Fund Concepts of Geometry
Fund Concepts of Geometry MATH 3181
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MATH 3181 SPRING 1999 Area in Hyperbolic Geometry Preliminaries A polygonal region is a plane gure which can be expressed as the union of a nite number of triangular regions in such a way that if two of the triangular regions intersect their intersection is an edge or a vertex of each ofthem Let R be a polygonal region A triangulation of R is a nite collection K T1T2Tn of triangular regions such that i the s intersect only at edges and vertices and ii their union is R Let R1 and R2 be polygonal regions Suppose that they have triangulations K1T1T2 Tn K2 such that for each i we have 2 Ti Then we say that R1 and R2 are equivalent by nite decomposition and we write R1 E Let 5 T defectT for any triangular region T Theorem If K1 and K2 are triangulations of the same polygonal region R then 6 K1 2 6K239 Theorem If two Saccheri quadrilaterals have the congruent summits and equal defects then their summit angles are congruent in which case the two Saccheri quadrilaterals are congruent Given AABC with BC considered as the base LetD and E be the midpoints of AB and AC let F G and H be the feet of the perpendiculars from B A and C respectively to E As you will proven in the homework oHCBF is a Saccheri quadrilateral It is known as the quadrilateral associated with AABC It depends on the choice of the base but it should be clear which base we mean Theorem Every triangular region is equivalent by finite decomposition to its associated quadrilateral region Theorem Every triangular region has the same defect as its associated quadrilateral region Theorem If AABC and ADEF have the same defect and a pair of congruent sides then the two triangular regions are equivalent by finite decomposition 1999 David Royster Hyperbolic Geometry For classroom use only MATH 3181 AREA 1N HYPERBOLIC GEOMETRY SPRING 1999 Theorem Bolyai s Theorem in the Hyperbolic Plane If Tl anal T2 are triangular regions anal 6T1 6T2 then T1 2 T2 Theorem If 5AABC gt5ADEF then there is a point P between A anal C such that 5AABP Requirements for an Area Function Now how do we de ne the area of a polygonal region in the hyperbolic plane Should we de ne the area in the same way that we do in the Euclidean plane If so what are the minimum requirements for an area function of the Euclidean plane Minimally it should satisfy the following Let R be the set of all polygonal regions in H 2 An area function should be a function a 9 R such that i 0R gt 0 for every R ii if R1 and R2 intersect only in edges and vertices then aR1uR2 aR1aR2 iii if Tl and T2 are triangular regions with the same base and altitude then a a T2 39 If there is such a function 06 then by ii and iii it will satisfy iV if R1 E R2 then 06 R1 06 because congruent triangles have the same bases and altitudes In Euclidean geometry we can show that this area function is unique and it must satisfy the formula 06 T bh for each triangular region T where b is the length of the base and h is the length of the altitude Theorem For hyperbolic geometry there is no such function 06 9 R satis ving i ii iii anal hence iv Proof Consider the right angle with APO 1 For each n let Pquot be the point of 130 131 such that l This gives a sequence of triangles AAPOPI AAPIPZ and a corresponding sequence of triangular regions Tl T2 By condition iii all the regions have the same area 06 A Now consider the defects ofthese triangles and let 6 5 For each n 1999 David Royster Hyperbolic Geometry For classroom use only MATH 3181 AREA 1N HYPERBOLIC GEOMETRY SPRING 1999 d1 d2 dquot 26AAPOPquot lt180 Since the partial sums 611 d2 dn are bounded we have that the in nite series idquot n0 limdn0 is convergent Therefore Hence 6quot lt oil for some n By Theorem 7 there is a point B betweenA and PO such that 6 ABPOPI 6 AAPHPn dquot Therefore by Bolyai39s Theorem the regions T and Tquot determined by these triangles are equivalent by nite decomposition By condition iv this means that 0T 0Tn But aT ocT1A 71 Therefore 0 T 0 T1 Because OCDAABqu gt 0 and 39 aT1aTaAABPl This must be impossible v The Uniqueness of Hyperbolic Area Theory Any reasonable area function 06 should have the following properties i 0R gt 0 for every R ii if R1 and R2 intersect only in edges and ve1tices then aR1uR2 aR1aR2 iii if R12 R2 then aR1 aR2 There is such a functionithe defect Recall that we know that the defect satisfies i ii and iii In fact we have the following theorem Theorem Let 0 9 R be an area function satis ving i ii anal iii Then there is a positive constant k gt 0 such that 727 a Rl ikz aRl l 80 for every R E Q We are not able to prove this at this time The proof is within your grasp but we do not have time to do it 1999 David Royster Hyperbolic Geometry For classroom use only MATH 3181 AREA 1N HYPERBOLIC GEOMETRY SPRING 1999 Corollary In H 2 the area of any triangle is at most kz There is no nite triangle whose area equals the maximal value kz although you can approach this area as closely as you wish and achieve it with a trebly asymptotic triangle J Bolyai proved that you can construct a circle of area kz and a regular 4sided polygon with a 727 4 angle that also has this area Angle of Parallelism In the Poincare model of hyperbolic geometry we found a formula relating the angle measurement of the angle of parallelism and the distance to which it is associated This formula is a special case of the general formula discovered independently by Bolyai and Lobachevsky Theorem Formula of BolyaiLobachevsky 06 tan 7 e M 2 where k is the proportionality constant from the area theorem We have proven this formula in the Poincare model where k 1 1999 David Royster Hyperbolic Geometry For classroom use only Neutral and Non Euclidean Geometries David C Royster UNC Charlotte1 September 67 2000 1Required Text Euclidean and NonEuclidean Geometn39es by Mel Greenberg Contents Introduction 1 Spherical Geometry 2 Logic and the Axiomatic Method 21 Introduction 22 Sets 23 Universal Sets and Compliments 24 Sentences and Statements 25 Sentence Connectives 26 Biconditionals and Combinations of Connectives 27 Quanti ers 28 Rules of Reasoning 29 Valid Arguments 3 Proof 31 Mathematical Systems 32 Proof 321 Proving Conditionals 322 Proving Biconditionals 323 ProvinszPm 324 Proof by Cases 325 Mathematical Induction 326 Proof by Contradiction 327 Proofs of Existence and Uniqueness 33 Proof Creativity 4 Euclid7s Mathematical System 5 Incidence Geometry 6 Betweenness Axioms 7 Congruence Theorems 8 Axioms of Continuity 38 43 47 51 55 C ON TEN TS 9 Neutral Geometry 91 Alternate lnterior Angles 92 Weak Exterior Angle Theorem 10 Theorems of Continuity 101 Elementary Continuity Principle 102 Measure of Angles and Segments 103 Saccheri Legendre Theorem 104 The Defect of a Triangle 11 The Work of Saccheri and Gauss 111 Saccheri 112 Gauss D Hyperbolic Geometry 121 The Hyperbolic Axiom and its Consequences 122 Angle Sums again 123 Similar Triangles CO Classi cation of Parallels 131 Fan Angles 132 Limiting Parallel Rays 133 Hyperparallel Lines 134 Classi cation of Parallels 135 Proof of Claim and Asymptotic Triangles 4 Strange New Triangles 5 Inversion in Euclidean Circles 16 Models of Hyperbolic Geometry 161 Consistency of Hyperbolic Geometry 162 The Beltrami Klein Model 163 The Poincare Half Plane Model 164 The Poincare Disk Model 165 lsomorphism of Models 166 Constructions in the Poincare Model 167 Return to the Klein Model 17 Area in Hyperbolic Geometry 171 Preliminaries 172 Requirements for an Area Function 173 The Uniqueness of Hyperbolic Area Theory 174 Angle of Parallelism 18 Hypercycles and Horocycles 19 The Pseudosphere 116 120 120 121 123 124 4 C ON TEN TS 20 Hyperbolic Trigonometry 133 201 Circumference and Area of a Circle 138 21 Hyperbolic Analytic Geometry 141 211 More on Quadrilaterals 141 212 Coordinate Geometry in the Hyperbolic Plane 143 The Origins of Geometry In the beginning geometry was a collection of rules for computing lengths areas and vol umes Many were crude approximations derived by trial and error This body of knowledge developed and used in construction navigation and surveying by the Babylonians and Egyptians was passed to the Greeks The Greek historian Herodotus 5th century BC credits the Egyptians with having originated the subject but there is much evidence that the Babylonians the Hindu civilization and the Chinese knew much of what was passed along to the Egyptians The Babylonians of 2000 to 1600 BC knew much about navigation and astronomy which required a knowledge of geometry They also considered the circumference of the circle to be three times the diameter Of course this would make 7139 3 7a small problem This value for 7139 carried along to later times The Roman architect Vitruvius took 7139 3 Prior to this it seems that the Chinese mathematicians had taken the same value for 7139 This value for 7139 was sancti ed by the ancient Jewish civilization and sanctioned in the scriptures In I Kings 723 we nd He then made the sea of cast metal it was round in shape the diameter from rim to rim being ten cubits it stood ue cubits high and it took a line thirty cubits long to go round itiThe New English Bible Rabbi Nehemiah attempted to change the value of 7139 to 2 but was rejected By 1800 BC the Egyptians according to the Rhind papyrus had the approximation 7139 N Z N 31604 To show how far we have progressed in the 1920 s the Indiana legislature passed a law mandating that 7139 It has since been appealed In 1789 Johann Lambert proved that 7139 is an irrational number and in 1882 F Lindemann proved that 7139 is transcendental ie it is not the solution to any algebraic equation with rational coef cients The ancient knowledge of geometry was passed on to the Greeks They seemed to be blessed with an inclination toward speculative thinking and the leisure to pursue this inclination They insisted that geometric statements be established by deductive reasoning rather than trial and error This began with Thales of Milete He was familiar with the computations right or wrong handed down from Egyptian and Babylonian mathematics In determining which ofthe computations were correct he developed the rst logical geometry This orderly development of theorems by proof was the distinctive characteristic of Greek mathematics and was new This new mathematics of Thales was continued over the next two centuries by Pythago ras and his disciples The Pythagoreans as a religious sect believed that the elevation of the soul and union with God were achieved by the study of music and mathematics Nonetheless they developed a large body of mathematics by using the deductive method Their foundation of plane geometry was brought to a conclusion around 400 BC in the El ements by the mathematician Hippocrates This treatise has been lost but many historians 5 C ON TEN TS agree that it probably covered most of Books IiIV of Euclid s Elements which appeared about a century later circa 300 BC Euclid was a disciple of the Platonic school Around 300 BC he produced the de nitive treatment of Greek geometry and number theory in his thirteen volume Elements In compiling this masterpiece Euclid built on the experience and achievements of his predecessors in preceding centuries on the Pythagore ans for Books lilV VII and IX on Archytas for Book Vlll on Eudoxus for Books V VI and Xll and on Theaetetus for Books X and Xlll So completely did Euclid s work supersede earlier attempts at presenting geometry that few traces remain of these efforts It s a pity that Euclid s heirs have not been able to collect royalties on his work for he is the most widely read author in the history of mankind His approach to geometry has dominated the teaching of the subject for over two thousand years Moreover the axiomatic method used by Euclid is the prototype for all of what we now call pure mathematics It is pure in the sense of pure thought no physical experiments need be performed to verify that the statements are correctionly the reasoning in the demonstrations need be checked In this treatise he organized a large body of known mathematics including discoveries of his own into the rst formal system of mathematics This formalness was exhibited by the fact that the Elements began with an explicit statement of assumptions called ax ioms or postulates together with de nitions The other statementsitheorems lemmae corollariesiwere then shown to follow logically from these axioms and de nitions Books IiIV VII and IX of the work dealt primarily with mathematics which we now classify as geometry and the entire structure is what we now call Euclidean geometry Euclidean geometry was certainly conceived by its creators as an idealization of physical geometry The entities of the mathematical system are concepts suggested by or abstracted from physical experience but differing from physical entities as an idea of an object differs from the object itself However a remarkable correlation existed between the two systems The angle sum of a mathematical triangle was stated to be 180 if one measured the angles of a physical triangle the angle sum did indeed seem to be 180 and so it went for a multitude of other relations Because of this agreement between theory and practice it is not surprising that many writers came to think of Euclid s axioms as self euident truths Centuries later the philosopher Immanuel Kant even took the position that the human mind is essentially Euclidean and can only conceive of space in Euclidean terms Thus almost from its inception Euclidean geometry had something of the character of dogma Euclid based his geometry on ve fundamental assumptions Postulate I For euery point P and for euery point Q not equal to P there exists a unique line 6 that passes through P and Postulate II For euery segment AB and for euery segment CD there exists a unique point E such that B is between A and E and segment CD is congruent to segment BE Postulate III For euery point 0 and euery point A not equal to 0 there exists a circle with center 0 and radius 04 Postulate IV All right angles are congruent to each other C ON TEN TS 7 Before we study the Fifth Postulate let me say a few words about his de nitions Euclid s methods are imperfect by modern standards He attempted to de ne everything in terms of a more familiar notion sometimes creating more confusion than he removed As an example A pointis that which has no part A line is breadthless length A straight line is a line which lies evenly with the points on itself A plane angle is the inclination to one another of two lines which meet When a straight line set upon a straight line makes adjacent angles equal to one another each of the equal angles is a right angle Euclid did not de ne length distance inclination or set upon Once having made the above de nitions Euclid never used them He used instead the rules of interaction between the de ned objects as set forth in his ve postulates and other postulates that he implicitly assumed but did not state Postulate V