quadrilateral
ABCD , we have ∡ ABC + ∡ BCD + ∡ CDA + ∡ DAB ≡ . Parallelograms
A quadrilateral
ABCD in the Euclidean plane is called nondegenerate ifany three points from
A, B, C, D do not lie on one line.A nondegenerate quadrilateral is called a parallelogram if its oppositesides are parallel.
ABC D If (cid:3) ABCD is a parallelogram,then(a) ∡ DAB = ∡ BCD ;(b) AB = CD .Proof. Since ( AB ) k ( CD ), the points C and D lie on the same side of ( AB ). Hence ∠ ABD and ∠ ABC have the same sign.The same way we get that ∠ CBD and ∠ CBA have the same sign.Since ∡ ABC ≡ − ∡ CBA , we get that the angles
DBA and
DBC haveopposite signs; that is, A and C lie on opposite sides of ( BD ).According to the transversal property (7.6), ∡ BDC ≡ − ∡ DBA and ∡ DBC ≡ − ∡ BDA.
By the ASA condition △ ABD ∼ = △ CDB . The latter implies both state-ments in the lemma.
Assume
ABCD is a quadrilateral such that AB = CD = BC = DA.
Show that
ABCD is a parallelogram.
A quadrilateral as in the exercise above is called a rhombus . Show that diagonals of a parallelogram intersect eachother at their midpoints.
A quadraliteral
ABCD is called a rectangle if the angles
ABC , BCD , CDA and
DAB are right. Note that according to the transversal property7.6, any rectangle is a parallelogram.A rectangle with equal sides is called a square . Show that the parallelogram
ABCD is a rectangle ifand only if AC = BD . Show that the parallelogram
ABCD is a rhombus if andonly if ( AC ) ⊥ ( BD ) . CHAPTER 7. PARALLEL LINES
Assume ℓ k m , and X, Y ∈ m . Let X ′ and Y ′ denote the foot pointsof X and Y on ℓ . Note that (cid:3) XY Y ′ X ′ is a rectangle. By Lemma 7.16, XX ′ = Y Y ′ . That is, any point on m lies on the same distance from ℓ .This distance is called the distance between ℓ and m . Method of coordinates
The following exercise is important; it shows that our axiomatic definitionagrees with the model definition described on page 11.
Let ℓ and m be perpendicular lines in the Euclideanplane. Given a point P , let P ℓ and P m denote the foot points of P on ℓ and m correspondingly.(a) Show that for any X ∈ ℓ and Y ∈ m there is a unique point P suchthat P ℓ = X and P m = Y .(b) Show that P Q = P ℓ Q ℓ + P m Q m for any pair of points P and Q .(c) Conclude that the plane is isometric to ( R , d ) ; see page 11. P QP ℓ Q ℓ P m Q m ℓ m Once this exercise is solved, we can apply themethod of coordinates to solve any problem in Eu-clidean plane geometry. This method is powerful,but it is often considered as a bad style.
Use the Exercise 7.21 to givean alternative proof of Theorem 3.17 in the Eu-clidean plane.That is, prove that given the real numbers a , b and c such that < a b c a + b, there is a triangle ABC such that a = BC , b = CA and c = AB . Consider two distinct points A = ( x A , y A ) and B == ( x B , y B ) on the coordinate plane. Show that the perpendicular bisectorto [ AB ] is described by the equation · ( x B − x A ) · x + 2 · ( y B − y A ) · y = x B + y B − x A − x B . Conclude that any line is described by an equation a · x + b · y = c for some constants a , b and c such that a = 0 or b = 0 . Show that for fixed real values a , b and c the equation x + y + a · x + b · y + c = 05 describes a circle, one-point set or empty set.Show that if it is a circle then it has center ( − a , − b ) and the radius r = ·√ a + b − · c . Use the previous exercise to show that given two distinctpoint A and B and positive real number k = 1 , the locus of points M suchthat AM = k · BM is a circle. The circle in the exercise above is an example of the so called
Apollo-nian circle . hapter 8 Triangle geometry
Triangle geometry is the study of the properties of triangles, includingassociated centers and circles.We discuss the most basic results in triangle geometry, mostly to showthat we have developed sufficient machinery to prove things.
Circumcircle and circumcenter
Perpendicular bisectors to the sides of any nondegener-ate triangle intersect at one point.
The point of intersection of the perpendicular bisectors is called cir-cumcenter . It is the center of the circumcircle of the triangle; that is, thecircle which passes thru all three vertices of the triangle. The circumcenterof the triangle is usually denoted by O . B A CO ℓ m Proof.
Let △ ABC be nondegenerate. Let ℓ and m beperpendicular bisectors to sides [ AB ] and [ AC ] corre-spondingly.Assume ℓ and m intersect, let O = ℓ ∩ m .Let us apply Theorem 5.2. Since O ∈ ℓ , we havethat OA = OB and since O ∈ m , we have that OA == OC . It follows that OB = OC ; that is, O lies on theperpendicular bisector to [ BC ].It remains to show that ℓ ∦ m ; assume the con-trary. Since ℓ ⊥ ( AB ) and m ⊥ ( AC ), we get that ( AC ) k ( AB ) (seeExercise 7.4). Therefore, by Theorem 5.5, ( AC ) = ( AB ); that is, △ ABC is degenerate — a contradiction. 567
There is a unique circle which passes thru the vertexesof a given nondegenerate triangle in the Euclidean plane.
Altitudes and orthocenter An altitude of a triangle is a line thru a vertex and perpendicular to theline containing the opposite side. The term altitude maybe also usedfor the distance from the vertex to its foot point on the line containingopposite side. The three altitudes of any nondegenerate triangle inter-sect in a single point.
The point of intersection of altitudes is called orthocenter ; it is usuallydenoted by H . Proof.
Let △ ABC be nondegenerate. B ′ A A ′ BCC ′ ℓ m n Consider three lines ℓ , m and n such that ℓ k ( BC ) , m k ( CA ) , n k ( AB ) ,ℓ ∋ A, m ∋ B, n ∋ C. Since △ ABC is nondegenerate, no pair of thelines ℓ , m and n is parallel. Set A ′ = m ∩ n, B ′ = n ∩ ℓ, C ′ = ℓ ∩ m. Note that (cid:3)
ABA ′ C , (cid:3) BCB ′ A and (cid:3) CBC ′ A are parallelograms. Ap-plying Lemma 7.16 we get that △ ABC is the median triangle of △ A ′ B ′ C ′ ;that is, A , B and C are the midpoints of [ B ′ C ′ ], [ C ′ A ′ ] and [ A ′ B ′ ] corre-spondingly.By Exercise 7.4, ( B ′ C ′ ) k ( BC ), the altitude from A is perpendicularto [ B ′ C ′ ] and from above it bisects [ B ′ C ′ ].Hence the altitudes of △ ABC are also perpendicular bisectors of △ A ′ B ′ C ′ . Applying Theorem 8.1, we get that altitudes of △ ABC in-tersect at one point.
Assume H is the orthocenter of an acute triangle ABC .Show that A is the orthocenter of △ HBC . Medians and centroid
A median of a triangle is the segment joining a vertex to the midpoint ofthe opposing side.8
CHAPTER 8. TRIANGLE GEOMETRY
The three medians of any nondegenerate triangle inter-sect in a single point. Moreover, the point of intersection divides eachmedian in the ratio 2:1.
The point of intersection of medians is called the centroid of the tri-angle; it is usually denoted by M . Proof.
Consider a nondegenerate triangle
ABC . Let [ AA ′ ] and [ BB ′ ] beits medians.According to Exercise 3.14, [ AA ′ ] and [ BB ′ ] are intersecting. Let usdenote the point of intersection by M .By SAS, △ B ′ A ′ C ∼ △ ABC and A ′ B ′ = · AB . In particular, ∡ ABC = ∡ B ′ A ′ C . A A ′ BB ′ C M
Note that ( AB ) k ( A ′ B ′ ). Indeed, since A ′ liesbetween B and C , we have that ∡ BA ′ B ′ + ∡ B ′ A ′ C == π . Therefore, ∡ B ′ A ′ B + ∡ A ′ BA = π ;it remains to apply the transversal property 7.6.Note that A ′ and A lie on opposite sidesfrom ( BB ′ ). Therefore, by the transversal property7.6, we get that ∡ B ′ A ′ M = ∡ BAM.
The same way we get that ∡ A ′ B ′ M = ∡ ABM.
By AA condition, △ ABM ∼ △ A ′ B ′ M . Since A ′ B ′ = · AB , we have A ′ MAM = B ′ MBM = 12 ;that is, M divides medians [ AA ′ ] and [ BB ′ ] in ratio 2:1.Note that M is a unique point on [ BB ′ ] such that B ′ MBM = 12 . Repeating the same argument for vertices B and C we get that all medians[ CC ′ ] and [ BB ′ ] intersect at M . Let (cid:3)
ABCD be a nondegenerate quadrilateral and X , Y , V , W be the midpoints of its sides [ AB ] , [ BC ] , [ CD ] and [ DA ] . Showthat (cid:3) XY V W is a parallelogram. Angle bisectors If ∡ ABX ≡ − ∡ CBX , then we say that the line ( BX ) bisects ∠ ABC , orline ( BX ) is the bisector of ∠ ABC . If ∡ ABX ≡ π − ∡ CBX , then theline ( BX ) is called the external bisector of ∠ ABC . AB C b i s e c t o r e x t e r n a l b i s ec t o r If ∡ ABA ′ = π ; that is, if B lies between A and A ′ , then bisector of ∠ ABC is the externalbisector of ∠ A ′ BC and the other way around.Note that the bisector and the externalbisector are uniquely defined by the angle. Show that for any angle, itsbisector and external bisector are perpendicu-lar.
The bisectors of ∠ ABC , ∠ BCA and ∠ CAB of a nondegenerate triangle
ABC are called bisectors of △ ABC at vertexes A , B and C correspondingly. Let △ ABC be a nondegenerate triangle. Assume that thebisector at the vertex A intersects the side [ BC ] at the point D . Then ➊ ABAC = DBDC . A BC DE ℓ
Proof.
Let ℓ be the line passing thru C that is parallelto ( AB ). Note that ℓ ∦ ( AD ); set E = ℓ ∩ ( AD ) . Also note that B and C lie on opposite sidesof ( AD ). Therefore, by the transversal property(7.6), ➋ ∡ BAD = ∡ CED.
Further, note that the angles
ADB and
EDC are vertical; in partic-ular, by 2.13 ∡ ADB = ∡ EDC.
By the AA similarity condition, △ ABD ∼ △
ECD . In particular, ➌ ABEC = DBDC .
Since ( AD ) bisects ∠ BAC , we get that ∡ BAD = ∡ DAC . Togetherwith ➋ , it implies that ∡ CEA = ∡ EAC . By Theorem 4.2, △ ACE isisosceles; that is, EC = AC. CHAPTER 8. TRIANGLE GEOMETRY
Together with ➌ , it implies ➊ . Formulate and prove an analog of Lemma 8.8 for theexternal bisector.
Equidistant property
Recall that distance from the line ℓ to the point P is defined as thedistance from P to its foot point on ℓ ; see page 39. X Assume △ ABC is not degenerate. Then a point X lies on the bisector or external bisector of ∠ ABC if and only if X isequidistant from the lines ( AB ) and ( BC ) .Proof. We can assume that X does not lie on the union of ( AB ) and( BC ). Otherwise the distance to one of the lines vanish; in this case X = B is the only point equidistant from the two lines. AB CXYZ
Let Y and Z be the reflections of X in ( AB ) and( BC ) correspondingly. Note that Y = Z. Otherwise both lines ( AB ) and ( BC ) are perpendicu-lar bisectors of [ XY ], that is, ( AB ) = ( BC ) which isimpossible since △ ABC is not degenerate.By Proposition 5.6, XB = Y B = ZB.
Note that X is equidistant from ( AB ) and ( BC ) if and only if XY == XZ . Applying SSS and then SAS, we get that XY = XZ. m△ BXY ∼ = △ BXZ. m ∡ XBY = ± ∡ BXZ.
Since Y = Z , we get that ∡ XBY = ∡ BXZ ; therefore ➍ ∡ XBY = − ∡ BXZ.
By Proposition 5.6, A lies on the bisector of ∠ XBY and B lies on thebisector of ∠ XBZ ; that is,2 · ∡ XBA ≡ ∡ XBY, · ∡ XBC ≡ ∡ XBZ. ➍ , 2 · ∡ XBA ≡ − · ∡ XBC.
The last identity means either ∡ XBA + ∡ XBC ≡ ∡ XBA + ∡ XBC ≡ π, and hence the result. Incenter X The angle bisectors of any nondegenerate triangleintersect at one point.
The point of intersection of bisectors is called the incenter of thetriangle; it is usually denoted by I . The point I lies on the same distancefrom each side. In particular, it is the center of a circle tangent to eachside of triangle. This circle is called the incircle and its radius is calledthe inradius of the triangle. A BCIZ A ′ XYB ′ Proof.
Let △ ABC be a nondegenerate triangle.Note that the points B and C lie on oppo-site sides of the bisector of ∠ BAC . Hence thisbisector intersects [ BC ] at a point, say A ′ .Analogously, there is B ′ ∈ [ AC ] such that( BB ′ ) bisects ∠ ABC .Applying Pasch’s theorem (3.12) twice for thetriangles AA ′ C and BB ′ C , we get that [ AA ′ ] and[ BB ′ ] intersect. Let I denotes the point of inter-section.Let X , Y and Z be the foot points of I on( BC ), ( CA ) and ( AB ) correspondingly. Apply-ing Proposition 8.10, we get that IY = IZ = IX.
From the same lemma, we get that I lies on the bisector or on the exteriorbisector of ∠ BCA .The line ( CI ) intersects [ BB ′ ], the points B and B ′ lie on oppositesides of ( CI ). Therefore, the angles ICB ′ and ICB have opposite signs.Note that ∠ ICA = ∠ ICB ′ . Therefore, ( CI ) cannot be the exterior bi-sector of ∠ BCA . Hence the result.2
CHAPTER 8. TRIANGLE GEOMETRY
More exercises
Assume that an angle bisector of a nondegenerate tri-angle bisects the opposite side. Show that the triangle is isosceles.
Assume that at one vertex of a nondegenerate trianglethe bisector coincides with the altitude. Show that the triangle is isosce-les.
A BC XY Z
Assume sides [ BC ] , [ CA ] and [ AB ] of △ ABC are tangent to the incircle at X , Y and Z correspondingly. Show that AY = AZ = · ( AB + AC − BC ) . By the definition, the vertexes of orthic triangle are the base pointsof the altitudes of the given triangle.
Prove that the orthocenter of an acute triangle coincideswith the incenter of its orthic triangle.What should be an analog of this statement for an obtuse triangle?
A BC D EF
Assume that the bisector at A ofthe triangle ABC intersects the side [ BC ] at thepoint D ; the line thru D and parallel to ( CA ) in-tersects ( AB ) at the point E ; the line thru E andparallel to ( BC ) intersects ( AC ) at F . Show that AE = F C . hapter 9 Inscribed angles
Angle between a tangent line and a chord
Let Γ be a circle with the center O . Assume the line ( XQ ) is tangent to Γ at X and [ XY ] is a chord of Γ . Then ➊ · ∡ QXY ≡ ∡ XOY.
Equivalently, ∡ QXY ≡ · ∡ XOY or ∡ QXY ≡ · ∡ XOY + π. QXY O Proof.
Note that △ XOY is isosceles.Therefore, ∡ Y XO = ∡ OY X .Let us apply Theorem 7.10 to △ XOY . We get π ≡ ∡ Y XO + ∡ OY X + ∡ XOY ≡≡ · ∡ Y XO + ∡ XOY.
By Lemma 5.16, ( OX ) ⊥ ( XQ ).Therefore, ∡ QXY + ∡ Y XO ≡ ± π . Therefore, 2 · ∡ QXY ≡ π − · ∡ Y XO ≡ ∡ XOY.
CHAPTER 9. INSCRIBED ANGLES
Inscribed angle
We say that a triangle is inscribed in the circle Γ if all its vertices lie on Γ.
Let Γ be a circle with the center O , and X, Y be twodistinct points on Γ . Then △ XP Y is inscribed in Γ if and only if ➋ · ∡ XP Y ≡ ∡ XOY.
Equivalently, if and only if ∡ XP Y ≡ · ∡ XOY or ∡ XP Y ≡ π + · ∡ XOY.P Q XY O
Proof; the “only if ” part.
Let (
P Q ) be the tangent lineto Γ at P . By Theorem 9.1,2 · ∡ QP X ≡ ∡ P OX, · ∡ QP Y ≡ ∡ P OY.
Subtracting one identity from the other, we get ➋ . “If ” part. Assume that ➋ holds for some P / ∈ Γ. Notethat ∡ XOY = 0. Therefore, ∡ XP Y = 0 nor π ; that is, △ P XY isnondegenerate.The line (
P X ) might be tangent to Γ at the point X or intersect Γ atanother point; it the latter case, let P ′ denotes this point of intersection. X YOP X YO P P ′ In the first case, by Theorem 9.1, we have2 · ∡ P XY ≡ ∡ XOY ≡ · ∡ XP Y.
Applying the transversal property (7.6), we get that ( XY ) k ( P Y ), whichis impossible since △ P XY is nondegenerate.In the second case, applying the “if” part and that P , X and P ′ lieon one line (see Exercise 2.11) we get that2 · ∡ XP P ′ ≡ · ∡ XP Y ≡ ∡ XOY ≡ · ∡ XP ′ Y ≡ · ∡ XP ′ P. Again, by transversal property, (
P Y ) k ( P ′ Y ), which is impossible since △ P XY is nondegenerate.5 Y ′ YP XX ′ O Let X , X ′ , Y and Y ′ bedistinct points on the circle Γ . Assume ( XX ′ ) meets ( Y Y ′ ) at the point P . Showthat(a) · ∡ XP Y = ∡ XOY + ∡ X ′ OY ′ ;(b) △ P XY ∼ △
P Y ′ X ′ ;(c) P X · P X ′ = | OP − r | , where O isthe center and r is the radius of Γ . Three chords [ XX ′ ] , [ Y Y ′ ] and [ ZZ ′ ] of the circle Γ intersect at onepoint. Show that XY ′ · ZX ′ · Y Z ′ = X ′ Y · Z ′ X · Y ′ Z. CA A ′ BB ′ Let Γ be a circumcircle of anacute triangle ABC . Let A ′ and B ′ denote thesecond points of intersection of the altitudes from A and B with Γ . Show that △ A ′ B ′ C is isosceles. Recall that the diameter of a circle is its chordwhich passes thru the center. Note that if [ XY ]is the diameter of a circle with center O , then ∡ XOY = π . Hence Theorem 9.2 implies the following: Let Γ be a circle with the diameter [ XY ] . Assume thatthe point P is distinct from X and Y . Then P ∈ Γ if and only if ∠ XP Y is right.
Given four points A , B , A ′ and B ′ , construct a point Z such that both angles AZB and A ′ ZB ′ are right. Let △ ABC be a nondegenerate triangle, A ′ and B ′ befoot points of altitudes from A and B respectfully. Show that the fourpoints A , B , A ′ and B ′ lie on one circle.What is the center of this circle? Assume a line ℓ , a circle with its center on ℓ and a point P / ∈ ℓ are given. Make a ruler-only construction of the perpendicular to ℓ from P . CHAPTER 9. INSCRIBED ANGLES
Inscribed quadrilaterals
A quadrilateral
ABCD is called inscribed if all the points A , B , C and D lie on a circle or a line. The quadrilateral
ABCD is inscribed if and only if ➌ · ∡ ABC + 2 · ∡ CDA ≡ . Equivalently, if and only if ∡ ABC + ∡ CDA ≡ π or ∡ ABC ≡ − ∡ CDA.A B CD
Proof of Theorem 9.10.
Assume △ ABC is degenerate.By Corollary 2.9, 2 · ∡ ABC ≡ · ∡ CDA ≡ D ∈ ( AB ); hence the result follows.It remains to consider the case if △ ABC is nondegenerate.Let O and Γ denote the circulcenter and circumcircle of △ ABC . Ac-cording to Theorem 9.2, ➍ · ∡ ABC ≡ ∡ AOC.
From the same theorem, D ∈ Γ if and only if ➎ · ∡ CDA ≡ ∡ COA.
Adding ➍ and ➎ , we get the result. Let [ XY ] and [ X ′ Y ′ ] be two parallel chords of a circle.Show that XX ′ = Y Y ′ . ABY X X ′ Y ′ Let Γ and Γ ′ be two circleswhich intersect at two distinct points: A and B .Assume [ XY ] and [ X ′ Y ′ ] are the chords of Γ and Γ ′ correspondingly, such that A lies between X and X ′ and B lies between Y and Y ′ . Show that ( XY ) k ( X ′ Y ′ ) . Method of additional circle
Assume that two chords [ AA ′ ] and [ BB ′ ] intersect atthe point P inside their circle. Let X be a point such that both angles XAA ′ and XBB ′ are right. Show that ( XP ) ⊥ ( A ′ B ′ ) . AB A ′ B ′ PX Y
Solution.
Set Y = ( A ′ B ′ ) ∩ ( XP ).Both angles XAA ′ and XBB ′ are right;therefore 2 · ∡ XAA ′ ≡ · ∡ XBB ′ . By Theorem 9.10, (cid:3)
XAP B is inscribed. Ap-plying this theorem again we get that2 · ∡ AXP ≡ · ∡ ABP.
Since (cid:3)
ABA ′ B ′ is inscribed,2 · ∡ ABB ′ ≡ · ∡ AA ′ B ′ . It follows that 2 · ∡ AXY ≡ · ∡ AA ′ Y. By the same theorem (cid:3)
XAY A ′ is inscribed, and therefore,2 · ∡ XAA ′ ≡ · ∡ XY A ′ . Since ∠ XAA ′ is right, so is ∠ XY A ′ . That is ( XP ) ⊥ ( A ′ B ′ ). Find an inaccuracy in the solution of the problem 9.13and try to fix it.
The method used in the solution is called method of additional cir-cle , since the circumcircles of the (cid:3)
XAP B and (cid:3)
XAP B above can beconsidered as additional constructions . LMNO X
Assume three lines ℓ, m and n in-tersect at point O and form six equal angles at O . Let X be a point distinct from O . Let L , M and N denotethe foot points of perpendiculars from X on ℓ, m and n correspondingly. Show that △ LM N is equilateral.
Assume that a point P lies on the circum-circle the triangle ABC . Show that three foot points of P on the lines ( AB ) , ( BC ) and ( CA ) lie on one line (which is called the Simson line of P ). Watch “Why is pi here? And why is it squared? A geo-metric answer to the Basel problem” by Grant Sanderson. (It is availableon YouTube.) CHAPTER 9. INSCRIBED ANGLES
Arcs
A subset of a circle bounded by two points is called a circle arc.More precisely, let Γ be a circle and A , B , C be distinct points on Γ.The subset which includes the points A , C as well as all the points on Γwhich lie with B on the same side of ( AC ) is called circle arc ABC . A BC X
Γ For the circle arc
ABC , the points A and C are called endpoints . There are two circlearcs of Γ with the given endpoints.A half-line [ AX ) is called tangent to thecircle arc ABC at A if the line ( AX ) is tan-gent to Γ, and the points X and B lie on thesame side of the line ( AC ).If B lies on the line ( AC ), then the arc ABC degenerates to one of two following a subsets of the line ( AC ): ⋄ If B lies between A and C , then we define the arc ABC as thesegment [ AC ]. In this case the half-line [ AC ) is tangent to the arc ABC at A . ⋄ If B ∈ ( AC ) \ [ AC ], then we define the arc ABC as the line ( AC )without all the points between A and C . If we choose points X and Y ∈ ( AC ) such that the points X , A , C and Y appear in the sameorder on the line, then the arc ABC is the union of two half-lines in[ AX ) and [ CY ). In this case, the half-line [ AX ) is tangent to thearc ABC at A .In addition, any half-line [ AB ) will be regarded as an arc. This degen-erate arc has only one end point A and it assumed to be tangent to itselfat A . The circle arcs together with the degenerate arcs will be called arcs . A point D lies on the arc ABC if and only if ∡ ADC = ∡ ABC or D coincides with A or C .Proof. If A , B and C lie on one line, then the statement is evident. A BC D
Assume Γ be the circle passing thru A , B and C .Assume D is distinct from A and C . According toTheorem 9.10, D ∈ Γ if and only if ∡ ADC = ∡ ABC or ∡ ADC ≡ ∡ ABC + π. By Exercise 3.13, if the first identity holds, then B and D lie on one side of ( AC ); in this case D belongsto the arc ABC . If the second identity holds, then the points B and9 D lie on opposite sides of ( AC ), in this case D does not belong to thearc ABC . The half-line [ AX ) is tangent to the arc ABC ifand only if ∡ ABC + ∡ CAX ≡ π. Proof.
For a degenerate arc
ABC , the statement is evident. Further weassume the arc
ABC is nondegenerate.Applying theorems 9.1 and 9.2, we get that2 · ∡ ABC + 2 · ∡ CAX ≡ . Therefore, either ∡ ABC + ∡ CAX ≡ π, or ∡ ABC + ∡ CAX ≡ .A BC X Since [ AX ) is the tangent half-line to the arc ABC , X and B lie on the same side of ( AC ). Therefore, theangles CAX , CAB and
ABC have the same sign. Inparticular, ∡ ABC + ∡ CAX
0; that is, we are leftwith the case ∡ ABC + ∡ CAX ≡ π. Assume that the half-lines [ AX ) and [ AY ) are tan-gent to the arcs ABC and
ACB correspondingly. Show that ∠ XAY isstraight.
Show that there is a unique arc with endpoints at thegiven points A and C , which is tangent at A to the given half line [ AX ) . Let [ AX ) be the tangent half-line to an arc ABC . As-sume Y is a point on the arc ABC which is distinct from A . Show that ∡ XAY → as AY → . A B B C Y Y X X Given two circle arcs AB C and AB C , let [ AX ) and [ AX ) be the half-lines tangentto the arcs AB C and AB C at A , and [ CY ) and [ CY ) be the half-lines tangent to the arcs AB C and AB C at C . Show that ∡ X AX ≡ − ∡ Y CY . Given an acute triangle
ABC make a compass-and-ruler construction of the point Z such that ∡ AZB = ∡ BZC = ∡ CZA = ± · π hapter 10 Inversion
Let Ω be the circle with center O and radius r . The inversion of a point P in Ω is the point P ′ ∈ [ OP ) such that OP · OP ′ = r . In this case the circle Ω will be called the circle of inversion and its center O is called the center of inversion .Ω O P P ′ T The inverse of O is undefined.Note that if P is inside Ω, then P ′ is outsideand the other way around. Further, P = P ′ ifand only if P ∈ Ω.Note that the inversion maps P ′ back to P . Let P be a point inside of thecircle Ω centered at O . Let T be a point wherethe perpendicular to ( OP ) from P intersects Ω .Let P ′ be the point where the tangent line to Ω at T intersects ( OP ) .Show that P ′ is the inverse of P in the cir-cle Ω . Let Γ be a circle with the center O . Assume A ′ and B ′ are the inverses of A and B in Γ . Then △ OAB ∼ △ OB ′ A ′ . Moreover, ➊ ∡ AOB ≡ − ∡ B ′ OA ′ , ∡ OBA ≡ − ∡ OA ′ B ′ , ∡ BAO ≡ − ∡ A ′ B ′ O. Proof.
Let r be the radius of the circle of the inversion. AA ′ BB ′ O From the definition of inversion, we get that OA · OA ′ = OB · OB ′ = r . Therefore,
OAOB ′ = OBOA ′ . Clearly, ➋ ∡ AOB = ∡ A ′ OB ′ ≡ − ∡ B ′ OA ′ . From SAS, we get that △ OAB ∼ △ OB ′ A ′ . Applying Theorem 3.7 and ➋ , we get ➊ . Let P ′ be the inverse of P in the circle Γ . Assume that P = P ′ . Show that the value P XP ′ X is the same for all X ∈ Γ . The converse to the exercise above also holds. Namely, given a positivereal number k = 1 and two distinct points P and P ′ the locus of points X such that P XP ′ X = k forms a circle which is called the Apollonian circle .In this case P ′ is inverse of P in the Apollonian circle. Let A ′ , B ′ and C ′ be the images of A , B and C underthe inversion in the incircle of △ ABC . Show that the incenter of △ ABC is the orthocenter of △ A ′ B ′ C ′ . Make a ruler-and-compass construction of the inverseof a given point in a given circle.
Cross-ratio
The following theorem gives some quantities expressed in distances orangles which do not change after inversion.
Let
ABCD and A ′ B ′ C ′ D ′ be two quadrilaterals suchthat the points A ′ , B ′ , C ′ and D ′ are the inverses of A , B , C , and D correspondingly.Then(a) AB · CDBC · DA = A ′ B ′ · C ′ D ′ B ′ C ′ · D ′ A ′ . CHAPTER 10. INVERSION (b) ∡ ABC + ∡ CDA ≡ − ( ∡ A ′ B ′ C ′ + ∡ C ′ D ′ A ′ ) . (c) If the quadrilateral ABCD is inscribed, then so is (cid:3) A ′ B ′ C ′ D ′ .Proof; (a). Let O be the center of the inversion. According to Lemma 10.2, △ AOB ∼ △ B ′ OA ′ . Therefore, ABA ′ B ′ = OAOB ′ . Analogously,
BCB ′ C ′ = OCOB ′ , CDC ′ D ′ = OCOD ′ , DAD ′ A ′ = OAOD ′ . Therefore,
ABA ′ B ′ · B ′ C ′ BC · CDC ′ D ′ · D ′ A ′ DA = OAOB ′ · OB ′ OC · OCOD ′ · OD ′ OA .
Hence (a) follows. (b).
According to Lemma 10.2, ➌ ∡ ABO ≡ − ∡ B ′ A ′ O, ∡ OBC ≡ − ∡ OC ′ B ′ , ∡ CDO ≡ − ∡ D ′ C ′ O, ∡ ODA ≡ − ∡ OA ′ D ′ . By Axiom IIIb, ∡ ABC ≡ ∡ ABO + ∡ OBC, ∡ D ′ C ′ B ′ ≡ ∡ D ′ C ′ O + ∡ OC ′ B ′ , ∡ CDA ≡ ∡ CDO + ∡ ODA, ∡ B ′ A ′ D ′ ≡ ∡ B ′ A ′ O + ∡ OA ′ D ′ . Therefore, summing the four identities in ➌ , we get that ∡ ABC + ∡ CDA ≡ − ( ∡ D ′ C ′ B ′ + ∡ B ′ A ′ D ′ ) . Applying Axiom IIIb and Exercise 7.15, we get that ∡ A ′ B ′ C ′ + ∡ C ′ D ′ A ′ ≡ − ( ∡ B ′ C ′ D ′ + ∡ D ′ A ′ B ′ ) ≡≡ ∡ D ′ C ′ B ′ + ∡ B ′ A ′ D ′ . Hence (b) follows. (c).