If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles the two straight lines if produced inde nitely meet on that side on which the angles are less than two right angles No one seemed to like this Fifth Postulate possibly not even Euclid himselfihe did not use it until Proposition 29 The reason that this statement seems out of place is that the rst 4 postulates seem to follow from experienceitry to draw more than one line through 2 different points The Fifth Postulate is unintuitive It does come from the study of parallel lines though Equivalent to this postulate is Playfair397s Postulate Giuen a line and a point not on that line there exists one and only one line through that point parallel to the giuen line Euclid s Fifth Postulate seemed to be too burdensome It should follow from the other axioms Since it is not intuitive we should be able to prove it as a theorem We should be able to prove that it is dependent in this Axiom system If we have a set of axioms A1 A2 An for our mathematical system and we can prove that Axiom An is derivable or provable from the other axioms then An is indeed redundant In some sense we are looking for a basis for this mathematical system Unlike vector spaces and linear algebra there is not a unique number of elements in this basis for it includes the axioms de nitions and the rules of logic that you use Many people have tried to prove the Fifth Postulate The rst known attempt to proue Euclid V as it became known was by Posidionius1st century BC He proposed to replaced the de nition of parallel lines those that do not intersect by de ning them as coplanar lines that are everywhere equidistant from one another It turns out that without Euclid V you cannot prove that such lines exist It is true that such a statement that parallel lines are equidistant from one another is equivalent to Euclid V Ptolemy followed with a proof that used the following assumption For euery line 6 and euery point P not on 5 there exists at most one line in through P such that m is parallel to Z We will show at a later date that this statement is equivalent to Euclid V and therefore this did not constitute a proof of Euclid V 8 C ON TEN TS Proclus 410485AD also attempted to prove Euclid V His argument used a limiting process He retained all of Euclid s de nitions all of his assumptions except Euclid V and hence all of his propositions which did not depend on Euclid V His plan was 1 to prove on this basis that a line which meets one of two parallels also meets the other and 2 to deduce Euclid V from this proposition His handling of step 2 was correctly handled The argument in step 1 runs substantially as follows Let g and h be parallel lines and let another line k meet h in P From Q a point of k situated between 9 and h drop a perpendicular to h As Q recedes inde nitely far from P its distance QR from h increases and exceeds any value however great In particular QR will exceed the distance between 9 and h For some position of Q then QR will equal the distance between 9 and h When this occurs k will meet 9 There are a number of assumptions here which go beyond those found in Euclid I will mention only the following two 0 the distance from one of two intersecting lines to the other increases beyond all bounds as we recede from their common point 0 the distance between two parallels never exceeds some nite value The rst of the two assumptions is not a grave error on the part of Proclus for it can be proved as a theorem on the basis of what he assumed from Euclid Unfortunately for Proclus his second assumption is equivalent to Euclid V Nasiraddin 120171274 John Wallis 161671703 Legendre 175271833 Wolfgang Bol yai Girolamo Saccheri 166771733 Johann Heinrich Lambert 172871777 and many oth ers tried to prove Euclid V and failed In these failures there developed a goodly number of substitutes for Euclid V 239e statements that were equivalent to the statement of Euclid V The following is a list of some of these that are more common 1 Through a point not on a given line there passes not more than one parallel to the line to Two lines that are parallel to the same line are parallel to each other OJ A line that meets one of two parallels also meets the other F If two parallels are cut by a transversal the alternate interior angles are equal 01 There exists a triangle whose angle sum is two right angles a Parallel lines are equidistant from one another1 1 There exist two parallel lines whose distance apart never exceeds some nite value 00 Similar triangles exist which are not congruent2 to Through any three noncollinear points there passes a circle H 0 Through any point within any angle a line can be drawn which meets both sides of the angle 1All perpendicular distances from either line to the other are equal 2Wallis7 Axiom C ON TEN TS 9 11 There exists a quadrilateral whose anglesum is four right angles 12 Any two parallel lines have a common perpendicular lt fell to three different mathematicians to independently show that Euclid V is not provable from the other axioms and what is derivable from them These were Carl Friedrich Gauss Janos Bolyai and Nicolai lvanovich Lobachevsky Once these men broke the ice the pieces of geometry began to fall into place More was learned about non Euclidean geometriesihyperbolic and elliptic or doubly elliptic The elliptic geometry was studied by Riemann gave rise to riemannian geometry and manifolds which gave rise to differential geometry which gave rise to relativity theory et al This gives us something to anticipate as we learn more about geometry We will spend our time studying hyperbolic geometry for it lends itself to better studyinot requiring major changes in the axiom system that we have chosen We may have an opportunity to see that hyperbolic geometry is now lending itself to considerations in the latest research areas of mathematics Chapter 1 Spherical Geometry Whereas basic plane geometry is concerned with points and lines and their interactions most of the early geometry of the Babylonians Arabs and Greeks was spherical geometryithe study of the Earth idealized as a sphere This early science was astronomy and the need to measure time accurately by the sun We have to come to some agreement on what lines and line segments on a sphere are going to be De nition 1 A great circle on a sphere is the intersection of that sphere with a plane passing through the center of the sphere Examples of great circles are the equator and the lines of constant longitude such as the Greenwich Mean Time Line 0 These are good choices to play the role of lines on our sphere For example given any to non antipodal1 there is a unique great circle joining those two points This is easy to see when you remember that three non collinear points determine a plane Take these two points and the center of the sphere as the three non collinear points The intersection of the plane determined by these points and the sphere is the great circle joining the two given points If great circles are to be lines then we can measure the angle between two intersecting great circles as the angle formed by the intersection of the two de ning planes with the plane tangent to the sphere at the point of intersection See the gure below Figure 3 With this de nition of angle we can form triangles 0n the sphere whose interior angle sum is greater than two right angles In fact we will show that the interior angle sum of all triangles 0n the sphere is greater than two right angles We will be working with radian measure for most of the semester to make many of the calculations neater Thus we are claiming that on a sphere the interior angle sums of any triangle is greater than 7139 1points that do not both lie on a line through the center of the sphere 10 5 Angles in Spherical Geometry In order to discuss how to measure the angle sums we will rst discuss the concept of the area of a triangle on the sphere We will discuss this more fully later but for the present time we will accept the following four properties of area for polygonal regions on the sphere C H The sphere of radius R has area 47rR2 to the area of a union of nonoverlapping regions is the sum of their areas OJ the areas of congruent regions are equal q the ratio of the area enclosed by two great circles to the area of the whole sphere is the same as the ratio of the angle between them to 27139 Theorem 11 On a sphere of radius R a triangle AABC with interior angles 04 B and V has area given by areaAABC 1321 B 39y 7 7r First some terminology a lune is one of the regions between two great circles and the antipodal points where they cross Antipodal points or gures are those that correspond to opposite ends of a diameter of the sphere A lune is given by the angle 9 at the center of the sphere between the two de ning planes Note that this is also the angle between the two great circles cf Figure 3 12 CHAPTER 1 SPHERICAL GEOMETRY Figure 2 A lune PROOF due to Euler 1781 From our assumptions above Assumption 4 tells us 9 2 2 Area of the lune 7 47rR 20R 11 7139 To make our proof easier we may assume that the triangle lies in one hemisphere If not you can use Assumption 2 to break the triangle up into smaller triangles that do lie in one hemisphere You can then use the following argument on each piece and then add them together gt gt A triangle is the intersection of three lunes the lune given by AB AC the lune given by AB BC and the lune given by BC AC Each of these three lunes has an antipodal lune and the intersection of these three antipodal lunes is an antipodal triangle AA B C Figure 3 A triangle and the associated lunes We will denote the lunes by the angle that includes them BOA BAC and ABC Starting with the triangle AABC and the lune ABC add to them the lune ACB This covers the triangle AABC three times Now add in the antipodal lune to lune CAB d the antipodal triangle AA B C This covers the hemisphere determined by the line BC counting the triangle three times So we get 2sz 2cm 23R 2sz 7 3 areaAABC areaAABC Thus we get our desired formula areaAABC 1321 B 39y 7 7139 We call the value of 04 B 39y 7 7139 the angle excess of the spherical triangle I Note that since every triangle on the sphere has area every triangle on the sphere has interior angle sum greater than 7139 If the radius of the sphere is very large and the triangle very small in area then the angle excess will be almost 0 Let ER denote the sphere of radius R centered at the origin in R3 ln rectangular coordinates ER 9671472 W2 y2 22 R2 14 CHAPTER 1 SPHERICAL GEOMETRY and in spherical coordinates ER R cos0 sin gtRsin0 sin gtRcos gt l 0 g 9 g 27139 0 g j g 7139 a39 Figure 4 Spherical coordinates The length of a path between two points along a great circle is found by measuring the angle in radians made by the radii at the two endpoints and multiply it by the radius of the sphere ie R0 Note that the length of a segment is unchanged when we retate the sphere around some axis or re ect it through a great circle Theorem 12 Spherical Pythagorean Theorem For a right triangle AABC on a sphere of radius R with a right angle at vertex C and sides of length a b and c then COSltgtCOSltgtCOSltgt Figure 5 Spherical right triangle PROOF Rotate the sphere so that the point A has coordinates R 00 and the point C lies in the my plane This will make 3 and g the angles determining the point B Here 3 is the central angle determined by side AC 04 is the angle determined by AC and 39y is the angle determined by AB This gives us the following spherical coordinates for the vertices of the triangle A R00 B ltRcos sing Rcos cosaRsin cosaRsina C Rcos Rsin O a Rsin sin 7 Oz Rcosg 7 04 Using what we know about the dot product in R3 we can nd the cosine of the angle A R2 cosa cos cos39y 7 W 7 T 7 cosa cos Now in radian measure a b c a E 5 E V E and we are done I Why is this called the spherical Pythagorean theorem First note that it does give a relationship of the hypotenuse of a right triangle with the two sides of this triangle To see how it is related to the classical Pythagorean theorem recall the Maclaurin series Taylor series at a 0 for the cosine function 2 4 6 cosz1imim m 16 CHAPTER 1 SPHERICAL GEOMETRY Using this in the formula from the spherical Pythagorean theorem gives us 1 02 c4 7 1 a2 a4 1 b2 4 72R224R47m 72R224R47m 72R224R47m a2 2 azbz W W t m t 39 quot terms in hi her owers of 0 terms in hi her owers of a and b C2 g 2 p a2b2 g 1132 Now as R gets arbitrarily large while a b and 0 stay small we get the classical Pythagorean theorem as a limiting case Theorem 13 Spherical Sine Theorem Let AABC be a spherical triangle on a sphere of radius B Let a b and 0 denote the lengths of the sides in radians and let ZA B and C denote the interior angles at each vertex Then suns suns suns sinZA T sinZB T sinZC PROOF We will rst prove this for a right triangle and then use that result to prove the general case In addition we can restrict our attention to triangles that lie entirely within a quarter of a hemisphere Cy Av Let AABC be a right triangle with right angle at C Extend the radius 0A to OA where A is the on the intersection of the tangent plane to the sphere at B with the line OA Likewise extend 00 to 00 This means that the triangles AOBC and AOBA are right triangles with right angle at B Additionally ZA BC ABC 1B Now we have other right angles as well First planes ABCquot and OBC are perpendic ular from above Now plane OCB is perpendiculare to plane OCA since ZACB is a right 17 gt gt angle Look at the two planesA BC and OAC They intersect in the line A C Since A C lies in the plane 00147 and OC lies in the plane OCB7 the two lines must be perpendicular Thus7 AOC A is a right triangle with right angle at C The classical Pythagorean theorem gives Combining these equations gives us ow ma ow wwfwBV wdfwBV AC2 BA2 R2 BC2 R2 mwwmhwmi and so AA BC is a triangle with right angle at C Let a B and 39y denote the angles at the center of the sphere determined by the arcs BC7 AC7 and AB7 respectively The de nition of the sine function gives us and so smm 7 AU 7 AU BA OA BA OA 7 SIDE i sinZB smm39 We can reverse the roles of A and B and this gives and so sin 1 R 7 i sinZA T SIHWL sin sin sinZB sin39y sinZA sinZB39 18 CHAPTER 1 SPHERICAL GEOMETRY C Altitudes Now if we have an arbitrary triangle we can construct the spherical analogue of the altitude to reduce the relation for two sides to the case of a right triangle which we just nished For example in the following gure inserting the altitudes from A and from B we get a R From the other altitude we have sin sinZC sin sinsinZA i b i i q i c i i c i s1n s1nZC 7 s1n 7 s1n s1n7r 7 1B 7 s1n s1nZB and the theorem follows I These theorems show that there are exactly six related pieces of information determining a spherical triangle 7 the three sides and the three angles Given any three of these pieces you can determine the other three pieces This means that similar triangles must be congruent This is not immediate We need one other piece which we are able to prove but will not do so now Theorem 14 Suppose that AABC is a right