Follows from (b) and Theorem 9.10.3
Inversive plane and circlines
Let Ω be a circle with the center O and the radius r . Consider theinversion in Ω.Recall that inverse of O is undefined. To deal with this problem it isuseful to add to the plane an extra point; it will be called the point atinfinity and we will denote it as ∞ . We can assume that ∞ is inverse of O and the other way around.The Euclidean plane with an added point at infinity is called the inversive plane .We will always assume that any line and half-line contains ∞ .It will be convenient to use the notion of circline , which means circleor line ; for instance we may say “if a circline contains ∞ , then it is aline” or “a circline which does not contain ∞ is a circle” .Note that according to Theorem 8.1, for any △ ABC there is a uniquecircline which passes thru A , B and C (if △ ABC is degenerate, then thisis a line and if not it is a circle).
In the inversive plane, inverse of a circline is a cir-cline.Proof.
Let O denotes the center of the inversion.Let Γ be a circline. Choose three distinct points A , B and C on Γ. (If △ ABC is nondegenerate, then Γ is the circumcircle of △ ABC ; if △ ABC is degenerate, then Γ is the line passing thru A , B and C .)Let A ′ , B ′ and C ′ denote the inverses of A , B and C correspondingly.Let Γ ′ be the circline which passes thru A ′ , B ′ and C ′ . According to 8.1,Γ ′ is well defined.Assume D is a point of the inversive plane which is distinct from A , C , O and ∞ . Let D ′ denotes the inverse of D .By Theorem 10.6 c , D ′ ∈ Γ ′ if and only if D ∈ Γ. Hence the result.It remains to prove that O ∈ Γ ⇔ ∞ ∈ Γ ′ and ∞ ∈ Γ ⇔ O ∈ Γ ′ .Since Γ is the inverse of Γ ′ , it is sufficient to prove that ∞ ∈ Γ ⇐⇒ O ∈ Γ ′ . Since ∞ ∈
Γ, we get that Γ is a line. Therefore, for any ε >
0, the lineΓ contains the point P with OP > r /ε . For the inversion P ′ ∈ Γ ′ of P ,we have that OP ′ = r /OP < ε . That is, the circline Γ ′ contains pointsarbitrary close to O . It follows that O ∈ Γ ′ . Q ′ Q Assume that the circle Γ ′ is the inverseof the circle Γ . Let Q denotes the center of Γ and Q ′ denotes the inverse of Q . Show that Q ′ is not the centerof Γ ′ . CHAPTER 10. INVERSION
Assume that a circumtool is a geometric construction tool which pro-duces a circline passing thru any three given points.
Show that with only a circumtool, it is impossible toconstruct the center of a given circle.
Show that for any pair of tangent circles in the inver-sive plane, there is an inversion which sends them to a pair of parallellines.
Consider the inversion of the inversive plane in thecircle Ω with the center O . Then(a) A line passing thru O is inverted into itself.(b) A line not passing thru O is inverted into a circle which passes thru O , and the other way around.(c) A circle not passing thru O is inverted into a circle not passingthru O .Proof. In the proof we use Theorem 10.7 without mentioning it. (a).
Note that if a line passes thru O , it contains both ∞ and O . There-fore, its inverse also contains ∞ and O . In particular, the image is a linepassing thru O . (b). Since any line ℓ passes thru ∞ , its image ℓ ′ has to contain O . If theline does not contain O , then ℓ ′
6∋ ∞ ; that is, ℓ ′ is not a line. Therefore, ℓ ′ is a circle which passes thru O . (c). If the circle Γ does not contain O , then its image Γ ′ does not con-tain ∞ . Therefore, Γ ′ is a circle. Since Γ
6∋ ∞ we get that Γ ′ O . Hencethe result. Method of inversion
Here is one application of inversion, which we include as an illustrationonly; we will not use it further in the book.
Let
ABCD be an inscribed quadrilateral.Assume that the points A , B , C and D appear on the circline in the sameorder. Then AB · CD + BC · DA = AC · BD.
Proof.
Assume the points
A, B, C, D lie on one line in this order.5
A B C Dx y z
Set x = AB , y = BC , z = CD . Note that x · z + y · ( x + y + z ) = ( x + y ) · ( y + z ) . Since AC = x + y , BD = y + z and DA = x + y + z , it proves the identity. A B C DA ′ B ′ C ′ D ′ It remains to consider the case when thequadrilateral
ABCD is inscribed in a circle,say Γ.The identity can be rewritten as AB · DCBD · CA + BC · ADCA · DB = 1 . On the left hand side we have two cross-ratios.According to Theorem 10.6 a , the left hand sidedoes not change if we apply an inversion to eachpoint.Consider an inversion in a circle centered at a point O which lies onΓ between A and D . By Theorem 10.11, this inversion maps Γ to a line.This reduces the problem to the case when A , B , C and D lie on one line,which was already considered.In the proof above, we rewrite Ptolemy’s iden-tity in a form which is invariant with respectto inversion and then apply an inversion whichmakes the statement evident. The solution ofthe following exercise is based on the same idea;one has to apply an inversion with center at A . Assume that four circles are mutually tangent to eachother. Show that there is four (among six) of their points of tangency lieon one circline. AB Assume thatthree circles tangent to each other and to twoparallel lines as shown on the picture.Show that the line passing thru A and B isalso tangent to the two circles at A . Perpendicular circles
Assume two circles Γ and Ω intersect at two points X and Y . Let ℓ and m be the tangent lines at X to Γ and Ω correspondingly. Analogously, ℓ ′ and m ′ be the tangent lines at Y to Γ and Ω.From Exercise 9.23, we get that ℓ ⊥ m if and only if ℓ ′ ⊥ m ′ .6 CHAPTER 10. INVERSION
We say that the circle Γ is perpendicular to the circle Ω (briefly Γ ⊥ Ω)if they intersect and the lines tangent to the circle at one point (andtherefore, both points) of intersection are perpendicular.Similarly, we say that the circle Γ is perpendicular to the line ℓ (brieflyΓ ⊥ ℓ ) if Γ ∩ ℓ = ∅ and ℓ perpendicular to the tangent lines to Γat one point (and therefore, both points) of intersection. According toLemma 5.16, it happens only if the line ℓ passes thru the center of Γ.Now we can talk about perpendicular circlines . Assume Γ and Ω are distinct circles. Then Ω ⊥ Γ ifand only if the circle Γ coincides with its inversion in Ω . XY QO
Proof.
Let Γ ′ denotes the inverse of Γ. “Only if ” part. Let O be the center of Ω and Q bethe center of Γ. Let X and Y denote the points ofintersections of Γ and Ω. According to Lemma 5.16,Γ ⊥ Ω if and only if ( OX ) and ( OY ) are tangentto Γ.Note that Γ ′ is also tangent to ( OX ) and ( OY )at X and Y correspondingly. It follows that X and Y are the foot pointsof the center of Γ ′ on ( OX ) and ( OY ). Therefore, both Γ ′ and Γ havethe center Q . Finally, Γ ′ = Γ, since both circles pass thru X . “If ” part. Assume Γ = Γ ′ .Since Γ = Ω, there is a point P which lies on Γ, but not on Ω. Let P ′ be the inverse of P in Ω. Since Γ = Γ ′ , we have that P ′ ∈ Γ. Inparticular, the half-line [ OP ) intersects Γ at two points. By Exercise5.12, O lies outside of Γ.As Γ has points inside and outside of Ω, the circles Γ and Ω intersect.The latter follows from Exercise 3.20.Let X be a point of their intersection. We need to show that ( OX ) istangent to Γ; that is, X is the only intersection point of ( OX ) and Γ.Assume Z is another point of intersection. Since O is outside of Γ, thepoint Z lies on the half-line [ OX ).Let Z ′ denotes the inverse of Z in Ω. Clearly, the three points Z, Z ′ , X lie on Γ and ( OX ). The latter contradicts Lemma 5.14.It is convenient to define the inversion in the line ℓ as the reflectionin ℓ . This way we can talk about inversion in an arbitrary circline . Let Ω and Γ be distinct circlines in the inversiveplane. Then the inversion in Ω sends Γ to itself if and only if Ω ⊥ Γ .Proof. By Theorem 10.15, it is sufficient to consider the case when Ω orΓ is a line.7Assume Ω is a line, so the inversion in Ω is a reflection. In this casethe statement follows from Corollary 5.7.If Γ is a line, then the statement follows from Theorem 10.11.
Let P and P ′ be two distinct points such that P ′ isthe inverse of P in the circle Ω . Assume that the circline Γ passes thru P and P ′ . Then Γ ⊥ Ω .Proof. Without loss of generality, we may assume that P is inside and P ′ is outside Ω. By Theorem 3.17, Γ intersects Ω. Let A denotes a point ofintersection.Let Γ ′ denotes the inverse of Γ. Since A is a self-inverse, the points A , P and P ′ lie on Γ ′ . By Exercise 8.2, Γ ′ = Γ and by Theorem 10.15,Γ ⊥ Ω. Let P and Q be two distinct points inside the circle Ω . Then there is a unique circline Γ perpendicular to Ω , which passesthru P and Q .Proof. Let P ′ be the inverse of the point P in the circle Ω. According toCorollary 10.17, the circline is passing thru P and Q is perpendicular toΩ if and only if it passes thru P ′ .Note that P ′ lies outside of Ω. Therefore, the points P , P ′ and Q aredistinct.According to Exercise 8.2, there is a unique circline passing thru P , Q and P ′ . Hence the result. Let Ω and Ω be two distinct circles in the Euclideanplane. Assume that the point P does not lie on Ω nor on Ω . Showthat there is a unique circline passing thru P which is perpendicular to Ω and Ω . Let P , Q , P ′ and Q ′ be points in the Euclidean plane.Assume P ′ and Q ′ are inverses of P and Q correspondingly. Show thatthe quadrilateral P QP ′ Q ′ is inscribed. Let Ω and Ω be two perpendicular circles with cen-ters at O and O correspondingly. Show that the inverse of O in Ω coincides with the inverse of O in Ω . Three distinct circles — Ω , Ω and Ω , intersect attwo points — A and B . Assume that a circle Γ is perpendicular to Ω and Ω . Show that Γ ⊥ Ω . CHAPTER 10. INVERSION
Assume you have two construction tools: the circum-tool which constructs a circline thru three given points, and a tool whichconstructs an inverse of a given point in a given circle.Assume that a point P does not lie on the two circles Ω , Ω . Usingonly the two given tools, construct a circline Γ that passes thru P , andperpendicular to both Ω and Ω . Angles after inversion
In the inversive plane, the inverse of an arc is anarc.Proof.
Consider four distinct points A , B , C and D ; let A ′ , B ′ , C ′ and D ′ be their inverses. We need to show that D lies on the arc ABC ifand only if D ′ lies on the arc A ′ B ′ C ′ . According to Proposition 9.18, thelatter is equivalent to the following: ∡ ADC = ∡ ABC ⇐⇒ ∡ A ′ D ′ C ′ = ∡ A ′ B ′ C ′ . The latter follows from Theorem 10.6 b .The following theorem states that the angle between arcs changes onlyits sign after the inversion. AA ′ B B ′ C C ′ X Y B C B ′ C ′ X Y Let AB C , AB C be two arcs in the inversive plane,and the arcs A ′ B ′ C ′ , A ′ B ′ C ′ be their inverses. Let [ AX ) and [ AX ) bethe half-lines tangent to AB C and AB C at A , and [ A ′ Y ) and [ A ′ Y ) be the half-lines tangent to A ′ B ′ C ′ and A ′ B ′ C ′ at A ′ . Then ∡ X AX ≡ − ∡ Y A ′ Y . angle between arcs can be defined as the angle between its tangenthalf-lines at the common endpoint. Therefore under inversion, the anglesbetween arcs are preserved up to sign. P A P From Exercise 5.23, it follows that the angle be-tween arcs with common endpoint A is the limit of ∡ P AP where P and P are points approaching A along the corresponding arcs. This observation can beused to define angle between a pair of curves emerg-ing from one point. It turns out that under inversion,angles between curves are also preserved up to sign. Proof.
Applying to Proposition 9.19, ∡ X AX ≡ ∡ X AC + ∡ C AC + ∡ C AX ≡≡ ( π − ∡ C B A ) + ∡ C AC + ( π − ∡ AB C ) ≡≡ − ( ∡ C B A + ∡ AB C + ∡ C AC ) ≡≡ − ( ∡ C B A + ∡ AB C ) − ( ∡ C B C + ∡ C AC ) . The same way we get that ∡ Y A ′ Y ≡ − ( ∡ C ′ B ′ A ′ + ∡ A ′ B ′ C ′ ) − ( ∡ C ′ B ′ C ′ + ∡ C ′ A ′ C ′ ) . By Theorem 10.6 b , ∡ C B A + ∡ AB C ≡ − ( ∡ C ′ B ′ A ′ + ∡ A ′ B ′ C ′ ) , ∡ C B C + ∡ C AC ≡ − ( ∡ C ′ B ′ C ′ + ∡ C ′ A ′ C ′ )and hence the result. Let P be the inverse of point Q in a circle Γ . Assumethat P ′ , Q ′ and Γ ′ are the inverses of P , Q and Γ in another circle Ω .Then P ′ is the inverse of Q ′ in Γ ′ . PQ P ′ Q ′ Γ Γ ′ Ω Proof. If P = Q , then P ′ = Q ′ ∈ Γ ′ .Therefore, P ′ is the inverse of Q ′ in Γ ′ .It remains to consider the case P = Q . Let ∆ and ∆ be two distinct cir-cles that intersect at P and Q . Ac-cording to Corollary 10.17, ∆ ⊥ Γ and∆ ⊥ Γ.Let ∆ ′ and ∆ ′ denote the inversesof ∆ and ∆ in Ω. Clearly, ∆ ′ meets∆ ′ at P ′ and Q ′ .From Theorem 10.25, the latter is equivalent to ∆ ′ ⊥ Γ ′ and ∆ ′ ⊥ Γ ′ .By Corollary 10.16, P ′ is the inverse of Q ′ in Γ ′ . hapter 11 Neutral plane
Let us remove Axiom V from our axiomatic system, see page 20. Thisway we define a new object called neutral plane or absolute plane . (In aneutral plane, the Axiom V may or may not hold.)Clearly, any theorem in neutral geometry holds in Euclidean geometry.In other words, the Euclidean plane is an example of a neutral plane. Inthe next chapter we will construct an example of a neutral plane that isnot Euclidean.In this book, the Axiom V was used for the first time in the proof ofuniqueness of parallel lines in Theorem 7.2. Therefore, all the statementsbefore Theorem 7.2 also hold in neutral geometry.It makes all the discussed results about half-planes, signs of angles,congruence conditions, perpendicular lines and reflections true in neutralgeometry. Recall that a statement above marked with “ X ”, for example“ Theorem. X ” if it holds in any neutral plane and the same proof works.Let us give an example of a theorem in neutral geometry, which admitsa simpler proof in Euclidean geometry. Assume that trian-gles
ABC and A ′ B ′ C ′ have right angles at C and C ′ correspondingly, AB = A ′ B ′ and AC = A ′ C ′ . Then △ ABC ∼ = △ A ′ B ′ C ′ .Euclidean proof. By the Pythagorean theorem BC = B ′ C ′ . Then thestatement follows from the SSS congruence condition.Note that the proof of the Pythagorean theorem used properties ofsimilar triangles, which in turn used Axiom V. Therefore this proof doesnot work in a neutral plane. 801 ABCD
Neutral proof.
Let D denotes the reflection of A in( BC ) and D ′ denotes the reflection of A ′ in ( B ′ C ′ ).Note that AD = 2 · AC = 2 · A ′ C ′ = A ′ D ′ ,BD = BA = B ′ A ′ = B ′ D ′ . By SSS congruence condition 4.4, we get that △ ABD ∼ = △ A ′ B ′ D ′ .The theorem follows, since C is the midpoint of [ AD ] and C ′ is themidpoint of [ A ′ D ′ ]. Give a proof of Exercise 8.12 which works in the neutralplane.
Let
ABCD be an inscribed quadrilateral in the neutralplane. Show that ∡ ABC + ∡ CDA ≡ ∡ BCD + ∡ DAB.
One cannot use the Theorem 9.10 to solve the exercise above, since ituses Theorems 9.1 and 9.2, which in turn uses Theorem 7.10.
Two angles of a triangle
In this section we will prove a weaker form of Theorem 7.10 which holdsin any neutral plane.
Let △ ABC be a nondegenerate triangle in the neu-tral plane. Then | ∡ CAB | + | ∡ ABC | < π. Note that according to 3.7, the angles
ABC , BCA and
CAB havethe same sign. Therefore, in the Euclidean plane the theorem followsimmediately from Theorem 7.10.
ABC X
In neutral geometry, we need to apply Lemma 7.7directly.
Proof.
By Lemma 7.7 there is a point X that lies with B on the same side of ( AC ) such that ∡ BAX = ∡ ABC .In particular, ➊ ∡ CAX ≡ ∡ CAB + ∡ ABC. CHAPTER 11. NEUTRAL PLANE
By 3.7, the angles
CAB , ABC have the same sign and since X and B lie on the same side of ( AC ), we also have that the angles CAB and
ABX have the same sign. Note that ∡ CAX π , otherwise X would lieon ( AC ). Therefore the identity ➊ implies | ∡ CAB | + | ∡ ABC | = | ∡ CAX | < π. Assume A , B , C and D are points in a neutral planesuch that · ∡ ABC + 2 · ∡ BCD ≡ . Show that ( AB ) k ( CD ) . Note that one cannot apply the transversal property (7.6).
Prove the side-angle-angle congruence condition in theneutral geometry.In other words, let △ ABC and △ A ′ B ′ C ′ be two triangles in a neutralplane. Show that △ ABC ∼ = △ A ′ B ′ C ′ if AB = A ′ B ′ , ∡ ABC = ± ∡ A ′ B ′ C ′ and ∡ BCA = ± ∡ B ′ C ′ A ′ . Note that in the Euclidean plane, the above exercise follows fromASA and the theorem on the sum of angles of a triangle (7.10). However,Theorem 7.10 cannot be used here, since its proof uses Axiom V. Later(Theorem 13.7) we will show that Theorem 7.10 does not hold in a neutralplane.
Assume that the point D lies between the vertices A and B of △ ABC in a neutral plane. Show that
CD < CA or CD < CB.
Three angles of triangle
Let △ ABC and △ A ′ B ′ C ′ be two triangles in theneutral plane such that AC = A ′ C ′ and BC = B ′ C ′ . Then AB < A ′ B ′ if and only if | ∡ ACB | < | ∡ A ′ C ′ B ′ | . AC BB ′ X Proof.
Without loss of generality, we mayassume that A = A ′ and C = C ′ and ∡ ACB, ∡ ACB ′ >
0. In this case we needto show that
AB < AB ′ ⇐⇒ ∡ ACB < ∡ ACB ′ . Choose a point X so that ∡ ACX = · ( ∡ ACB + ∡ ACB ′ ) . Note that ⋄ ( CX ) bisects ∠ BCB ′ . ⋄ ( CX ) is the perpendicular bisector of [ BB ′ ]. ⋄ A and B lie on the same side of ( CX ) if and only if ∡ ACB < ∡ ACB ′ . From Exercise 5.3, A and B lie on the same side of ( CX ) if and only if AB < AB ′ . Hence the result. Let △ ABC be a triangle in the neutral plane. Then | ∡ ABC | + | ∡ BCA | + | ∡ CAB | π. The following proof is due to Legendre [12], earlier proofs were due toSaccheri [17] and Lambert [11].
Proof.
Let △ ABC be the given triangle. Set a = BC, b = CA, c = AB,α = ∡ CAB, β = ∡ ABC, γ = ∡ BCA.
Without loss of generality, we may assume that α, β, γ > B A A . . . A n C C . . . C n c c c ca b d a b d a b d a bαβγ δ αβγ Fix a positive integer n . Consider the points A , A , . . . , A n on thehalf-line [ BA ), such that BA i = i · c for each i . (In particular, A == B and A = A .) Let us construct the points C , C , . . . , C n , so that ∡ A i A i − C i = β and A i − C i = a for each i .4 CHAPTER 11. NEUTRAL PLANE
By SAS, we have constructed n congruent triangles △ ABC = △ A A C ∼ = ∼ = △ A A C ∼ = . . . ∼ = △ A n A n − C n . Set d = C C and δ = ∡ C A C . Note that ➋ α + β + δ = π. By Proposition 11.4, we get that δ > △ A C C ∼ = △ A C C ∼ = . . . ∼ = △ A n − C n − C n . In particular, C i C i +1 = d for each i .By repeated application of the triangle inequality, we get that n · c = A A n A C + C C + · · · + C n − C n + C n A n == a + ( n − · d + b. In particular, c d + n · ( a + b − d ) . Since n is arbitrary positive integer, the latter implies c d. From Proposition 11.8 and SAS, the latter is equivalent to γ δ. From ➋ , the theorem follows.The defect of triangle △ ABC is defined asdefect( △ ABC ) := π − | ∡ ABC | − | ∡ BCA | − | ∡ CAB | . Note that Theorem 11.9 sates that, the defect of any triangle in aneutral plane has to be nonnegative. According to Theorem 7.10, anytriangle in the Euclidean plane has zero defect.5
A C BD
Let △ ABC be a nondegenerate trianglein the neutral plane. Assume D lies between A and B .Show that defect( △ ABC ) = defect( △ ADC ) + defect( △ DBC ) . How to prove that somethingcannot be proved
Many attempts were made to prove that any theorem in Euclidean geom-etry holds in neutral geometry. The latter is equivalent to the statementthat Axiom V is a theorem in neutral geometry.Some of these attempts were accepted as proofs for long periods oftime, until a mistake was found.There is a number of statements in neutral geometry which are equiv-alent to the Axiom V. It means that if we exchange the Axiom V to anyof these statements, then we will obtain an equivalent axiomatic system.The following theorem provides a short list of such statements. Weare not going to prove it in the book.
A neutral plane is Euclidean if and only if one of thefollowing equivalent conditions holds:(a) There is a line ℓ and a point P / ∈ ℓ such that there is only one linepassing thru P and parallel to ℓ .(b) Every nondegenerate triangle can be circumscribed.(c) There exists a pair of distinct lines that lie on a bounded distancefrom each other.(d) There is a triangle with an arbitrary large inradius.(e) There is a nondegenerate triangle with zero defect.(f ) There exists a quadrilateral in which all the angles are right. It is hard to imagine a neutral plane that does not satisfy some of theproperties above. That is partly the reason for the large number of falseproofs; each used one of such statements by accident.Let us formulate the negation of (a) above as a new axiom; we labelit h-V as a hyperbolic version of Axiom V on page 20.h-V. For any line ℓ and any point P / ∈ ℓ there are at least two linesthat pass thru P and parallel to ℓ .6 CHAPTER 11. NEUTRAL PLANE
By Theorem 7.2, a neutral plane that satisfies Axiom h-V is not Eu-clidean. Moreover, according to the Theorem 11.11 (which we do notprove) in any non-Euclidean neutral plane, Axiom h-V holds.It opens a way to look for a proof by contradiction. Simply exchangeAxiom V to Axiom h-V and start to prove theorems in the obtained ax-iomatic system. In the case if we arrive to a contradiction, we prove theAxiom V in neutral plane. This idea was growing since the 5th century;the most notable results were obtained by Saccheri in [17].The system of axioms I–IV and h-V defines a new geometry which isnow called hyperbolic or Lobachevskian geometry . The more this geome-try was developed, it became more and more believable that there is nocontradiction; that is, the system of axioms I–IV and h-V is consistent .In fact, the following theorem holds true:
The hyperbolic geometry is consistent if and only ifso is the Euclidean geometry.
The claims that hyperbolic geometry has no contradiction can befound in private letters of Gauss, Schweikart and Taurinus. They allseem to be afraid to state it in public. For instance, in 1818 Gauss writesto Gerling: . . . I am happy that you have the courage to express yourselfas if you recognized the possibility that our parallels theoryalong with our entire geometry could be false. But the waspswhose nest you disturb will fly around your head.
Lobachevsky came to the same conclusion independently. Unlike theothers, he had the courage to state it in public and in print (see [13]). Thatcost him serious troubles. A couple of years later, also independently,Bolyai published his work (see [7]).It seems that Lobachevsky was the first who had a proof of Theo-rem 11.12 altho its formulation required rigorous axiomatics, which werenot developed at his time. Later, Beltrami gave a cleaner proof of the“if” part of the theorem. It was done by modeling points, lines, distancesand angle measures of one geometry using some other objects in anothergeometry. The same idea was used earlier by Lobachevsky; in [14, 34] hemodeled the Euclidean plane in the hyperbolic space.The proof of Beltrami is the subject of the next chapter.The discovery of hyperbolic geometry was one of the main scientificdiscoveries of the 19th century, on the same level are Mendel’s laws andthe law of multiple proportions. The oldest surviving letters were the Gauss letter to Gerling in 1816 and yet moreconvincing letter dated 1818 of Schweikart sent to Gauss via Gerling. Curvature
In a letter from 1824 Gauss writes:
The assumption that the sum of the three angles is lessthan π leads to a curious geometry, quite different from oursbut completely consistent, which I have developed to my en-tire satisfaction, so that I can solve every problem in it withthe exception of a determination of a constant, which cannotbe designated a priori. The greater one takes this constant,the nearer one comes to Euclidean geometry, and when it ischosen indefinitely large the two coincide. The theorems ofthis geometry appear to be paradoxical and, to the uninitiated,absurd; but calm, steady reflection reveals that they containnothing at all impossible. For example, the three angles of atriangle become as small as one wishes, if only the sides aretaken large enuf; yet the area of the triangle can never exceeda definite limit, regardless how great the sides are taken, norindeed can it ever reach it. In modern terminology, the constant which Gauss mentions, can beexpressed as 1 / √− k , where k
0, is the so called curvature of the neutralplane, which we are about to introduce.The identity in Exercise 11.10 suggests that the defect of a triangleshould be proportional to its area. In fact, for any neutral plane, there is a nonpositive real number k such that k · area( △ ABC ) + defect( △ ABC ) = 0for any △ ABC . This number k is called the curvature of the plane.For example, by Theorem 7.10, the Euclidean plane has zero curvature.By Theorem 11.9, the curvature of any neutral plane is nonpositive.It turns out that up to isometry, the neutral plane is characterized byits curvature; that is, two neutral planes are isometric if and only if theyhave the same curvature.In the next chapter, we will construct a hyperbolic plane ; this is, anexample of neutral plane with curvature k = − c , the area changes by factor c , while the defect staysthe same. Therefore, taking c = √− k , we can get the neutral plane ofthe given curvature k <
0. In other words, all the non-Euclidean neutralplanes become identical if we use r = 1 / √− k as the unit of length. The area in the neutral plane is discussed briefly in the end of Chapter 20, but thereader could also refer to an intuitive understanding of area measurement. CHAPTER 11. NEUTRAL PLANE
In Chapter 16, we discuss the geometry of the unit sphere. Althospheres are not neutral planes, the spherical geometry is a close relativeof Euclidean and hyperbolic geometries.Nondegenerate spherical triangles have negative defects. Moreover, if R is the radius of the sphere, then R · area( △ ABC ) + defect( △ ABC ) = 0for any spherical triangle
ABC . In other words, the sphere of the radius R has the curvature k = R . hapter 12 Hyperbolic plane
In this chapter, we use inversive geometry to construct the model ofhyperbolic plane — a neutral plane which is not Euclidean.Namely, we construct the so called conformal disc model of the hy-perbolic plane. This model was discovered by Beltrami in [5] and oftencalled the
Poincar´e disc model .The figure above shows the conformal disc model of the hyperbolicplane which is cut into congruent triangles with angles π , π and π .890 CHAPTER 12. HYPERBOLIC PLANE
Conformal disc model
In this section, we give new names for some objects in the Euclidean planewhich will represent lines, angle measures and distances in the hyperbolicplane.
Hyperbolic plane.
Let us fix a circle on the Euclidean plane and call it absolute . The set of points inside the absolute will be called the hyperbolicplane (or h-plane ).Note that the points on the absolute do not belong to the h-plane.The points in the h-plane will be also called h-points .Often we will assume that the absolute is a unit circle.
Hyperbolic lines.
The intersections of the h-plane with circlines per-pendicular to the absolute are called hyperbolic lines or h-lines . PQ AB
Γh-plane By Corollary 10.18, there is aunique h-line which passes thruthe given two distinct h-points P and Q . This h-line will be de-noted by ( P Q ) h .The arcs of hyperbolic lineswill be called hyperbolic seg-ments or h-segments . An h-segment with endpoints P and Q will be denoted by [ P Q ] h .The subset of an h-line onone side from a point will becalled a hyperbolic half-line (or h-half-line ). More precisely, an h-half-line is an intersection of the h-plane with arc which perpendicular to theabsolute with only one endpoint in the h-plane. An h-half-line startingat P and passing thru Q will be denoted by [ P Q ) h .If Γ is the circle containing the h-line ( P Q ) h , then the points of in-tersection of Γ with the absolute are called ideal points of ( P Q ) h . (Notethat the ideal points of an h-line do not belong to the h-line.)An ordered triple of h-points, say ( P, Q, R ) will be called h-triangle
P QR and denoted by △ h P QR .So far, (
P Q ) h is just a subset of the h-plane; below we will introduceh-distance and later we will show that ( P Q ) h is a line for the h-distancein the sense of the Definition 1.8. Show that an h-line is uniquely determined by its idealpoints.
Show that an h-line is uniquely determined by one ofits ideal points and one h-point on it. Show that the h-segment [ P Q ] h coincides with the Eu-clidean segment [ P Q ] if and only if the line ( P Q ) pass thru the center ofthe absolute. Hyperbolic distance.
Let P and Q be distinct h-points; let A and B denote the ideal points of ( P Q ) h . Without loss of generality, we mayassume that on the Euclidean circle containing the h-line ( P Q ) h , thepoints A, P, Q, B appear in the same order.Consider the function δ ( P, Q ) := AQ · P BAP · QB .
Note that the right hand side is a cross-ratio; by Theorem 10.6 it isinvariant under inversion. Set δ ( P, P ) = 1 for any h-point P . Let usdefine h-distance as the logarithm of δ ; that is, P Q h := ln[ δ ( P, Q )] . The proof that
P Q h is a metric on the h-plane will be given below.For now it is just a function which returns a real value P Q h for any pairof h-points P and Q . Let O be the center of the absolute and the h-points O , X and Y lie on one h-line in the same order. Assume OX = XY . Provethat OX h < XY h . Hyperbolic angles.
Consider three h-points P , Q and R such that P = Q and R = Q . The hyperbolic angle P QR (briefly ∠ h P QR ) is anordered pair of h-half-lines [ QP ) h and [ QR ) h .Let [ QX ) and [ QY ) be (Euclidean) half-lines which are tangent to[ QP ] h and [ QR ] h at Q . Then the hyperbolic angle measure (or h-anglemeasure ) of ∠ h P QR denoted by ∡ h P QR and defined as ∡ XQY . Let ℓ be an h-line and P be an h-point which does not lieon ℓ . Show that there is a unique h-line passing thru P and perpendicularto ℓ . Plan of the proof
We defined all the h-notions needed in the formulation of the axioms I–IVand h-V. It remains to show that all these axioms hold; this will be doneby the end of this chapter.Once we are done with the proofs, we get that the model provides anexample of a neutral plane; in particular, Exercise 12.5 can be proved thesame way as Theorem 5.5.2
CHAPTER 12. HYPERBOLIC PLANE
Most importantly we will prove the “if”-part of Theorem 11.12.Indeed, any statement in hyperbolic geometry can be restated in theEuclidean plane using the introduced h-notions. Therefore, if the systemof axioms I–IV and h-V leads to a contradiction, then so does the systemaxioms I–V.
Auxiliary statements
Consider an h-plane with a unit circle as the absolute.Let O be the center of the absolute and P be another h-point. Let P ′ denotes the inverse of P in the absolute.Then the circle Γ with the center P ′ and radius √ − OP OP is perpendic-ular to the absolute. Moreover, O is the inverse of P in Γ . Γ O P P ′ T Proof.
Follows from Exercise 10.21.Assume Γ is a circline which is perpen-dicular to the absolute. Consider the in-version X X ′ in Γ, or if Γ is a line, set X X ′ to be the reflection in Γ.The following observation says that themap X X ′ respects all the notions in-troduced in the previous section. Togetherwith the lemma above, it implies that in any problem which is formulatedentirely in h-terms we can assume that a given h-point lies in the centerof the absolute. The map X X ′ described above is a bi-jection of the h-plane to itself. Moreover, for any h-points P , Q , R suchthat P = Q and Q = R , the following conditions hold:(a) The h-line ( P Q ) h , h-half-line [ P Q ) h and h-segment [ P Q ] h are trans-formed into ( P ′ Q ′ ) h , [ P ′ Q ′ ) h and [ P ′ Q ′ ] h correspondingly.(b) δ ( P ′ , Q ′ ) = δ ( P, Q ) and P ′ Q ′ h = P Q h .(c) ∡ h P ′ Q ′ R ′ ≡ − ∡ h P QR . It is instructive to compare this observation with Proposition 5.6.
Proof.
According to Theorem 10.15, the map sends the absolute to itself.Note that the points on Γ do not move, it follows that points inside of theabsolute remain inside after the mapping and the other way around.Part (a) follows from 10.7 and 10.25.Part (b) follows from Theorem 10.6.Part (c) follows from Theorem 10.25.3
Let Γ be a circle which is perpendicular to the absoluteand let Q be an h-point lying on Γ . Assume P is an h-point and P ′ is itsinversion in Γ . Show that P Q h = P ′ Q h . Assume that the absolute is a unit circle centered at O .Given an h-point P , set x = OP and y = OP h . Then y = ln 1 + x − x and x = e y − e y + 1 .A O P B Proof.
Note that the h-line ( OP ) h forms adiameter of the absolute. If A and B are theideal points as in the definition of h-distance,then OA = OB = 1 ,P A = 1 + x,P B = 1 − x. In particular, y = ln AP · BOP B · OA = ln 1 + x − x . Taking the exponential function of the left and the right hand sideand applying obvious algebra manipulations, we get that x = e y − e y + 1 . Assume the points P , Q and R appear on one h-linein the same order. Then P Q h + QR h = P R h . Proof.
Note that
P Q h + QR h = P R h is equivalent to ➊ δ ( P, Q ) · δ ( Q, R ) = δ ( P, R ) . CHAPTER 12. HYPERBOLIC PLANE
Let A and B be the ideal points of ( P Q ) h . Without loss of generality,we can assume that the points A , P , Q , R and B appear in the sameorder on the circline containing ( P Q ) h . Then δ ( P, Q ) · δ ( Q, R ) = AQ · BPQB · P A · AR · BQRB · QA == AR · BPRB · P A == δ ( P, R ) . Hence ➊ follows.Let P be an h-point and ρ >
0. The set of all h-points Q such that P Q h = ρ is called an h-circle with the center P and the h-radius ρ . Any h-circle is a Euclidean circle which lies completelyin the h-plane.More precisely for any h-point P and ρ > there is a ˆ ρ > and apoint ˆ P such that P Q h = ρ ⇐⇒ ˆ P Q = ˆ ρ for any h-point Q .Moreover, if O is the center of the absolute, then1. ˆ O = O for any ρ and2. ˆ P ∈ ( OP ) for any P = O . O Q P ˆ P ∆ ′ ρ Proof.
According to Lemma 12.9, OQ h == ρ if and only if OQ = ˆ ρ = e ρ − e ρ + 1 . Therefore, the locus of h-points Q suchthat OQ h = ρ is a Euclidean circle, de-note it by ∆ ρ .If P = O , applying Lemma 12.6 andthe main observation (12.7) we get a circleΓ perpendicular to the absolute such that P is the inverse of O in Γ.Let ∆ ′ ρ be the inverse of ∆ ρ in Γ. Since the inversion in Γ preservesthe h-distance, P Q h = ρ if and only if Q ∈ ∆ ′ ρ .According to Theorem 10.7, ∆ ′ ρ is a Euclidean circle. Let ˆ P and ˆ ρ denote the Euclidean center and radius of ∆ ′ ρ .Finally, note that ∆ ′ ρ reflects to itself in ( OP ); that is, the center ˆ P lies on ( OP ). Assume P , ˆ P and O are as in the Lemma 12.11 and P = O . Show that ˆ P ∈ [ OP ] . Axiom I
Evidently, the h-plane contains at least two points. Therefore, to showthat Axiom I holds in the h-plane, we need to show that the h-distancedefined on page 91 is a metric on h-plane; that is, the conditions (a) – (d) in Definition 1.1 hold for h-distance.The following claim says that the h-distance meets the conditions (a) and (b) . Given the h-points P and Q , we have P Q h > and P Q h = 0 if and only if P = Q .Proof. According to Lemma 12.6 and the main observation (12.7), wemay assume that Q is the center of the absolute. In this case δ ( Q, P ) = 1 + QP − QP > QP h = ln[ δ ( Q, P )] > . Moreover, the equalities holds if and only if P = Q .The following claim says that the h-distance meets the condition 1.1 c . For any h-points P and Q , we have P Q h = QP h .Proof. Let A and B be ideal points of ( P Q ) h and A, P, Q, B appear onthe circline containing (
P Q ) h in the same order. PQ AB
Then
P Q h = ln AQ · BPQB · P A == ln BP · AQP A · QB == QP h . The following claim shows, in particular, that thetriangle inequality (which is condition 1.1 d ) holds for h -distance. Given a triple of h-points P , Q and R , we have P Q h + QR h > P R h . Moreover, the equality holds if and only if P , Q and R lie on one h-linein the same order. CHAPTER 12. HYPERBOLIC PLANE
Proof.
Without loss of generality, we may assume that P is the center ofthe absolute and P Q h > QR h > Q and h-radius ρ = QR h .Let S and T be the points of intersection of ( P Q ) and ∆.By Lemma 12.10,
P Q h > QR h . Therefore, we can assume that thepoints P , S , Q and T appear on the h-line in the same order.According to Lemma 12.11, ∆ is a Euclidean circle; let ˆ Q denotes itsEuclidean center. Note that ˆ Q is the Euclidean midpoint of [ ST ]. P Q ˆ Q ∆ S TR
By the Euclidean triangle inequality ➋ P T = P ˆ Q + ˆ QR > P R and the equality holds if and only if T = R .By Lemma 12.9, P T h = ln 1 + P T − P T ,P R h = ln 1 + P R − P R .
Since the function f ( x ) = ln x − x is increasing for x ∈ [0 , ➋ implies P T h > P R h and the equality holds if and only if T = R .Finally, applying Lemma 12.10 again, we get that P T h = P Q h + QR h . Hence the claim follows.
Axiom II
Note that once the following claim is proved, Axiom II follows from Corol-lary 10.18.
A subset of the h-plane is an h-line if and only if itforms a line for the h-distance in the sense of Definition 1.8.Proof.
Let ℓ be an h-line. Applying the main observation (12.7) we canassume that ℓ contains the center of the absolute. In this case, ℓ is anintersection of a diameter of the absolute and the h-plane. Let A and B be the endpoints of the diameter.Consider the map ι : ℓ → R defined as ι ( X ) = ln AXXB . ι : ℓ → R is a bijection.Further, if X, Y ∈ ℓ and the points A , X , Y and B appear on [ AB ] inthe same order, then ι ( Y ) − ι ( X ) = ln AYY B − ln AXXB = ln AY · BXY B · XB = XY h . We proved that any h-line is a line for h-distance. The converse followsfrom Claim 12.15.
Axiom III
Note that the first part of Axiom III follows directly from the definitionof the h-angle measure defined on page 91. It remains to show that ∡ h satisfies the conditions IIIa, IIIb and IIIc on page 20.The following two claims say that ∡ h satisfies IIIa and IIIb. Given an h-half-line [ OP ) h and α ∈ ( − π, π ] , there is aunique h-half-line [ OQ ) h such that ∡ h P OQ = α . For any h-points P , Q and R distinct from an h-point O , we have ∡ h P OQ + ∡ h QOR ≡ ∡ h P OR.
Proof of 12.17 and 12.18.
Applying the main observation, we may assumethat O is the center of the absolute. In this case, for any h-point P = O , [ OP ) h is the intersection of [ OP ) with h-plane. Hence the claims12.17 and 12.18 follow from the corresponding axioms of the Euclideanplane.The following claim says that ∡ h satisfies IIIc. The function ∡ h : ( P, Q, R ) ∡ h P QR is continuous at any triple of points ( P, Q, R ) such that Q = P , Q = R and ∡ h P QR = π .Proof. Let O denotes the center of the absolute. We can assume that Q is distinct from O .Let Z denotes the inverse of Q in the absolute; let Γ denotes thecircle perpendicular to the absolute which is centered at Z . According toLemma 12.6, the point O is the inverse of Q in Γ.8 CHAPTER 12. HYPERBOLIC PLANE
Let P ′ and R ′ denote the inversions in Γ of the points P and R cor-respondingly. Note that the point P ′ is completely determined by thepoints Q and P . Moreover, the map ( Q, P ) P ′ is continuous at anypair of points ( Q, P ) such that Q = O . The same is true for the map( Q, R ) R ′ According to the main observation ∡ h P QR ≡ − ∡ h P ′ OR ′ . Since ∡ h P ′ OR ′ = ∡ P ′ OR ′ and the maps ( Q, P ) P ′ , ( Q, R ) R ′ are continuous, the claim follows from the corresponding axiom of theEuclidean plane. Axiom IV
The following claim says that Axiom IV holds in the h-plane.
In the h-plane, we have △ h P QR ∼ = △ h P ′ Q ′ R ′ if andonly if Q ′ P ′ h = QP h , Q ′ R ′ h = QR h and ∡ h P ′ Q ′ R ′ = ± ∡ P QR.
Proof.
Applying the main observation, we can assume that both Q and Q ′ coincide with the center of the absolute. In this case ∡ P ′ QR ′ = ∡ h P ′ QR ′ = ± ∡ h P QR = ± ∡ P QR.
Since QP h = QP ′ h and QR h = QR ′ h , Lemma 12.9 implies that the same holds for the Euclidean distances; thatis, QP = QP ′ and QR = QR ′ . By SAS, there is a motion of the Euclidean plane which sends Q to itself, P to P ′ and R to R ′ Note that the center of the absolute is fixed by the correspondingmotion. It follows that this motion gives also a motion of the h-plane; inparticular, the h-triangles △ h P QR and △ h P ′ QR ′ are h-congruent. Axiom h-V
Finally, we need to check that the Axiom h-V on page 85 holds; that is,we need to prove the following claim.
For any h-line ℓ and any h-point P / ∈ ℓ there are at leasttwo h-lines which pass thru P and have no points of intersection with ℓ . ABP m n ℓ Instead of proof.
Applying the main obser-vation we can assume that P is the centerof the absolute.The remaining part of the proof can beguessed from the picture Show that in the h-plane there are 3 mutually parallel h-linessuch that any pair of these three lines lieson one side of the remaining h-line.
Hyperbolic trigonometry
In this section we give formulas for h-distance using hyperbolic functions .One of these formulas will be used in the proof of the hyperbolic Pythag-orean theorem (13.13).Recall that ch, sh and th denote hyperbolic cosine , hyperbolic sine and hyperbolic tangent ; that is, the functions defined bych x := e x + e − x , sh x := e x − e − x , th x := sh x ch x . These hyperbolic functions are analogous to sine and cosine and tan-gent.
Prove the following identities: ch ′ x = sh x ; sh ′ x = ch x ; (ch x ) − (sh x ) = 1 . The identities ch(2 · x ) = (ch x ) + (sh x ) and sh(2 · x ) = 2 · sh x · ch x hold for any real value x .Proof. (sh x ) + (ch x ) = ( e x − e − x ) + ( e x + e − x ) == e · x + e − · x == ch(2 · x );2 · sh x · ch x = 2 · ( e x − e − x ) · ( e x + e − x ) == e · x − e − · x == sh(2 · x ) . CHAPTER 12. HYPERBOLIC PLANE
Let P and Q be two h-poins distinct fromthe center of absolute. Denote by P ′ and Q ′ the inverses of P and Q inthe absolute. P QP ′ Q ′ Show that(a) ch[ · P Q h ] = s P Q ′ · P ′ QP P ′ · QQ ′ ; (b) sh[ · P Q h ] = s P Q · P ′ Q ′ P P ′ · QQ ′ ; (c) th[ · P Q h ] = s P Q · P ′ Q ′ P Q ′ · P ′ Q ; (d) ch P Q h = P Q · P ′ Q ′ + P Q ′ · P ′ QP P ′ · QQ ′ . hapter 13 Geometry of the h-plane
In this chapter, we study the geometry of the plane described by theconformal disc model. For briefness, this plane will be called the h-plane .We can work with this model directly from inside of the Euclideanplane. We may also use the axioms of neutral geometry since they allhold in the h-plane; the latter proved in the previous chapter.
Angle of parallelism
Let P be a point off an h-line ℓ . Drop a perpendicular ( P Q ) h from P to ℓ ; let Q be its foot point. Let ϕ be the smallest value such that the h-line( P Z ) h with | ∡ h QP Z | = ϕ does not intersect ℓ .The value ϕ is called the angle of parallelism of P to ℓ . Clearly, ϕ depends only on the h-distance s = P Q h . Further, ϕ ( s ) → π/ s → ϕ ( s ) → s → ∞ . (In the Euclidean geometry, the angle ofparallelism is identically equal to π/ P QZ ℓ ϕ If ℓ , P and Z are as above, then theh-line m = ( P Z ) h is called asymptoticallyparallel to ℓ . In other words, two h-linesare asymptotically parallel if they shareone ideal point. (In hyperbolic geome-try, the term parallel lines is often usedfor asymptotically parallel lines ; we do notfollow this convention.)Given P ℓ , there are exactly twoasymptotically parallel lines thru P to ℓ ;the remaining parallel lines are called ultraparallel . 10102 CHAPTER 13. GEOMETRY OF THE H-PLANE
On the diagram, the two solid h-lines passing thru P are asymptoti-cally parallel to ℓ ; the dashed h-line is ultra parallel to ℓ . Let Q be the foot point of P on h-line ℓ . Then P Q h = · ln ϕ − cos ϕ , where ϕ is the angle of parallelism of P to ℓ . ABP X ZQ ϕ
Proof.
Applying a motion ofthe h-plane if necessary, wemay assume P is the centerof the absolute. Then the h-lines thru P are the intersec-tions of Euclidean lines withthe h-plane.Let A and B denote theideal points of ℓ . Without lossof generality, we may assumethat ∠ AP B is positive. Inthis case ϕ = ∡ QP B = ∡ AP Q = · ∡ AP B.
Let Z be the center of the circle Γ containing the h-line ℓ . Set X to be the point of intersection of the Euclidean segment [ AB ] and theline ( P Q ).Note that,
P X = cos ϕ . Therefore, by Lemma 12.9, P X h = ln ϕ − cos ϕ . Note that both angles
P BZ and
BXZ are right. Since the angle
P ZB is shared, △ ZBX ∼ △
ZP B . In particular, ZX · ZP = ZB ;that is, X is the inverse of P in Γ.The inversion in Γ is the reflection of the h-plane in ℓ . Therefore P Q h = QX h == · P X h == · ln ϕ − cos ϕ . Let
ABC be an h-triangle and β = | ∡ h ABC | . Thenthe h-distance from B to ( AC ) h is less than · ln 1 + cos β − cos β . Proof.
Let ϕ denotes the angle of parallelism of B to ( AC ) h . Note that ϕ > β ; therefore · ln 1 + cos ϕ − cos ϕ < · ln 1 + cos β − cos β . It remains to apply Proposition 13.1.
Let
ABC be an equilateral h-triangle with side .Show that | ∡ h ABC | < − . Inradius of h-triangle
The inradius of any h-triangle is less than · ln 3 .Proof. Let I and r be the h-incenter and h-inradius of △ h XY Z .Note that the h-angles
XIY , Y IZ and
ZIX have the same sign.Without loss of generality, we can assume that all of them are positiveand therefore ∡ h XIY + ∡ h Y IZ + ∡ h ZIX = 2 · π We can assume that ∡ h XIY > · π ; if not relabel X , Y and Z . X YZI
Since r is the h-distance from I to ( XY ) h ,Corollary 13.2 implies that r < · ln π − cos π == · ln 1 + − == · ln 3 . Let (cid:3) h ABCD be a quadrilateral in the h-plane suchthat the h-angles at A , B and C are right and AB h = BC h . Find theoptimal upper bound for AB h . Circles, horocycles and equidistants
Note that according to Lemma 12.11, any h-circle is a Euclidean circlewhich lies completely in the h-plane. Further, any h-line is an intersectionof the h-plane with the circle perpendicular to the absolute.In this section we will describe the h-geometric meaning of the inter-sections of the other circles with the h-plane.04
CHAPTER 13. GEOMETRY OF THE H-PLANE
You will see that all these intersections have a perfectly round shape in the h-plane.One may think of these curves as trajectories of a car which drives inthe plane with a fixed position of the steering wheel.In the Euclidean plane, this way you either run along a circle or alonga line.In the hyperbolic plane, the picture is different. If you turn the steeringwheel to the far right, you will run along a circle. If you turn it less, at acertain position of the wheel, you will never come back to the same point,but the path will be different from the line. If you turn the wheel furthera bit, you start to run along a path which stays at some fixed distancefrom an h-line. m gAB PP ′ Γ ∆
Equidistants of h-lines.
Consider theh-plane with the absolute Ω. Assumea circle Γ intersects Ω in two distinctpoints, A and B . Let g denotes the in-tersection of Γ with the h-plane.Let us draw an h-line m with the idealpoints A and B . According to Exer-cise 12.1, m is uniquely defined.Consider any h-line ℓ perpendicularto m ; let ∆ be the circle containing ℓ .Note that ∆ ⊥ Γ. Indeed, accordingto Corollary 10.16, m and Ω invert to themselves in ∆. It follows that A is the inverse of B in ∆. Finally, by Corollary 10.17, we get that ∆ ⊥ Γ.Therefore, inversion in ∆ sends both m and g to themselves. For anytwo points P ′ , P ∈ g there is a choice of ℓ and ∆ as above such that P ′ is the invese of P in ∆. By the main observation (12.7) the inversion in∆ is a motion of the h-plane. Therefore, all points of g lie on the samedistance from m .In other words, g is the set of points which lie on a fixed h-distanceand on the same side of m . Γ A Such a curve g is called equidistant to h-line m . In Euclidean geometry, the equidis-tant from a line is a line; apparently in hy-perbolic geometry the picture is different. Horocycles.
If the circle Γ touches theabsolute from inside at one point A , thenthe complement h = Γ \{ A } lies in the h-plane. This set is called a horocycle . It alsohas a perfectly round shape in the sensedescribed above.05Horocycles are the border case between circles and equidistants to h-lines. A horocycle might be considered as a limit of circles thru a fixedpoint with the centers running to infinity along a line. The same horocycleis a limit of equidistants which pass thru fixed point to the h-lines runningto infinity. Find the leg of an isosceles right h-triangle inscribed ina horocycle.
Hyperbolic triangles
Any nondegenerate hyperbolic triangle has a positivedefect.
A CB D
Proof.
Fix an h-triangle
ABC . According toTheorem 11.9, ➊ defect( △ h ABC ) > . It remains to show that in the case of equality, △ h ABC degenerates.Without loss of generality, we may assumethat A is the center of the absolute; in this case ∡ h CAB = ∡ CAB . Yetwe may assume that ∡ h CAB, ∡ h ABC, ∡ h BCA, ∡ ABC, ∡ BCA > . Let D be an arbitrary point in [ CB ] h distinct from B and C . FromProposition 9.19, we have ∡ ABC − ∡ h ABC ≡ π − ∡ CDB ≡ ∡ BCA − ∡ h BCA.
From Exercise 7.12, we get thatdefect( △ h ABC ) = 2 · ( π − ∡ CDB ) . Therefore, if equality holds in ➊ , then ∡ CDB = π . In particular, theh-segment [ BC ] h coincides with the Euclidean segment [ BC ]. By Exer-cise 12.3, the latter can happen only if the h-line ( BC ) h passes thru thecenter of the absolute ( A ); that is, if △ h ABC degenerates.The following theorem states, in particular, that nondegenerate hyper-bolic triangles are congruent if their corresponding angles are equal. Inparticular, in hyperbolic geometry, similar triangles have to be congruent.06
CHAPTER 13. GEOMETRY OF THE H-PLANE
Two nondegenerate h-triangles
ABC and A ′ B ′ C ′ are congruent if ∡ h ABC = ± ∡ h A ′ B ′ C ′ , ∡ h BCA == ± ∡ h B ′ C ′ A ′ and ∡ h CAB = ± ∡ h C ′ A ′ B ′ .Proof. Note that if AB h = A ′ B ′ h , then the theorem follows from ASA. A ′ B ′ C ′ B ′′ C ′′ Assume the contrary. Without loss of general-ity, we may assume that AB h < A ′ B ′ h . Therefore,we can choose the point B ′′ ∈ [ A ′ B ′ ] h such that A ′ B ′′ h = AB h .Choose an h-half-line [ B ′′ X ) so that ∡ h A ′ B ′′ X = ∡ h A ′ B ′ C ′ . According to Exercise 11.5, ( B ′′ X ) h k ( B ′ C ′ ) h .By Pasch’s theorem (3.12), ( B ′′ X ) h inter-sects [ A ′ C ′ ] h . Let C ′′ denotes the point of intersection.According to ASA, △ h ABC ∼ = △ h A ′ B ′′ C ′′ ; in particular, ➋ defect( △ h ABC ) = defect( △ h A ′ B ′′ C ′′ ) . Applying Exercise 11.10 twice, we get that ➌ defect( △ h A ′ B ′ C ′ ) = defect( △ h A ′ B ′′ C ′′ )++ defect( △ h B ′′ C ′′ C ′ ) + defect( △ h B ′′ C ′ B ′ ) . By Theorem 13.7, all the defects have to be positive. Thereforedefect( △ h A ′ B ′ C ′ ) > defect( △ h ABC ) . On the other hand,defect( △ h A ′ B ′ C ′ ) = | ∡ h A ′ B ′ C ′ | + | ∡ h B ′ C ′ A ′ | + | ∡ h C ′ A ′ B ′ | == | ∡ h ABC | + | ∡ h BCA | + | ∡ h CAB | == defect( △ h ABC )— a contradiction.Recall that a bijection from a h-plane to itself is called angle preserving if ∡ h ABC = ∡ h A ′ B ′ C ′ for any △ h ABC and its image △ h A ′ B ′ C ′ . Show that any angle-preserving transformation of theh-plane is a motion. Conformal interpretation
Let us give another interpretation of the h-distance.
Consider the h-plane with the unit circle centered at O as the absolute. Fix a point P and let Q be another point in the h-plane.Set x = P Q and y = P Q h . Then lim x → yx = 21 − OP . The above formula tells us that the h-distance from P to a nearby point Q is almost proportional to the Euclidean distance with the coefficient − OP . The value λ ( P ) = − OP is called the conformal factor of theh-metric.The value λ ( P ) = · (1 − OP ) can be interpreted as the speed limit atthe given point P . In this case the h-distance is the minimal time neededto travel from one point of the h-plane to another point. Γ O PQQ ′ P ′ Proof. If P = O , then according toLemma 12.9 ➍ yx = ln x − x x → x → P = O , let P ′ denotes the inverseof P in the absolute. Let Γ denotes thecircle with the center P ′ perpendicular tothe absolute.According to the main observation(12.7) and Lemma 12.6, the inversion in Γ is a motion of the h-planewhich sends P to O . In particular, if Q ′ denotes the inverse of Q in Γ,then OQ ′ h = P Q h .Set x ′ = OQ ′ . According to Lemma 10.2, x ′ x = OP ′ P ′ Q .
Since P ′ is the inverse of P in the absolute, we have that P O · OP ′ = 1.Therefore, x ′ x → OP ′ P ′ P = 11 − OP as x → ➍ , it implies that yx = yx ′ · x ′ x → − OP CHAPTER 13. GEOMETRY OF THE H-PLANE as x → The circumference of an h-circle of the h-radius r is · π · sh r, where sh r denotes the hyperbolic sine of r ; that is, sh r := e r − e − r . Before we proceed with the proof, let us discuss the same problem inthe Euclidean plane.The circumference of a circle in the Euclidean plane can be defined asthe limit of perimeters of regular n -gons inscribed in the circle as n → ∞ .Namely, let us fix r >
0. Given a positive integer n , consider △ AOB such that ∡ AOB = · πn and OA = OB = r . Set x n = AB . Note that x n is the side of a regular n -gon inscribed in the circle of radius r . Therefore,the perimeter of the n -gon is n · x n . ABO · π n r r x n The circumference of the circle with theradius r might be defined as the limit ➎ lim n →∞ n · x n = 2 · π · r. (This limit can be taken as the definitionof π .)In the following proof, we repeat thesame construction in the h-plane. Proof.
Without loss of generality, we canassume that the center O of the circle isthe center of the absolute.By Lemma 12.9, the h-circle with the h-radius r is the Euclidean circlewith the center O and the radius a = e r − e r + 1 . Let x n and y n denote the side lengths of the regular n -gons inscribedin the circle in the Euclidean and hyperbolic plane correspondingly.Note that x n → n → ∞ . By Lemma 13.10,lim n →∞ y n x n = 21 − a . ➎ , we get that the circumference of the h-circle can be foundthe following way: lim n →∞ n · y n = 21 − a · lim n →∞ n · x n == 4 · π · a − a == 4 · π · (cid:16) e r − e r +1 (cid:17) − (cid:16) e r − e r +1 (cid:17) == 2 · π · e r − e − r · π · sh r. Let circum h ( r ) denotes the circumference of the h-circle of the h-radius r . Show that circum h ( r + 1) > · circum h ( r ) for all r > . Pythagorean theorem
Recall that ch denotes hyperbolic cosine ; that is, the function defined bych x := e x + e − x . Assume that
ACB is anh-triangle with right angle at C . Set a = BC h , b = CA h and c = AB h . Then ➏ ch c = ch a · ch b. The formula ➏ will be proved by means of direct calculations. Let usdiscuss the limit cases of this formula.Note that ch x can be written using the Taylor expansionch x = 1 + · x + · x + . . . . CHAPTER 13. GEOMETRY OF THE H-PLANE
It follows that if a , b and c are small, then1 + · c ≈ ch c = ch a · ch b ≈≈ (1 + · a ) · (1 + · b ) ≈≈ · ( a + b ) . In other words, the original Pythagorean theorem (6.4) is a limit case ofthe hyperbolic Pythagorean theorem for small triangles.For large a and b the terms e − a and e − b are neglectable. In this casewe have the following approximations:ch a · ch b ≈ e a · e b == e a + b − ln 2 ≈≈ ch( a + b − ln 2) . Therefore c ≈ a + b − ln 2. Assume that
ACB is an h-triangle with right angleat C . Set a = BC h , b = CA h and c = AB h . Show that c + ln 2 > a + b. In the proof of the hyperbolic Pythagorean theorem we use the fol-lowing formula from Exercise 12.25:ch
P Q h = P Q · P ′ Q ′ + P Q ′ · P ′ QP P ′ · QQ ′ , here P , Q are h-points distinct from the absolute and P ′ , Q ′ are theirinversions in the absolute. A complete proof of this formula is given inthe hints. AA ′ B B ′ C Proof of the hyperbolic Pythagorean theorem.
Weassume that absolute is a unit circle. By the mainobservation (12.7) we can assume that C is thecenter of absolute. Let A ′ and B ′ denote the in-verses of A and B in the absolute.Set x = BC , y = AC . By Lemma 12.9 a = ln x − x , b = ln y − y . Therefore ➐ ch a = · ( x − x + − x x ) = ch b = · ( y − y + − y y ) == x − x , = y − y . B ′ C = x , A ′ C = y . Therefore BB ′ = x − x, AA ′ = y − y. Since the triangles
ABC , A ′ BC , AB ′ C , A ′ B ′ C are right, the originalPythagorean theorem (6.4) implies AB = p x + y , AB ′ = q x + y ,A ′ B = q x + y , A ′ B ′ = q x + y . According to Exercise 12.25, ➑ ch c = AB · A ′ B ′ + AB ′ · A ′ BAA ′ · BB ′ == p x + y · q x + y + q x + y · q x + y ( y − y ) · ( x − x ) == x + y + 1 + x · y (1 − y ) · (1 − x )= 1 + x − x · y − y . Finally note that ➐ and ➑ imply ➏ . hapter 14 Affine geometry
Affine transformations
Affine geometry studies the so called incidence structure of the Euclideanplane. The incidence structure says which points lie on which lines andnothing else; we cannot talk about distances, angle measures and so on.In other words, affine geometry studies the properties of the Euclideanplane which preserved under affine transformations defined below.A bijection of Euclidean plane to itself is called affine transformation if it maps any line to a line.Recall that three points are collinear if they lie on one line. Notethat affine transformation sends collinear points to collinear; the followingexercise gives a converse.
Assume f is a bijection from Euclidean plane to itselfwhich sends collinear points to collinear points. Show that f is an affinetransformation. (In other words, show that f maps noncollinear points tononcollinear.) Show that affine transformation sends a pair of parallellines to a pair of parallel lines.
Constructions
Let us consider geometric constructions with a ruler and a parallel tool ;the latter makes possible to draw a line thru a given point parallel toa given line. By Exercisers 14.2, any construction with these two toolsare invariant with respect to affine transformation. For example, to solve11213the following exercise, it is sufficient to prove that midpoint of a givensegment can be constructed with a ruler and a parallel tool.
Let M be the midpoint of segment [ AB ] in the Euclideanplane. Assume that an affine transformation sends the points A , B and M to A ′ , B ′ and M ′ correspondingly. Show that M ′ is the midpointof [ A ′ B ′ ] . The following exercise will be used in the proof of Theorem 14.7.
Assume that the points with the coordinates (0 , , (1 , , ( a, and ( b, are given. Using a ruler and a parallel tool, construct thepoints with the coordinates ( a · b, and ( a + b, . Use ruler and parallel tool to construct the center ofthe given circle.
Matrix form
Since the lines are defined in terms of metric; any motion of Euclideanplane is also an affine transformation.Let us describe an example of an affine transformation which is not amotion.Fix a coordinate system on the Euclidean plane. Let us use the columnnotation for the coordinates; that is, we will write ( xy ) instead of ( x, y ).As it follows from the theorem below, the so called shear mapping ( xy ) (cid:0) x + k · yy (cid:1) is an affine transformation. The shear mapping canchange the angle between vertical and horizontal lines almost arbitrary.The latter can be used to prove impossibility of some constructions witha ruler and a parallel tool; here is one example. Show that with a ruler and a parallel tool one cannotconstruct a line perpendicular to a given line.