triangle on the sphere of radius R with a right angle at C Then cosA sinB cos Homework 1 Prove that two great circles intersect 2 Prove that two great circles bisect each other 3 The sphere of radius one can be considered as 21 zyz lm2y222 1 10 M 0 g a 2m g M W In these two coordinate systems determine the distance along great circles between two arbitrary points on the sphere as a function of the coordinates 4 A pole of a great circle is one ofthe endpoints of a diameter ofthe sphere perpendicular to the plane of the great circle For example7 the North and South poles are both poles for the equator Prove that through a given point7 not on a given great circle and not the pole of the great circle7 there is a unique great circle ghrough the given point perpendicular to the given great circle Chapter 2 Logic and the Axiomatic Method 21 Introduction Mathematicians use a large number of methods to discover new resultsitrial and error computation of special cases inspired guessing pulling results from thin air The difference in this and an astrologer for example is that we have an accepted method called the axiomatic method for proving that these results are correct Proofs give us assurance that results are correct In many cases they also give more general results For example the Egyptians and Hindus knew by experiment that if a triangle has sides of lengths 3 4 and 5 it is then a right triangle But the Greeks proved that if a triangle has sides of lengths a b and c and if a2 b2 02 then the triangle is a right triangle There is no amount of checking by experiment that could give this general result Proofs give us insight into relationships among different things that we are studying forcing us to organize our thoughts in a coherent way If you gain nothing else from the course than this you have still gained the greatest gift that mathematics has to offer I wish to persuade you that a certain statement is true or false by pure reasoning I could do this by showing you that the statement follows logically from some other statement that you may already believe I may have to convince you that that statement is also true and follows from another statement This process may continue until I reach a statement which you are willing to believe one which does not need justi cation That statement plays the role of an axiom If no such statement exists then I will be caught in an in nite regress giving one proof after another ad in nitum There are three requirements that must be met before we can agree that a proof is correct Requirement 1 There must be mutual understanding of the words and symbols used in the discourse Requirement 2 There must be acceptance of certain statements called axioms without justi cation Requirement 3 There must be agreement on how and when one statement follows logically from another ie agreement on certain rules of reasoning There should be no problem in reaching mutual understanding so long as we use terms familiar to both and use them consistently If I use an unfamiliar term you have the right to demand a de nition of this term De nitions cannot be given arbitrarily they are 20 2 2 SETS 21 subject to the rules of reasoning referred to in Requirement 2 Also we cannot de ne every term that we use In order to de ne one term we must use other terms and to de ne these terms we must use still other terms and so on If we were not allowed to leave some terms unde ned we would get involved in in nite regress Let us begin with this 22 Sets We need some basic information about sets in order to study the logic and the axiomatic method This is not a formal study of sets but consists only of basic de nitions and notation Braces and are used to name or enumerate sets The roster method for naming sets is simply to list all of the elements of a set between a pair of braces For example the set of integers 1 2 3 and 4 could be named 1 2 34 This does not work well for sets containing a large number of elements though it can be used The more common method for this is known as the set builder notation A property is speci ed which is held by all objects in a set Pz read P of I will denote a sentence referring to the variable m For example m 23 z is an odd integer 1 lt z lt 4 The set of all objects x such that z satis es Pm is denoted by x l Pltzgt The set 1 23 4 can be named zl1 x 4 xEZm Zl1 m 4 From hence forth the words object element and member mean the same thing when referring to sets Sets will be denoted mainly by capital Roman letters and elements of the sets by small letters The following have the same meaning a E A a is in set A a is a member of set A a is an element of set A Likewise a A means that a is not an element of set A A is a subset of B if every element of A is also an element of B The following have the same meaning 22 CHAPTER 2 LOGIC AND THE AXIOMATIC METHOD A C B Every element of A is an element of B lfaEAthena B A is included in B B contains A A is a subset of B Note that a set is always a subset of itself If A and B are sets then we say that A B if A and B represent the same set A B A and B are the same set A and B have the same members A C B and B C A The set which contains no elements is known as the empty set and is denoted by 0 Note that for each set A Q C A The intersection of two sets A and B is the set of all elements common to both sets The intersection is symbolized by A O B or x l x E A and z E B The union of two sets A and B is the set of elements which are in A or B or both The union is symbolized by AUBorzlz Aorz B 23 Universal Sets and Compliments When we are working in an area or on a certain problem we always have a frame of reference in which we are working called a universal set In our geometry course it will be the set of points that lie on a plane In calculus we consider the set of real numbers the set of real functions the set of differentiable functions and the set of continuous functions as universal sets The complement of a set A is de ned to be the set of all elements of the universal set which are not in A and is symbolized by CA A Ac Note that A U A0 is always the universal set while A O A0 0 24 Sentences and Statements Logic and mathematical proof can be studies just like algebra In fact much of symbolic logic is just that A declarative sentence which is true or false but not both is called a statement The following are statements Babe Ruth hit 714 home runs Jack Nicklaus has won 20 major golf titles 236 24 SEN TEN CES AND STATEMENTS 23 The 25000th digit of 7r is 7 The following are not statements He is a golfer Why is a duck m 1 0 m 7 y a The sentence He is a golfer cannot be judged true or false because we do not know who He is If the word He is replaced by Bobby Jones forming the sentence Bobby Jones is a golfer the sentence becomes a true statement Similarly if z in the sentence m 1 0 is replaced by 2 forming the sentence 2 1 0 the sentence then becomes a false statement The letter z is a variable in the sentence z 1 O A letter or other symbol that can represent various elements of a universal set is called a variable We can make a sentence a statement by replacing its variables by elements of the universal set or by attaching phrases such as For every or There exists to the sentence For example m lt 3 is not a statement but each of the following is a statement 1 lt 3 5 lt 3 For every real number m z lt 3 There exists a real number m such that z lt 3 Replacements for variables of a sentence are always chosen from some universal set Any replacement which makes a sentence true is called a solution The set of all solutions is called the solution set of the sentence 24 CHAPTER 2 LOGIC AND THE AXIOMATIC METHOD 25 Sentence Connectives If P and Q are sentences then the sentence P and Q is called the conjunction of P and Q denoted by P A Q For any statement there are just two possible truth values true T or false If P and Q are both true then P A Q is true If one or both of P and Q are false then P A Q is false The truth table below de nes the truth values of P A Q for all possible truth value combinations of P and Q If P and Q are sentences then the sentence P or Q is called the disjunction of P and Q denoted by P V Q In mathematics we use an inclusive or That is P V Q is true when P is true or Q is true or both are true P V Q is false only when P and Q are false the truth table for P V Q is thus de ned below PlQHPAQ PlQHPVQ TT T TT T P NP TF F TF T T F FT F FT T F T F F F F F F A negation or denial of a sentence is formed in many ways For example the negation of P 2 is rational is represented by each of the following N P It is false that 2 is rational 2 is not rational 2 is irrational The truth table for negation is obvious You need to realize that there are other symbols besides N for negations that are in common usage ay b meansab azb meansaltb a A meansa A If P and Q are sentences the sentence If P then Q is denoted by P a Q or P a Q We construct a truth table for P a Q just as for the the other connectives A V and N However the de nition is not at all obvious Consider the sentence If I get an A in mathematics then I will take the next course 25 SENTENCE CONNECTIVES 25 Suppose a fellow student says this When is the sentence true and when is it false Let P denote the statement I get an A in mathematics and let Q denote the statement I will take the next course Consider the following four cases 1 P true He gets an A in mathematics true He takes the next course true He gets an A in mathematics false He does not take the next course He does not get an A in mathematics true He takes the next course He does not get an A in mathematics false He does not take the next course It is easy to see that 1 is true and that 2 is false You cannot claim that the original statement was false in 3 since he takes the next course even though he did not get an A Likewise7 in 4 you cannot claim that the statement was false7 since he did not get an A and he did not take the next course The truth table for this sentence is as below The sentence P a Q is called a conditional with P the antecedent and Q the consequent ln mathematics the conditional is encountered in many forms The following have the same meaning P a Q If P7 then Q P implies Q Q if P7 P only ifQ Q provided P Q whenever P7 Q when P P is a sufficient condition for Q Q is a necessary condition for P 26 CHAPTER 2 LOGIC AND THE AXIOMATIC METHOD 26 Biconditionals and Combinations of Connectives A sentence of the type P 5 Q A Q 5 P is called a biconditionall7 denoted P gt Q When P and Q are sentences7 the truth table for P ltgt Q is ln mathematics the biconditional is encountered in many forms The following have the same meaning P gt Q P is equivalent to Q P if and only if Q Q if and only if P P iff Q If P7 then Q and conversely lf Q7 then P and conversely P is a necessary and sufficient condition for Q Q is a necessary and sufficient condition for P Combinations of N ltgt7 A and V often occur A facility at recognizing them is essential for mathematical reading and proof Consider the following statement lfp is prime7 then ifp is even p must be smaller than 7 This breaks up into three statements P p is prime Q p is even R p must be smaller than 7 We can then translate the original statement into P a Q a R If k is perpendicular to Z and Z is perpendicular to m7 then k is parallel to m 2 7 QUANTIFIERS 27 Let P k is perpendicular to 6 Q l is perpendicular to m R k is parallel to m Then the sentence translates as P A Q a R 27 Quanti ers Sentences involving the phrases For every and There exists also play a very impor tant role in the structure of mathematical sentences The symbol V called the universal quanti er denotes phrases such as For each For every For all A sentence such as For every x Pm can be translated symbolically into Vz Pz or Va The following sentences have the same meaning Va z is an integer z E Q For every x if z is an integer then x E Q For all x if z is an integer then x E Q For each x if z is an integer then x E Q Every integer belongs to Q Every integer is a rational number If x is an integer then x E Q Note that in the last sentence the universal quanti er is understood and not written The symbol 3 called the existential quanti er symbolizes phrases such as There exists There is at least one For at least one and Some A sentence such as There exists an x such that Pm translates symbolically to HzPm or 3m The following have the same meaning 3m x is a natural number There exists an x such that z is a natural number Some number is a natural number There is at least one natural number 28 CHAPTER 2 LOGIC AND THE AXIOMATIC METHOD Quanti ers often appear together Consider the following examples Vsz x y 0 For every x and for every y z y O Vz y z y 0 For every x there exists a y so that z y O Hsz z y 0 There exists an x such that for all y z y O Hz y z y 0 There exists an x and there exists a y such that z y 0 The following sentence For every x if z is even then there exists a y such that z 2y translates as Vzz is even a y z 2y These quanti ers refer to some universal set which if not explicitly given must be easily inferred from the context We will be interested only in nonempty universal sets De nition 2 The sentence VzPm is true if and only if the solution set of Pm equals the universal set This sentence is false if the solution set is a proper subset of the universal set 239e if there is at least one element of the universal set for which Pm is false De nition 3 The sentence HzPm is true if the solution set of Pm is nonempty This sentence is false if the solution set of Pm is empty 239e if for every replacement of x by a member a of the universal set Pa is false For more complicated mathematical sentences containing more quanti ers let us look at a few examples Example 271 Suppose Pmy is a sentence with variables x and y The sentence VszPmy is true if and only if for every replacement of z and y by members a and b from the universal set the statement Pa b is true The sentence is false if there is a replacement for z or a replacement for y for which the statement is false Example 272 The sentence HmVy Pmy is true if there exists a replacement a for x such that Vy Pa y is true This same a makes the sentence Pa 1 true for every I in the universal set Note that the sentence Hsz z gt y is false There is no replacement a for x which makes the sentence Vy a gt y true 28 Rules of Reasoning Mathematicians assume a certain class of sentences to be true before we ever prove any theorems in a mathematical system We call these sentences mles of reasoning They could be called reasoning axioms An important class of these rules of reasoning are known as tautologies A tautology is a sentence which is true no matter what the truth value of its constituent parts 28 RULES OF REASONING 29 Example 281 The sentence P a P V Q is a tautology7 where P and Q represent arbitrary mathematical sentences We can show that this is a tautology from a truth table PVQ PgtPVQ mmaaw HHHH The way in which we do this is to compute the truth values for P V Q in the third