A map β from the plane to itself is an affine transfor-mation if and only if ➊ β : ( xy ) (cid:0) a bc d (cid:1) · ( xy ) + ( vw ) = (cid:16) a · x + b · y + vc · x + d · y + w (cid:17) for a fixed invertible matrix (cid:0) a bc d (cid:1) and a vector (cid:0) vw (cid:1) .In particular, any affine transformation of Euclidean plane is contin-uous. In the proof of the “only if” part, we will use the following algebraiclemma:14
CHAPTER 14. AFFINE GEOMETRY
Assume f : R → R is a function such that forany x, y ∈ R we have(a) f (1) = 1 ,(b) f ( x + y ) = f ( x ) + f ( y ) ,(c) f ( x · y ) = f ( x ) · f ( y ) .Then f is the identity function; that is, f ( x ) = x for any x ∈ R . Note that we do not assume that f is continuous.The function f satisfying these three conditions is called field auto-morphism . Therefore, the lemma states that the identity function is theonly automorphism of the field of real numbers. For the field of com-plex numbers, the conjugation z ¯ z (see page 142) gives an example ofnontrivial automorphism. Proof. By (b) we have f (0) + f (1) = f (0 + 1) . By (a) f (0) + 1 = 1;whence ➋ f (0) = 0 . Applying (b) again, we get that0 = f (0) = f ( x ) + f ( − x ) . Therefore, ➌ f ( − x ) = − f ( x ) for any x ∈ R . Applying (b) recurrently, we get that f (2) = f (1) + f (1) = 1 + 1 = 2; f (3) = f (2) + f (1) = 2 + 1 = 3; . . . Together with ➌ , the latter implies that f ( n ) = n for any integer n. By (c) f ( m ) = f ( mn ) · f ( n ) . Therefore ➍ f ( mn ) = mn mn .Assume a >
0. Then the equation x · x = a has a real solution x = √ a .Therefore, [ f ( √ a )] = f ( √ a ) · f ( √ a ) = f ( a ). Hence f ( a ) >
0. That is, ➎ a > ⇒ f ( a ) > . Applying ➌ , we also get ➏ a ⇒ f ( a ) . Now we are ready to the final step in the proof. Assume f ( a ) = a forsome a ∈ R . Then there is a rational number mn which lies between a and f ( a ); that is, the numbers x = a − mn and y = f ( a ) − mn have opposite signs.By ➍ , y + mn = f ( a ) == f ( x + mn ) == f ( x ) + f ( mn ) == f ( x ) + mn ;that is, f ( x ) = y . By ➎ and ➏ the values x and y can not have oppositesigns — a contradiction. Assume γ is an affine transformation which fix threepoints ( ) , ( ) and ( ) on the coordinate plane. Then γ is the identitymap; that is, γ ( xy ) = ( xy ) for any point ( xy ) .Proof. Since an affine transformation sends lines to lines, we get that eachaxes is mapped to itself.According to Exercise 14.2, parallel lines are mapped to parallel lines.Therefore, we get that horizontal lines mapped to horizontal lines andvertical lines mapped to vertical. In other words, γ ( xy ) = (cid:16) f ( x ) h ( y ) (cid:17) . for some functions f, h : R → R .Note that f (1) = h (1) = 1 and according to Exercise 14.4, both f and h satisfies the other two conditions of the algebraic lemma (14.8).Applying the lemma, we get that f and h are identity functions and sois γ .16 CHAPTER 14. AFFINE GEOMETRY
Proof of Theorem 14.7; “if ” part.
Recall that matrix (cid:0) a bc d (cid:1) is invertibleif det (cid:0) a bc d (cid:1) = a · d − b · c = 0;in this case the matrix a · d − b · c · (cid:0) d − b − c a (cid:1) is the inverse of (cid:0) a bc d (cid:1) .Assume that the map β is described by ➊ . Note that ➐ ( xy ) a · d − b · c · (cid:0) d − b − c a (cid:1) · (cid:0) x − vy − w (cid:1) . is inverse of β . In particular, β is a bijection.Any line in the plane is given by equation ➑ p · x + q · y + r = 0 , where p = 0 or q = 0 (see Exercise 7.23). Let us find ( xy ) from its β -imageby formula ➐ and substitute the result in ➑ . We will get the equation ofthe image of the line. The equation has the same type as ➑ , with differentconstants; in particular, it describes a line.Therefore we proved that β is an affine transformation. “Only if ” part. Fix an affine transformation α . Set( vw ) = α ( ) , ( ac ) = α ( ) − α ( ) , (cid:0) bd (cid:1) = α ( ) − α ( ) . Note that the points α ( ), α ( ), α ( ) do not lie on one line. There-fore, the matrix (cid:0) a bc d (cid:1) is invertible.For the affine transformation β defined by ➊ we have β ( ) = α ( ) ,β ( ) = α ( ) ,β ( ) = α ( ) . Consider the composition γ = α ◦ β − is the identity map. Notethat γ is an affine transformation which fix points ( ), ( ) and ( ). ByLemma 14.9, γ is the identity map or equivalently α = β . On inversive transformations
Recall that inversive plane is Euclidean plane with added a point at in-finity, denoted by ∞ . We assume that every line passes thru ∞ . Recallthat the term circline stands for circle or line .17An inversive transformation is a bijection from inversive plane to itselfthat sends circlines to circlines. Inversive geometry studies the circlineincidence structure of inversive plane; it says which points lie on whichcirclines.
A map from inversive plane to itself is an inversivetransformation if and only if it can be presented as a composition of in-versions and reflections.
Exercise 18.15 gives another description of inversive transformationswhich use complex coordinates.
Proof.
According to Theorem 10.7 any inversion is an inversive trans-formation. Therefore, the same holds for composition of inversions andreflection.To prove the converse, fix an inversive transformation α .Assume α ( ∞ ) = ∞ . Recall that any circline passing thru ∞ is a line.If follows that α maps lines to lines; that is, it is an affine transformation.Note that α is not an arbitrary affine transformation — it maps circlesto circles.Composing α with a reflection, say ρ , we can assume that α ′ = ρ ◦ α maps the unit circle with center at the origin to a concentric circle.Composing the obtained map α ′ with a homothety χ : ( xy ) (cid:0) k · xk · y (cid:1) , we can assume that α ′′ = χ ◦ α ′ sends the unit circle to itself.Composing the obtained map α ′′ with a reflection ρ in a line thru theorigin, we can assume that α ′′′ = ρ ◦ α ′′ maps the point (1 ,
0) to itself.By Exercise 14.5, α ′′′ fixes the center of the circle; that is, it fixes theorigin.The obtained map α ′′′ is an affine transformation. Applying Theo-rem 14.7, together with the properties of α ′′ described above we get that α ′′′ : ( xy ) (cid:0) b d (cid:1) · ( xy )for an invertible matrix (cid:0) b d (cid:1) . Since the point (0 ,
1) maps to the unitcircle we get that b + d = 1 . Since the point ( √ , √ ) maps to the unit circle we get that( b + d ) = 1 . It follows α ′′′ : ( xy ) (cid:0) ± (cid:1) · ( xy ) ;18 CHAPTER 14. AFFINE GEOMETRY that is, either α ′′′ is the identity map or the reflection in the x -axis.Note that the homothety χ is a composition of two inversions in con-centric circles. Therefore, α is a composition of inversions and reflectionsif and only are so is α ′ , α ′′ and α ′′′ .In the remaining case α ( ∞ ) = ∞ , set P = α ( ∞ ). Consider an in-version β in a circle with center at P . Note that β ( P ) = ∞ ; therefore, β ◦ α ( ∞ ) = ∞ . Since β is inversive, so is β ◦ α . From above we get that β ◦ α is a composition of reflections and inversions; therefore, so is α . Show that inversive transformations preserve the anglebetween arcs up to sign.More precisely, assume A ′ B ′ C ′ , A ′ B ′ C ′ are the images of two arcs AB C , AB C under an inversive transformation. Let α and α ′ denotethe angle between the tangent half-lines to AB C and AB C at A andthe angle between the tangent half-lines to A ′ B ′ C ′ and A ′ B ′ C ′ at A ′ correspondingly. Then α ′ = ± α. Show that any reflection can be presented as a com-position of three inversions.
The exercise above together with Theorem 14.10, implies that anyinversive map is a composition of inversions — no reflections needed. hapter 15
Pro jective geometry
Real projective plane
In the Euclidean plane, two distinct lines might have one or zero pointsof intersection (in the latter case the lines are parallel). Our aim is toextend Euclidean plane by ideal points so that any two distinct lines willhave exactly one point of intersection.A collection of lines in the Euclidean plane iscalled concurrent if they all intersect at a singlepoint or all of them pairwise parallel. A maximalset of concurrent lines in the plane is called pen-cil . There are two types of pencils: central pencils contain all lines passing thru a fixed point calledthe center of the pencil and parallel pencil containpairwise parallel lines.Each point in the Euclidean plane uniquely de-fines a central pencil with the center in it. Notethat any two lines completely determine the pencilcontaining both.Let us add one ideal point for each parallel pen-cil, and assume that all these ideal points lie on one ideal line . We also assume that the ideal line be-longs to each parallel pencil.We obtain the so called real projective plane . It comes with an inci-dence structure — we say that three points lie on one line if the corre-sponding pencils contain a common line.Projective geometry studies this incidence structure on the projectiveplane. Loosely speaking, any statement in projective geometry can beformulated using only terms collinear points , concurrent lines .11920 CHAPTER 15. PROJECTIVE GEOMETRY
Euclidean space
Let us repeat the construction of metric d (page 11) in the space.Let R denotes the set of all triples ( x, y, z ) of real numbers. Assume A = ( x A , y A , z A ) and B = ( x B , y B , z B ) are arbitrary points in R . Definethe metric on R the following way: AB := p ( x A − x B ) + ( y A − y B ) + ( z A − z B ) . The obtained metric space is called
Euclidean space .Assume at least one of the real numbers a , b or c is distinct from zero.Then the subset of points ( x, y, z ) ∈ R described by equation a · x + b · y + c · z + d = 0is called plane ; here d is a real number.It is straightforward to show the following: ⋄ Any plane in the Euclidean space is isometric to the Euclidean plane. ⋄ Any three points in the space lie on a plane. ⋄ An intersection of two distinct planes (if it is nonempty) is a line ineach of these planes.These statements make possible to generalize many notions and resultsfrom Euclidean plane geometry to the Euclidean space by applying planegeometry in the planes of the space.
Perspective projection
Consider two planes Π and Π ′ in the Euclidean space. Let O be a pointwhich does not belong neither to Π nor Π ′ .ΠΠ ′ O Let us define the perspective projection from Π to Π ′ with center at O . The projection of a point P ∈ Π is defined as the intersection point P ′ = Π ′ ∩ ( OP ).21Note that the perspective projection sends collinear points to collinear.Indeed, assume three points P , Q , R lie on one line ℓ in Π and P ′ , Q ′ , R ′ are their images in Π ′ . Let Θ be the plane containing O and ℓ . Then allthe points P , Q , R , P ′ , Q ′ , R ′ lie on Θ. Therefore, the points P ′ , Q ′ , R ′ lie on the intersection line ℓ ′ = Θ ∩ Π ′ .The perspective projection is not a bijection between the planes. In-deed, if the line ( OP ) is parallel to Π ′ (that is, if ( OP ) ∩ Π ′ = ∅ ) thenthe perspective projection is not defined. Also, if ( OP ′ ) k Π for P ′ ∈ Π ′ ,then the point P ′ is not an image of the perspective projection.Let us remind that a similar story happened with inversion. An inver-sion is not defined at its center; moreover, the center is not an inverse ofany point. To deal with this problem we passed to inversive plane whichis Euclidean plane extended by one ideal point.A similar strategy works for perspective projection Π → Π ′ , but thistime real projective plane is the right choice of extension.Let ˆΠ and ˆΠ ′ denote the corresponding real projective planes. Let usdefine a bijection between points in the real projective plane ˆΠ and theset Λ of all the lines passing thru O . If P ∈ Π, then take the line ( OP ); if P is an ideal point of ˆΠ, so it is defined by a parallel pencil of lines, thentake the line thru O parallel to the lines in this pencil.The same construction gives a bijection between Λ and ˆΠ ′ . Composingthese two bijections ˆΠ ↔ Λ ↔ ˆΠ ′ , we get a bijection between ˆΠ andˆΠ ′ which coincides with the perspective projection P P ′ where it isdefined.Note that the ideal line of ˆΠ maps to the intersection line of Π ′ andthe plane thru O parallel to Π. Similarly the ideal line of ˆΠ ′ is the imageof the intersection line of Π and the plane thru O parallel to Π ′ .Strictly speaking we described a transformation from one real projec-tive plane to another, but if we identify the two planes, say by fixing acoordinate system in each, we get a projective transformation from theplane to itself. Let O be the origin of ( x, y, z ) -coordinate space and theplanes Π and Π ′ are given by the equations x = 1 and y = 1 correspond-ingly. The perspective projection from Π to Π ′ with center at O sends P to P ′ . Assume P has coordinates (1 , y, z ) , find the coordinates of P ′ .For which points P ∈ Π the perspective projection is undefined? Whichpoints P ′ ∈ Π ′ are not images of points under perspective projection? Projective transformations
A bijection from the real projective plane to itself which sends lines tolines is called projective transformation .22
CHAPTER 15. PROJECTIVE GEOMETRY
Projective geometry studies the properties of real projective planewhich preserved under projective transformations.Note that any affine transformation defines a projective transforma-tion on the corresponding real projective plane. We will call such pro-jective transformations affine ; these are projective transformations whichsend the ideal line to itself.The perspective projection discussed in the previous section gives anexample of projective transformation which is not affine.
Given a line ℓ in the real projective plane, there is aperspective projection which sends ℓ to the ideal line.Moreover, any projective transformation can be obtained as a compo-sition of an affine transformation and a perspective projection.Proof. Identify the projective plane with a plane Π in the space. Fixa point
O / ∈ Π and choose a plane Π ′ which is parallel to the planecontaining ℓ and O . The corresponding perspective projection sends ℓ tothe ideal line.Assume α is a projective transformation.If α sends ideal line to itself, then it has to be affine. It proves thetheorem in this case.If α sends the ideal line to the line ℓ , choose a perspective projection β which sends ℓ back to the ideal line. The composition β ◦ α sends theideal line to itself. That is, β ◦ α is affine, hence the result. Moving points to infinity
Theorem 15.2 makes possible to take any line in the projective plane anddeclare it to be ideal. In other words, we can choose preferred affineplane by removing one line from the projective plane. This constructionprovides a method for solving problems in projective geometry which willbe illustrated by the following classical example:
Consider three concurrent lines ( AA ′ ) , ( BB ′ ) and ( CC ′ ) in the real projective plane. Set X = ( BC ) ∩ ( B ′ C ′ ) , Y = ( CA ) ∩ ( C ′ A ′ ) , Z = ( AB ) ∩ ( A ′ B ′ ) . Then the points X , Y and Z are collinear.Proof. Without loss of generality, we may assume that the line ( XY ) isideal. If not, apply a perspective projection which sends the line ( XY )to the ideal line.23 A A ′ B B ′ C C ′ ZXY
That is, we can assume that( BC ) k ( B ′ C ′ ) and ( CA ) k ( C ′ A ′ )and we need to show that( AB ) k ( A ′ B ′ ) . Assume that the lines ( AA ′ ), ( BB ′ )and ( CC ′ ) intersect at point O . Since( BC ) k ( B ′ C ′ ), the transversal property(7.6) implies that ∡ OBC = ∡ OB ′ C ′ and ∡ OCB = ∡ OC ′ B ′ . By the AA similaritycondition, △ OBC ∼ △ OB ′ C ′ . In partic-ular, OBOB ′ = OCOC ′ .A A ′ B B ′ C C ′ O The same way we get that △ OAC ∼∼ △ OA ′ C ′ and OAOA ′ = OCOC ′ . Therefore,
OAOA ′ = OBOB ′ . By the SAS similarity condition, we getthat △ OAB ∼ △ OA ′ B ′ ; in particular, ∡ OAB = ± ∡ OA ′ B ′ .Note that ∡ AOB = ∡ A ′ OB ′ .Therefore, ∡ OAB = ∡ OA ′ B ′ . By the transversal property 7.6, ( AB ) k ( A ′ B ′ ).The case ( AA ′ ) k ( BB ′ ) k ( CC ′ ) is done similarly. In this case thequadrilaterals B ′ BCC ′ and A ′ ACC ′ are parallelograms. Therefore, BB ′ = CC ′ = AA ′ . Hence (cid:3) B ′ BAA ′ is a parallelogram and ( AB ) k ( A ′ B ′ ).Here is another classical theorem of projective geometry. Assume that two triples of points A , B , C ,and A ′ , B ′ , C ′ are collinear. Set X = ( BC ′ ) ∩ ( B ′ C ) , Y = ( CA ′ ) ∩ ( C ′ A ) , Z = ( AB ′ ) ∩ ( A ′ B ) . CHAPTER 15. PROJECTIVE GEOMETRY AA ′ BB ′ CC ′ A A ′ BB ′ CC ′ XYZ
Then the points X , Y , Z are collinear. Pappus’ theorem can be proved the same way as Desargues’ theorem.
Idea of the proof.
Applying a perspective projection, we can assume that X and Y lie on the ideal line. It remains to show that Z lies on the idealline.In other words, assuming that ( AB ′ ) k ( A ′ B ) and ( AC ′ ) k ( A ′ C ), weneed to show that ( BC ′ ) k ( B ′ C ). Finish the proof of Pappus’ theorem using the idea de-scribed above.
Duality
Assume that a bijection P ↔ p between the set of lines and the set ofpoints of a plane is given. That is, given a point P , we denote by p thecorresponding line; and the other way around, given a line s we denoteby S the corresponding point.The bijection between points and lines is called duality if P ∈ s ⇐⇒ p ∋ S. for any point P and line s . p q r sADCB EF P QRSa b c def Dual configurations. The standard definition of duality is more general; we consider a special case whichis also called polarity . Show that Euclidean plane does not admit a duality.
The real projective plane admits a duality.Proof.
Consider a plane Π and a point
O / ∈ Π in the space; let ˆΠ denotesthe corresponding real projective plane.Recall that there is a natural bijection ˆΠ ↔ Λ between ˆΠ and the setΛ of all the lines passing thru O . Denote it by P ↔ ˙ P ; that is, ⋄ if P ∈ Π, then ˙ P = ( OP ); ⋄ if P is an ideal point of ˆΠ, so P is defined as a parallel pencil oflines, set ˙ P to be the line thru O which is parallel to each lines inthis pencil.Similarly there is a natural bijection s ↔ ˙ s between lines in ˆΠ andall the planes passing thru O . If s is a line in Π, then ˙ s is the planecontaining O and s ; if s is the ideal line of ˆΠ, take ˙ s is the plane thru O parallel to Π.It is straightforward to check that ˙ P ⊂ ˙ s if and only if P ∈ s ; that is,the bijections P ↔ ˙ P and s ↔ ˙ s remember all the incidence structure ofthe real projective plane ˆΠ.It remains to construct a bijection ˙ s ↔ ˙ S between the set of planesand the set of lines passing thru O such that ➊ ˙ r ⊂ ˙ S ⇐⇒ ˙ R ⊃ ˙ s for any two lines ˙ r and ˙ s passing thru O .Set ˙ S to be the plane thru O which is perpendicular to ˙ s . Note thatboth conditions ➊ are equivalent to ˙ r ⊥ ˙ s ; hence the result follows. Consider the Euclidean plane with ( x, y ) -coordinates;let O denotes the origin. Given a point P = O with coordinates ( a, b ) consider the line p given by the equation a · x + b · y = 1 .Show that the correspondence P to p extends to a duality of the realprojective plane.Which line corresponds to O ?Which point of the real projective plane corresponds to the line a · x ++ b · y = 0 ? The existence of duality in the real projective planes makes possibleto formulate an equivalent dual statement to any statement in projectivegeometry. For example, the dual statement for “the points X , Y and Z lie on one line ℓ ” would be the “lines x , y and z intersect at one point L ”.Let us formulate the dual statement for Desargues’ theorem 15.3.26 CHAPTER 15. PROJECTIVE GEOMETRY
Consider the collinear points X , Y and Z . Assume that X = ( BC ) ∩ ( B ′ C ′ ) , Y = ( CA ) ∩ ( C ′ A ′ ) , Z = ( AB ) ∩ ( A ′ B ′ ) . Then the lines ( AA ′ ) , ( BB ′ ) and ( CC ′ ) are concurrent. In this theorem the points X , Y and Z are dual to the lines ( AA ′ ),( BB ′ ) and ( CC ′ ) in the original formulation, and the other way around.Once Desargues’ theorem is proved, applying duality (15.7) we get thedual Desargues’ theorem. Note that the dual Desargues’ theorem is theconverse to the original Desargues’ theorem 15.3. Formulate the dual Pappus’ theorem (see 15.4). (a) Given two parallel lines, construct with a ruler only a third parallelline thru a given point.(b) Given a parallelogram, construct with a ruler only a line parallel toa given line thru a given point.
Axioms
Note that the real projective plane described above satisfies the followingset of axioms:p-I. Any two distinct points lie on a unique line.p-II. Any two distinct lines pass thru a unique point.p-III. There exist at least four points of which no three are collinear.Let us take these three axioms as a definition of the projective plane ;so the real projective plane discussed above becomes a particular exampleof projective plane.There is an example of projective plane which containsexactly 3 points on each line. This is the so called
Fanoplane which you can see on the diagram; it contains 7points and 7 lines. This is an example of finite projectiveplane ; that is, projective plane with finitely many points.
Show that any line in projective planecontains at least three points.
Consider the following analog of Axiom p-III:27p-III ′ . There exist at least four lines of which no three are concurrent. Show that Axiom p-III ′ is equivalent to Axiom p-III .That is, p-I, p-II and p-III imply p-III ′ ,and p-I, p-II and p-III ′ imply p-III . The exercise above shows that in the axiomatic system of projectiveplane, lines and points have the same rights. In fact, one can switch every-where words “point” with “line”, “pass thru” with “lies on”, “collinear”with “concurrent” and we get an equivalent set of axioms — Axioms p-Iand p-II convert into each other, and the same happens wit the pair p-IIIand p-III ′ . Assume that one of the lines in a finite projectiveplane contains exactly n + 1 points.(a) Show that each line contains exactly n + 1 points.(b) Show that the number of the points in the plane has to be n + n + 1 . (c) Show that there is no projective plane with exactly 10 points.(d) Show that in any finite projective plane the number of points coin-cides with the number of lines. The number n in the above exercise is called order of finite projectiveplane. For example Fano plane has order 2. Here is one of the mostfamous open problem in finite geometry. The order of any finite projective plane is a powerof a prime number. hapter 16
Spherical geometry
Spherical geometry studies the surface of a unit sphere. This geometryhas applications in cartography, navigation and astronomy.The spherical geometry is a close relative of the Euclidean and hy-perbolic geometries. Most of the theorems of hyperbolic geometry havespherical analogs, but spherical geometry is easier to visualize.
Euclidean space
Recall that Euclidean space is the set R of all triples ( x, y, z ) of realnumbers such that the distance between a pair of points A = ( x A , y A , z A )and B = ( x B , y B , z B ) is defined by the following formula: AB := p ( x A − x B ) + ( y A − y B ) + ( z A − z B ) . The planes in the space are defined as the set of solutions of equation a · x + b · y + c · z + d = 0for real numbers a , b , c and d such that at least one of the numbers a , b or c is not zero. Any plane in the Euclidean space is isometric to theEuclidean plane.A sphere in the space is the direct analog of circle in the plane. For-mally, sphere with center O and radius r is the set of points in the spacethat lie on the distance r from O .Let A and B be two points on the unit sphere centered at O . The spherical distance from A to B (briefly AB s ) is defined as | ∡ AOB | .In spherical geometry, the role of lines play the great circles ; that is,the intersection of the sphere with a plane passing thru O .12829Note that the great circles do not form lines in the sense of Defini-tion 1.8. Also, any two distinct great circles intersect at two antipodalpoints. In particular, the sphere does not satisfy the axioms of the neutralplane. Pythagorean theorem
Here is an analog of the Pythagorean theorems (6.4 and 13.13) in sphericalgeometry.
Let △ s ABC be a sphericaltriangle with a right angle at C . Set a = BC s , b = CA s and c = AB s .Then cos c = cos a · cos b. In the proof, we will use the notion of the scalar product which we areabout to discuss.Let v A = ( x A , y A , z A ) and v B = ( x B , y B , z B ) denote the positionvectors of points A and B . The scalar product of the two vectors v A and v B in R is defined as ➊ h v A , v B i := x A · x B + y A · y B + z A · z B . Assume both vectors v A and v B are nonzero; let ϕ denotes the anglemeasure between them. Then the scalar product can be expressed thefollowing way: ➋ h v A , v B i = | v A |·| v B |· cos ϕ, where | v A | = q x A + y A + z A , | v B | = q x B + y B + z B .OCB A xy z Now, assume that the points A and B lie onthe unit sphere Σ in R centered at the origin. Inthis case | v A | = | v B | = 1. By ➋ we get that ➌ cos AB s = h v A , v B i . Proof of the spherical Pythagorean Theorem.
Since the angle at C is right, we can choose thecoordinates in R so that v C = (0 , , v A lies in the xz -plane, so v A == ( x A , , z A ), and v B lies in yz -plane, so v B = (0 , y B , z B ).30 CHAPTER 16. SPHERICAL GEOMETRY
Applying, ➌ , we get that z A = h v C , v A i = cos b,z B = h v C , v B i = cos a. Applying, ➊ and ➌ , we get thatcos c = h v A , v B i == x A · · y B + z A · z B == cos b · cos a. Show that if △ s ABC is a spherical triangle with a rightangle at C , and AC s = BC s = π , then AB s = π . Inversion of the space
The inversion in a sphere is defined the same way as we define the inversionin a circle.Formally, let Σ be the sphere with the center O and radius r . The inversion in Σ of a point P is the point P ′ ∈ [ OP ) such that OP · OP ′ = r . In this case, the sphere Σ will be called the sphere of inversion and itscenter is called the center of inversion .We also add ∞ to the space and assume that the center of inversionis mapped to ∞ and the other way around. The space R with the point ∞ will be called inversive space .The inversion of the space has many properties of the inversion of theplane. Most important for us are the analogs of theorems 10.6, 10.7, 10.25which can be summarized as follows: The inversion in the sphere has the following proper-ties:(a) Inversion maps a sphere or a plane into a sphere or a plane.(b) Inversion maps a circle or a line into a circle or a line.(c) Inversion preserves the cross-ratio; that is, if A ′ , B ′ , C ′ and D ′ arethe inverses of the points A , B , C and D correspondingly, then AB · CDBC · DA = A ′ B ′ · C ′ D ′ B ′ C ′ · D ′ A ′ . (d) Inversion maps arcs into arcs. (e) Inversion preserves the absolute value of the angle measure betweentangent half-lines to the arcs. We do not present the proofs here, but they nearly repeat the corre-sponding proofs in plane geometry. To prove (a) , you will need in additionthe following lemma; its proof is left to the reader.
Let Σ be a subset of the Euclidean space which containsat least two points. Fix a point O in the space.Then Σ is a sphere if and only if for any plane Π passing thru O , theintersection Π ∩ Σ is either empty set, one point set or a circle. The following observation helps to reduce part (b) to part (a) . Any circle in the space is an intersection of twospheres.
Let us define a circular cone as a set formed by linesegments from a fixed point, called the tip of the cone, to allthe points on a fixed circle, called the base of the cone; wealways assume that the base does not lie in the same planeas the tip. We say that the cone is right if the center ofthe base circle is the foot point of the tip on the base plane;otherwise we call it oblique . Let K be an oblique circular cone. Show that thereis a plane Π that is not parallel to the base plane of K such that theintersection Π ∩ K is a circle. Stereographic projection
O PSN P ′ ΣΥThe plane thru P , O and S .Consider the unit sphere Σ centered atthe origin (0 , , x + y + z == 1.Let Π denotes the xy -plane; it is de-fined by the equation z = 0. Clearly, Πruns thru the center of Σ.Let N = (0 , ,
1) and S = (0 , , − ∩ Π.For any point P = S on Σ, consider the line ( SP ) in the space. Thisline intersects Π in exactly one point, denoted by P ′ . Set S ′ = ∞ .32 CHAPTER 16. SPHERICAL GEOMETRY
The map ξ s : P P ′ is called the stereographic projection from Σ to Π with respect to the south pole . The inverse of this map ξ − s : P ′ P iscalled the stereographic projection from Π to Σ with respect to the southpole .The same way, one can define the stereographic projections ξ n and ξ − n with respect to the north pole N .Note that P = P ′ if and only if P ∈ Ω.Note that if Σ and Π are as above, then the composition of the stere-ographic projections ξ s : Σ → Π and ξ − s : Π → Σ are the restrictions ofthe inversion in the sphere Υ with the center S and radius √ the absolute valueof the angle measure between arcs on the sphere.This makes it particularly useful in cartography. A map of a big regionof earth cannot be done in a constant scale, but using a stereographicprojection, one can keep the angles between roads the same as on earth.In the following exercises, we assume that Σ, Π, Υ, Ω, O , S and N are as above. Show that ξ n ◦ ξ − s , the composition of stereographicprojections from Π to Σ from S , and from Σ to Π from N is the inverseof the plane Π in Ω . Show that a stereographic projection Σ → Π sends thegreat circles to circlines on the plane which intersects Ω at two oppositepoints. The following exercise is analogous to Lemma 13.10.
Fix a point P ∈ Π and let Q be another point in Π . Let P ′ and Q ′ denote their stereographic projections to Σ . Set x = P Q and y = P ′ Q ′ s . Show that lim x → yx = 21 + OP . Central projection
The central projection is analogous to the projective model of hyperbolicplane which is discussed in Chapter 17.Let Σ be the unit sphere centered at the origin which will be denotedby O . Let Π + denotes the plane defined by the equation z = 1. This planeis parallel to the xy -plane and it passes thru the north pole N = (0 , , O PP ′ N Σ + Π + Recall that the northern hemisphere of Σ, isthe subset of points ( x, y, z ) ∈ Σ such that z > + .Given a point P ∈ Σ + , consider the half-line[ OP ). Let P ′ denotes the intersection of [ OP )and Π + . Note that if P = ( x, y, z ), then P ′ =( xz , yz , P ↔ P ′ is a bijection between Σ + and Π + .The described bijection Σ + ↔ Π + is called the central projection ofthe hemisphere Σ + .Note that the central projection sends the intersections of the greatcircles with Σ + to the lines in Π + . The latter follows since the greatcircles are intersections of Σ with planes passing thru the origin as wellas the lines in Π + are the intersection of Π + with these planes.The following exercise is analogous to Exercise 17.4 in hyperbolic ge-ometry. Let △ s ABC be a nondegenerate spherical triangle.Assume that the plane Π + is parallel to the plane passing thru A , B and C . Let A ′ , B ′ and C ′ denote the central projections of A , B and C .(a) Show that the midpoints of [ A ′ B ′ ] , [ B ′ C ′ ] and [ C ′ A ′ ] are centralprojections of the midpoints of [ AB ] s , [ BC ] s and [ CA ] s correspond-ingly.(b) Use part (a) to show that the medians of a spherical triangle inter-sect at one point. hapter 17 Pro jective model
The projective model is another model of hyperbolic plane discovered byBeltrami; it is often called
Klein model . The projective and conformalmodels are saying exactly the same thing but in two different languages.Some problems in hyperbolic geometry admit simpler proof using theprojective model and others have simpler proof in the conformal model.Therefore, it worth to know both.