column rst7 and then use columns one and three to compute the truth values in column four Logical Axiom 1 Every tautology is a rule of reasoning The following are tautologies that we commonly use You will nd these listed in the Rules of Logic that you have been given 1 PWNQWFWN P i Q lt2 Q w P contrdpositiie P A P a Q a Q Modus ponens P P Q A Q a R a P a B Law of Syllogism PAQltgtPVQ PVQltgtPAQ1 PQltgtPAQ PQ gtPVQ P A Q a P P gt P P A Q s P v Q PgtQgtQgtP PRARP PA Q A RA R a P a 22 PV N P Law of the Excluded Middle P gt P PP PQVRPAQR 1Items 4 and 5 comprise De Morgan s Laws 2Items 12 and 13 comprise Proof by Contradiction 30 CHAPTER 2 LOGIC AND THE AXIOMATIC METHOD 18 P SlASl SZA ASn1 SnASn 13 a P a R Law of Syllogism 19 P a R A a 13 a P V a R Proof by Cases 2 O PAQWMQAP 21 PVQltgtQVP3 22 PRQ gtPARQ 23 PAQARltgtPAQAR 24 PVQVR lt2 PVQVR4 25 PAQVRltgtPAQVPAR 26 PVQARltgtPVQAPVR5 27 PeolmmmomonMQ 29 Valid Arguments The tautologies in the preceding section are not all that there are If you want to make a deduction based on a sentence check its truth table If it is a tautology use it Tautologies provide lots of reasoning theorems before we ever start deduction within a mathematical system There are actually two branches of formal logic the statement calculus involving state ments and reasoning by tautology and the predicate calculus involving quanti ed sen tences All we are doing is taking a quick guided tour through informal logic and so we will not study these areas in great detail However from the predicate calculus we get another collection of reasoning sentences some of which are listed below These cannot be veri ed by tautology Logical Axiom 2 Let U be a universal set Each of the following is a rule of reasoning 1 Wm Pm VzPm Va 2 VmPm ltgt Pa for any a E U 3 3mPx ltgt Pa for some a E U An argument is an assertion that from a certain set of sentences 1 5 called premises or assumptions one can deduce another sentence Q called an inference or con Clvsion Such an argument can be denoted by 51sro Arguments are either valid correct or invalid incorrect 3Items 20 and 21 are the Commutative Laws 4Items 23 and 24 are the Associative Laws 5Items 25 and 26 are the Distributive Laws 2 9 VALID ARG UMEN TS 31 De nition 4 1 5 F Q is a valid argument if and only if 51 A A Sn Q is a rule of reasoning Logical Axiom 3 Rule of Substitution Suppose P ltgt Q Then P and Q may be substituted for one another in any sentence Logical Axiom 4 Euery sentence of the type N 3mPm ltgt Vm N Pm is true Logical Axiom 5 Euery sentence of the type N VmPm ltgt 3m N Pm is true To prove a sentence of the type VzPm false7 one could try to prove Hz N Pm true This is referred to as prouz39dz39ng a counterexample Chapter 3 Proof 31 Mathematical Systems A mathematical system consists of the following 1 QWFWN 00 T 9 a set of unde ned concepts a universal set a set of relations a set of operations a set of logical axioms a set of non logical axiomsithese axioms pertain to the elements being studied the relations and the operations and not to the logic being used a set of theorems a set of de nitions an underlying set theory In plane geometry the unde ned concepts were those of point and line The universal set was the set of points in the plane The relations were such concepts as equality per pendicularity and parallelism We have mentioned the logical axioms A non logical axiom would be of the form Two different points are on exactly one line 32 Proof De nition 5 Suppose A1A2 Ak are all the axioms and previously proved theorems of a mathematical system A formal proof or deduction of a sentence P is a sequence of statements 51 2 Sn where 1 Sn is P and one of the following holds 1 S is one of A1A2 Ak or 32 PROOF 33 2 5 follows form the previous statements by a valid argument using the rules of reasoning A theorem is any sentence deduced from the axioms andor the previous theorems The same is true of lemma and proposition For some mathematicians there is a hierarchy of lemma7 proposition7 and theorem with lemma being the easiest to prove and theorem the most difficult7 or longest Other mathematicians make little or no distinction between these objects7 and will call everything a theorem Example 321 Suppose a mathematical system contains just the following axioms A1 abcmltyA23 A2 a b c The following is a formal proof of z lt y 51 abcmltyA23 7byAl Sgt a b C 7by A2 53 m lt y A 2 3 by modus ponens on 5152 S4 z lt y by the tautology P A Q a P In practice mathematicians do not write formal proofs They write informal proofs An informal proof is an argument which shows the existence of a formal proof As such it gives enough of the formal proof so that another person becomes conuinced Thus we might call an informal proof a conuincing argument Mathematicians try to convince other mathematicians You will try to convince your fellow students and me7 your professor An informal proof of the above example runs as follows From A1 and A2 it follows that z lt y A 2 3 Thus7 z lt y Henceforth7 we will be writing only informal proofs The art of mathematics is creating proofs Just as every other artisan7 the mathematician has some basic modes of proof We will now consider a few of these 321 Proving Conditionals You usually proved a sentence of the type P a Q in plane geometry by assuming P and deducing Q You considered Q the conclusion In actually7 P a Q was the conclusion it was what you were trying to prove To prove P a Q rst assume P to be true Then using P and all other theorems and axioms try to deduce Q Once Q is deduced in this manner you have completed a proof of P a Q You have not shown that Q is true you have only shown that Q is true if P is true Whether P is true is another question whether Q is true is another question What you have shown to be true is P a Q This technique is called the Rule of Conditional Proof or the Deduction Theorem More formally7 suppose that A1 Ak are the axioms and previously proved theorems To prove P a Q is to show that From A1 Ak we can deduce P a Q 34 CHAPTER 3 PROOF is a valid argument To do this temporarily assume P to be an axiom and show that From A17Ak7P we can deduce Q is a valid argument A second technique of proving P a Q is by the contrapositive We can prove P a Q by proving N Q N P Often the rule of conditional proof is used to prove the contrapositive 322 Proving Biconditionals There are three modes of proof for biconditional sentences 1 Prove P a Q and Q P 2 Prove P a Q and N P N 5 Provide an iff string A word about the i strmg We produce a string of equivalent sentences from P to Q This is the Law of Syllogism from the list of tautologies 323 Proving Vm Pz To prove Vz Pm let x represent an arbitrary element of the universal set and prove that Pm is true Then since x was arbitrary element of the universal set7 we may generalize that VzPm is true The justi cation is Logical Axiom 2 324 Proof by Cases Proof by cases is used several ways and involves the connective or We will be trying to prove a sentence of the type P V R a Q This type of proof utilizes the tautology P QMU QH PVR Ql The proof is accomplished by proving the antecedent of this sentence7 PimMRin Hence7 P a Q and R a Q must be proved Any mode of proof for conditional sentences can be used Similarly7 a proof by cases of PlngvangtQ is accomplished by proving P1 Q A A Pn Q The art of producing a proof by cases lies in the discovery of what set of exhaustive cases is appropriate 32 PROOF 35 325 Mathematical Induction This is a technique that is all too often overlooked in geometry I include it here for completeness Suppose Pn is a sentence which is a statement for any n E N then the Principle of Mathematical Induction is 131 AVk Pk Pk 1 VnPn If we can prove the antecedent of this statement P1 A Vk Pk Pk 1 then by Modus ponens we can deduce Vn Thus there are two steps in the proof of VnPn Basic Step Prove P1 Inductive Step Prove Vk Pk Pk 1 Note that its name is misleading Mathematical induction is deductive reasoning not induc tive reasoning Inductive reasoning is making a conjecture or guess based on observations and your previous mathematical experience 326 Proof by Contradiction A contradiction is a statement which is false no matter what the truth value of its con stituent parts It can usually be expressed symbolically in the form RA N R A proof by contradiction of a statement P is a proof that assumes N P and yields a sentence of the type RA N R where R is any sentence including P an axiom or any previously proved theorem This is justi ed by the tautology NPRANRP Intuitively P can only be true or false since we are assuming only a two valued logic If we assume its negation true and this yields another sentence both true and false then N P cannot be true so P must be true The phrases reductio ad absurdum and indirect proof both refer to proof by contradiction The importance of being able to form sentence negations is realized when doing proofs by contradiction To begin such proofs you must know how to form negations Comparing proof techniques we see that with the Rule of Conditional Proof we assume P with the explicit intention of deducing Q With the contrapositive we assume N Q with the explicit intention of deducing N P But in using Proof by Contradiction we assume both P and N Q and try to deduce any sentence R and its negation N R 327 Proofs of Existence and Uniqueness The sentence There exists an x such that Px is denoted by 3x The sentence There exists exactly one x such that Px 36 CHAPTER 3 PROOF is denoted by Elm There are two parts to proving a sentence of this form H Existence Part Prove that there is an x such that Pm is true to Uniqueness Part Here you must prove that if there are two elements z and 2 such that Pm and Pz are each true then x 2 Symbolically Vsz A Z 33 Proof Creativity In the previous part of this chapter you learned several modes of proof The intent is that these will become part of your mathematical toolbox Just because you have the tools does not guarantee that you can create a proof There are some helpful procedures to follow as aids in creating a proof Translate to Symbolic Logic A typical comment made when proofs are attempted is I do not know where to start This statement is made with a great gnashing of teeth and wringing of hands One procedure to follow is comparable to that of solving a problem in basic algebra First translate what you are requested to prove into symbolic logic Then seeing the structure of the translated sentence you can select a mode of proof Still knowing a mode of proof that could be used does not guarantee success Suppose you want to attempt to prove a sentence of the type P a Q by using the Rule of Conditional Proof You want to assume P and deduce Q A question often asked is How do get from P to Q There is no certain way No one way will always work Certainly knowing to assume P and deduce Q is a step in the right direction The mode of proof provides the structure for the proof building this structure is usually a more creative task I can give a few hints Analogy An important aid in carrying out proofs is to get ideas from other proofs This is supposed by comments of mathematicians who argue that to be good at mathematics you need lots of practice lots of exposure to different proofs Analytic Process This is known as working backwards You want to prove P a Q Start with Q and try to nd an R such that R a Q Then try to nd and S such that S a R Then look to see if P a S If not try to ll in another step Continue this until you nd a sentence Rn such that P Rn and RniRn71gtgtRgtQ Do not be surprised if you do not see this process outlined in a text or reference book It is rare that if this processed is used it is then explicitly mentioned Usually the proof will be given as PgtRngtRn1gtgtRgtQ 33 PROOF CREATIVITY 37 Do Something Approach This is simply trial and error You want to prove P a Q by assuming P and deducing Q You have no particular way to get from P to Q but start out get involved do something try different approaches prove all that you can You do not have to show all of this in your nal version of your proof but it can help you get started When reading proofs in mathematics texts and journals you are not aware of the blind alleys and unsuccessful attempts preceding a successful proof This leads you to think the established mathematician never follows a wrong path or makes a mistake Trial and error is very much a part of mathematical creativity Use of De nitions Another helpful procedure is to recall all relevant de nitions It is a tendency to read a de nition and ignore its importance in later proofs Use of Previously Proved Theorems It is helpfulinay it is essential in starting a proof to examine all previously proved theorems for results which might be relevant to the proof Chapter 4 Euclid s Mathematical System We will accept the Rules of Reasoning that we have been discussing as our logic rules in our mathematical system There are several ways in which we can go about setting forth our axioms and unde ned terms for geometry We can keep the number of axioms down to a very few in which case we need a large number of unde ned terms relations and operations We can keep the number of unde ned terms relations and operations to a small number thus requiring a larger number of axioms We will actually work with the Axiom system set up by David Hilbert in 1899 in his book Foundations of Geometry He not only clari ed Euclid s de nitions but also lled in the gaps in some of Euclid s proofs Hilbert recognized that Euclid s proof for the sideangle side criterion of congruence in triangles was based on an unstated assumption the principle of superposition and that this criterion had to be treated as an axiom He built on the earlier work of Moritz Pasch who in 1882 published the rst rigorous treatise on geometry Pasch made explicit Euclid s unstated assumptions about betweenness At the same time we could have used the axioms set forth by Garret Birkhoff These axioms are the ones with which you are more familiar being the basis for the texts in High School Geometry l have included both sets of axioms so that you can compare the two We will work exclusively with the Hilbert axioms An interesting exercise is to prove everything that we will prove using Birkhoff s axioms Some proofs are easier and some are harder First let us set our unde ned terms They will be 0 point 0 line 0 incidence a point being incident with a line 0 betweenness o congmence These objects are to be taken as unde ned terms We will not attempt to make a de nition of any of them However when we wish to look at a speci c example of our geometry called a model we will have to establish how each of these unde ned terms is to be interpreted For example if we wish to look at the model where points are ordered pairs coming from the usual Cartesian plane and lines are the usual straight lines in the Cartesian plane then we would be able to easily state what it means for a point to be incident with 38 39 a particular line what it means for one point to lie between two other points and what it means for two line segments to be congruent We could change our example though to be a set of seven points where lines would be ordered pairs of points incidence would be set inclusion and betweenness and congruence would be meaningless The particular model that we choose will determine our interpretation of the terms but will not de ne them At this point we need to make some de nitions In order to do this we need our axiom system I shall include all of the axioms here though we will discuss them in more detail later Hilbert7s Axioms for Neutral Geometry GROUP 1 INCIDENCE AXIOMs 171 For every point P and for every point Q not equal to P there exists a unique line 6 that passes through P and Q 172 For every line 6 there exist at least two distinct points incident with Z IE3 There exist three distinct points with the property that no line is incident with all three of them GROUP 11 BETWEENESS AXIOMs B71 If A B C then A B and C are three distinct points all lying on the same line and C B A gt B72 Given any two distinct points B and D there exist points A C and E lying on BD suchthatABDBCDandBDE B73 If A B and C are three distinct points lying on the same line then one and only one of the points is between the other two B74 PLANE SEPARATION AXIOM For every line 6 and for any three points A B and C not lying on 6 i if A and B are on the same side of Z and B and C are on the same side of 6 then A and C are on the same side of 6 ii if A and B are on opposite sides of Z and B and C are on opposite sides of 6 then A and C are on the same side of Z GROUP 111 CONGRUENCE AXIOMs CEI If A and B are distinct points and if A is any point then for each ray r emanating from A there is a unique point B on r such that B 7 A and AB E A B 072 If AB E CD and AB E EF then CD E EF Moreover every segment is congruent to itself 073 HA B C A B C AB EA B and BCE B C then ACEA C 40 CHAPTER 4 EUCLID S MATHEMATICAL SYSTEM 074 Given any ZBAC and given any ray ABe11anating from a point A then there is a unique ray A C on a given side of line A B such that ZB A C E ZBAC 075 If A E B and A E 10 then B E 10 Moreover every angle is congruent to itself 076 SAS If two sides and the included angle of one triangle are congruent respectively to two sides and the included angle of another triangle then the two triangles are congruent GROUP IV CONTINUITY AXIOMs ARCHIMEDES7 AXIOM lf AB and CD are any segments then there is a number n such that if segment CD is laid off 71 times on the ray A emanating from A then a point E is reached where 71 CD E AE and B is between A and E DEDEKIND S AXIOM Suppose that the set of all points on a line 6 is the union 21 U 22 of two nonempty subsets such that no point of 21 is between two points of 22 and vice versa Then there is a unique point 0 lying on 6 such that P1 O P2 if and only If P1 6 21 and P2 6 22 and 0 7g P1 P2 The following two Principles follow from Dedekind s Axiom yet are at times more useful CIRCULAR CONTINUITY PRINCIPLE lfa circle V has one point inside and one point outside another circle 39y then the two circles intersect in two points ELEMENTARY CONTINUITY PRINCIPLE If one endpoint of a segment is inside a circle and the other outside the circle then the segment intersects the circle Birkhoff7s Axioms for Neutral Geometry The setting for these axioms is the Absolute or Neutral Plane It is universal in the sense that all points belong to this plane It is denoted by A2 AXIOM 1 There exist nonempty subsets of A2 called lines with the property that each two points belong to exactly one line AXIOM 2 Corresponding to any two points A B E A2 there exists a unique number dAB dBA E R the distance from A to B which is 0 if and only if A B AXIOM 3 Birkhoff Ruler Axiom If k is a line and R denotes the set of real numbers there exists a one to one correspondence X lt gt z between the points X E k and the numbers z E R such that dA B la 7 bl where A lt gt a and B lt gt b AXIOM 4 For each line k there are exactly two nonempty conuer sets R and R satisfying 239 A2R OkoR n R O R 0 O k Q and R O k 0 That is they are pairwise disjoint m le E R and Y E R then XY k 7 Q AXIOM 5 To each angle ABC there exists a unique real number x with 0 g x g 180 which is the degree measure of the angle 35 1ABC AXIOM 6 If E c lnt iABC then ZABDo ZDBC39o ZABCo AXIOM 7 If E is a ray in the edge k of an open half plane HkP then there exist a one to one correspondence between the open rays in HkP emanating from A and the set of real numbers between 0 and 180 so that if H lt gt x then BAXo z AXIOM 8 SAS If a correspondence of two triangles or a triangle with itself is such that two sides and the angle between them are respectively congruent to the corresponding two sides and the angle between them the correspondence is a congruence of triangles 42 CHAPTER 4 EUCLID S MATHEMATICAL SYSTEM De nition 6 The plane is the collection of all points and lines De nition 7 The segment gt AB C 6 AB l C A7 C B7 or C lies between A and B A and B are the endpoints of the segment AB De nition 8 Given points 0 and A in the plane The set of points P such that OP E 0A is a circle with center 0 Each segment OP is called a radius De nition 9 The ray gt E C 6 AB l C 6 AB or B lies between A and C gt E emanates from A and is part of AB De nition 10 The rays E m we opposite if they are distinct7 emanate from A7 and are part of the same line AB AC De nition 11 An angle with vertex A is a point A together with 2 non opposite rays E and E called the sides7 emanating from A Denote this angle by A CAB BAC De nition 12 If BAD and CAD have a common side E and the other two sides E and E form opposite rays7 the angles are supplements or supplementary angles De nition 13 An angle BAD is a right angle if it has a supplementary angle to which it is congruent De nition 14 Two lines 6 and m are parallel if they do not intersect 226 if no point lies on both of them Denote this by Z m De nition 15 Two lines 6 and m are perpendicular Z L m7 if they have a point A in common and there exist rays E a part of Z and E a part of m7 such that BAC is a right angle Chapter 5 Incidence Geometry We are now ready to begin our study of geometries in earnest We will study neutral geometry based on the axioms of Hilbert This means that we will study all that we can almost without the introduction of a Parallel Postulate of any sort At the appropriate time we will add a parallel postulate and see where we will be led For an ease of notation let A B C denote the statement that the point B lies between the point A and the point C De nition 16 Let t be any line and let A and B be any points that do not lie on Z If A B or if the segment AB contains no point lying on t we say that A and B are on the same side of Z If if A 7 B and if AB contains a point of l we say that A and B are on opposite sides of 6 Let us quickly review the incidence axioms Incidence Axiom 1 For euery point P and for euery point Q not equal to P there exists a unique line 6 that passes through P and Incidence Axiom 2 For euery line 6 there exists at least two distinct points incident with Incidence Axiom 3 There exist three distinct points with the property that no line is incident with all three of them This does not seem like much but already we can prove several easy properties that any set satisfying these three axioms must have Theorem 51 If 6 and m are distinct lines that are not parallel then 6 and m have a unique point in common Let s be brave and give a formal and an informal proof of this theorem Having done that I think that you will see how an informal proof is really a rigorous proof just not a formal proof PROOF First the formal proof We shall break the statement into its three constituent parts Pl m 44 CHAPTER 5 INCIDENCE GEOMETRY Q 6 Wm R l and m have a unique point in common We are to prove then that P A Q a R 51 6 Wm and l 7 m Assuming P and Q in an RC P proof 52 l and m have at least one point in common negation of the condition of being parallel 3 Assume l and m have more than one point in common Proof by Contradiction S4 From 1 and 53 l and m have at least two points A and B in common S5 There exists a unique line through A and B Axiom 1 6 Thus7 l m S4 and S5 S7 We have Q and N Q7 a contradiction 5391 and 56 SS Thus 6 and m have a unique point in common the other case of 52 An informal proof of this result follows much the same line7 but is easier to read PROOF Since 6 and m are not parallel and since 6 7 m7 they must have at least one point in common Assume that they have more than one point in common They then have at least two points in common Axiom I1 says that two points determine a unique line7 so 6 m7 which is contrary to the hypothesis Thus7 l and m have a unique point in common I De nition 17 Two or more lines are concurrent if they intersect in one common point De nition 18 Two or more points are collinear if they are all incident with the same line We have four other results to mention Theorem 52 For every line there is at least one point not incident with it Theorem 53 For every point there is at least one line not incident with it Theorem 54 For every point there exist at least two lines incident with it Theorem 55 There exist three distinct lines which are not concurrent 45 To introduce you to the concept of a model for geometry let us look at a simple example of some mathematical object which satis es the three axioms of incidence based on our interpretation of the unde ned concepts Example 501 Consider the set U A B C We shall interpret a point to be a singleton subset of U Thus A B and C are points We shall interpret a line to be a doubleton subset of U The lines are then A B A C and B C We shall agree that a point is incident with a line if it is a subset of the line Now before we continue we must verify that each of the Incidence Axioms is valid in this particular example Axiom I 1 le and Y are any of the points of this geometry then X Y is the unique line which contains them for there are only three possible lines Axiom I 2 If XY is any line in this geometry then X and Y are two distinct points incident with it Axiom I f The points A B and C are three distinct points which are not collinear Thus this is a model of a geometry which satis es the Incidence Axioms Such a geometry is called an incidence geometry There are a number of different ways of visualizing this geometry Example 502 Again consider the set U A B C We shall interpret a point to be a doubleton subset of U The points are then A B A C and B C We shall interpret a line to be a singleton subset of U Thus A B and C are lines We shall agree that a point is incident with a line if it contains the line as a subset Now we must verify that each of the Incidence Axioms is valid in this example Axiom I 1 If X and Y are any of the points of this geometry then X O Y is the unique line which contains them There are three possible points and each must contain two elements from U Thus the intersection in the set U will be nonempty Axiom I 2 If X is any line in this geometry then X Y and X Z are two distinct points incident with it Axiom I f The points A B B C and A C are three distinct points which are not collinear since their intersection is empty Thus this too is a model of an incidence geometry It is sometimes referred to as the dual geometry to the previous example Note that since the three incidence axioms hold for each of these two examples the ve theorems must also hold One further item to note about these geometriesithere are no parallel lines Given a line and a point not on that line in each of these two examples there is no line through that point parallel to the given line We say that these two models exhibit the elliptic parallel property This property is not inherent in the incidence axioms but is inherent in the examples There are other examples of incidence geometries which do not exhibit the elliptic parallel property This implies that we cannot prove the Euclidean Parallel Postulate based only on the incidence axioms In fact we cannot prove that parallel lines even exist based solely on the incidence axioms Furthermore we cannot prove that they do not exist Example 503 If we take U A B C D and take the same interpretation for point and line as in Example 1 then we will have an incidence geometry which exhibits the Euclidean parallel propertyiunique parallels 46 CHAPTER 5 INCIDENCE GEOMETRY Example 504 If we take U A B C D E and take the same interpretation for point and line as in Example 1 then we will have an incidence geometry which exhibits the hyperbolic parallel propertyimultiple parallels De nition 19 We say that two models for incidence geometry are isomorphic if there is a one to one correspondence between the points of the models P lt gt P and the lines of the models 6 lt gt 6 which preserves the incidence relations 239e P is incident with 6 if and only if P is incident with Z Note that we will not be able to have a model with the elliptic parallel property isomor phic to a model with the hyperbolic or Euclidean parallel property for incidence would not be preserved Chapter 6 Betweenness Axioms You should review the Betweenness Axioms in our list of axioms There are numerous Propositions proved in the text based on the Betweenness Axioms We shall list most of them but shall prove only a few Proposition 61 For any two points A and B 239 E o El AB gt a E u El AB C A B Figure 61 It seems clear from Figure 61 that every point P lying on the line 6 through A B and C must either belong to ray E or to an opposite ray This statement seems similar to the second assertion of Proposition 61 but it is actually much more complicated You are now discussing four points and not the three of Proposition 61 You will prove this assertion in your homework with the addition of another axiom Let us call the assertion 0 AB and apointP is collinear with A B and C implies thatP E as the line separation property This is something that you will prove but knowing that it can be proven we shall use it as we need Recall the de nitions of same side and opposite sides Also recall the last Between ness Axiom It is called the plane separation axiom Note that it indirectly guarantees that our geometry will be only two dimensional for such an axiom will not hold in three dimensional space The points A and C may lie in the same plane with 6 yet the point B need not In this case the conditions would be satis ed yet the conclusion would not hold De nition 20 Let 6 be any line A side of the line 6 containing A is the set of all points that are on the same side of l as A This is also called the halfplane bounded by 6 containing A 48 CHAPTER 6 BETWEENNESS AXIOMS Proposition 62 Every line bounds exactly two half planes and these half planes have no point in common Proposition 63 GivenA B C andA C D thenB C D andA B D Figure 62 PROOF Compare Figure 62 From Exercise 1 A B C and D are four distinct nts on hne From Incidence Axiom 5 there is at least one int ltLgot on the line AD Line EC exists by Incidence iom 1 and by hypothesis AD EC TthA