Special bijection on the h-plane
Consider the conformal disc model with the absolute at the unit circle Ωcentered at O . Choose a coordinate system ( x, y ) on the plane with theorigin at O , so the circle Ω is described by the equation x + y = 1. O PNS P ′ ˆ P Σ ΠThe plane thru P , O and S . Let us think that our plane is thecoordinate xy -plane in the Euclideanspace; denote it by Π. Let Σ be the unitsphere centered at O ; it is described bythe equation x + y + z = 1 . Set S = (0 , , −
1) and N = (0 , , → Σ from S ; given point P ∈ Πdenote its image in Σ. Note that theh-plane is mapped to the north hemi-sphere ; that is, to the set of points( x, y, z ) in Σ described by the inequality z > P ′ ∈ Σ consider its foot point ˆ P on Π; this is the closestpoint to P ′ .Note that the composition P ↔ P ′ ↔ ˆ P of these two maps gives abijection from the h-plane to itself. Further note that P = ˆ P if and onlyif P ∈ Ω or P = O . Show that the bijection P ↔ ˆ P described above can bedescribed the following way: set ˆ O = O and for any other point P take ˆ P ∈ [ OP ) such that O ˆ P = 2 · x x , where x = OP . Let ( P Q ) h be an h-line with the ideal points A and B .Then ˆ P , ˆ Q ∈ [ AB ] .Moreover, ➊ A ˆ Q · B ˆ P ˆ QB · ˆ P A = (cid:18) AQ · BPQB · P A (cid:19) . In particular, if
A, P, Q, B appear in the same order, then
P Q h = · ln A ˆ Q · B ˆ P ˆ QB · ˆ P A .
Proof.
Consider the stereographic projection Π → Σ from the south pole S . Note that it fixes A and B ; denote by P ′ and Q ′ the images of P and Q ; A B ˆ PP ′ ΓThe plane Λ.According to Theorem 16.3 c , ➋ AQ · BPQB · P A = AQ ′ · BP ′ Q ′ B · P ′ A .
By Theorem 16.3 e , each circline in Π whichis perpendicular to Ω is mapped to a circle in Σwhich is still perpendicular to Ω. It follows thatthe stereographic projection sends ( P Q ) h to theintersection of the north hemisphere of Σ with aplane perpendicular to Π.Let Λ denotes the plane; it contains thepoints A , B , P ′ , ˆ P and the circle Γ = Σ ∩ Λ. (It also contains Q ′ and ˆ Q but we will not use these points for a while.)Note that36 CHAPTER 17. PROJECTIVE MODEL ⋄ A, B, P ′ ∈ Γ, ⋄ [ AB ] is a diameter of Γ, ⋄ ( AB ) = Π ∩ Λ, ⋄ ˆ P ∈ [ AB ] ⋄ ( P ′ ˆ P ) ⊥ ( AB ).Since [ AB ] is the diameter of Γ, by Corollary 9.6, the angle AP ′ B isright. Hence △ A ˆ P P ′ ∼ △ AP ′ B ∼ △ P ′ ˆ P B . In particular AP ′ BP ′ = A ˆ PP ′ ˆ P = P ′ ˆ PB ˆ P .
Therefore ➌ A ˆ PB ˆ P = (cid:18) AP ′ BP ′ (cid:19) . The same way we get that ➍ A ˆ QB ˆ Q = (cid:18) AQ ′ BQ ′ (cid:19) . Finally, note that ➋ + ➌ + ➍ imply ➊ .The last statement follows from ➊ and the definition of h-distance.Indeed, P Q h := ln AQ · BPQB · P A == ln A ˆ Q · B ˆ P ˆ QB · ˆ P A ! == · ln A ˆ Q · B ˆ P ˆ QB · ˆ P A . A B A B A B Γ Γ Γ Ω Let Γ , Γ and Γ be three circles perpendicular to thecircle Ω . Let [ A B ] , [ A B ] and [ A B ] denote the common chords of Ω and Γ , Γ , Γ correspondingly.Show that the chords [ A B ] , [ A B ] and [ A B ] intersect at one point in-side Ω if and only if Γ , Γ and Γ intersect at two points. Projective model
The following picture illustrates the map P ˆ P described in the previoussection — if you take the picture on the left and apply the map P ˆ P ,Conformal model Projective modelyou get the picture on the right. The pictures are conformal and projectivemodel of the hyperbolic plane correspondingly. The map P ˆ P is a“translation” from one to another.In the projective model things look different; some become simpler,other things become more complicated. Lines.
The h-lines in the projective model are chords of the absolute;more precisely, chords without its endpoints.
Circles and equidistants.
The h-circles and equidistants in the projec-tive model are certain type of ellipses and their open arcs.It follows since the stereographic projection sends circles on the planeto circles on the unit sphere and the foot point projection of circle back tothe plane is an ellipse. (One may define ellipse as a foot point projectionof a circle.)
A BP Q
Distance.
Consider a pair of h-points P and Q . Let A and B be the ideal pointof the h-line in projective model; that is, A and B are the intersections of the Eu-clidean line ( P Q ) with the absolute.Then by Lemma 17.2, ➎ P Q h = · ln AQ · BPQB · P A , assuming the points
A, P, Q, B appear onthe line in the same order.38
CHAPTER 17. PROJECTIVE MODEL
Angles.
The angle measures in the projective model are very differentfrom the Euclidean angles and it is hard to figure out by looking on thepicture. For example all the intersecting h-lines on the picture above areperpendicular. There are two useful exceptions: ⋄ If O is the center of the absolute, then ∡ h AOB = ∡ AOB. ⋄ If O is the center of the absolute and ∡ OAB = ± π , then ∡ h OAB = ∡ OAB = ± π . To find the angle measure in the projective model, you may apply amotion of the h-plane which moves the vertex of the angle to the centerof the absolute; once it is done the hyperbolic and Euclidean angles havethe same measure.
Motions.
The motions of the h-plane in the conformal and projectivemodels are relevant to inversive transformations and projective transfor-mation in the same way. Namely: ⋄ Inversive transformations that preserve the h-plane describe motionsof the h-plane in the conformal model. ⋄ Projective transformations that preserve h-plane describe motionsof the h-plane in the projective model. The following exercise is a hyperbolic analog of Exercise 16.10. Thisis the first example of a statement which admits an easier proof using theprojective model.
Let P and Q be the points in h-plane which lie on thesame distance from the center of the absolute. Observe that in the projec-tive model, h-midpoint of [ P Q ] h coincides with the Euclidean midpoint of [ P Q ] h .Conclude that if an h-triangle is inscribed in an h-circle, then its me-dians meet at one point.Recall that an h-triangle might be also inscribed in a horocycle or anequidistant. Think how to prove the statement in this case. m ℓ st Z Let ℓ and m are h-lines in theprojective model. Let s and t denote the Euclideanlines tangent to the absolute at the ideal points of ℓ . Show that if the lines s , t and the extension of m intersect at one point, then ℓ and m are perpendicular h-lines. The idea described in the solution of Exercise 16.6 and in the sketch of proof ofTheorem 19.13 can be used to construct many projective transformations of this type. Use the projective model to derive the formula for angleof parallelism (Proposition 13.1).
Use projective model to find the inradius of the idealtriangle.
The projective model of h-plane can be used to give another proof ofthe hyperbolic Pythagorean theorem (13.13).
A BCst u XY First let us recall its statement: ➏ ch c = ch a · ch b, where a = BC h , b = CA h and c = AB h and △ h ACB is an h-triangle with right angle at C .Note that we can assume that A is the center of theabsolute. Set s = BC , t = CA , u = AB . According tothe Euclidean Pythagorean theorem (6.4), we have ➐ u = s + t . It remains to express a , b and c using s , u and t and show that ➐ implies ➏ . Finish the proof of hyperbolic Pythagoreantheorem (13.13) indicated above.
Bolyai’s construction
Assume we need to construct a line thru P asymptotically parallel to thegiven line ℓ in the h-plane.If A and B are ideal points of ℓ in the projective model, then we couldsimply draw the Euclidean line ( P A ). However the ideal points do not liein the h-plane, therefore there is no way to use them in the construction.In the following construction we assume that you know a compass-and-ruler construction of the perpendicular line; see Exercise 5.21.
1. Drop a perpendicular from P to ℓ ; denote it by m . Let Q be the footpoint of P on ℓ .2. Erect a perpendicular from P to m ; denote it by n .3. Mark by R a point on ℓ distinct from Q .4. Drop a perpendicular from R to n ; denote it by k .5. Draw the circle Γ with center P and the radius QR . Mark by T apoint of intersection of Γ with k .6. The line ( P T ) h is asymptotically parallel to ℓ . CHAPTER 17. PROJECTIVE MODEL
Explain what happens if one performs the Bolyai con-struction in the Euclidean plane.
To prove that Bolyai’s construction gives the asymptotically parallelline in the h-plane, it is sufficient to show the following:
Assume P , Q , R , S , T be points in h-plane suchthat ⋄ S ∈ ( RT ) h , ⋄ ( P Q ) h ⊥ ( QR ) h , ⋄ ( P S ) h ⊥ ( P Q ) h , ⋄ ( RT ) h ⊥ ( P S ) h and ⋄ ( P T ) h and ( QR ) h are asymptotically parallel.Then QR h = P T h .Proof. We will use the projective model. Without loss of generality, wemay assume that P is the center of the absolute. As it was noted on page138, in this case the corresponding Euclidean lines are also perpendicular;that is, ( P Q ) ⊥ ( QR ), ( P S ) ⊥ ( P Q ) and ( RT ) ⊥ ( P S ).Let A be the common ideal point of ( QR ) h and ( P T ) h . Let B and C denote the remaining ideal points of ( QR ) h and ( P T ) h correspondingly.Note that the Euclidean lines ( P Q ), (
T R ) and ( CB ) are parallel. PQRS T ABC ℓ m n k Therefore, △ AQP ∼ △
ART ∼ △
ABC.
In particular,
ACAB = ATAR = APAQ .
It follows that
ATAR = APAQ = BRCT = BQCP .
In particular, AT · CPT C · P A = AR · BQRB · QA .
Applying the formula for h-distance in the projective model ➎ , we getthat QR h = P T h . hapter 18 Complex coordinates
In this chapter, we give an interpretation of inversive geometry usingcomplex coordinates. The results of this chapter will not be used in thisbook, but they lead to deeper understanding of both concepts.
Complex numbers
Informally, a complex number is a number that can be put in the form ➊ z = x + i · y, where x and y are real numbers and i = − C . If x , y and z are as in ➊ , then x is called the real part and y the imaginary part ofthe complex number z . Briefly it is written as x = Re z and y = Im z. On the more formal level, a complex number is a pair of real numbers( x, y ) with the addition and multiplication described below. The formula x + i · y is only a convenient way to write the pair ( x, y ). ➋ ( x + i · y ) + ( x + i · y ) := ( x + x ) + i · ( y + y );( x + i · y ) · ( x + i · y ) := ( x · x − y · y ) + i · ( x · y + y · x ) . Complex coordinates
Recall that one can think of the Euclidean plane as the set of all pairs ofreal numbers ( x, y ) equipped with the metric AB = p ( x A − x B ) + ( y A − y B ) , CHAPTER 18. COMPLEX COORDINATES where A = ( x A , y A ) and B = ( x B , y B ).One can pack the coordinates ( x, y ) of a point in one complex number z = x + i · y . This way we get a one-to-one correspondence between pointsof the Euclidean plane and C . Given a point Z = ( x, y ), the complexnumber z = x + i · y is called the complex coordinate of Z .Note that if O , E and I are points in the plane with complex coordi-nates 0, 1 and i , then ∡ EOI = ± π . Further, we assume that ∡ EOI = π ;if not, one has to change the direction of the y -coordinate. Conjugation and absolute value
Let z be a complex number with real part x and imaginary part y . If y = 0, we say that the complex number z is real and if x = 0 we saythat z is imaginary . The set of points with real (imaginary) complexcoordinates is a line in the plane, which is called real (correspondingly imaginary ) line. The real line will be denoted as R .The complex number ¯ z := x − i · y is called the complex conjugate of z .Let Z and ¯ Z be the points in the plane with the complex coordinates z and ¯ z correspondingly. Note that the point ¯ Z is the reflection of Z inthe real line.It is straightforward to check that ➌ x = Re z = z + ¯ z , y = Im z = z − ¯ zi · , x + y = z · ¯ z. The last formula in ➌ makes it possible to express the quotient wz oftwo complex numbers w and z = x + i · y : wz = z · ¯ z · w · ¯ z = x + y · w · ¯ z. Note that z + w = ¯ z + ¯ w, z − w = ¯ z − ¯ w, z · w = ¯ z · ¯ w, z/w = ¯ z/ ¯ w. That is, the complex conjugation respects all the arithmetic operations.The value | z | := p x + y = p ( x + i · y ) · ( x − i · y ) = √ z · ¯ z is called the absolute value of z . If | z | = 1, then z is called a unit complexnumber . Show that | v · w | = | v |·| w | for any v, w ∈ C . Z and W are points in the Euclidean plane, z and w aretheir complex coordinates, then ZW = | z − w | . The triangle inequality for the points with complex coordinates 0, v and v + w implies that | v + w | | v | + | w | for any v, w ∈ C ; this inequality is also called triangle inequality . Use the identity u · ( v − w ) + v · ( w − u ) + w · ( u − v ) = 0 for u, v, w ∈ C and the triangle inequality to prove the Ptolemy’s inequality(6.8). Euler’s formula
Let α be a real number. The following identity is called Euler’s formula : ➍ e i · α = cos α + i · sin α. In particular, e i · π = − e i · π = i . 0 1 ie i · α α Geometrically, Euler’s formula means the fol-lowing: Assume that O and E are the pointswith complex coordinates 0 and 1 correspond-ingly. Assume OZ = 1 and ∡ EOZ ≡ α, then e i · α is the complex coordinate of Z . In particular, the complexcoordinate of any point on the unit circle centered at O can be uniquelyexpressed as e i · α for some α ∈ ( − π, π ]. Why should you think that ➍ is true? The proof of Euler’s identitydepends on the way you define the exponential function. If you never hadto apply the exponential function to an imaginary number, you may takethe right hand side in ➍ as the definition of the e i · α .In this case, formally nothing has to be proved, but it is better tocheck that e i · α satisfies familiar identities. Mainly, e i · α · e i · β = e i · ( α + β ) . CHAPTER 18. COMPLEX COORDINATES
The latter can be proved using ➋ and the following trigonometric formu-las, which we assume to be known:cos( α + β ) = cos α · cos β − sin α · sin β, sin( α + β ) = sin α · cos β + cos α · sin β. If you know the power series for the sine, cosine and exponential func-tion, the following might convince that the identity ➍ holds: e i · α = 1 + i · α + ( i · α )
2! + ( i · α )
3! + ( i · α )
4! + ( i · α )
5! + · · · == 1 + i · α − α − i · α
3! + α
4! + i · α − · · · == (cid:18) − α
2! + α − · · · (cid:19) + i · (cid:18) α − α
3! + α − · · · (cid:19) == cos α + i · sin α. Argument and polar coordinates
As before, we assume that O and E are the points with complex coordi-nates 0 and 1 correspondingly.Let Z be the a point distinct form O . Set ρ = OZ and θ = ∡ EOZ .The pair ( ρ, θ ) is called the polar coordinates of Z .0 1 i z θ = a r g z ρ = | z | If z is the complex coordinate of Z , then ρ = | z | . The value θ is called the argument of z (briefly, θ = arg z ). In this case, z = ρ · e i · θ = ρ · (cos θ + i · sin θ ) . Note thatarg( z · w ) ≡ arg z + arg w and arg zw ≡ arg z − arg w if z = 0 and w = 0. In particular, if Z , V , W are points with complexcoordinates z , v and w correspondingly, then ➎ ∡ V ZW = arg (cid:18) w − zv − z (cid:19) ≡≡ arg( w − z ) − arg( v − z )45if ∡ V ZW is defined.
Use the formula ➎ to show that ∡ ZV W + ∡ V W Z + ∡ W ZV ≡ π for any △ ZV W in the Euclidean plane.
Assume that points V , W and Z have complex coordi-nates v , w and z = v · w correspondingly and the point O and E are asabove. Show that △ OEV ∼ △
OW Z.
The following theorem is a reformulation of Theorem 9.10 which usescomplex coordinates.
Let (cid:3)
U V W Z be a quadrilateral and u , v , w and z bethe complex coordinates of its vertices. Then (cid:3) U V W Z is inscribed if andonly if the number ( v − u ) · ( z − w )( v − w ) · ( z − u ) is real. The value ( v − u ) · ( w − z )( v − w ) · ( z − u ) is called the complex cross-ratio ; it will be dis-cussed in more details below. Observe that the complex number z = 0 is real if andonly if arg z = 0 or π ; in other words, · arg z ≡ .Use this observation to show that Theorem 18.5 is indeed a reformu-lation of Theorem 9.10. U VWZ U ′ V ′ W ′ Z ′ Let U , V , W , Z , U ′ , V ′ , W ′ and Z ′ be points on the plane with com-plex coordinates u , v , w , z , u ′ , v ′ , w ′ and z ′ correspondingly.Assume that (cid:3) U V W Z , (cid:3) U V V ′ U ′ , (cid:3) V W W ′ V ′ , (cid:3) W ZZ ′ W ′ and (cid:3) ZU U ′ Z ′ areinscribed.a) Express it using Theorem 18.5.b) Use (a) to show that (cid:3) U ′ V ′ W ′ Z ′ isinscribed. CHAPTER 18. COMPLEX COORDINATES
Method of complex coordinates
The following problem illustrates the method of complex coordinates.
Let △ OP V and △ OQW be isosceles right trianglessuch that ∡ V P O = ∡ OQW = π and M be the midpoint of [ V W ] . Assume P , Q and M are distinct points.Show that △ P M Q is an isosceles right triangle.Solution.
Choose the complex coordinates so that O is the origin; denoteby v, w, p, q, m the complex coordinates of the corresponding points.Note that v = (1 + i ) · p, w = (1 − i ) · q.O WVP QM Therefore m = · ( v + w ) == i · p + − i · q. By a straightforward computations, we getthat p − m = i · ( q − m ) . In particular, | p − m | = | q − m | and arg p − mq − m = π ; that is, P M = QM and ∡ QM P = π . A B CEO
Consider three squares withcommon sides as on the diagram. Use themethod of complex coordinates to show that ∡ EOA + ∡ EOB + ∡ EOC = ± π . Fractional linear transformations
Watch video “M¨obius transformations tevealed” byDouglas Arnold and Jonathan Rogness. (It is available on YouTube.)
The complex plane C extended by one ideal number ∞ is called the extended complex plane . It is denoted by ˆ C , so ˆ C = C ∪ {∞} A fractional linear transformation or M¨obius transformation of ˆ C is afunction of one complex variable z which can be written as f ( z ) = a · z + bc · z + d , a , b , c , d are complex numbers satisfying a · d −− b · c = 0. (If a · d − b · c = 0 the function defined above is a constant andis not considered to be a fractional linear transformation.)In case c = 0, we assume that f ( − d/c ) = ∞ and f ( ∞ ) = a/c ;and if c = 0 we assume f ( ∞ ) = ∞ . Elementary transformations
The following three types of fractional linear transformations are called elementary :1. z z + w, z w · z for w = 0 , z z . The geometric interpretations.
Let O denotes the point with thecomplex coordinate 0.The first map z z + w, corresponds to the so called parallel trans-lation of the Euclidean plane, its geometric meaning should be evident.The second map is called the rotational homothety with the centerat O . That is, the point O maps to itself and any other point Z maps toa point Z ′ such that OZ ′ = | w |· OZ and ∡ ZOZ ′ = arg w .The third map can be described as a composition of the inversion inthe unit circle centered at O and the reflection in R (the composition canbe taken in any order). Indeed, arg z ≡ − arg z . Therefore,arg z = arg(1 / ¯ z );that is, if the points Z and Z ′ have complex coordinates z and 1 / ¯ z , then Z ′ ∈ [ OZ ). Clearly, OZ = | z | and OZ ′ = | / ¯ z | = | z | . Therefore, Z ′ isthe inverse of Z in the unit circle centered at O .Finally, the reflection of Z ′ in R , has complex coordinate z = (1 / ¯ z ). The map f : ˆ C → ˆ C is a fractional linear transfor-mation if and only if it can be expressed as a composition of elementarytransformations.Proof; the “only if ” part. Fix a fractional linear transformation f ( z ) = a · z + bc · z + d . CHAPTER 18. COMPLEX COORDINATES
Assume c = 0. Then f ( z ) = ac − a · d − b · cc · ( c · z + d ) == ac − a · d − b · cc · z + dc . That is, ➏ f ( z ) = f ◦ f ◦ f ◦ f ( z ) , where f , f , f and f are elementary transformations of the followingform: ⋄ f ( z ) = z + dc , ⋄ f ( z ) = z , ⋄ f ( z ) = − a · d − b · cc · z , ⋄ f ( z ) = z + ac .If c = 0, then f ( z ) = a · z + bd . In this case f ( z ) = f ◦ f ( z ) , where f ( z ) = ad · z and f ( z ) = z + bd . “If ” part. We need to show that by composing elementary transforma-tions, we can only get fractional linear transformations. Note that it issufficient to check that the composition of a fractional linear transforma-tions f ( z ) = a · z + bc · z + d . with any elementary transformation z z + w , z w · z and z z is afractional linear transformations.The latter is done by means of direct calculations. a · ( z + w ) + bc · ( z + w ) + d = a · z + ( b + a · w ) c · z + ( d + c · w ) ,a · ( w · z ) + bc · ( w · z ) + d = ( a · w ) · z + b ( c · w ) · z + d ,a · z + bc · z + d = b · z + ad · z + c . The image of a circline under a fractional lineartransformation is a circline. Proof.
By Proposition 18.11, it is sufficient to check that each elementarytransformation sends a circline to a circline.For the first and second elementary transformation, the latter is evi-dent.As it was noted above, the map z z is a composition of inversionand reflection. By Theorem 10.11, the inversion sends a circline to acircline. Hence the result. Show that the inverse of a fractional linear transfor-mation is a fractional linear transformation.
Given distinct values z , z , z ∞ ∈ ˆ C , construct a frac-tional linear transformation f such that f ( z ) = 0 , f ( z ) = 1 and f ( z ∞ ) = ∞ . Show that such a transformation is unique.
Show that any inversion is a composition of the com-plex conjugation and a fractional linear transformation.Use Theorem 14.10 to conclude that any inversive transformation iseither fractional linear transformation or a complex conjugate to a frac-tional linear transformation.
Complex cross-ratio
Given four distinct complex numbers u , v , w and z , the complex number( u − w ) · ( v − z )( v − w ) · ( u − z )is called the complex cross-ratio ; it will be denoted by ( u, v ; w, z ).If one of the numbers u , v , w , z is ∞ , then the complex cross-ratio hasto be defined by taking the appropriate limit; in other words, we assumethat ∞∞ = 1. For example,( u, v ; w, ∞ ) = ( u − w )( v − w ) . Assume that U , V , W and Z are the points with complex coordinates u , v , w and z correspondingly. Note that U W · V ZV W · U Z = | ( u, v ; w, z ) | , ∡ W U Z + ∡ ZV W = arg u − wu − z + arg v − zv − w ≡≡ arg( u, v ; w, z ) . CHAPTER 18. COMPLEX COORDINATES
It makes it possible to reformulate Theorem 10.6 using the complex co-ordinates the following way:
Let
U W V Z and U ′ W ′ V ′ Z ′ be two quadrilaterals suchthat the points U ′ , W ′ , V ′ and Z ′ are inverses of U , W , V , and Z corre-spondingly. Assume u , w , v , z , u ′ , w ′ , v ′ and z ′ are the complex coordi-nates of U , W , V , Z , U ′ , W ′ , V ′ and Z ′ correspondingly.Then ( u ′ , v ′ ; w ′ , z ′ ) = ( u, v ; w, z ) . The following exercise is a generalization of the theorem above. Itadmits a short and simple solution which uses Proposition 18.11.
Show that complex cross-ratios are invariant underfractional linear transformations.That is, if a fractional linear transformation maps four distinct com-plex numbers u, v, w, z to complex numbers u ′ , v ′ , w ′ , z ′ respectively, then ( u ′ , v ′ ; w ′ , z ′ ) = ( u, v ; w, z ) . Schwarz–Pick theorem
The following theorem shows that the metric in the conformal disc modelnaturally appears in other branches of mathematics. We do not give aproof, but it can be found in any textbook on geometric complex analysis.Let D denotes the unit disc in the complex plane centered at 0; thatis, a complex number z belongs to D if and only if | z | < D as a h-plane in the conformal disc model; theh-distance between z, w ∈ D will be denoted by d h ( z, w ).A function f : D → C is called holomorphic if for every z ∈ D there isa complex number s such that f ( z + w ) = f ( z ) + s · w + o ( | w | ) . In other words, f is complex-differentiable at any z ∈ D . The number s above is called the derivative of f at z and is denoted by f ′ ( z ). Assume f : D → D is a holomorphicfunction. Then d h ( f ( z ) , f ( w )) d h ( z, w ) for any z, w ∈ D .If the equality holds for one pair of distinct numbers z, w ∈ D , then itholds for any pair. In this case f is a fractional linear transformation aswell as a motion of the h-plane. Show that if a fractional linear transformation f ap-pears in the equality case of Schwarz–Pick theorem, then it can be writtenas f ( z ) = v · z + ¯ ww · z + ¯ v . where | w | < | v | . Show that th[ · d h ( z, w )] = (cid:12)(cid:12)(cid:12)(cid:12) z − w − z · ¯ w (cid:12)(cid:12)(cid:12)(cid:12) . Conclude the inequality in Schwarz–Pick theorem can be rewritten as (cid:12)(cid:12)(cid:12)(cid:12) z ′ − w ′ − z ′ · ¯ w ′ (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) z − w − z · ¯ w (cid:12)(cid:12)(cid:12)(cid:12) , where z ′ = f ( z ) and w ′ = f ( w ) . Show that the Schwarz lemma stated below followsfrom Schwarz–Pick theorem.
Let f : D → D be a holomorphic function and f (0) = 0 . Then | f ( z ) | | z | for any z ∈ D .Moreover, if equality holds for some z = 0 , then there is a unit complexnumber u such that f ( z ) = u · z for any z ∈ D . hapter 19 Geometric constructions
Geometric constructions have great pedagogical value as an introductionto mathematical proofs. We were using construction problems everywherestarting from Chapter 5.In this chapter we briefly discuss the classical results in geometricconstructions.
Classical problems
In this section we list a couple of classical construction problems; eachknown for more than a thousand years. The solutions of the followingtwo problems are quite nontrivial.
Construct an inscribed quadrilat-eral with given sides.
Construct a circle which is tangent tothree given circles.
The following exercise is a simplified version ofthe problem of Apollonius, which is still nontrivial.
Construct a circle which passesthru a given point and is tangent to two intersectinglines.
The following three problems cannot be solved in principle; that is,the needed compass-and-ruler construction does not exist.
Doubling the cube.
Construct the side of a new cube, which has thevolume twice as big as the volume of a given cube. a , one needs to constructa segment of length √ · a . Squaring the circle.
Construct a square with the same area as a givencircle. If r is the radius of the given circle, we need to construct a segmentof length √ π · r . Angle trisection.
Divide the given angle into three equal angles.
In fact, there is no compass-and-ruler construction which trisects an-gle with measure π . Existence of such a construction would imply con-structability of a regular 9-gon which is prohibited by the following famousresult:A regular n -gon inscribed in a circle with center O is a sequence ofpoints A . . . A n on the circle such that ∡ A n OA = ∡ A OA = · · · = ∡ A n − OA n = ± n · π. The points A , . . . , A n are vertexes , the segments [ A A ] , . . . , [ A n A ] are sides and the remaining segments [ A i A j ] are diagonals of the n -gon.A construction of a regular n -gon, therefore, is reduced to the con-struction of an angle with size n · π . A regular n -gon can be constructedwith a ruler and a compass if and only if n is the product of a power of and any number of distinct Fermat primes. A Fermat prime is a prime number of the form 2 k +1 for some integer k .Only five Fermat primes are known today:3 , , , , . For example, ⋄ one can construct a regular 340-gon since 340 = 2 · ·
17 and 5 aswell as 17 are Fermat primes; ⋄ one cannot construct a regular 7-gon since 7 is not a Fermat prime; ⋄ one cannot construct a regular 9-gon; altho 9 = 3 · Constructible numbers
In the classical compass-and-ruler constructions initial configuration canbe completely described by a finite number of points; each line is defined54
CHAPTER 19. GEOMETRIC CONSTRUCTIONS by two points on it and each circle is described by its center and a pointon it (equivalently, you may describe a circle by three points on it).The same way the result of construction can be described by a finitecollection of points.We may always assume that the initial configuration has at least twopoint; if not add one or two points to the configuration. Moreover, ap-plying a rescaling to whole plane, we can assume that the first two pointsin the initial configuration lie on distance 1 from each other.In this case we can choose a coordinate system, such that one of theinitial points is the origin (0 ,
0) and yet another initial point has thecoordinates (1 , n points is described by 2 · n − x , y , . . .. . . , x n , y n .It turns out that the coordinates of any point constructed with acompass and ruler can be written thru the numbers x , y , . . . , x n , y n usingthe four arithmetic operations “+”, “ − ”, “ · ”, “ / ” and the square root“ √ ”.For example, assume we want to find the points X = ( x , y ) and X = ( x , y ) of the intersections of a line passing thru A = ( x A , y A ) and B = ( x B , y B ) and the circle with center O = ( x O , y O ) which passes thruthe point W = ( x W , y W ). Let us write the equations of the circle and theline in the coordinates ( x, y ): ( ( x − x O ) + ( y − y O ) = ( x W − x O ) + ( y W − y O ) , ( x − x A ) · ( y B − y A ) = ( y − y A ) · ( x B − x A ) . Expressing y from the second equation and substituting the result in thefirst one, gives us a quadratic equation in x , which can be solved using“+”, “ − ”, “ · ”, “ / ” and “ √ ” only.The same can be performed for the intersection of two circles. Theintersection of two lines is even simpler; it is described as a solution oftwo linear equations and can be expressed using only four arithmeticoperations; the square root “ √ ” is not needed.On the other hand, it is easy to produce compass-and-ruler construc-tions which produce segments of the lengths a + b and a − b from twogiven segments of lengths a > b .To perform “ · ”, “ / ” and “ √ ” consider the following diagram: let[ AB ] be a diameter of a circle; fix a point C on the circle and let D bethe foot point of C on [ AB ]. Note that △ ABC ∼ △
ACD ∼ △
BDC.