and D are on opposite sides of EC We claim that A and B lie on the we side 0 ltAssume not ie assume that A and B lie on opposite sides of EC Then EC intersects AB in a point between A and B By Proposition 51 that point must be C Thus we have that A C B and ltAB C which contradicts Betweenness Axio Thus A and B lie on the same side of EC Since A and D are on opposite sides of ltEC as we have proven earlier by Betweenness Axiom ABand DE on opposite sides of EC Hence the unique point of intersection of the lines EC and BD lies between B and D This unique point is C by an earlier step Thus B C D This proves half of Proposition 63 The other half is left as an exercise I Proposition 64 Line Separation Property If C A B and l is the line through A B and C then for every point P incident with l P is incident with either ray E or the opposite ray A The next theorem is very interesting Theorem 61 Pasch7s Theorem If AABC is any triangle and l is any line intersect ing side AB in a point between A and B then 6 also intersects either side AC or side BC If C is not incident with t then 6 does not intersect both AC and BC PROOF Compare Figure 63 Either C E l or C Z If C E t then we are done Thus let us assume that C 6 By hypothesis A B 6 and l O AB 7 0 Thus A and B are on opposite sides of t Then since there are only two sides to any line either C is on the same side of l as A or it is on the opposite side If A and C are on the same side of t then B and C are on opposite sides of 6 Thus 6 BC 7 0 Likewise if A and C are on opposite sides then 6 AC 7 0 and we are done I The following propositions are left to you to prove Proposition 65 Given A B C then AC ABUBC and B is the only point common to the segments AB and BC Proposition 66 Given A gtk B gtk C then B is the only point common to the rays El and W andzTBE 7 De nition 21 Given an angle ZCABHwe shall say that a point D is in the interior of the angle if D is on the same side of AB as C and on the same side of AC as B Note that this implies that the interior of an angle is the intersection of two half planes lt gt Proposition 67 Let CAB be any angle and let D be any point incident with line BC D is in the interior of CAB if and only ifB gtk D gtk C Proposition 68 IfD is in the interior of angle ZCAB then i so is every other point on the ray E except A ii no point on the opposite ray to E is in the interior of ZCAB iii ifC gtk A gtk E then B is in the interior of ZDAE De nition 22 Ray 1TD is between rays E and E if E and E are not opposite rays and D is in the interior of ZCAB Theorem 62 Crossbar Theorem IszD is between E and R then 1TD intersects the segment BC PROOF We are given a ray 1TD between rays E and Let us use proof by contradiction and assume that A O BC 0 Let U be the opposite ray to If H O BC P7 then B P C and by Proposition 67 we have that P lies in the interior of ZCAB However7 this contradicts ii A C Figure 63 Pasch s Triangle 50 CHAPTER 6 BETWEENNESS AXIOMS Figure 64 of Proposition 68 which says that no point on the opposimay can be interior to the angle Thus we have that HO BC 0 Now this means that AD BC by Proposition 61 part ii It follows that B and Clie on the same side of the line AD Let me a point on the line AC so that C A E Then C and E agon opposite sides of AD and by Betweenness Axiom 4 B and E are on opposite sides of AD Applying Proposition 68 part iii to this situation we know tha is in the interior of ZDAE which means that B and E are on the same side of the line AD We now have a contradiction Thus we have that E O BC 7 0 Lastly we need to consider the incidence of rays and triangles De nition 23 The interior of a triangle is the intersection of the interiors of the three angles A point is in the exterior of the triangle if it is not in the interior or on any of the three sides Proposition 69 i If a ray ITQ emanating from a point exterior to AABC intersects side AB in a point between A and B then ITQ also intersects AC or BC ii If a ray emanates from an interior point of AABC then it intersects on of the sides and if it does not pass through a vertex it intersects only one side Chapter 7 Congruence Theorems According to our development7 congruence has been a relationship between segments or a relationship between angles ln Geometry we are accustomed to seeing congruence as a relationship between triangles We can make this so by a de nition De nition 24 Two triangles are congruent if there exists a one tO one correspondence between their vertices sot that the corresponding sides and corresponding angles are con gruent Proposition 71 If in AABC we have AB E AC then B E 10 A Figure 71 PROOF This simple proof is due to Pappus Consider the correspondence of vertices A lt gt A7 B lt gt C7 and C lt gt B Under this correspondence7 two sides and the included angle of AABC are congruent respectively to the corresponding sides and included angle of AACB Hence7 by SAS the triangles are congruent Therefore7 the corresponding angles are congruent and B E 10 I Proposition 72 Segment Subtraction IfA B C D E F AB E DE and ACgDF thenBCgEF 52 CHAPTER 7 CON GRUEN CE THEOREMS C D E B A F Figure 72 PROOF From what we are given let us assume that BC E EF By Axiom 0 there exists a unique point C on ray W so that BC E EG By our hypothesis we have that G E F Since AB E DE and BC E EG by Axiom C 3 AC E DG But then by Axiom 0 2 we have that DF E DC It follows from Axiom 0 1 that F G We have reached a contradiction Thus it must be that BC E EF and we are done I Proposition 73 Giuen AC E DF then for any point B between A and C there exists a unique point E D gtk E gtk F such that AB E DE PROOF The proof of this proposition is in the text I We can use this result to help us establish a partial ordering on the line segments in the plane De nition 25 AB lt CD or CD gt AB means that there exists a point E between C and D such that AB E CE Proposition 74 Segment Ordering i exactly one of the following condi tions holds trichotomy AB lt CD AB B CD or AB gt CD ii if AB lt CD and CD E EF then AB lt EF iii if AB gt CD and CD E EF then AB gt EF iu ifAB lt CD and CD lt EF then AB lt EF PROOF We will prove the rst item only The proofs of the remaining three items will be left to you as homework We are given segments AB and CD If AB E CD then we are done So let us assume that AB E CD By Axiom 0 there is a unique point E E E5 so that AB E CE E E D else AB E CE E CD which is impossible under our assumption Thus we must have that C E D or C D E from Axiom B 2 If C E D then by de nition AB lt CD and we are done Suppose then that C D E By Proposition 73 there exists a unique point F 6 AB so that AF E CD In this case by the de nition AB gt CD and we are done I Proposition 75 Supplements of congruent angles are congruent Proposition 76 i Vertical angles are congurent to each other Figure 73 ii An angle congruent to a right angle is a right angle Proposition 77 For every line 6 and every point P there exists a line through P perpen dicular to t PROOF Either P E t or P 6 First let us assume that P 6 Let A B E 6 Such points exist by Axiom I 2 The ray E lies on one side of Z By Axiom C 4 there exists a ray H on the opposite side of 6 from P so that XAB E PAB By Axiom C l there is a point P E H so that AP E AP Since P and P are on opposite sides of t the segment PP intersects the line 6 Let Q PP Z If Q A then PQB PAB E P AB P QB Thus PP T Z If Q 7 A then APAQ E AP AQ by SAS Thus from the de nition of congruent triangles PQA E P AQ and PP T 6 Now if P E 6 there is an X 6 From this point X apply the previous technique to construct a perpendicular line to 6 through X By Axiom C 4 we can copy this angle on one side of t at P From the second part of Proposition 76 the other side of this angle is part of a line through P perpendicular to t I Proposition 78 ASA Given AABC and ADEF with A E D C E F and AC E DF Then AABC E ADEF Proposition 79 If in AABC we have that B E C then AB E AC and AABC Proposition 710 Angle Addition Given H3 between a and E E7 between FD and 13 CBC FEH and angleGBA HED Then ABC DEF Proposition 711 Angle Subtraction Given H3 between and W E between E and W CBG E FEH and GBA E HED then ABC E DEF 54 CHAPTER 7 CON CRUEN CE THEOREMS P B C E F Figure 74 PROOF Since H3 lies between and R we may apply the Crossbar Theorem to nd that W intersects AC Without loss of generality we may assume that this point of intersection is in fact the point C Then we have that A C C Assume that the points D F and H are chosen so that AB ED BC EH and BC EF This is nothing but a relabeling of the points Let us ass e then that ABC 7 DEH Then there exists a unique ray W on the same side of EH as FD so that HEP ABC By our assumption 7 ET so by Proposition 710 ABC PEF Since ABC DEF the uniqueness of W implies that E73 E a contradiction Thus ABC E DEH I As with line segments there is a natural method for de ning an ordering on angles De nition 26 ABC lt DEF means that there exists a ray E between FD and W so that ABC E CEF This gives us the following results which are completely analogous to those for segments Proposition 712 Ordering of Angles 239 Exactly one of the following con ditions holds P lt 62 P 62 or P gt 62 it If P lt Q and Q g R then P lt R m If P gt Q and Q g R then P gt R a If P lt Q and Q lt R then P lt R Proposition 713 SSS Cluen triangles AABC and ADEF If AB DE BC EF and AC DF then AABC ADEF Proposition 714 Euclid7s Fourth Postulate All right angles are congruent to each other Chapter 8 Axioms of Continuity These axioms are the axioms which give us our correspondence between the real line and a Euclidean line These are necessary to guarantee that our Euclidean plane is complete The rst axiom gives us some information about the relative sizes of segments as compared one to another ARCHIMEDES7 AXIOM If AB and CD are any segments then there is a number n such that if segment CD is laid o n times on the ray 1E emanating from A then a point E is reached where n CD AE and B is between A and E This is derived from the Archimedean Axiom in the real number system This should not be surprising for we wish to have a one to one correspondence between each euclidean line and the set of real numbers In the real line the Archimedean Postulate takes on the avor Archimedean Postulate Let M and e be any tow positive numbers Then there is a positive integer n such that ne gt M The main point for geometry is that if you choose any segment to be of unit length then every segment has nite length with respect to this measure Nothing can be too big n CD gt AB and nothing can be too small n EF gt CD where CD was chosen as our unit length This is still not enough for the purposes of geometry for the set of rational numbers Q satis es this property but causes trouble in another situation Let us consider the set of rational points in the cartesian plane Call this Q x Q Consider the ray passing from the origin through the point 1 1 This segment has length Now on the ray making up the positive real axis we are unable to nd a point satisfying Congruence Axiom 1 No point exists whose distance to the origin is We require a stronger property Dedekind s Axiom Suppose that the set of all points on a line 6 is the union 21 U 22 of two nonempty subsets such that no point of 21 is between two points of 22 and vice uersa Then there is a unique point 0 lying on t such that P1 gtk O gtk P2 if and only if P1 6 El and P2 6 22 and O 7 P1 P2 In order to better understand this axiom we need to study the concept in the real line Suppose that ab E R with a lt 1 Suppose that 5152 C ab satisfying 56 CHAPTER 8 AXIOMS OF CONTINUITY c 1651 andbCSg d lszSl andaltyltmtheny Sl The fourth item above indicates that we have the property that if X C 51 then a m C 51 Let 2 C SQ CLAIM lf 2 lt w lt b then w E SQ Assume not so that w C 51 This means that a w C 51 This places 2 C 51 which then means that 2 C 51 SQ a contradiction to item Thus w C 52 and in fact 212 C 52 There are three different situations possible here 1 1 1 and SQ 1 2 1 1 b and 2 b 3 If neither of the above two conditions occurs then we classify numbers as Sl numbers or Sg numbers We are willing to believe that there exists an element 3 C a b so that if z lt c then x C 51 and if z gt c then x C 52 The existence of such a real number is guaranteed by the Least Upper Bound Axiom for the real numbers Since 51 is bounded above by b it has a least upper bound Does this least upper bound separate the sets 51 and 52 as above Let c lub 1 and let x C 51 Then by the de nition of a least upper bound z lt c and for any 6 gt O of e C 51 Thus if y lt c then y C 51 If y gt c then y cannot be in 1 for c is bigger than every element in 51 Thus y C 52 The sets 51 and 52 form a Dedekind cut of the set a b and c is called the Dedekind number of the cut Our Dedekind s Axiom is a translation of this phenomenon to a line Without Dedekind s Axiom there is no guarantee that there is a segment of length 7139 or of length e or of certain other non constructible lengths eg It is Dedekind s Axiom that allows us to make the correspondence of the line in our geometry and the real line There are well de ned geometries that exist without the Dedekind Axiom such as the geometry of the surd plane They do not have all of the properties which we wish to have or to which we are accustomed to having It is with Dedekind s Axiom that we are able to introduce a coordinate system and do geometry analytically in the fashion of Fermat and Descartes To see why we will want Dedekind s Axiom consider the manner in which you construct the perpendicular to a given line at a given point First using the given point as a center draw a circle of positive radius The circle intersects the line in two points At each of these points you then construct a circle of larger radius and these two circles intersect in two points Drawing the line between the two points of intersection gives a line perpendicular to the given line at the given point There are two problems with this proof and they are both very subtle Why does the line intersect the circle at all Why do the two circles then intersect In the surd plane these are not necessarily true The rst of these problems is addressed by the following principle which follows from Dedekind s Axiom We de ne a point A to be inside a circle centered at O with radius OR if OA lt OR A point B is outside the circle if OB gt OR Elementary Continuity Principle If one endpoint of a segment is inside a circle and the other outside then the segment intersects the circle We shall prove this after we have developed some theorems in geometry that we shall need The second problem above is that of the intersection of two circles It is addressed by the following principle7 which is again a corollary of Dedekind s Axiom Circular Continuity Principle If a circle V has one point inside and one point outside another circle 39y then the two circles intersect in two points We shall prove this later also Chapter 9 Neutral Geometry How much of our geometry does not depend on a parallel axiom How much is indepen dent of this particular