It follows that AD · DC = BD .55 A BDC
Using this diagram, one shouldguess the compass-and-ruler construc-tions which produce segments of lengths √ a · b and a b .For example, to construct √ a · b , dothe following: (1) construct points A , B and D ∈ [ AB ] such that AD = a and BD = b ; (2) construct the circle Γ onthe diameter [ AB ]; (3) draw the line ℓ thru D perpendicular to ( AB ); (4) let C be an intersection of Γ and ℓ . Then DC = √ a · b .Taking 1 for a or b above, we can produce √ a , a , b . Combining theseconstructions we can produce a · b = ( √ a · b ) , ab = a · b . In other wordswe produced a compass-and-ruler calculator , which can do “+”, “ − ”, “ · ”,“ / ” and “ √ ”.The discussion above gives a sketch of the proof of the following the-orem: Assume that the initial configuration of geometric con-struction is given by the points A = (0 , , A = (1 , , A = ( x , y ) , . . .. . . , A n = ( x n , y n ) . Then a point X = ( x, y ) can be constructed usinga compass-and-ruler construction if and only if both coordinates x and y can be expressed from the integer numbers and x , y , x , y , . . . , x n , y n using the arithmetic operations “ + ”, “ − ”, “ · ”, “ / ” and the square root“ √ ”. The numbers which can be expressed from the given numbers using thearithmetic operations and the square root “ √ ” are called constructible ;if the list of given numbers is not given, then we can only use the integers.The theorem above translates any compass-and-ruler constructionproblem into a purely algebraic language. For example: ⋄ The impossibility of a solution for doubling the cube problem statesthat √ √ − ”, “ · ”, “ / ” and “ √ ”. ⋄ The impossibility of a solution for squaring the circle states that √ π , or equivalently π , is not a constructible number. ⋄ The Gauss–Wantzel theorem says for which integers n the numbercos · πn is constructible.Some of these statements might look evident, but rigorous proofs requiresome knowledge of abstract algebra (namely, field theory) which is out ofthe scope of this book.In the next section, we discuss similar but simpler examples of impos-sible constructions with an unusual tool.56 CHAPTER 19. GEOMETRIC CONSTRUCTIONS (a) Show that diagonal or regular pentagon is √ times larger thanits side.(b) Use (a) to make a compass-and-ruler construction of a regular pen-tagon. Constructions with a set square
A set square is a construction tool shown on the pic-ture — it can produce a line thru a given point whichmakes the angles π or ± π to a given line and it canbe also used as a ruler; that is, it can produce a linethru a given pair of points. Trisect a given segment with a set square.
Let us consider set-square constructions. Following the same lines asin the previous section, we can define set-square constructible numbers and prove the following analog of Theorem 19.5:
Assume that the initial configuration of a geomet-ric construction is given by the points A = (0 , , A = (1 , , A == ( x , y ) , . . . , A n = ( x n , y n ) . Then a point X = ( x, y ) can be constructedusing a set-square construction if and only if both coordinates x and y canbe expressed from the integer numbers and x , y , x , y , . . . , x n , y n usingthe arithmetic operations “ + ”, “ − ”, “ · ”, “ / ”. Let us apply this theorem to show the impossibility of some construc-tions with a set square.Note that if all the coordinates x , y , . . . , x n , y n are rational numbers,then the theorem above implies that with a set square, one can onlyconstruct the points with rational coordinates. A point with both rationalcoordinates is called rational , and if at least one of the coordinates isirrational, then the point is called irrational . Show that an equilateral triangle in the Euclidean planehas at least one irrational point.Conclude that with a set square, one cannot construct an equilateraltriangle.
Make a set-square construction which verifies if thegiven triangle is equilateral. (We assume that we can “verify” if twoconstructed points coincide.) More impossible constructions
In this section we discuss yet another source of impossible constructions.Recall that a circumtool produces a circle passing thru any given threepoints or a line if all three points lie on one line. Let us restate Exer-cise 10.9.
Exercise.
Show that with a circumtool only, it is impossible to constructthe center of a given circle Γ . Remark.
In geometric constructions, we allow to choose some free points,say any point on the plane, or a point on a constructed line, or a pointwhich does not lie on a constructed line and so on.In principle, when you make such a free choice it is possible to mark thecenter of Γ by accident. Nevertheless, we do not accept such a coincidenceas true construction; we say that a construction produces the center if itproduces it for any free choices.
Solution.
Arguing by contradiction, assume we have a construction of thecenter.Apply an inversion in a circle perpendicular to Γ to the whole con-struction. According to Corollary 10.16, the circle Γ maps to itself. Sincethe inversion sends a circline to a circline, we get that the whole construc-tion is mapped to an equivalent construction; that is, a constriction witha different choice of free points.According to Exercise 10.8, the inversion sends the center of Γ toanother point. That is, following the same construction, we can end upat a different point — a contradiction.
Show that there is no circumtool-only constructionwhich verifies if the given point is the center of a given circle. (We assumethat we can only “verify” if two constructed points coincide.)
A similar example of impossible constructions for a ruler and a paralleltool is given in Exercise 14.6.Let us discuss yet another example for a ruler-only construction. Notethat ruler-only constructions are invariant with respect to the projectivetransformations. In particular, to solve the following exercise, it is suf-ficient to construct a projective transformation which fixes two points A and B and moves its midpoint. Show that the midpoint of a given segment cannot beconstructed with only a ruler.
The following theorem is a stronger version of the exercise above.58
CHAPTER 19. GEOMETRIC CONSTRUCTIONS
The center of a given circle cannot be constructedwith only a ruler.Sketch of the proof.
It is sufficient to construct a projective transformationwhich sends the given circle Γ to a circle Γ ′ such that the center of Γ ′ is not the image of the center of Γ. (Compare the construction withExercise 16.6.)Let Γ be a circle which lies in the plane Π in the Euclidean space.By Theorem 16.3, the inverse of a circle in a sphere is a circle or aline. Fix a sphere Σ with the center O so that the inversion Γ ′ of Γ is acircle and the plane Π ′ containing Γ ′ is not parallel to Π; any sphere Σ ina general position will do.Let Z and Z ′ denote the centers of Γ and Γ ′ . Note that Z ′ / ∈ ( OZ ).It follows that the perspective projection Π → Π ′ with center at O sendsΓ to Γ ′ , but Z ′ is not the image of Z . Construction of a polar
In this section we describe a powerful trick that can be used in the con-structions only with ruler.Assume Γ is a circle in the plane and
P / ∈ Γ. Draw two lines x and y thru P which intersect Γ at two pairs of points X , X ′ and Y , Y ′ .Let Z = ( XY ) ∩ ( X ′ Y ′ ) and Z ′ = ( XY ′ ) ∩ ( X ′ Y ). Consider the line p = ( ZZ ′ ). Γ Ppx y X X ′ Y Y ′ ZZ ′ The following claim will be used in the constructions without a proof.
The constructed line p = ( ZZ ′ ) does not depend on thechoice of the lines x and y . Moreover, P p is a duality (see page 124). The line p is called the polar of P with respect to Γ. The same waythe point P is called the polar of the line p with respect to Γ.59 Let p be the polar line of point P with respect to thecircle Γ . Assume that p intersects Γ at points V and W . Show that thelines ( P V ) and ( P W ) are tangent to Γ .Come up with a ruler-only construction of the tangent lines to thegiven circle Γ thru the given point P / ∈ Γ . Assume two concentric circles Γ and Γ ′ are given.Construct the common center of Γ and Γ ′ with a ruler only. hapter 20 Area
The area functional will be defined by Theorem 20.7. This theorem isgiven without proof, but it follows immediately from the properties of
Lebesgue measure on the plane. The construction of Lebesgue measuretypically use the method of coordinates and it is included in any textbookin real analysis. Based on this theorem, we develop the concept of areawith no cheating.We choose this approach since any rigorous introduction to area istedious. We do not want to cheat and at the same time we do not wantto waste your time; soon or later you will have to learn Lebesgue measureif it is not done already.
Solid triangles
We say that the point X lies inside a nondegenerate triangle ABC if thefollowing three condition hold: ⋄ A and X lie on the same side of the line ( BC ); ⋄ B and X lie on the same side of the line ( CA ); ⋄ C and X lie on the same side of the line ( AB ). A BCX
The set of all points inside △ ABC and on its sides[ AB ], [ BC ], [ CA ] will be called solid triangle ABC anddenoted by ▲ ABC . Show that any solid triangle is convex ;that is, if
X, Y ∈ ▲ ABC , then [ XY ] ⊂ ▲ ABC . The notations △ ABC and ▲ ABC look similar, they also have closebut different meanings, which better not to confuse. Recall that △ ABC is an ordered triple of distinct points (see page 17), while ▲ ABC is aninfinite set of points. 16061In particular, ▲ ABC = ▲ BAC for any triangle
ABC . Indeed, anypoint which belong to the set ▲ ABC also belongs to the set ▲ BAC andthe other way around. On the other hand, △ ABC = △ BAC simplybecause the ordered triple of points (
A, B, C ) is distinct from the orderedtriple (
B, A, C ).Note that ▲ ABC ∼ = ▲ BAC even if △ ABC ≇ △ BAC , where congru-ence of the sets ▲ ABC and ▲ BAC is understood the following way:
Two sets S and T in the plane are called congruent (briefly S ∼ = T ) if T = f ( S ) for some motion f of the plane. If △ ABC is not degenerate and ▲ ABC ∼ = ▲ A ′ B ′ C ′ , then after relabeling the vertices of △ ABC we will have △ ABC ∼ = △ A ′ B ′ C ′ . Indeed it is sufficient to show that if f is a motion which maps ▲ ABC to ▲ A ′ B ′ C ′ , then f maps each vertex of △ ABC to a vertex △ A ′ B ′ C ′ .The latter follows from the characterization of vertexes of solid trianglesgiven in the following exercise: Let △ ABC be nondegenerate and X ∈ ▲ ABC . Showthat X is a vertex of △ ABC if and only if there is a line ℓ which intersects ▲ ABC at the single point X . Polygonal sets
Elementary set on the plane is a set of one of the following three types: ⋄ one-point set; ⋄ segment; ⋄ solid triangle.A set in the plane is called polygonal if it can bepresented as a union of finite collection of elementarysets.Note that according to this definition, the empty set ∅ is a polygonalset. Indeed, ∅ is a union of an empty collection of elementary sets.A polygonal set is called degenerate if it can be presented as union offinite number of one-point sets and segments.If X and Y lie on opposite sides of the line ( AB ), then the union ▲ AXB ∪ ▲ BY A is a polygonal set which is called solid quadrilateral
AXBY and denoted by (cid:4)
AXBY . In particular, we can talk about solidparallelograms , rectangles and squares .62 CHAPTER 20. AREA
Typically a polygonal set admits many pre-sentation as union of a finite collection of el-ementary sets. For example, if (cid:3)
AXBY is aparallelogram, then (cid:4)
AXBY = ▲ AXB ∪ ▲ BY A = ▲ XAY ∪ ▲ Y BX.
Show that a solid square is not degenerate.
Show that a circle is not a polygonal set.
Definition of area
For any two polygonal sets P and Q , the union P ∪ Q aswell as the intersection
P ∩ Q are also polygonal sets.
A class of sets which closed with respect to union and intersection iscalled a ring of sets . The claim above, therefore, states that polygonalsets in the plane form a ring of sets.
Informal proof.
Let us present P and Q as a union of finite collection ofelementary sets P , . . . , P k and Q , . . . , Q n correspondingly.Note that P ∪ Q = P ∪ · · · ∪ P k ∪ Q ∪ · · · ∪ Q n . Therefore,
P ∪ Q is polygonal.Note that
P ∩ Q is the union of sets P i ∩ Q j for all i and j . Therefore,in order to show that P ∩ Q is polygonal, it is sufficient to show that each P i ∩ Q j is polygonal for any pair i , j .The diagram should suggest an idea for the proof ofthe latter statement in case if P i and Q j are solid trian-gles. The other cases are simpler; a formal proof can bebuilt on Exercise 20.1The following theorem defines the area as a functionwhich returns a real number for any polygonal set andsatisfying certain conditions. We omit the proof of this theorem. It followsfrom the construction of Lebesgue measure which can be found in any textbook on real analysis. For each polygonal set P in the Euclidean plane thereis a real number s called area of P (briefly s = area P ) such that(a) area ∅ = 0 and area K = 1 where K a solid square with unit side; (b) the conditions P ∼ = Q ⇒ area P = area Q ; P ⊂ Q ⇒ area P area Q ;area P + area Q = area( P ∪ Q ) + area(
P ∩ Q ) hold for any two polygonal sets P and Q .Moreover, the area function P 7→ area P is uniquely defined by the above conditions. For any polygonal set P in the Euclidean plane, wehave area P > . Proof.
Since ∅ ⊂ P , we get thatarea ∅ area P . Since area ∅ = 0 the result follows. Vanishing area and subdivisions
Any one-point set as well as any segment in theEuclidean plane have vanishing area.Proof.
Fix a line segment [ AB ]. Consider a sold square (cid:4) ABCD .Note that given a positive integer n , there are n disjoint segments[ A B ] , . . . , [ A n B n ] in (cid:4) ABCD , such that each [ A i B i ] is congruent to[ AB ] in the sense of the Definition 20.2. A B . . .. . . A n B n Applying the last identity in Theorem 20.7 fewtimes, we get that n · area[ AB ] = area ([ A B ] ∪ · · · ∪ [ A n B n ]) area( (cid:4) ABCD )That is, area[ AB ] n · area( (cid:4) ABCD )for any positive integer n . Therefore, area[ AB ] CHAPTER 20. AREA
On the other hand, by Proposition 20.8,area[ AB ] > { A } we have that ∅ ⊂ { A } ⊂ [ AB ]. Therefore,0 area { A } area[ AB ] = 0 . Hence area { A } = 0. Any degenerate polygonal set in the Euclidean planehas vanishing area.Proof.
Let P be a degenerate set, say P = [ A B ] ∪ · · · ∪ [ A n B n ] ∪ { C , . . . , C k } . Applying Theorem 20.7 together with Proposition 20.8, we get thatarea P area[ A B ] + · · · + area[ A n B n ]++ area { C } + · · · + area { C k } . By Proposition 20.9, the right hand side vanish.On the other hand, area P > Q Q R We say that polygonal set P is subdivided into twopolygonal sets Q and Q if P = Q ∪ Q and the inter-section Q ∩ Q is degenerate. (Recall that accordingto Claim 20.6, the set Q ∩ Q is polygonal.) Assume polygonal sets P is sub-divided into two polygonal set Q and Q . Then area P = area Q + area Q . Proof.
By Theorem 20.7,area P = area Q + area Q − area( Q ∩ Q ) . Since Q ∩ Q is degenerate, by Corollary 20.10,area( Q ∩ Q ) = 0and hence the result.65 Area of solid rectangles
The solid rectangle in the Euclidean plane with sides a and b has area a · b . Assume that a function s returns a nonneg-ative real number s ( a, b ) for any pair of positive real numbers ( a, b ) andit satisfies the following identities: s (1 ,
1) = 1; s ( a, b + c ) = s ( a, b ) + s ( a, c ) s ( a + b, c ) = s ( a, c ) + s ( b, c ) for any a, b, c > . Then s ( a, b ) = a · b for any a, b > . The proof is similar to the proof of Lemma 14.8.
Proof.
Note that if a > a ′ and b > b ′ then ➊ s ( a, b ) > s ( a ′ , b ′ ) . Indeed, since s returns nonnegative numbers, we get that s ( a, b ) = s ( a ′ , b ) + s ( a − a ′ , b ) >> s ( a ′ , b ) = > s ( a ′ , b ′ ) + s ( a ′ , b − b ′ ) >> s ( a ′ , b ′ ) . Applying the second and third identity few times we get that s ( a, m · b ) = s ( m · a, b ) = m · s ( a, b )for any positive integer m . Therefore s ( kl , mn ) = k · s ( l , mn ) == k · m · s ( l , n ) == k · m · l · s (1 , n ) == k · m · l · n · s (1 ,
1) == kl · mn for any positive integers k , l , m and n . That is, the needed identity holdsfor any pair of rational numbers a = kl and b = mn .66 CHAPTER 20. AREA
Arguing by contradiction, assume s ( a, b ) = a · b for some pair of pos-itive real numbers ( a, b ). We will consider two cases: s ( a, b ) > a · b and s ( a, b ) < a · b .If s ( a, b ) > a · b , we can choose a positive integer n such that ➋ s ( a, b ) > ( a + n ) · ( b + n ) . Set k = ⌊ a · n ⌋ + 1 and m = ⌊ b · n ⌋ + 1; equivalently, k and m are positiveintegers such that a < kn a + n and b < mn b + n . By ➊ , we get that s ( a, b ) s ( kn , mn ) == kn · mn ( a + n ) · ( b + n ) , which contradicts ➋ .The case s ( a, b ) < a · b is similar. Fix a positive integer n such that a > n , b > n and ➌ s ( a, b ) < ( a − n ) · ( b − n ) . Set k = ⌈ a · n ⌉ − m = ⌈ b · n ⌉ −
1; that is, a > kn > a − n and b > mn > b − n . Applying ➊ again, we get that s ( a, b ) > s ( kn , mn ) == kn · mn >> ( a − n ) · ( b − n ) , which contradicts ➌ . Proof of Theorem 20.12.
Let R a,b denotes the solid rectangle with sides a and b . Set s ( a, b ) = area R a,b . By theorem 20.7, s (1 ,
1) = 1. That is, the first identity in the algebraiclemma holds.67 R a,c R b,c R a + b,c Note that the rectangle R a + b,c can be subdividedinto two rectangle congruent to R a,c and R b,c . There-fore, by Proposition 20.11,area R a + b,c = area R a,c + area R b,c That is, the second identity in the algebraic lemma holds. The proof ofthe third identity is analogues.It remains to apply the algebraic lemma.
Area of solid parallelograms
Let (cid:3)
ABCD be a parallelogram in the Euclideanplane, a = AB and h be the distance between the lines ( AB ) and ( CD ) .Then area( (cid:4) ABCD ) = a · h. Proof.
Let A ′ and B ′ denote the foot points of A and B on the line ( CD ).Note that ABB ′ A ′ is a rectangle with sides a and h . By Proposi-tion 20.12, ➍ area( (cid:4) ABB ′ A ′ ) = h · a.A B CDA ′ B ′ Without loss of generality, we may as-sume that (cid:4)
ABCA ′ contains (cid:4) ABCD and (cid:4)
ABB ′ A ′ .In this case (cid:4) ABB ′ D admits two sub-divisions. First into (cid:4) ABCD and ▲ AA ′ D .Second into (cid:4) ABB ′ A ′ and ▲ BB ′ C .By Proposition 20.11, ➎ area( (cid:4) ABCD ) + area( ▲ AA ′ D ) == area( (cid:4) ABB ′ A ′ ) + area( ▲ BB ′ C ) . Note that ➏ △ AA ′ D ∼ = △ BB ′ C. Indeed, since the quadrilaterals
ABB ′ A ′ and ABCD are parallelograms,by Lemma 7.16, we have that AA ′ = BB ′ , AD = BC and DC = AB == A ′ B ′ . It follows that A ′ D = B ′ C . Applying the SSS congruencecondition, we get ➏ .In particular, ➐ area( ▲ BB ′ C ) = area( ▲ AA ′ D ) . CHAPTER 20. AREA
A B CD B ′ C ′ D ′ Subtracting ➐ from ➎ , we get thatarea( (cid:4) ABCD ) = area( (cid:4)
ABB ′ D ) . From ➍ , the statement follows. Assume (cid:3)
ABCD and (cid:3) AB ′ C ′ D ′ are two parallelograms such that B ′ ∈ [ BC ] and D ∈∈ [ C ′ D ′ ] . Show that area( (cid:4) ABCD ) = area( (cid:4) AB ′ C ′ D ′ ) . Area of solid triangles
Let a = BC and h A to be the altitude from A in △ ABC . Then area( ▲ ABC ) = · a · h A . Remark.
It is acceptable to write area( △ ABC ) for area( ▲ ABC ), since △ ABC completely determines the solid triangle ▲ ABC . Proof.
Draw the line m thru A which is parallel to ( BC ) and line n thru C parallel to ( AB ). Note that m ∦ n ; set D = m ∩ n . By construction, (cid:3) ABCD is a parallelogram.
A BCD m n Note that (cid:4)
ABCD admits a subdivision into ▲ ABC and ▲ CDA . Therefore,area( (cid:4)
ABCD ) = area( ▲ ABC ) + area( ▲ CDA )Since (cid:3)
ABCD is a parallelogram, Lemma 7.16 im-plies that AB = CD and BC = DA.
Therefore, by the SSS congruence condition, we have △ ABC ∼ = △ CDA .In particular area( ▲ ABC ) = area( ▲ CDA ) . From above and Proposition 20.14, we get thatarea( ▲ ABC ) = · area( (cid:4) ABCD ) == · h A · a Let h A , h B and h C denote the altitudes of △ ABC from vertices A , B and C correspondingly. Note that from Theorem 20.16,it follows that h A · BC = h B · CA = h C · AB.
Give a proof of this statement without using area.
Assume M lies inside the parallelogram ABCD ; thatis, M belongs to the solid parallelogram (cid:4) ABCD , but does not lie on itssides. Show that area( ▲ ABM ) + area( ▲ CDM ) = · area( (cid:4) ABCD ) . Assume that diagonals of a nondegenerate quadrilat-eral
ABCD intersect at point M . Show that area( ▲ ABM ) · area( ▲ CDM ) = area( ▲ BCM ) · area( ▲ DAM ) . Let r be the inradius of △ ABC and p be its semipe-rimeter ; that is, p = · ( AB + BC + CA ) . Show that area( ▲ ABC ) = p · r. Show that for any affine transformation β there is a constant k > such that the equality area[ β ( ▲ )] = k · area ▲ . holds for any solid triangle ▲ .Moreover, if β has the matrix form ( xy ) (cid:0) a bc d (cid:1) · ( xy ) + ( vw ) , then k = | det (cid:0) a bc d (cid:1) | = | a · d − b · c | . Area method
In this section we will give examples of slim proofs using the propertiesof area. Note that these proofs are not truly elementary since the priceone pays to introduce the area function is high.We start with the proof of the Pythagorean theorem. In the Elementsof Euclid, the Pythagorean theorem was formulated as equality ➑ belowand the proof used a similar technique.70 CHAPTER 20. AREA
Proof.
We need to show that if a and b are legs and c is the hypotenuseof a right triangle, then a + b = c . Let T denotes the right solid triangle with legs a and b and by Q x bethe solid square with side x .Let us construct two subdivisions of Q a + b . TT T T Q a Q b T TT T Q c
1. Subdivide Q a + b into two solidsquares congruent to Q a and Q b and 4solid triangles congruent to T , see the leftdiagram.2. Subdivide Q a + b into one solidsquare congruent to Q c and 4 solid righttriangles congruent to T , see the right diagram.Applying Proposition 20.11 few times, we get thatarea Q a + b = area Q a + area Q b + 4 · area T == area Q c + 4 · area T . Therefore, ➑ area Q a + area Q b = area Q c . By Theorem 20.12, area Q x = x , for any x >
0. Hence the statement follows.
Build another proof of the Pythagoreantheorem based on the diagram.(In the notations above it shows a subdivision of Q c into Q a − b and four copies of T if a > b .) Show that the sum of distances from a point to thesides of an equilateral triangle is the same for all points inside the trian-gle.
Let us prove Lemma 8.8 using the area method. That is, we need toshow that if △ ABC is nondegenerate and its angle bisector at A intersects[ BC ] at the point D . Then ABAC = DBDC .
In the proof we will use the following claim:71
Assume that two triangles
ABC and A ′ B ′ C ′ in the Eu-clidean plane have equal altitudes dropped from A and A ′ correspondingly.Then area( ▲ A ′ B ′ C ′ )area( ▲ ABC ) = B ′ C ′ BC .
In particular, the same identity holds if A = A ′ and the bases [ BC ] and [ B ′ C ′ ] lie on one line.Proof. Let h be the altitude. By Theorem 20.16,area( ▲ A ′ B ′ C ′ )area( ▲ ABC ) = · h · B ′ C ′ · h · BC = B ′ C ′ BC . A BC D
Proof of Lemma 8.8.
Applying Claim 20.24, we getthat area( ▲ ABD )area( ▲ ACD ) =
BDCD .
By Proposition 8.10 the triangles
ABD and
ACD have equal altitudes from D . ApplyingClaim 20.24 again, we get thatarea( ▲ ABD )area( ▲ ACD ) =
ABAC and hence the result.
Assume that the point X lies inside a nondegeneratetriangle ABC . Show that X lies on the median from A if and only if area( ▲ ABX ) = area( ▲ ACX ) . Build a proof of Theorem 8.5 based on the Exer-cise 20.25.Namely, show that medians of nondegenerate triangle intersect at onepoint and the point of their intersection divides each median in the ratio1:2.
Area in the neutral planes and spheres
An analog of Theorem 20.7 holds in the neutral planes and spheres. In theformulation of this theorem, the solid unit square K has to be exchangedto a fixed nondegenerate polygonal set. One has to make such change forgood reason — hyperbolic plane and sphere have no unit squares.The set K in this case plays role of the unit measure for the area andchanging K will require conversion of area units.72 CHAPTER 20. AREA
AB CD K n n n According to the standard convention, the set K istaken so that on small scales area behaves like area in theEuclidean plane. Say, if K n denotes the solid quadrilat-eral (cid:4) ABCD with right angles at A , B and C such that AB = BC = n , then we may assume that ➒ n · area K n → n → ∞ . This convention works equally well for spheres and neutral planes,including Euclidean plane. In spherical geometry equivalently we mayassume that if r is the radius of the sphere, then the area of whole sphereis 4 · π · r .Recall that defect of triangle △ ABC is defined asdefect( △ ABC ) := π − | ∡ ABC | − | ∡ BCA | − | ∡ CAB | . It turns out that any neutral plane or sphere there is a real number k such that ➓ k · area( ▲ ABC ) + defect( △ ABC ) = 0for any △ ABC .This number k is called curvature ; k = 0 for the Euclidean plane, k = − k = 1 for the unit sphere and k = r for thesphere of radius r .In particular, it follows that any ideal triangle in h-plane has area π .Similarly in the unit sphere the area of equilateral triangle with rightangles has area π ; since whole sphere can be subdivided in eight suchtriangles, we get that the area of unit sphere is 4 · π .The identity ➓ can be used as an alternative way to introduce areafunction; it works on spheres and all neutral planes, except for the Eu-clidean plane. Quadrable sets
A set S in the plane is called quadrable if for any ε > P and Q such that P ⊂ S ⊂ Q and area
Q − area P < ε. If S is quadrable, its area can be defined as the necessarily unique realnumber s = area S such that the inequalityarea Q s area P P and Q such that P ⊂ S ⊂ Q . Let D be the unit disc; that is, D is a set which con-tains the unit circle Γ and all the points inside Γ .Show that D is a quadrable set. Since D is quadrable, the expression area D makes sense and the con-stant π can be defined as π = area D .It turns out that the class of quadrable sets is the largest class for whichthe area can be defined in such a way that it satisfies all the conditionsin Theorem 20.7 incliding uniqueness.There is a way to define area for all bounded sets which satisfies allthe conditions in the Theorem 20.7 excluding uniqueness. (A set in theplane is called bounded if it lies inside of a circle.)In the hyperbolic plane and in the sphere there is no similar construc-tion. If you wonder why, read about doubling the ball paradox of FelixHausdorff, Stefan Banach and Alfred Tarski. ints Exercise 1.2.
Only the triangle inequality requires a proof — the rest of conditions inDefinition 1.1 are evident. Let A = ( x A , y A ), B = ( x B , y B ) and C = ( x C , y C ). Set x = x B − x A , y = y B − y A ,x = x C − x B , y = y C − y B . (a). The inequality d ( A, C ) d ( A, B ) + d ( B, C )can be written as | x + x | + | y + y | | x | + | y | + | x | + | y | . The latter follows since | x + x | | x | + | x | and | y + y | | y | + | y | . (b). The inequality ➊ d ( A, C ) d ( A, B ) + d ( B, C )can be written as q(cid:0) x + x (cid:1) + (cid:0) y + y (cid:1) q x + y + q x + y . Take the square of the left and the right hand sides, simplify, take the square again andsimplify again. You should get the following inequality:0 ( x · y − x · y ) , which is equivalent to ➊ and evidently true. (c). The inequality d ∞ ( A, C ) d ∞ ( A, B ) + d ∞ ( B, C )can be written as ➋ max {| x + x | , | y + y |} max {| x | , | y |} + max {| x | , | y |} . Without loss of generality, we may assume thatmax {| x + x | , | y + y |} = | x + x | . Further, | x + x | | x | + | x | max {| x | , | y |} + max {| x | , | y |} . Hence ➋ follows. Exercise 1.4. If A = B , then d X ( A, B ) >
0. Since f is distance-preserving, d Y ( f ( A ) , f ( B )) = d X ( A, B ) . Therefore, d Y ( f ( A ) , f ( B )) >
0; hence f ( A ) = f ( B ). Exercise 1.5.
Set f (0) = a and f (1) = b . Note that b = a + 1 or a −
1. Moreover, f ( x ) = a ± x and at the same time, f ( x ) = b ± ( x −
1) for any x .If b = a + 1, it follows that f ( x ) = a + x for any x .The same way, if b = a −
1, it follows that f ( x ) = a − x for any x . Exercise 1.6.
Show that the map ( x, y ) ( x + y, x − y ) is an isometry ( R , d ) →→ ( R , d ∞ ). That is, you need to check if this map is bijective and distance-preserving. Exercise 1.7.
First prove that two points A = ( x A , y A ) and B = ( x B , y B ) on the Man-hattan plane have a unique midpoint if and only if x A = x B or y A = y B ; compare withthe example on page 17.Then use above statement to prove that any motion of the Manhattan plane can bewritten in one of the following two ways:( x, y ) ( ± x + a, ± y + b ) or ( x, y ) ( ± y + b, ± x + a ) , for some fixed real numbers a and b . (In each case we have 4 choices of signs, so for fixedpair ( a, b ) we have 8 distinct motions.) Exercise 1.9.
Assume three points A , B and C lie on one line. Note that in this caseone of the triangle inequalities with the points A , B and C becomes an equality.Set A = ( − , B = (0 ,
0) and C = (1 , d and d all the triangleinequalities with the points A , B and C are strict. It follows that the graph is not a line.For d ∞ show that ( x, | x | ) x gives the isometry of the graph to R . Conclude thatthe graph is a line in ( R , d ∞ ). Exercise 1.10.
Applying the definition of lines, the problems are reduced to the following:Assume that a = b , find the number of solutions for each of the following two equations: | x − a | = | x − b | and | x − a | = 2 ·| x − b | . Each can be solved by taking the square of the left and the right hand sides. Thenumbers of solutions are 1 and 2 correspondingly.
Exercise 1.11.
Fix an isometry f : ( P Q ) → R such that f ( P ) = 0 and f ( Q ) = q > f ( X ) = x . By the definition of the half-line X ∈ [ P Q ) if and only if x >
0. Show that the latter holds if and only if | x − q | = (cid:12)(cid:12) | x | − | q | (cid:12)(cid:12) . Hence the result will follow. HINTS
Exercise 1.12.
The equation 2 · α ≡ · α = 2 · k · π for some integer k .Therefore, α = k · π for some integer k .Equivalently, α = 2 · n · π or α = (2 · n + 1) · π for some integer n . The first identitymeans that α ≡ α ≡ π . Exercise 1.13. (a).