postulate As you will see7 a large portion of our knowledge does not depend on any parallel postulate This study will then help us to see the true role of a parallel postulate in Geometry 91 Alternate Interior Angles De nition 27 Let E be a set of lines in the plane A line k is transversal of E if 1 k E and 2 k my wforallme Let 6 be transversal to m and n at points A and B7 respectively We say that each of the angles of intersection of Z and m and of Z and n has a transversal side in Z and a non transversal side not contained in Z A De nition 28 An angle of intersection of m and k and one of n and k are alternate interior angles if their transversal sides are opposite directed and intersecting7 and if their non transversal sides lie on opposite sides of 6 Two of these angles are corresponding angles if their transversal sides have like directions and their non transversal sides lie on the same side of 6 De nition 29 If k and Z are lines so that k 6 0 we shall call these lines non intersecting 91 ALTERNATE INTERIOR ANGLES 59 We want to reserve the word parallel for later Theorem 91 Alternate Interior Angle Theorem If two lines out by a transversal have a pair of congruent alternate interior angles then the two lines are non intersecting Figure 91 PROOF Let m and n be two lines cut by the transversal 6 Let the points of intersection be B and B respectively Choose a point A on m on one side of t and choose A E n on the same side of t as A Likewise choose C E m on the opposite side of 6 from A Choose Cquot E n on the same side of t as C Hence it is on the opposite side of 6 from A by the Plane Separation Axiom We are given that ZA B B E ZOBB Assume that the lines m and n are not non intersecting ie they have a nonempty intersection Let us denote this point of intersection by D D is on one side of 6 so by changing the labeling if necessary we may assume that D lie gt the same side of t as C and C By Congruence Axiom 1 there is a unique point E 6 HA so that HE E BD Since BB E BB by Axiom 0 2 we may apply the SAS Axiom to prove that AEBB E ADBB From the de nition of congruent triangles it follows that DBB E ZEBB Now the supplement of ZDBB is congruent to the supplement of EBB by Proposition 75 The supplement of EBB is DBB and DBB E ZEBB Therefore ZEBB is congruent to the supplement of ZDBB Since the angles share a side they are themselves supplementary Thus E E n and we have shown that DE C n or that m O n is more that one point contradicting Proposition 51 Thus m and n must be non intersecting Corollary 1 Ifm and n are distinct lines both perpendicular to the line 5 then m and n are non intersecting PROOF t is the transversal to m and n The alternate interior angles are right angles By Proposition 7 14 all right angles are congruent so the Alternate Interior Angle Theorem applies m and n are non intersecting Corollary 2 UP is a point not on 5 then the perpendicular dropped from P to Z is unique PROOF Assume that m is a perpendicular to 6 through P intersecting t at Q If n is another perpendicular to 6 through P intersecting t at R then m and n are two distinct lines perpendicular to Z By the above corollary they are non intersecting but each contains P Thus the second line cannot be distinct and the perpendicular is unique 60 CHAPTER 9 NEUTRAL GEOMETRY U Q D Figure 92 The point at which this perpendicular intersects the line 6 is called the foot of the perpendicular Corollary 3 If 6 is any line and P is any point not on 6 there exists at least one line in through P which does not intersect l PROOF By Corollary 2 there is a unique line m through P perpendicular to Z By Proposition 77 there is a unique line n through P perpendicular to m By Corollary 1 l and n are non intersecting I Note that while we have proved that there is a line through P which does not intersect l we have not and cannot proved that it is unique 92 Weak Exterior Angle Theorem Let AABC be any triangle in the plane This triangle gives us not just three segments but in fact three lines De nition 30 An angle supplementary to an angle of a triangle is called an exterior angle of the triangle The two angles of the triangle not adjacent to this exterior angle are called the remote interior angles Theorem 92 Exterior Angle Theorem An exterior angle of a triangle is greater than either remote interior angle See Figure 92 PROOF We shall show that ACD gt A In a like manner you can show that ACD gt 13 Then by using the same techniques you can prove the same for the other two exterior angles By Proposition 712 part i either A lt ACD A ACD or A gt ACD If A BAC ACD then by the Alternate Interior Angle Theorem E and E are non intersecting This is impossible since they both contain B Assume then that A gt ACD By the de nition of this ordering on angles there exists a ray TE between E and E so that CAE ACD 92 WEAK EXTERIOR ANGLE THEOREM 61 By the Crossbar TheoremQ AE intersects BC in a point G Again by the Alternate Interior Angle Theorem AE and CD are non intersecting This is a contradiction Thus A lt ZACD I Proposition 91 SAA Congruence In triangles AABC and ADEF given that AC E DF A 1D and B e E then AABC ADEF C Figure 93 PROOF lf AB E DE we are done by Angle Side Angle Thus let us assume that AB E DE Then by Proposition 74 we must have that either AB lt DE or AB gt DE lf AB lt DE then there is a point H 6 DE so that AB E DH Then by the SAS Axiom AABC E ADHF Thus B E ZDHF But ZDHF is exterior to AFHE so by the Exterior Angle Theorem ZDHF gt E E 1B Thus ZDHF gt B by Proposition 712 and we have a contradiction Therefore AB is not less than DE By a similar argument we can show that assuming that AB gt DE leads to a similar contradiction Thus our hypothesis that AB E DE cannot be valid Thus AB E DE and AABC E ADEF by ASA I Proposition 92 Two right triangles are congruent if the hypotenuse and a leg of one are congruent respectiuely to the hypotenuse and a leg of the other Proposition 93 Euery segment has a unique midpoint Q Figure 94 PROOF Let AB be any segment in the plane and let C be any point not on ltESuch a point exists by Axiom 173 There exists a unique ray BX on the opposite side of AB from P such that ZPAB E ZXBA by Axiom 074 There Munique point Q E BX so that AP E BQ by Axiom 071 Q is on the opposite side of AB from P by meworlgP roblem 9 Chapter 3 and Axiom B74 Since P and Q are on opposite sides of AB PQ AB 7 0 Let M denote this point of intersection By Axiom B72 either AMB MAB ABM MA orMB 62 CHAPTER 9 NEUTRAL GEOMETRY We want to show that A M B so let us assume N A M B Since MB g BA by construction we have from the Alternate erior Mle Thigrem that AP andiQ are non intersecting lfM A then A PM 6 AP and AP AB which intersects BQ We can dispose of the case M B similarly lt Thus assume that M A B This will mean that the line PA will intersect side MB of AMBQ at a point between M and B ThusltbyPasch heorem it must intersect mer MQ or BQ It cannot intersect side BQ as AP and BQ are non intersecting lf AP intersects MQ then it must contain MQ for P Q and M are collinear Thus M A which we have already shown is impossible Thus we have shown that M A B is not possible In the same manner we can show that A B M is impossible Thus we have that A M B This means that AMP E ZBMQ since they are vertical angles By Angle Angle Side we have that AAMP E ABMQ Thus AM E MB and M is the midpoint of AB I Proposition 94 i Euery angle has a unique bisector ii Euery segment has a unique perpendicular bisector Proposition 95 In a triangle AABC the greater angle lies opposite the greater side and the greater side lies opposite the greater angle ie AB gt BC if and only if C gt A Proposition 96 Giuen AABC and AABC ifAB AB and BC B C then B lt B if and only if AC lt A C Chapter 10 Theorems of Continuity 101 Elementary Continuity Principle We will now take up the Axioms of Continuity First we will prove the two Continuity Principles and then discuss some of the different uses of the Continuity Axioms in our work First we shall need the famous Triangle Inequality It is usually proved after we have given a measure to line segments but that is not necessary Proposition 101 Triangle Inequality If A B and C are three noncollinedr points then AC lt AB BC where the sum is segment addition 0 A B D Figure 101 PROOF There is a unique point D such that A B D and BD E BC by the rst congruence axiom Then ZBCD ZBDC by Proposition 71 Now AD AB BD and since BD E BC we have that AD E AB BC By Proposition 67 is between the rays E1 and Then by de nition ZACD gt ZBCD By Proposition 712 and what we have just shown ZACD gt ZADC Thus by Proposition 95 AD gt AC and we are done Theorem 101 Elementary Continuity Principle If one endpoint of a segment is inside a circle and the other outside then the segment intersects the circle PROOF Let F be a circle centered at O with radius OR so that R E F Let AB be the above segment Let IntF A OA lt OR and ErtF B OB gt OR Let 21 XlXeABandOXltOR 22 XlXeABandOXZOR 63 64 CHAPTER 10 THEOREMS OF CONTINUITY Note that neither of 21 or 22 is empty We wish to show that these two sets form a Dedekind cut of the segment AB There are then several things to show 1 If Y 6 AB then by Proposition 74 we must have that OX lt OR OX E OR or OX gt OR This puts X in one of the two sets Therefore 21 U 22 AB to Clearly 21 O 22 0 OJ We have that A E 21 and B E 22 BE Let X E 21 and A C X We need to show then that C E 21 If OA E OX then A E X OCX is exterior to AOAC so OCX gt A E X Thus by Proposition 95 00 lt OX Since OX lt OR by Proposition 74 00 lt OR and C E 21 Now assume that OA E OX Then for the rst case assume that OA lt OX Hence X lt A Again A lt OCX which implies 00 lt OX lt OR Thus C E 21 If OA gt OX reverse the labels in the previous argument and we again nd that C E 21 Thus 21 and 22 form a Dedekind cut of AB Let M be the cut point CLAIM OM E OR Assume OM lt OR This puts M E 21 and so M 7 B Now there is a segment which is less than the difference between OM and OR Let M be a point outside P and on AB so that MM is congruent to that segment Thus we have that M E 22 By the Triangle lnequality we have that OM lt OMMM OMMM lt OR OM lt OR But M E 22 so that OM 2 OR a contradiction Assuming that OM gt OR we use the same technique to arrive at a contradiction Thus OMEORandMEP 102 Measure of Angles and Segments To avoid some of the dif culties that we faced in the previous two proofs and to facilitate matters at a later time we will introduce a measure for angles and for segments The proof of the Theorem requires the axioms of continuity for the rst time The axioms of continuity are not needed if one merely wants to de ne the addition for congruence classes of segments and then prove the triangle inequality for these congruence classes It is in order to prove several of our theorems that we need the measurement of angles and segments by real numbers and for such measurement Archimedes s axiom is required However the fourth and eleventh parts of Theorem 102 the proofs for which require Dedekind s axiom are never used in proofs in the text It is possible to introduce coordinates without the continuity axioms as in discussed in Appendix B of the text The notation A will be used for the number of degrees in A and the length of segment AB will be denoted by E 102 MEASURE OF ANGLES AND SEGMENTS Theorem 102 A There is a unique way of assigning a degree measure to each angle such that the following properties hold i A is a real number such that 0 lt A lt 180 ii A 90 if and only if A is a right angle iii A B if and only if A B iu UR is interior ot DAB then DAB u For euery real number m between 0 and 180 there exists an angle A DAC CAB such that A z ui If B is supplementary to A then A B 1800 uii A gt B if and only if A gt B B Giuen a segment 0 called the unit segment Then there is a unique way of assigning a length AB to each segment AB such that the following properties 0 39 i E is a positiue real number and W 1 ii E W if and only ifAB CD iii A gtk B gtk C ifand only if m iu Elt W if and only ifAB lt CD u E euery positiue real number m there exists a segment AB such that Bm De nition 31 An angle A is acute if A lt 90 7 and is obtuse if A gt 90 Corollary 1 The sum of the degree measures of any two angles of a triangle is less than 180 This follows from the Exterior Angle Theorem and Theorem 102 PROOF We want to show that A B lt 180 From the Exterior Angle Theorem and Theorem 1027 A lt CBD A B lt CBD B 180 7 since they are supplementary angles Corollary 2 Triangle Inequality If A B and C are three noncollinear points then E lt E W Theorem 102 offers an easier proof of this than the one that we gave 66 CHAPTER 10 THEOREMS OF CONTINUITY A B 103 Saccheri Legendre Theorem This theorem gives us a setting for our later exploration into noneuclidean geometry Theorem 103 Saccheri Legendre Theorem The sum of the degree measures of the three angles in any triangle is less than or equal to 180 14 1B 10quot 180 PROOF Let us assume not 239e7 assume that we have a triangle AABC in which A0 B0 CO gt 180 So there is an x E RJr so that 14quot 1B 10quot 1800 z A C Figure 102 Saccheri Legendre Theorem Compare Figure 102 Let D be the midpoint of BC and let E be the unique point on E so that DE E AD Then by SAS ABAD E AGED This makes 13quot ZDC39Eo ZEquot ZBADquot Thus7 A4quot 1B 10 ZBAD ZEACquot 13 ZACBquot ZEquot ZEACquot ZDCE ZACDquot ZEquot 14 10 So7 AABC and AACE have the same angle sum7 even though they need not be congruent Note that ZBAEo ZC AEo ZBAC 7 hence ZC EAo ZC AEo ZBACquot 104 THE DEFECT OF A TRIANGLE 67 It is impossible for both of the angles 1CE14 and 1C 14E to have angle measure greater than 121BAC so at least one of the angles has angle measure greater than or equal to 121B14C39 Therefore there is a triangle AACE so that the angle sum is 180 1 x but in which one angle has measure less than or equal to 12114 Repeat this construction to get another triangle with angle sum 180 1 x but in which one angle has measure less than or equal to 14114 Now there is an n E ZJr so that 1 o 27L114 x by the Archimedean property of the real numbers Thus after a nite number of iterations of the above construction we obtain a triangle with angle sum 180 1 z in which one angle has measure less than or equal to 1 o 27L114 z Then the other two angles must sum to a number greater than 180 contradicting Corollary 1 to Theorem 102 Corollary 1 In AABC the sum of the degree measures of two angles is less than or equal to the degree measure of their remote exterior angle 104 The Defect of a Triangle Since the angle sum of any triangle in neutral geometry is not more than 180 we can compute the difference between the number 180 and the angle sum of a given triangle De nition 32 The defect of a triangle AABC is the number defectA14BC 180 7 9114 1B 10 ln euclidean geometry we are accustomed to having triangles whose defect is zero ls this always the case The Saccheri Legrendre Theorem indicates that it may not be so However what we wish to see is that the defectiueness of triangles is preserved That is if we have one defective triangle then