By the triangle inequality, | f ( A ′ ) − f ( A ) | d ( A ′ , A ) . Therefore, we can take δ = ε . (b). By the triangle inequality, | f ( A ′ , B ′ ) − f ( A, B ) | | f ( A ′ , B ′ ) − f ( A, B ′ ) | + | f ( A, B ′ ) − f ( A, B ) | d ( A ′ , A ) + d ( B ′ , B ) . Therefore, we can take δ = ε . Exercise 1.14.
Fix A ∈ X and B ∈ Y such that f ( A ) = B .Fix ε >
0. Since g is continuous at B , there is a positive value δ such that d Z ( g ( B ′ ) , g ( B )) < ε if d Y ( B ′ , B ) < δ . Since f is continuous at A , there is δ > d Y ( f ( A ′ ) , f ( A )) < δ if d X ( A ′ , A ) < δ . Since f ( A ) = B , we get that d Z ( h ( A ′ ) , h ( A )) < ε if d X ( A ′ , A ) < δ . Hence the result.
Exercise 2.1.
By Axiom I, there are at least two points in the plane. Therefore, byAxiom II, the plane contains a line. To prove (a) , it remains to note that line is an infiniteset of points. To prove (b) apply in addition Axiom III.
Exercise 2.3.
By Axiom II, ( OA ) = ( OA ′ ). Therefore, the statement boils down two thefollowing: Assume f : R → R is a motion of the line which sends → and one positive numberto a positive number, then f is an identity map. The latter follows from Exercise 1.5.
Exercise 2.6.
By Proposition 2.5, ∡ AOA = 0. It remains to apply Axiom IIIa.
Exercise 2.10.
Apply Proposition 2.5, Theorem 2.8 and Exercise 1.12.
Exercise 2.11.
By Axiom IIIb,2 · ∡ BOC ≡ · ∡ AOC − · ∡ AOB ≡ . By Exercise 1.12, it implies that ∡ BOC is either 0 or π . It remains to apply Exercise 2.6and Theorem 2.8 correspondingly in these two cases. Exercise 2.12.
Fix two points A and B provided by Axiom I.Fix a real number 0 < α < π . By Axiom IIIa there is a point C such that ∡ ABC = α .Use Proposition 2.2 to show that △ ABC is nondegenerate.
Exercise 2.14.
Applying Proposition 2.13, we get that ∡ AOC = ∡ BOD . It remains toapply Axiom IV.
Exercise 3.1.
Set α = ∡ AOB and β = ∡ BOA . Note that α = π if and only if β = π .Otherwise α = − β . Hence the result. Exercise 3.3.
Set α = ∡ AOB , β = ∡ BOC and γ = ∡ COA . By Axiom IIIb andProposition 2.5, we have ➊ α + β + γ ≡ . Note that 0 < α + β < · π and | γ | π . If γ >
0, then ➊ implies α + β + γ = 2 · π and if γ <
0, then ➊ implies α + β + γ = 0 . Exercise 3.11.
Note that O and A ′ lie on the same side of ( AB ). Analogously O and B ′ lie on the same side of ( AB ). Hence the result. Exercise 3.13.
Apply Theorem 3.7 for △ P QX and △ P QY and then apply Corol-lary 3.10 a . Exercise 3.14.
Note that it is sufficient to consider the cases when A ′ = B, C and B ′ = A, C .Apply Pasch’s theorem (3.12) twice: (1) for △ AA ′ C and ( BB ′ ), and (2) for △ BB ′ C and ( AA ′ ). Exercise 3.15.
Assume that Z is the point of intersection.Note that Z = P and Z = Q . Therefore, Z / ∈ ( P Q ).Show that Z and X lie on one side of ( P Q ). Repeat the argument to show that Z and Y lie on one side of ( P Q ). It follows that X and Y lie on the same side of ( P Q ) — acontradiction.
Exercise 3.20.
The “only-if” part follows from the triangle inequality. To prove “if” partapply Theorem 3.17 for the values r , r and d . Exercise 4.3.
Apply Theorem 4.2 twice.
Exercise 4.5.
Consider the points D and D ′ , such that M is the midpoint of [ AD ] and M ′ is the midpoint of [ A ′ D ′ ]. Show that △ ABD ∼ = △ A ′ B ′ D ′ and use it to prove that △ A ′ B ′ C ′ ∼ = △ ABC . Exercise 4.6. (a)
Apply SAS. (b)
Use (a) and apply SSS.
Exercise 4.7.
Choose B ′ ∈ [ AC ] such that AB = AB ′ . Note that BC = B ′ C . By SSS, △ ABC ∼ = △ AB ′ C . HINTS
Exercise 4.8.
Without loss of generality, we may assume that X is distinct from A , B and C . Set f ( X ) = X ′ ; assume X ′ = X .Note that AX = AX ′ , BX = BX ′ and CX = CX ′ . By SSS we get that ∡ ABX == ± ∡ ABX ′ . Since X = X ′ , we get that ∡ ABX ≡ − ∡ ABX ′ . The same way we get that ∡ CBX ≡ − ∡ CBX ′ . Subtracting these two identities from each other, we get that ∡ ABC ≡ − ∡ ABC.
Conclude that ∡ ABC = 0 or π . That is, △ ABC is degenerate — a contradiction.
Exercise 5.1.
By Axiom IIIb and Theorem 2.8, we have ∡ XOA − ∡ XOB ≡ π. Since | ∡ XOA | , | ∡ XOB | π , we get that | ∡ XOA | + | ∡ XOB | = π. Hence the statement follows.
Exercise 5.3.
Assume X and A lie on the same side of ℓ .Note that A and B lie on opposite sides of ℓ . Therefore, by Corollary 3.10, [ AX ] doesnot intersect ℓ and [ BX ] intersects ℓ ; let Y denotes the intersection point. A BX Yℓ
Note that
Y / ∈ [ AX ]. By Exercise 4.7, BX = AY + Y X > AX.
This way we proved the “if” part. To prove the “only if” part,it remains to switch A and B , repeat the above argument and applyTheorem 5.2. Exercise 5.4.
Apply Exercise 5.3, Theorem 4.1 and Exercise 3.3.
Exercise 5.8.
Choose an arbitrary nondegenerate triangle
ABC . Let △ ˆ A ˆ B ˆ C denotes itsimage after the motion.If A = ˆ A , apply the reflection in the perpendicular bisector of [ A ˆ A ]. This reflectionsends A to ˆ A . Let B ′ and C ′ denote the reflections of B and C correspondingly.If B ′ = ˆ B , apply the reflection in the perpendicular bisector of [ B ′ ˆ B ]. This reflectionsends B ′ to ˆ B . Note that ˆ A ˆ B = ˆ AB ′ ; that is, ˆ A lies on the perpendicular bisector.Therefore, ˆ A reflects to itself. Let C ′′ denotes the reflection of C ′ .Finally, if C ′′ = ˆ C , apply the reflection in ( ˆ A ˆ B ). Note that ˆ A ˆ C = ˆ AC ′′ and ˆ B ˆ C == ˆ BC ′′ ; that is, ( AB ) is the perpendicular bisector of [ C ′′ ˆ C ]. Therefore, this reflectionsends C ′′ to ˆ C .Apply Exercise 4.8 to show that the composition of the constructed reflections coincideswith the given motion. Exercise 5.9.
Note that ∡ XBA = ∡ ABP , ∡ P BC = ∡ CBY . Therefore, ∡ XBY ≡ ∡ XBP + ∡ P BY ≡ · ( ∡ ABP + ∡ P BC ) ≡ · ∡ ABC. ABC DX
Exercise 5.11. If ∠ ABC is right, the statement follows fromLemma 5.10. Therefore, we can assume that ∠ ABC is obtuse.Draw a line ( BD ) perpendicular to ( BA ). Since ∠ ABC is obtuse,the angles
DBA and
DBC have opposite signs.By Corollary 3.10, A and C lie on opposite sides of ( BD ). In partic-ular, [ AC ] intersects ( BD ) at a point; denote it by X .Note that AX < AC and by Lemma 5.10, AB AX . Exercise 5.12.
Let O be the center of the circle. Note that we can assume that O = P .Assume P lies between X and Y . By Exercise 5.1, we can assume that ∠ OP X is rightor obtuse. By Exercise 5.11,
OP < OX ; that is, P lies inside Γ.If P does not lie between X and Y , we can assume that X lies between P and Y .Since OX = OY , Exercise 5.11 implies that ∠ OXY is acute. Therefore, ∠ OXP is obtuse.Applying Exercise 5.11 again we get that
OP > OX ; that is, P lies outside Γ. Exercise 5.13.
Apply Theorem 5.2.
Exercise 5.15.
Use Exercise 5.13 and Theorem 5.5.
Exercise 5.17.
Let P ′ be the reflection of P in ( OO ′ ). Note that P ′ lies on both circlesand P ′ = P if and only if P / ∈ ( OO ′ ). Exercise 5.18.
Apply Exercise 5.17.
Exercise 5.19.
Let A and B be the points of intersection. Note that the centers lie onthe perpendicular bisector of the segment [ AB ]. Exercise 5.21. Exercise 5.22. HINTS
Exercise 5.23. Exercise 5.24.Exercise 6.3.
By the AA similarity condition, the transformation multiplies the sides ofany nondegenerate triangle by some number which may depend on the triangle.Note that for any two nondegenerate triangles which share one side this number is thesame. Applying this observation to a chain of triangles leads to a solution.
Exercise 6.5.
Apply that △ ADC ∼ △
CDB . Exercise 6.6.
Apply the Pythagorean theorem (6.4) and the SSS congruence condition.
Exercise 6.7.
By the AA similarity condition (6.2), △ AY C ∼ △
BXC .Conclude that
Y CAC = XCBC .
Apply the SAS similarity condition to show that △ ABC ∼ △
Y XC .Use AA and equality of vertical angles to prove that △ AZX ∼ △
BZY . Exercise 7.4.
Apply Proposition 7.1 to show that k k m . By Corollary 7.3, k k n ⇒⇒ m k n . The latter contradicts that m ⊥ n . Exercise 7.5.
Repeat the construction in Exercise 5.21 twice.
Exercise 7.8.
By the transversal property 7.6, ∡ B ′ BC ≡ π − ∡ C ′ B ′ B. Since B ′ lies between A and B , we get that ∡ ABC = ∡ B ′ BC and ∡ AB ′ C ′ + ∡ C ′ B ′ B ≡≡ π . Hence ∡ ABC = ∡ AB ′ C ′ .The same way we can prove that ∡ BCA = ∡ B ′ C ′ A . It remains to apply the AAsimilarity condition. Exercise 7.9.
Assume we need to trisect segment [ AB ]. Construct a line ℓ = ( AB ) withfour points A, C , C , C such that C and C trisect [ AC ]. Draw the line ( BC ) anddraw parallel lines thru C and C . The points of intersections of these two lines with( AB ) trisect the segment [ AB ]. Exercise 7.11.
Apply twice Theorem 4.2 and twice Theorem 7.10. Exercise 7.12. If △ ABC is degenerate, then one of the angle measures is π and the othertwo are 0. Hence the result.Assume △ ABC is nondegenerate. Set α = ∡ CAB , β = ∡ ABC and γ = ∡ BCA .By Theorem 3.7, we may assume that 0 < α, β, γ < π . Therefore, ➊ < α + β + γ < · π. By Theorem 7.10, ➋ α + β + γ ≡ π. From ➊ and ➋ the result follows. Exercise 7.13.
Apply twice Theorem 4.2 and once Theorem 7.10.
A B CO X
Exercise 7.14.
Let O denotes the center of the circle.Note that △ AOX is isosceles and ∠ OXC is right. Applying 7.10and 4.2 and simplifying, you should get4 · ∡ CAX ≡ π. Show that ∠ CAX has to be acute. It follows then that ∡ CAX == ± π . Exercise 7.15.
Apply Theorem 7.10 to △ ABC and △ BDA . Exercise 7.17.
Since △ ABC is isosceles, ∡ CAB = ∡ BCA .By SSS, △ ABC ∼ = △ CDA . Therefore, ± ∡ DCA = ∡ BCA = ∡ CAB.
Since D = C , we get “ − ” in the last formula. Use the transversal property (7.6) toshow that ( AB ) k ( CD ). Repeat the argument to show that ( AD ) k ( BC ) Exercise 7.18.
Apply Lemma 7.16 together with the transversal property (7.6) to thediagonals and a pair of opposite sides. After that use the ASA-congruence condition (4.1).
Exercise 7.19.
By Lemma 7.16 and SSS, AC = BD ⇐⇒ ∡ ABC = ± ∡ BCD.
By the transversal property (7.6), ∡ ABC + ∡ BCD ≡ π. Therefore, AC = BD ⇐⇒ ∡ ABC = ∡ BCD = ± π . Exercise 7.20.
Fix a parallelogram
ABCD . By Exercise 7.18, its diagonals [ AC ] and[ BD ] have a common midpoint; denote it by M .Use SSS and Lemma 7.16 to show that AB = CD ⇐⇒ △ AMB ∼ = △ AMD ⇐⇒ ∡ AMB = ± π . HINTS
Exercise 7.21. (a).
Use the uniqueness of the parallel line (Theorem 7.2). (b)
Use Lemma 7.16 and the Pythagorean theorem (6.4).
Exercise 7.22.
Set A = (0 , B = ( c,
0) and C = ( x, y ). Clearly, AB = c , AC = x + y and BC = ( c − x ) + y .It remains to show that there is a pair of real numbers ( x, y ) which satisfy the followingsystem of equations: ( b = x + y a = ( c − x ) + y if 0 < a b c a + c . Exercise 7.23.
Let M = ( x, y ). Note that MA = MB if and only if( x − x A ) + ( y − y A ) = ( x − x B ) + ( y − y B ) . To prove the first, simplify this equation. For the second part use that any line is aperpendicular bisector to some line segment.
Exercise 7.24.
Rewrite the equation as( x + a ) + ( y + b ) = ( a ) + ( b ) − c and think. Exercise 7.25.
We can choose the coordinates so that B = (0 ,
0) and A = ( a,
0) for some a >
0. If M = ( x, y ), then the equation AM = k · BM can be written in coordinates as k · ( x + y ) = ( x − a ) + y . It remains to rewrite this equation as in Exercise 7.24.
Exercise 8.2.
Apply Theorem 8.1 and Theorem 5.2.
Exercise 8.4.
Note that ( AC ) ⊥ ( BH ) and ( BC ) ⊥ ( AH ) and apply Theorem 8.3. Exercise 8.6.
Use the idea from the proof of Theorem 8.5 to show that ( XY ) k ( AC ) kk ( V W ) and ( XV ) k ( BD ) k ( Y W ). Exercise 8.7.
Let ( BX ) and ( BY ) be the internal and external bisectors of ∠ ABC . Then2 · ∡ XBY ≡ · ∡ XBA + 2 · ∡ ABY ≡≡ ∡ CBA + π + 2 · ∡ ABC ≡≡ π + ∡ CBC = π and hence the result. Exercise 8.9. If E is the point of intersection of ( BC ) with the external bisector of ∠ BAC , then
ABAC = EBEC .
It can be proved along the same lines as Lemma 8.8. Exercise 8.12.
Apply Lemma 8.8. Also see the solution of Exercise 11.2.
Exercise 8.13.
Apply ASA for the two triangles which the bisector cuts from the originaltriangle.
Exercise 8.14.
Let I be the incenter. By SAS, we get that △ AIZ ∼ = △ AIY . Therefore, AY = AZ . The same way we get that BX = BZ and CX = CY . Hence the result. Exercise 8.15.
Let △ ABC be the given acute triangle and △ A ′ B ′ C ′ be its orthic triangle.Note that △ AA ′ C ∼ △ BB ′ C . Use it to show that △ A ′ B ′ C ∼ △ ABC .The same way we get that △ AB ′ C ′ ∼ △ ABC . It follows that ∡ A ′ B ′ C = ∡ AB ′ C ′ .Conclude that ( BB ′ ) bisects ∠ A ′ B ′ C ′ .If △ ABC is obtuse, then its orthocenter coincides with one of the excenters of △ ABC ;that is, the point of intersection of two external and one internal bisectors of △ ABC . Exercise 8.16.
Apply 4.2, 7.6 and 7.16.
Exercise 9.3. (a).
Apply Theorem 9.2 for ∠ XX ′ Y and ∠ X ′ Y Y ′ and Theorem 7.10 for △ P Y X ′ . (b) If P is inside of Γ then P lies between X and X ′ and between Y and Y ′ in this case ∠ XP Y is vertical to ∠ X ′ P Y ′ . If P is outside of Γ then [ P X ) = [
P X ′ ) and [ P Y ) = [
P Y ′ ).In both cases we have that ∡ XP Y = ∡ X ′ P Y ′ .Applying Theorem 9.2 and Exercise 2.11, we get that2 · ∡ Y ′ X ′ P ≡ · ∡ Y ′ X ′ X ≡ · ∡ Y ′ Y X ≡ · ∡ P Y X.
According to Theorem 3.7, ∠ Y ′ X ′ P and ∠ P Y X have the same sign; therefore ∡ Y ′ X ′ P = ∡ P Y X.
It remains to apply the AA similarity condition. (c)
Apply (b) assuming [
Y Y ′ ] is the diameter of Γ. Exercise 9.4.
Apply Exercise 9.3 b three times. Exercise 9.5.
Let X any Y be the foot points of the altitudes from A and B . Let O denotes the circumcenter.By AA condition, △ AXC ∼ △
BY C . Thus ∡ A ′ OC ≡ · ∡ A ′ AC ≡ − · ∡ B ′ BC ≡ − ∡ B ′ OC.
By SAS, △ A ′ OC ∼ = △ B ′ OC . Therefore, A ′ C = B ′ C . Exercise 9.7.
Construct the circles Γ and Γ ′ on the diameters [ AB ] and [ A ′ B ′ ] corre-spondingly. By Corollary 9.6, any point Z in the intersection Γ ∩ Γ ′ will do. Exercise 9.8.
Note that ∡ AA ′ B = ± π and ∡ AB ′ B = ± π . Then apply Theorem 9.10 to (cid:3) AA ′ BB ′ .If O is the center of the circle, then ∡ AOB ≡ · ∡ AA ′ B ≡ π. That is, O is the midpoint of [ AB ]. HINTS
Exercise 9.9.
Guess the construction from the diagram. Toprove it, apply Theorem 8.3 and Corollary 9.6.
Exercise 9.11.
Apply the transversal property (7.6) and thetheorem on inscribed angles (9.2).
Exercise 9.12.
Apply Theorem 9.10 twice for (cid:3)
ABY X and (cid:3)
ABY ′ X ′ and use the transversal property (7.6). Exercise 9.14.
One needs to show that the lines ( A ′ B ′ ) and( XP ) are not parallel, otherwise the first line in the proof doesnot make sense.In addition, the following identities:2 · ∡ AXP ≡ · ∡ AXY, · ∡ ABP ≡ · ∡ ABB ′ , · ∡ AA ′ B ′ ≡ · ∡ AA ′ Y. Exercise 9.15.
By Corollary 9.6, the points L , M and N lie on the circle Γ with diame-ter [ OX ]. It remains to apply Theorem 9.2 for the circle Γ and two inscribed angles withvertex at O . A BC XY Z P
Advanced exercise 9.16.
Let X , Y and Z denote the foot pointsof P on ( BC ), ( CA ) and ( AB ) correspondingly.Notice that (cid:3) AZP Y , (cid:3) BXP Z , (cid:3) CY P X and (cid:3)
ABCP areinscribed. Therefore2 · ∡ CXY ≡ · ∡ CP Y, · ∡ BXZ ≡ · ∡ BP Z, · ∡ Y AZ ≡ · ∡ Y P Z, · ∡ CAB ≡ · ∡ CP B.
Conclude that 2 · ∡ CXY ≡ · ∡ BXZ and hence X , Y and Z lie on one line. Exercise 9.20.
By Theorem 7.10, ∡ ABC + ∡ BCA + ∡ CAB ≡ π. It remains to apply Proposition 9.19 twice.
Exercise 9.21. If C ∈ ( AX ), then the arc is the line segment [ AC ] or the union of twohalf-lines in ( AX ) with vertices at A and C .Assume C / ∈ ( AX ). Let ℓ be the perpendicular line dropped from A to ( AX ) and m be the perpendicular bisector of [ AC ].Note that ℓ ∦ m ; set O = ℓ ∩ m . Note that the circle with center O passing thru A isalso passing thru C and tangent to ( AX ).Note that one the two arcs with endpoints A and C is tangent to [ AX ).The uniqueness follow from the propositions 9.18 and 9.19. Exercise 9.22.
Use 9.19 and 7.10 to show that ∡ XAY = ∡ ACY.
By Axiom IIIc, ∡ ACY → AY →
0; hence the result.
Exercise 9.23.
Apply Proposition 9.19 twice. Exercise 9.24.
Guess the construction from the diagram. Toshow that it produces the needed point, apply Theorem 9.2.
Exercise 10.1.
By Lemma 5.16, ∠ OT P ′ is right. Therefore, △ OP T ∼ △
OT P ′ and in particular OP · OP ′ = OT and hence the result. Exercise 10.3.
Let O denotes the center of Γ. Suppose X, Y ∈∈ Γ; in particular, OX = OY .Note that the inversion sends X and Y to themselves. ByLemma 10.2, △ OP X ∼ △
OXP ′ and △ OP Y ∼ △
OY P ′ . Therefore,
P XP ′ X = OPOX = OPOY = P YP ′ Y and hence the result. Exercise 10.4.
By Lemma 10.2, ∡ IA ′ B ′ ≡ − ∡ IBA, ∡ IB ′ C ′ ≡ − ∡ ICB, ∡ IC ′ A ′ ≡ − ∡ IAC, ∡ IB ′ A ′ ≡ − ∡ IAB, ∡ IC ′ B ′ ≡ − ∡ IBC, ∡ IA ′ C ′ ≡ − ∡ ICA.
It remains to apply the theorem on the sumof angles of triangle (7.10) to show that ( A ′ I ) ⊥⊥ ( B ′ C ′ ), ( B ′ I ) ⊥ ( C ′ A ′ ) and ( C ′ I ) ⊥ ( B ′ A ′ ). Exercise 10.5.
Guess the construction from thediagram (the two nonintersecting lines on the di-agram are parallel).
Exercise 10.8.
First show that for any r > x, y distinct from 0, r ( x + y ) / (cid:18) r x + r y (cid:19) / x = y .Let ℓ denotes the line passing thru Q , Q ′ and the center of the inversion O . Choose anisometry ℓ → R which sends O to 0; assume x, y ∈ R are the values of ℓ for the two pointsof intersection ℓ ∩ Γ; note that x = y . Assume r is the radius of the circle of inversion. Thenthe left hand side above is the coordinate of Q ′ and the right hand side is the coordinateof the center of Γ ′ . Exercise 10.9.
A solution is given on page 157. HINTS
P Q Q ′ A A ′ B B ′ X X ′ Y Y ′ Exercise 10.10.
Apply an inversion in acircle with the center at the only point ofintersection of the circles; then use Theo-rem 10.11.
Exercise 10.13.
Label the points of tan-gency by X , Y , A , B , P and Q as on thediagram above. Apply an inversion withthe center at P . Observe that the twocircles which tangent at P become par-allel lines and the remaining two circlesare tangent to each other and these twoparallel lines.Note that the points of tangency A ′ , B ′ , X ′ and Y ′ with the parallel lines are vertexesof a square; in particular they lie on one circle. These points are images of A , B , X and Y under the inversion. By Theorem 10.7, the points A , B , X and Y also lie on one circline. Advanced exercise 10.14.
Apply the inversion in a circle with center A . The point A will go to infinity, the two circles tangent at A will become parallel lines and the twoparallel lines will become circles tangent at A ; see the diagram. B ′ A It remains to show that the dashed line AB ′ isparallel to the other two lines. Exercise 10.19.
Let P and P be the inverses of P in Ω and Ω . Note that the points P , P and P aremutually distinct.Use Theorem 8.1, to show that there is uniquecircline Γ which passes thru P , P and P . Use Corol-lary 10.17 to show that Γ ⊥ Ω and Γ ⊥ Ω . Use Theorem 10.15 to prove uniqueness. Exercise 10.20.
Apply Theorem 10.6 b , Exercise 7.15 and Theorem 9.2. Exercise 10.21.
Let T denotes a point of intersection of Ω and Ω . Let P be the footpoint of T on ( O O ). Show that △ O P T ∼ △ O T O ∼ △ T P O . Conclude that P coincides with the inverses of O in Ω and of O in Ω . Exercise 10.22.
Since Γ ⊥ Ω and Γ ⊥ Ω , Corollary 10.16 implies that the circles Ω and Ω are inverted in Γ to themselves.Therefore, the points A and B are inverses of each other.Since Ω ∋ A, B , Corollary 10.17 implies that Ω ⊥ Γ. Exercise 10.23.
Follow the solution of Exercise 10.19.
Exercise 11.2 . Let D denotes the midpoint of [ BC ]. Assume ( AD ) is the angle bisectorat A .Let A ′ ∈ [ AD ) be the point distinct from A such that AD = A ′ D . Note that △ CAD ∼ = ∼ = △ BA ′ D . In particular, ∡ BAA ′ = ∡ AA ′ B . It remains to apply Theorem 4.2 for △ ABA ′ . Exercise 11.3.
The statement is evident if A , B , C and D lie on one line. In the remaining case, let O denotes the circumcenter. Apply theorem about isoscelestriangle (4.2) to the triangles AOB , BOC , COD , DOA . (Note that in the Euclidean plane the statement follows from Theorem 9.10 and Exer-cise 7.15, but one cannot use these statements in the neutral plane.) Exercise 11.5.
Arguing by contradiction, assume2 · ( ∡ ABC + ∡ BCD ) ≡ , but ( AB ) ∦ ( CD ). Let Z be the point of intersection of ( AB ) and ( CD ).Note that 2 · ∡ ABC ≡ · ∡ ZBC, · ∡ BCD ≡ · ∡ BCZ.
Apply Proposition 11.4 to △ ZBC and try to arrive to a contradiction. B ′ A ′ C ′ C ′′ Exercise 11.6.
Let C ′′ ∈ [ B ′ C ′ ) be the point such that B ′ C ′′ = BC .Note that by SAS, △ ABC ∼ = △ A ′ B ′ C ′′ . Conclude that ∡ B ′ C ′ A ′ = ∡ B ′ C ′′ A ′ .Therefore, it is sufficient to show that C ′′ = C ′ . If C ′ = C ′′ applyProposition 11.4 to △ A ′ C ′ C ′′ and try to arrive to a contradiction. Exercise 11.7.
Use Exercise 5.4 and Proposition 11.4.Alternatively, use the same argument as in the solution of Exer-cise 5.12.
Exercise 11.10.
Note that | ∡ ADC | + | ∡ CDB | = π . Then applythe definition of the defect. Exercise 12.1.
Let A and B be the ideal points of the h-line ℓ . Notethat the center of the Euclidean circle containing ℓ lies at the intersection of the linestangent to the absolute at the ideal points of ℓ . Exercise 12.2.
Assume A is an ideal point of the h-line ℓ and P ∈ ℓ . Let P ′ denotes theinverse of P in the absolute. By Corollary 10.16, ℓ lies in the intersection of the h-planeand the (necessarily unique) circline passing thru P , A and P ′ . Exercise 12.3.
Let Ω and O denote the absolute and its center.Let Γ be the circline containing [ P Q ] h . Note that [ P Q ] h = [ P Q ] if and only if Γ is aline.Let P ′ denotes the inverse of P in Ω. Note that O , P and P ′ lie on one line.By the definition of h-line, Ω ⊥ Γ. By Corollary 10.16, Γ passes thru P and P ′ .Therefore, Γ is a line if and only if it pass thru O . Exercise 12.4.
Assume that the absolute is a unit circle.Set a = OX = OY . Note that 0 < a < and OX h = ln a − a , XY h = ln (1+2 · a ) · (1 − a )(1 − · a ) · (1+ a ) . It remains to check that the inequalities1 < a − a < (1+2 · a ) · (1 − a )(1 − · a ) · (1+ a ) HINTS hold if 0 < a < . Exercise 12.5.
Spell the meaning of terms “perpendicular” and “h-line” and then applyExercise 10.19.
Exercise 12.8.
Apply the main observation (12.7 b ). Exercise 12.12.
Let X and Y denote the points of intersections of ( OP ) and ∆ ′ ρ . Consideran isometry ( OP ) → R such that O corresponds to 0. Let x , y , p and ˆ p denote the realnumbers corresponding to X , Y , P and ˆ P .We can assume that p > x < y . Note that ˆ p = x + y and(1 + x ) · (1 − p )(1 − x ) · (1 + p ) = (1 + p ) · (1 − y )(1 − p ) · (1 + y ) . It remains to show that all this implies 0 < ˆ p < p . Exercise 12.22.
Look at the diagram and think.
Advanced exercise 12.25.
By Corollary 10.26 andTheorem 10.6, the right hand sides in the identities sur-vive under an inversion in a circle perpendicular to theabsolute.As usual we assume that the absolute is a unit circle.Let O denotes the h-midpoint of [ P Q ] h . By the mainobservation (12.7) we can assume that O is the centerof the absolute. In this case O is also the Euclideanmidpoint of [ P Q ].Set a = OP = OQ ; in this case we have P Q = 2 · a, P P ′ = QQ ′ = a − a,P ′ Q ′ = 2 · a , P Q ′ = QP ′ = a + a. and P Q h = ln (1+ a ) (1 − a ) = 2 · ln a − a . Thereforech[ · P Q h ] = · ( a − a + − a a ) = r P Q ′ · P ′ QP P ′ · QQ ′ = a + a a − a == a − a ; = a − a . Hence the part (a) follows. Similarly,sh[ · P Q h ] = · (cid:16) a − a − − a a (cid:17) = r P Q · P ′ Q ′ P P ′ · QQ ′ = 2 a − a == · a − a ; = · a − a . Hence the part (b) follows. The parts (c) and (d) follow from (a) , (b) , the definition of hyperbolic tangent andthe double-argument identity for hyperbolic cosine, see 12.24. Exercise 13.3.
By triangle inequality, the h-distance from B to ( AC ) h is at least 50. Itremains to estimate | ∡ h ABC | using Corollary 13.2. The inequalities cos ϕ − · ϕ for | ϕ | < π and e >
10 should help to finish the proof.
Exercise 13.5.
Note that the angle of parallelism of B to ( CD ) h is bigger than π , andit converges to π as CD h → ∞ .Applying Proposition 13.1, we get that BC h < · ln 1 + √ − √ = ln (cid:16) √ (cid:17) . The right hand side is the limit of BC h if CD h → ∞ . Therefore, ln (cid:0) √ (cid:1) is theoptimal upper bound. Exercise 13.6.
As usual, we assume that the absolute is a unit circle.Let
P QR be a hyperbolic triangle with a right angle at Q , such that P Q h = QR h andthe vertices P , Q and R lie on a horocycle.Without loss of generality, we may assume that Q is the center of the absolute. In thiscase ∡ h P QR = ∡ P QR = ± π and P Q = QR . Q PRA B
Note that Euclidean circle passing thru P , Q and R is tangent to the absolute. Conclude that P Q = √ .Apply 12.9 to find P Q h . Exercise 13.9.
Apply AAA-congruence condition(13.8).
Exercise 13.12.
Apply Proposition 13.11. Use that e >> r e − r is decreasing. Exercise 13.14.
Apply the hyperbolic Pythagoreantheorem and the definition of hyperbolic cosine.
Exercise 14.1.
Assume that a triple of noncollinear points P , Q and R are mapped toone line ℓ . Note that all three lines ( P Q ), ( QR ) and ( RP ) are mapped to ℓ . Therefore,any line which connects two points on these three lines is mapped to ℓ .Note that any point in the plane lies on a line passing thru two distinct points on thesethree lines. Therefore, the whole plane is mapped to ℓ . The latter contradicts that themap is a bijection. Exercise 14.2.
Assume the two distinct lines ℓ and m are mapped to the intersectinglines ℓ ′ and m ′ . Let P ′ denotes their point of intersection. A BM
Let P be the inverse image of P ′ . By the definition of affine map, ithas to lie on both ℓ and m ; that is, ℓ and m are intersecting. Hence theresult. Exercise 14.3.
According to the remark before the exercise, it is suffi-cient to construct the midpoint of [ AB ] with a ruler and a parallel tool.Guess a construction from the diagram. HINTS
Exercise 14.4.
Let O , E , A and B denote the points with the coordinates(0 , , a,
0) and ( b,
0) correspondingly.To construct a point W with the coordinates (0 , a + b ), try to construct two parallel-ograms ABP Q and
BW P Q .To construct Z with coordinates (0 , a · b ) choose a line ( OE ′ ) = ( OE ) and try toconstruct the points A ′ ∈ ( OE ′ ) and Z ∈ ( OE ) so that △ OEE ′ ∼ △ OAA ′ and △ OE ′ B ∼∼ △ OA ′ Z . Exercise 14.5.
Draw two parallel chords [ XX ′ ] and [ Y Y ′ ]. Set Z = ( XY ) ∩ ( X ′ Y ′ ) and Z ′ = ( XY ′ ) ∩ ( X ′ Y ). Note that ( ZZ ′ ) passes thru the center.Repeat the same construction for another pair of parallel chords. The center lies inthe intersection of the obtained lines. Exercise 14.6.
Assume a construction produces two perpendicular lines. Apply a shearmapping which changes the angle between the lines.Note that it transforms the construction to the same construction for other free choicespoints. Therefore, this construction does not produce perpendicular lines in general (itmight be the center only by coincidence).
Exercise 14.11.
Apply 10.25 and 14.10.
Exercise 14.12.
Fix a line ℓ . Choose a circle Γ with its center not on ℓ . Let Ω be theinverse of ℓ in Γ; note that Ω is a circle.Let ι Γ and ι Ω denote the inversions in Γ and Ω. Apply 10.26 to show that the compo-sition ι Γ ◦ ι Ω ◦ ι Γ is the reflection in ℓ . Exercise 15.1.
Since O , P and P ′ lie on one line we have that the coordinates of P ′ are proportional to the coordinates of P . The y coordinate of P ′ has to be equal to 1.Therefore, P ′ has coordinates ( y , , zy ). Exercise 15.5.
Assume that ( AB ) meets ( A ′ B ′ ) at O . Since ( AB ′ ) k ( BA ′ ), we get that △ OAB ′ ∼ △ OBA ′ and OAOB = OB ′ OA ′ . Similarly, since ( AC ′ ) k ( CA ′ ), we get that OAOC = OC ′ OA ′ . Therefore
OBOC = OC ′ OB ′ . Applying the SAS similarity condition, we get that △ OBC ′ ∼ △ OCB ′ . Therefore,( BC ′ ) k ( CB ′ ).The case ( AB ) k ( A ′ B ′ ) is similar. Exercise 15.6.
Assume the contrary. Choose two parallel lines ℓ and m . Let L and M be their dual points. Set s = ( ML ), then its dual point S has to lie on both ℓ and m — acontradiction. Exercise 15.8.
Assume M = ( a, b ) and the line s is given by the equation p · x + q · y = 1.Then M ∈ s is equivalent to p · a + q · b = 1. The latter is equivalent to m ∋ S where m is the line given by the equation a · x + b · y = 1and S = ( p, q ).To extend this bijection to the whole projective plane, assume that (1) the ideal linecorresponds to the origin and (2) the ideal point given by the pencil of the lines b · x − a · y = c for different values of c corresponds to the line given by the equation a · x + b · y = 0. PP ′ Q Q ′ RR ′ Exercise 15.10.
Assume one set of concurrent lines a , b , c , andanother set of concurrent lines a ′ , b ′ , c ′ are given. Set P = b ∩ c ′ , Q = c ∩ a ′ , R = a ∩ b ′ ,P ′ = b ′ ∩ c, Q ′ = c ′ ∩ a, R ′ = a ′ ∩ b. Then the lines (
P P ′ ), ( QQ ′ ) and ( RR ′ ) are concurrent. Exercise 15.11.
To solve (a) , assume ( AA ′ ) and ( BB ′ ) are the givenlines and C is the given point. Apply the dual Desargues’ theorem(15.9) to construct C ′ so that ( AA ′ ), ( BB ′ ) and ( CC ′ ) are concurrent.Since ( AA ′ ) k ( BB ′ ), we get that ( AA ′ ) k ( BB ′ ) k ( CC ′ ).A similar solution can be build on the dual Pappus’ theorem, see Exercise 15.10.For part (b) , apply one of the discussed theorems to construct a line parallel to thegiven one and then apply part (a) . Exercise 15.12.
Let A , B , C and D be the point provided by Axiom p-III. Given aline ℓ , we can assume that A / ∈ ℓ , otherwise permute the labels of the points. Then byaxioms p-I and p-II, the three lines ( AB ), ( AC ) and ( AD ) intersect ℓ at distinct points.In particular, ℓ contains at least three points. Exercise 15.13.
Let A , B , C and D be the point provided by Axiom p-III. Show thatthe lines ( AB ), ( BC ), ( CD ) and ( DA ) satisfy Axiom p-III ′ . The proof of the converse issimilar. Exercise 15.14.
Let ℓ be a line with n + 1 points on it.By Axiom p-III, given any line m there is a point P which does not lie on ℓ nor on m .By axioms p-I and p-II, there is a bijection between the lines passing thru P and thepoints on ℓ . In particular, there are exactly n + 1 lines passing thru P .The same way there is bijection between the lines passing thru P and the points on m .Hence (a) follows.Fix a point X . By Axiom p-I, any point Y in the plane lies in a unique line passingthru X . From part (a) , each such line contains X and yet n point. Hence (b) follows.To solve (c) , show that the equation n + n + 1 = 10 does not admit an integer solutionand then apply part (b) .To solve (d) , count the number of lines crossing a given line using the part (a) andapply (b) . Exercise 16.2.
Applying the Pythagorean theorem, we get thatcos AB s = cos AC s · cos BC s = . Therefore, AB s = π . HINTS
A BC
Alternatively, look at the tessellation of the sphere on the pic-ture; it is made from 24 copies of △ s ABC and yet 8 equilateraltriangles. From the symmetry of this tessellation, it follows that[ AB ] s occupies of the equator. Exercise 16.6.
Consider the inversion of the base in a spherewith the center at the tip of the cone and apply Theorem 16.3.
Exercise 16.7.
Note that points on Ω do not move. Moreover,the points inside Ω are mapped outside of Ω and the other wayaround.Further, note that this map sends circles to circles; moreover, the perpendicular circlesare mapped to perpendicular circles. In particular, the circles perpendicular to Ω aremapped to themselves.Consider arbitrary point
P / ∈ Ω. Let P ′ denotes the inverse of P in Ω. Choose twodistinct circles which pass thru P and P ′ . According to Corollary 10.17, Γ ⊥ Ω andΓ ⊥ Ω.Therefore, the inversion in Ω sends Γ to itself and the same holds for Γ .The image of P has to lie on Γ and Γ . Since the image of P is distinct from P , weget that it has to be P ′ . Exercise 16.8.
Apply Theorem 16.3 b . Exercise 16.9.
Set z = P ′ Q ′ . Note that yz → x → x → zx = 21 + OP . Recall that the stereographic projection is the inversion in the sphere Υ with the centerat the south pole S restricted to the plane Π. Show that there is a plane Λ passing thru S , P , Q , P ′ and Q ′ . In the plane Λ, the map Q Q ′ is an inversion in the circle Υ ∩ Λ.This reduces the problem to Euclidean plane geometry. The remaining calculations inΛ are similar to those in the proof of Lemma 13.10.
Exercise 16.10. (a) . Observe and use that OA ′ = OB ′ = OC ′ . (b). Note that the medians of spherical triangle
ABC map to the medians of Euclidean atriangle A ′ B ′ C ′ . It remains to apply Theorem 8.5 for △ A ′ B ′ C ′ . Exercise 17.1.
Let N , O , S , P , P ′ and ˆ P be as on the diagram on page 134.Notice that △ NOP ∼ △ NP ′ S ∼ △ P ′ ˆ P P and2 · NO = NS . It remains to do some algebraic manipu-lations. Exercise 17.3.
Consider the bijection P ↔ ˆ P of the h-plane with absolute Ω. Note that ˆ P ∈ [ A i B i ] if and onlyif P ∈ Γ i . Exercise 17.4.
The observation follows since the reflec-tion in the perpendicular bisector of [
P Q ] is a motion ofthe Euclidean plane, and a motion of the h-plane as well.Without loss of generality, we may assume that thecenter of the circumcircle coincides with the center of the absolute. In this case the h-medians of the triangle coincide with the Euclidean medians.It remains to apply Theorem 8.5. Exercise 17.5.
Let ˆ ℓ and ˆ m denote the h-lines in the conformal model which correspondto ℓ and m . We need to show that ˆ ℓ ⊥ ˆ m as arcs in the Euclidean plane.The point Z , where s meets t , is the center of the circle Γ containing ˆ ℓ .If ¯ m is passing thru Z , then the inversion in Γ exchanges the ideal points of ˆ ℓ . Inparticular, ˆ ℓ maps to itself. Hence the result. P Qϕ
Exercise 17.6.
Let Q be the foot point of P on the line and ϕ be theangle of parallelism. We can assume that P is the center of the absolute.Therefore P Q = cos ϕ and P Q h = · ln 1 + cos ϕ − cos ϕ . Exercise 17.7.
Apply Exercise 17.6 for ϕ = π . Exercise 17.8.
Note that b = · ln 1 + t − t ;therefore ➊ ch b = · (cid:16)q t − t + q − t t (cid:17) = 1 √ − t . The same way we get that ➋ ch c = 1 √ − u . Let X and Y are the ideal points of ( BC ) h . Applying the Pythagorean theorem (6.4)again, we get that CX = CY = p − t . Therefore, a = · ln √ − t + s √ − t − s , and ➌ ch a = · s √ − t + s √ − t − s + s √ − t − s √ − t + s == √ − t √ − t − s == √ − t √ − u . Finally, note that ➊ , ➋ and ➌ imply the theorem. Exercise 17.10.
In the Euclidean plane, the circle Γ is tangent to k ; that is, the point T of intersection of Γ and k is unique. It defines a unique line ( P T ) parallel to ℓ . Exercise 18.1.
Use that | z | = z · ¯ z for z = v , w and v · w . HINTS
Exercise 18.2.
Given a quadrilateral
ABCD , we can choose the complex coordinates sothat A has complex coordinate 0. Rewrite the terms in the Ptolemy’s inequality in termsof the complex coordinates u , v and w of B , C and D ; apply the identity and the triangleinequality. Exercise 18.3.
Let z , v and w denote the complex coordinates of Z , V and W corre-spondingly. Then ∡ ZV W + ∡ V W Z + ∡ W ZV ≡ arg w − vz − v + arg z − wv − w + arg v − zw − z ≡≡ arg ( w − v ) · ( z − w ) · ( v − z )( z − v ) · ( v − w ) · ( w − z ) ≡≡ arg( − ≡≡ π. Exercise 18.4.
Note and use that ∡ EOV = ∡ W OZ = arg v, OWOZ = OZOW = | v | . Exercise 18.6.
Note thatarg ( v − u ) · ( z − w )( v − w ) · ( z − u ) ≡ arg v − uz − u + arg z − wv − w == ∡ ZUV + ∡ V W Z.
The statement follows since the value ( v − u ) · ( z − w )( v − w ) · ( z − u ) is real if and only if2 · arg ( v − u ) · ( z − w )( v − w ) · ( z − u ) ≡ . Exercise 18.7.
Check the following identity:( v − u ) · ( z − w )( v − w ) · ( z − u ) · ( v ′ − u ′ ) · ( z ′ − w ′ )( v ′ − w ′ ) · ( z ′ − u ′ ) = ( v − u ) · ( u ′ − v ′ )( v − v ′ ) · ( u ′ − u ) · ( z − w ) · ( w ′ − z ′ )( z − z ′ ) · ( w ′ − w ) ·· ( v − v ′ ) · ( w ′ − w )( v − w ) · ( w ′ − v ′ ) · ( z − z ′ ) · ( u ′ − u )( z − u ) · ( u ′ − z ′ ) . By Theorem 18.5, five from six cross ratios in the this identity are real. Therefore so isthe sixth cross ratio; it remains to apply the theorem again.
Exercise 18.9.
We can choose the complex coordinates so that the points O , E , A , B and C have coordinates 0, 1, 1 + i , 2 + i and 3 + i correspondingly.Set ∡ EOA = α , ∡ EOB = β and ∡ EOC = γ . Note that α + β + γ ≡ arg(1 + i ) + arg(2 + i ) + arg(3 + i ) = ≡ arg[(1 + i ) · (2 + i ) · (3 + i )] = ≡ arg[10 · i ] = ≡ π . Note that these three angles are acute and conclude that α + β + γ = π . Exercise 18.13.
Show that the inverse of each elementary transformation is elementaryand use Proposition 18.11.
Exercise 18.14.
The fractional linear transformation f ( z ) = ( z − z ∞ ) · ( z − z )( z − z ) · ( z − z ∞ )meets the conditions.To show the uniqueness, assume there is another fractional linear transformation g ( z )which meets the conditions. Then the composition h = g ◦ f − is a fractional lineartransformation; set h ( z ) = a · z + bc · z + d .Note that h ( ∞ ) = ∞ ; therefore, c = 0. Further, h (0) = 0 implies b = 0. Finally, since h (1) = 1, we get that ad = 1. Therefore, h is the identity ; that is, h ( z ) = z for any z . Itfollows that g = f . Exercise 18.15.
Let Z ′ be the inverse of the point Z . Assume that the circle of theinversion has center W and radius r . Let z , z ′ and w denote the complex coordinate ofthe points Z , Z ′ and W correspondingly.By the definition of inversion:arg( z − w ) = arg( z ′ − w ) , | z − w |·| z ′ − w | = r It follows that (¯ z ′ − ¯ w ) · ( z − w ) = r . Equivalently, z ′ = (cid:18) ¯ w · z + [ r − | w | ] z − w (cid:19) . Exercise 18.17.
Check the statement for each elementary transformation. Then applyProposition 18.11.
Exercise 18.19.
Note that f = a · z + bc · z + d preserves the unit circle | z | = 1. Use Corollary 10.26and Proposition 18.11 to show that f commutes with the inversion z / ¯ z . In otherwords, 1 /f ( z ) = f (1 / ¯ z ) or ¯ c · ¯ z + ¯ d ¯ a · ¯ z + ¯ b = a/ ¯ z + bc/ ¯ z + d for any z ∈ ˆ C . The latter identity leads to the required statement. The condition | w | < | v | follows since f (0) ∈ D . Exercise 18.20.
Note that the inverses of the points z and w have complex coordinates1 / ¯ z and 1 / ¯ w . Apply Exercise 12.25 and simplify.The second part follows since the function x th( · x ) is increasing. Exercise 18.21.
Apply Schwarz–Pick theorem for a function f such that f (0) = 0 andthen apply Lemma 12.9. HINTS
O PXA B CX
Exercise 19.3.
Let O be the point of intersection of the lines.Draw a line ℓ thru the given point P and O . Construct a circleΓ, tangent to both lines, which crosses ℓ . Let X denotes one ofthe points of intersections.Consider the homothety with the center at O which sends X to P . The image of Γ is the needed circle. Exercise 19.6, (a) . Show that the angles marked the same wayon the on the diagram of regular pentagon are equal to each other.Conclude that XC = BC and △ ABC ∼ △
AXB . Therefore
ABAC = AXAB = AC − ABAB = ACAB − . It remains to solve for
ACAB . (b). Choose two points P and Q and use the compass-and-rulercalculator to construct two points V and W such that V W == √ · P Q . Then construct a pentagon with the sides
P Q anddiagonals
V W . Exercise 19.7.
Note that with a set square we can construct a line parallel to given linethru the given point. It remains to modify the construction in Exercise 14.3.
Exercise 19.9.
Assume that two vertices have rational coordinates, say ( a , b ) and( a , b ). Find the coordinates of the third vertex. Use that the number √ Exercise 19.10.
Guess the construction from the diagram. Provethat it verifies that the triangle is equilateral.
Exercise 19.11.
Apply the same argument as in Exercise 10.9.
Exercise 19.12.
Consider the perspective projection as in Exer-cise 15.1. Let A = (1 , , B = (1 , ,
1) and M = (1 , , M is the midpoint of [ AB ].Their images are A ′ = (1 , , B ′ = ( , , ) and M ′ = ( , , ).Clearly, M ′ is not the midpoint of [ A ′ B ′ ]. Exercise 19.15.
The line v polar to V is tangent to Γ. Since V ∈ p ,by Claim 19.14, we get that P ∈ v ; that is, ( P V ) = v . Hence the statement follows. PX ′ XYY ′ Z Exercise 19.16.
Choose a point P outside of the bigger circle.Construct the lines dual to P for both circles. Note that thesetwo lines are parallel.Assume that the lines intersect the bigger circle at two pairsof points X , X ′ and Y, Y ′ . Set Z = ( XY ) ∩ ( X ′ Y ′ ). Note thatthe line ( P Z ) passes thru the common center.The center is the intersection of (
P Z ) and another line con-structed the same way.
Exercise 20.1.
Assume the contrary; that is, there is a point W ∈ [ XY ] such that W / ∈ ▲ ABC . Without loss of generality, we may assume that W and A lie on the opposite sides ofthe line ( BC ).It imples that both segments [ W X ] and [
W Y ] intersect ( BC ). By Axiom II, W ∈ ( BC )— a contradiction. Exercise 20.3.
To prove the “only if” part, consider the line passing thru the vertexwhich is parallel to the opposite side.To prove the “if” part, use Pasch’s theorem (3.12).
Exercise 20.4.
Assume the contrary; that is, a solid square Q can be presented as a unionof a finite collection of segments [ A B ] , . . . , [ A n B n ] and one-point sets { C } , . . . , { C k } .Note that Q contains an infinite number of mutually nonparallel segments. There-fore, we can choose a segment [ P Q ] in Q which is not parallel to any of the segments[ A B ] , . . . , [ A n B n ].It follows that [ P Q ] has at most one common point with each of the sets [ A i B i ]and { C i } . Since [ P Q ] contains infinite number of points, we arrive to a contradiction.
A B CD B ′ C ′ D ′ E Exercise 20.5.
First note that among elementary sets only one-pointsets can be subsets of the a circle. It remains to note that any circlecontains an infinite number of points.
Exercise 20.15.
Let E denotes the point of intersection of the lines( BC ) and ( C ′ D ′ ).Use Proposition 20.14 to prove the following two identities:area( (cid:4) AB ′ ED ) = area( (cid:4) ABCD ) , area( (cid:4) AB ′ ED ) = area( (cid:4) AB ′ C ′ D ′ ) . Hence the statement follows.
Exercise 20.17.
Without loss of generality, we may assume that the angles
ABC and
BCA are acute.Let A ′ and B ′ denote the foot points of A and B on ( BC ) and ( AC ) correspondingly.Note that h A = AA ′ and h B = BB ′ . A BC A ′ B ′ Note that △ AA ′ C ∼ △ BB ′ C ; indeed the angle at C is shared andthe angles at A ′ and B ′ are right. In particular AA ′ BB ′ = ACBC or, equivalently, h A · BC = h B · AC.
Along the same lines, we get that h C · AB = h B · AC.
Hence the statement follows. HINTS
A B CD EF M
Exercise 20.18.
Draw the line ℓ thru M parallel to [ AB ] and [ CD ]; itsubdivides (cid:4) ABCD into two solid parallelograms which will be denotedby (cid:4)
ABEF and (cid:4)
CDF E . In particular,area( (cid:4)
ABCD ) = area( (cid:4)
ABEF ) + area( (cid:4)
CDF E ) . By Proposition 20.14 and Theorem 20.16 we get thatarea( ▲ ABM ) = · area( (cid:4) ABEF ) , area( ▲ CDM ) = · area( (cid:4) CDF E )and hence the result.
Exercise 20.19.
Let h A and h C denote the distances from A and C to the line ( BD )correspondingly. According to Theorem 20.16,area( ▲ ABM ) = · h A · BM ; area( ▲ BCM ) = · h C · BM ;area( ▲ CDM ) = · h C · DM ; area( ▲ ABM ) = · h A · DM.
Therefore area( ▲ ABM ) · area( ▲ CDM ) = · h A · h C · DM · BM == area( ▲ BCM ) · area( ▲ DAM ) . Exercise 20.20.
Let I be the incenter of △ ABC . Note that ▲ ABC can be subdividedinto ▲ IAB , ▲ IBC and ▲ ICA .It remains to apply Theorem 20.16 to each of these triangles and sum up the results.
Exercise 20.22.
Assuming a > b , we subdivided Q c into Q a − b and four triangles con-gruent to T . Therefore ➊ area Q c = area Q a − b + 4 · area T . According to Theorem 20.16, area T = · a · b . Therefore, the identity ➊ can be writtenas c = ( a − b ) + 2 · a · b. Simplifying, we get the Pythagorean theorem.The case a = b is yet simpler. The case b > a can be done the same way. Exercise 20.23. If X is a point inside of △ ABC , then ▲ ABC is subdivided into ▲ ABX , ▲ BCX and ▲ CAX . Thereforearea( ▲ ABX ) + area( ▲ BCX ) + area( ▲ CAX ) = area( ▲ ABC ) . Set a = AB = BC = CA . Let h , h and h denote the distances from X to the sides[ AB ], [ BC ] and [ CA ]. Then by Theorem 20.16,area( ▲ ABX ) = · h · a, area( ▲ BCX ) = · h · a, area( ▲ CAX ) = · h · a. Therefore, h + h + h = a · area( ▲ ABC ) . Exercise 20.25.
Let X be a point inside △ ABC . Let Y denotes the point of intersectionof ( AX ) and [ BC ].Let b and c denote the distances from B and C to the line ( AX ). XA BC Y b c By Theorem 20.16, we get the following equivalences:area( ▲ AXB ) = area( ▲ AXC ) , m b = c, m area( ▲ AY B ) = area( ▲ AY C ) , m BY = CY.
Exercise 20.26.
Let M denotes the intersection of two medians [ AA ′ ] and [ BB ′ ]. FromExercise 20.25 we have area( ▲ ABM ) = area( ▲ ACM ) and area( ▲ ABM ) = area( ▲ CBM ).Therefore, area( ▲ BCM ) = area( ▲ ACM ).According to Exercise 20.25, M lies on the median [ CC ′ ]. That is, medians [ AA ′ ],[ BB ′ ] and [ CC ′ ] intersect at one point M . MA BC A ′ C ′ B ′ By Theorem 20.16, we get thatarea( ▲ C ′ AM ) = · area( ▲ BAM )= · area( ▲ CAM )Applying Claim 20.24, we get that MC ′ MC = area( ▲ C ′ AM )area( ▲ CAM ) = . Exercise 20.27.
Let P n and Q n be the solid regular n -gons so that Γ is inscribed in Q n and circumscribed around P n . Clearly, P n ⊂ D ⊂ Q n . Show that area P n area Q n = (cos πn ) ; in particular,area P n area Q n → n → ∞ . Next show that area Q n < n > Q n − area P n ) →
0. Hence the result. ndex ∠ , 14 ∠ h , 91 ∡ , 14 ∡ h , 91 △ , 17 △ h , 90 ▲ , 160 (cid:3) , 46 (cid:4) , 161 ∼ , 43 ∼ =, 17, 161 ≡ , 14 k , 48 ⊥ , 35 ∞ , 73( P Q ), [
P Q ), [
P Q ], 13(
P Q ) h , [ P Q ) h ,[ P Q ] h , 90( u, v ; w, z ), 149AA similarity condition, 44absolute, 90absolute plane, 80absolute value, 142acute angle, 35affine transformation, 112, 122altitude, 57angle, 14acute and obtuse angles, 35angle of parallelism, 101positive and negative angles, 24right angle, 35straight angle, 22vertical angles, 23angle between arcs, 79angle measure, 20hyperbolic angle measure, 91angle preserving transformation, 44, 106Apollonian circle, 55, 71arc, 68area, 162argument, 144ASA congruence condition, 31asymptotically parallel lines, 101base, 32base of cone, 131between, 22 bijection, 12bisectorangle bisector, 59external bisector, 59perpendicular bisector, 35bounded set, 173center, 30center of the pencil, 119central projection, 133centroid, 58ch, 99, 109chord, 39circle, 30circle arc, 68circline, 73circular cone, 131circumcenter, 56circumcircle, 56circumtool, 78, 157collinear, 112collinear points, 13compass-and-ruler calculator, 155complex conjugate, 142complex coordinate, 142concurrent, 119conformal disc model, 89conformal factor, 107congruentsets, 161congruent triangles, 17consistent, 86constructible numbers, 155continuous, 16convex set, 160cross-ratio, 71complex cross-ratio, 145, 149curvature, 87, 172 d , d , d ∞ , 11defect of triangle, 84degeneratetriangle, 22polygonal set, 161quadrilateral, 53diagonaldiagonal of a regular n -gon, 153diagonals of quadrilateral, 46 NDEX diameter, 39direct motion, 38discrete metric, 11distance, 10between parallel lines, 54from a point to a line, 39distance-preserving map, 12doubling the ball, 173duality, 124elementary set, 161elementary transformation, 147ellipse, 137endpoint of arc, 68equidistant, 104equilateral triangle, 32equivalence relation, 17, 43, 49Euclidean metric, 11Euclidean plane, 20Euclidean space, 120Euler’s formula, 143excenter, 183extended complex plane, 146Fano plane, 126Fermat prime, 153field automorphism, 114finite projective plane, 126foot point, 36fractional linear transformation, 146great circle, 128h-angle measure, 91h-circle, 94h-half-line, 90h-line, 90h-plane, 90h-point, 90h-radius, 94h-segment, 90half-line, 13half-plane, 27holomorphic function, 150horocycle, 104hyperbolic angle, 91hyperbolic angle measure, 91hyperbolic cosine, 99, 109hyperbolic functions, 99hyperbolic geometry, 86hyperbolic plane, 90hyperbolic sine, 99, 108hyperbolic tangent, 99hypotenuse, 45ideal line, 119ideal point, 90, 119identity map, 195imaginary complex number, 142imaginary line, 142imaginary part, 141incenter, 61incidence structure, 112 incircle, 61indirect motion, 38injective map, 12inradius, 61inscribed triangle, 64insideinside a circle, 39inside a triangle, 160intersecting lines, 48inverse, 12inversion, 70center of inversion, 70, 130circle of inversion, 70inversion in a sphere, 130inversion in the circline, 76inversion in the line, 76sphere of inversion, 130inversive plane, 73inversive space, 130inversive transformation, 117irrational point, 156isometry, 12isosceles triangle, 32leg, 45line, 12Lobachevskian geometry, 86M¨obius transformation, 146Manhattan plane, 11metric, 10metric space, 10motion, 12neutral plane, 80oblique circular cone, 131obtuse angle, 35order of finite projective plane, 127orthic triangle, 62orthocenter, 57outside a circle, 39parallel lines, 48ultra parallel lines, 101parallel tool, 112parallel translation, 147parallelogram, 53solid parallelogram, 161pencil, 119perpendicular, 35perpendicular bisector, 35perpendicular circlines, 76perspective projection, 120plane absolute plane, 80Euclidean plane, 20h-plane, 90hyperbolic plane, 90inversive plane, 73neutral plane, 80plane in the space, 120point, 10 INDEX ideal point, 90point at infinity, 73polar coordinates, 144polarity, 124polygonal set, 161degenerate polygonal set, 161projective model, 137projective plane, 126projective transformation, 121quadrable set, 172quadrilateral, 46degenerate quadrilateral, 53inscribed quadrilateral, 66solid quadrilateral, 161radius, 30rational point, 156real complex number, 142real line, 11, 142real part, 141real projective plane, 119rectangle, 53solid rectangle, 161reflection, 37regular n -gon, 153rhombus, 53righ circular cone, 131rotational homothety, 147ruler-and-compass construction, 41SAA congruence condition, 82SAS congruence condition, 31SAS similarity condition, 44scalar product, 129secant line, 40segment, 13sh, 99, 108side side of a regular n -gon, 153side of quadrilateral, 46side of the triangle, 27similar triangles, 43Simson line, 67solid quadrilateral, parallelogram,rectangle, square, 161triangle, 160sphere, 128spherical distance, 128square, 53solid square, 161SSS congruence condition, 33SSS similarity condition, 44stereographic projection, 132subdivision of polygonal set, 164tangentcircles, 40half-line, 68line, 40th, 99tip of cone, 131transversal, 50 triangle, 17congruent triangles, 17degenerate triangle, 22orthic triangle, 62right triangle, 45similar triangles, 43solid triangle, 160triangle inequality, 143unit complex number, 142vertexvertex of a regular n -gon, 153vertex of quadrilateral, 46vertex of the angle, 14vertical angles, 23 sed resources [1] Akopyan, A. V. Geometry in figures,
Siberian Math. J.
Aufbau der Geometrie aus dem Spiegelungsbegriff,
M¨obius transformations revealed [5] Beltrami, E.
Teoria fondamentale degli spazii di curvatura costante,
Annali. diMat., ser II, 2 (1868), 232–255.[6] Birkhoff, G. D. A set of postulates for plane geometry, based on scale andprotractors,
Annals of Mathematics
33 (1932), 329–345.[7] Bolyai, J.
Appendix.
Euclid’s Elements. [9] Greenberg, M. J.
Euclidean and non-Euclidean geometries: Development andhistory,
Kiselev’s geometry. Book I. Planimetry,
Adapted from Russianby Alexander Givental.[11] Lambert, J. H.
Theorie der Parallellinien,
El´ements de g´eom´etrie , 1794.[13]
РЊР«РҫРөСЉРҳРўСҒРәРҷРє, Р№. РҶ. Р¤ Р„РөСЉРөРњРөСҖРүРҳЫиРҳСЂСҐРҷРҷ,
РӘРөРҹРөР„СҒРәРҷРє РўРҳСҒСЂР„РҷРә,
РўСҜР». 25–28 (1829–1830 РүРү.). [14] Lobachevsky, N. I.
Geometrische Untersuchungen zur Theorie der Paral-lellinien,
Elementary geometry from an advanced standpoint,
Problems in plane and solid geometry, translated and edited byDimitry Leites. 2006.[17] Saccheri, G. G.
Euclides ab omni nævo vindicatus,
РїРөСҐСҜРүРҷР„ , РҶ. Рһ.
РҮРҳЫиРҳСЂСҐРҷСЅ 7–9 , 1997. [19] Tarski, A. What is elementary geometry? in