all of the triangles are defective By defective we mean that the triangles have positive defect Theorem 104 Additivity of Defect Let AABC be any triangle and let D be a point between A and B Then defect AABC defect AACD defect ABCD PROOF Since W lies in 1140B we know that 1ACB 1ACD 1BCD and since 114DC and 1BDC are supplementary angles 114DC 1BDC 180 There fore defectA14BC 180 7 114 4 1B 10 180 7 114 1B 1ACD 1BCD 180 180 7 114 1B 1ACD 1BCD 114DC39 1BDC defectA14C39D defectABCD 68 CHAPTER 10 THEOREMS OF CONTINUITY C A D B Figure 103 Corollary 1 defect AABC 0 if and only if defect AACD defect ABCD 0 A rectangle is a quadrilateral all of whose angles are right angles We cannot prove the existence or non existence of rectangles in Neutral Geometry Nonetheless the following result is extremely useful Theorem 105 If there exists a triangle of defect 0 then a rectangle exists If a rectangle exists then every triangle has defect 0 Let me rst outline the proof in ve steps 1 Construct a right triangle having defect 0 2 From a right triangle of defect 0 construct a rectangle 3 From one rectangle construct arbitrarily large rectangles 4 Prove that all right triangles have defect 0 5 If every right triangle has defect 0 then every triangle has defect 0 Having outlined the proof each of the steps is relatively straightforward 1 Construct a right triangle having defect 0 Let us assume that we have a triangle AABC so that defectAABC 0 We may assume that AABC is not a right triangle or we are done Now at least two angles are acute since the angle sum of any two angles is always less th 180quot Let us assume that A and B are acute Also let D be the foot of C on AB We need to know that A D B C u 3 Figure 104 Assume not ie assume that D A B See Figure 104 This means that CAB is exterior to ACAD and therefore A gt CODA 90 This makes A obtuse 104 THE DEFECT OF A TRIANGLE to 03 BE 69 a contradiction Similarly if A B D we can show that B is obtuse Thus we must have that A D B This makes AADC and ABDC right triangles By Corollary 1 above since AABC has defect 0 each of them has defect O and we have two right triangles with defect 0 From a right triangle of defect 0 construct a rectangle We now have a right triangle of defect 0 Take ACBD from Steluy which has a right angle at D There is a unique ray W on the opposite side of BC from D so that ZDBC E ZBCX from Axiom C 4 From Axiom C 1 there is a unique point E on W such that CE E BD E X Figure 105 Thus ACDB E ABEC by SAS Then BOEC 90 and ABEC must also have defect 0 Now clearly since defectACDB 0 ZDBCo BODo 90 and hence ZEC39BO BODo ZEC39Do 90 Likewise ZEBDo 90 and DCDBE is a rectangle From one rectangle construct arbitrarily large rectangles Given any right triangle AXYZ we can construct a rectangle DPQRS so that PS gt XZ and RS gt YZ By applying Archimedes Axiom we can nd a number n so that we lay off segment BD in the above rectangle on the ray W to reach the point P so that 71 BD E PZ and P X Z We make 71 copies of our rectangle sitting on PZ PS This gives us a rectangle with vertices P Z S Y and some other point Now using the same technique we can nd a number m and a point B on W so that m BE E RZ and R Y Z NOw constructing m copies of the long rectangle gives us the requisite rectangle containing AXYZ Prove that all right triangles have defect 0 Let AXYZ be an arbitrary right triangle By Step 3 we can embed it in a rectangle DPQRS Since APQR E APSE we have that RPSo 1 PBS0 90 and then APRS has defect 0 Using Corollary 1 to Theorem 104 we nd defectAPtXY 0 and thus defectAXYZ 0 Therefore each triangle has defect O 70 CHAPTER 10 THEOREMS OF CONTINUITY R P X S Y Figure 106 5 If every right triangle has defect 0 then every triangle has defect 0 As in the rst step use the foot of a vertex to decompose the triangle into two right triangles each of which has defect 0 from Step 4 Thus the original triangle has defect 0 Corollary 1 If there is a triangle with positive defect then all triangles have positive defect Chapter 1 1 The Work of Saccheri and Gauss 1 11 Saccheri Girolamo Saccheri was a Jesuit priest living from 1667 to 1733 Before he died he published a book entitled Euclides ab omni nceuo uindicatus Euclid Freed of Every Flaw lt sat unnoticed for over a century and a half until rediscovered by the Italian mathematician Beltrami He wished to prove Euclid s Fifth Postulate from the other axioms To do so he decided to use a reductio ad absurdum argument He assumed the negation of the Parallel Postulate and tried to arrive at a contradiction He studied a family of quadrilaterals that have come to be called Saccheri quadrilaterals Let S be a convex quadrilateral in which two adjacent angles are right angles The segment joining these two vertices is called the base The side opposite the base is the summit and the other two sides are called the sides If the sides are congruent to one another then this is called a Saccheri quadrilateral The angles containing the summit are called the summit angles Theorem 111 In a Saccheri quadrilateral i the summit angles are congruent and ii the line joining the midpoints of the base and the summiticalled the alti tudeiis perpendicular to both D N C A M 3 PROOF Let M be the midpoint of AB and let N be the midpoint of CD 1 We are given that DABO ABCO 90 Now7 AD BC and AB AB7 so that by SAS ADAB ACBA7 which implies that BD E AC Also7 since CD E CD then we may apply the SSS criterion to see that ACDB g ADCA Then7 it is clear that D E 10 71 72 CHAPTER 11 THE WORK OF SACCHERI AND GAUSS gt gt gt 2 We need to show that MN is perpendicular to both AB and CD Now DN E CN AD E BC and D E 0 Thus by SAS AADN E ABCN This means then that AN E BN Also AM E BM and MN E MN By SSS AANM E ABNM and it follows that ZAMN E MN They are supplewtary angles hence they must be right angles Therefore MN is perpendicular to AB gt Using the analogous proof and triangles ADMN and ACMN we can show that MN is perpendicular to CD Thus we are done I Saccheri considered all possible cases of such a quadrilateral They are ADC is a right angle HRA the hypothesis of the right angle ADC is an obtuse angle HOA the hypothesis of the obtuse angle ADC is an acute angle HAA the hypothesis of the acute angle We shall see that HRA is equivalent to Euclid s Postulate V so we may take HOA or HAA as negations of Postulate V The Three Musketeers Theorem implies that if one of HRA HOA or HAA holds for one quadrilateral then it holds for all Theorem 112 In a Saccheri quadrilateral DABCD on the base AB under the assumption HRA HOA or HAA we haue AB N CD AB gt CD or AB lt CD respectiuely and the angle sum of a triangle is equal to greater than or less than two right angles respectiuely PROOF Let M and N denote the midpoints of AB andCD respectively We will work with the assumption HAA since the others are similar or already known We wish to show that AB lt CD Suppose CD lt AB then CN lt BM There is a point T E 1 so that NT N BM and N C T Thus ZMBT must be greater than a right angle since W is between W and DNMBT is a Saccheri quadrilateral on the base NM which implies that ZBTN N ZTBM which implies that ZBTN is greater than a right angle In ABTC ZBCN is an exterior angle and hence is greater than ZBTN Recall that by hypothesis ZBCN is an acute angle Thus we have an acute angle greater than an angle that is greater than a right angle our contradiction Therefore AB lt CD Now suppose we have a right triangle AABC with right angle at B Construct AD perpendicular to AB with AD N BC Assuming HAA CD gt AB and ZDAC gt ZACB since the greater side subtends the greater angle Now CAB ZDAC is a right angle Thus the angle sum of AABC is less than two right angles I gt gt gt Consider two parallel lines k d l and let k AB and l CD Let X Y g and constuct lines perpendicular to AB through X and Y Call these lines XU and YV We ar terestg h the angles ZUXY and ZVYX If both angles are acute then the two lines AB and CD have a common perpendicular between the segments UX and VY If one of the angles is a right angle and the other is acute then the two lines already have a common perpendiculASuppose that ZUXY is acute and ZVYX is obtuse If we move V away from U along AB then ZVYX may change to a right angle or remain obtuse lt gt Theorem 113 As aboue anMssuming HAA if AV moues away from U along AB ZVYX remains obtuse then CD is asymptotic to AB 112 GAUSS 73 A U V B Saccheri now shows the following Theorem 114 Giuen a point P not on a line b there are three classes of lines through P 1 lines meeting 5 2 lines with a common perpendicular to l and 3 lines withour a common perpendicular to l and hence asymptotic to l Saccheri now concludes the hypothesis of the acute angle is absolutely false because it is repugnant to the nature of straight lines77 The aw in Saccheri s work was observed by the Swiss mathematician JH Lambert in 1786 He himself contributed to the work in non euclidean geometries with the following A convex quadrilateral three of whose angles are right angles is called a Lambert quadrilateral Under the HAA assumption the following are true Theorem 115 The fourth angle of a Lambert quadrilateral is acute Theorem 116 The side adjacent to the acute angle of a Lambert quadrilateral is greater than its opposite side Theorem 117 In a Saccheri quadrilateral the summit is greater than the base and the sides are greater than the altitude PROOF Using Theorem 111 if M is the midpoint of AB and N is the midpoint of CD7 then DAMND is a Lambert quadrilateral Thus7 AB gt MN and7 since BC E AB7 both sides are greater than the altitude Also7 applying Theorem 111 DN gt AM Since CD 2DN and AB 2AM it follows that CD gt AB7 so that the summit is greater than the base I 112 Gauss The contributions of Carl Friedrich Gauss 177771855 are found in two brief unpublished letters and notes He begins with a different de nition of parallel 74 CHAPTER 11 THE WORK OF SACCHERI AND GAUSS gt gt De nition 33 Given a e AB and a poirLP not on AB let AP be the perpendicular from P to AB A line PQ is parallel to AB thmugh P if fomy line PS ltwith S in the interior of ZAPQ and 0 lt ZAPSo lt ZAPQquot it follows that PS intersects AB Soniwould seem tth we consider the set of lines tmugh the point P sweeping up from AP the parallel PQ is the rst line not meeting AB Unless we assume Euclid s Mulate V we do not know that this line is unique Thus it may depend on the side of AP we choose Therefore we speak of parallel lines in a direction These lines will be called nonintersecting lines to distinguish them from parallel lines Theorem 118 In a given direction being parallel is an equivalence relation PROOF We need to show three things gt gt a First we must show that it is well de nithat is if PQ is parallel to AB and S is any point between P and Q then SQ is parallel to AB gt gt b We must also show that the relation is re egive that is if PQ is parallel to AB in a gven direction then AB is parallel to PQ in the same direction gt c We must show that the relation is transitive if AB is parallel to amp in a given direction and PQ is parallel to BS in the same direction then AB is parallel to BS in that direction To begin gt a Let P S Q and consider any line through S SC suc hat C is int r to 1Aampand ZBAP Construct PC By our assumption PC intersects Thus SC has entered a triangle so by Pasch s Theorem it must intersect AB gt b Let R be the foot of A in PQ Let S be a point in the interiors of BAR and ZARQ and construct the line AS Chapter 12 Hyperbolic Geometry 121 The Hyperbolic Axiom and its Consequences What is hyperbolic geometry It is an example of one of many non euclidean geometries It is often called Bolyai Lobacheuskiian geometry after two of its discouers Janos Bolyai and Nikolai lvanovich Lobachevsky Bolyai rst announced his discoveries in a 26 page appendix to a book by his father7 the Tentamen7 in 1831 You can read about what happened to him and his work in the text Another of the great mathematicians who seems to have preceeded Bolyai in his work is Carl F redrich Gauss He seems to have done some work in the area dating from 17927 but never published it The rst to publish a complete account of non euclidean geometry was Lobachevsky in 1829 It was rst published in Russian and was not widely read In 1840 he published a treatise in German We shall add one more axiom to the list of axioms for Neutral Geometry HYPERBOLIC AXIOM In hyperbolic geometry there exists a line b and a point P not on 6 such that at least two distinct lines parallel to 6 pass through P IgtEltI t We shall denote the set of all points in the plane by H27 and call this the hyperbolic plane Lemma 121 There exists a triangle whose angle sum is less than 180 P Y m X X 71 Q R 5 7 6 CHAPTER 12 H YPERB OLIC GEOMETRY PROOF Let 6 be a line and P a point not on 6 such that two parallels to 6 pass through P We can construct one of these parallels as previously done using perpendiculagLet Q be the foot of the perpendicular to 6 through P Let in be the perpendicular to PQ through P Then in and Z are non intersecting Let n be another line through P which does not intersect Z This line exists by the Hyperbolic Axiom Let W be a ray of n lying between mandaray ofm lt CLAIM There is a point B E Z on the same side of PQ as X and Y so that ZQRP lt ZXPY PROOF OF CLAIM The idea is to construct a sequence of angles romp ZQRZP iQRnP so that ZQRHlPquot lt ZQR739P We will then apply Archimedes Axiom for real numbers to complete the proof There is a point R1 6 6 so that QR1 PQ Then AQRlP is isosceles and ZQRlPquot 45 Also there is a point R2 6 6 so that F R1 R2 and Ptle E PR1 Then APP le is isosceles and ZRlPRZ E ZQRZP Since ZQRlP is exterior to APP le it follows that 131133 ZQRZPquot ZQRlPquot so then ZQRZPquot 22 Continuing with this construction we nd a point Rn E 6 so that Q gtk RnA Rn and o 45 ZQPWP Applying the Archimedean axiom we see that for any positive real mber for example ZXPYquot there is a point B E 6 so that R is on the same side of PQ as X and Y and ZQRPquot lt ZXPYquot Thus we have proved our claim Now 17R lies in the interior of ZQPX for if not then W is in the interior of ZQRP By the Crossbar Theorem it follows that 1 6 7 0 which implies that n and Z are not non intersectingia contradiction Thus ZRPQquot lt ZXPQquot Then ZRPQquot ZQRPquot lt ZXPQquot ZQRPquot lt ZXPQquot ZXPYquot 90 Therefore ZPquot if ZRquot lt 180 and defeetAPQR gt 0 I The Hyperbolic Axiom only hypothesizes the existence of one line and one point not on that line for which there are two non intersecting lines With the above theorem we can now prove a much stronger theorem Theorem 121 Universal Hyperbolic Theorem In H2 for every line 6 and for every point P not on 6 there pass through P at least two distinct lines neither of which intersect Z PRM Drop a perpendicular to Z and construct a line in through P perpendicular to PQ Let R be any other point on 6 and construct a perpend ar t to 6 through R Now let S be the foot of the perpendicular to t throrrghsP Now PS does not intersect 6 since both are perpendicular to t At the same time PS 7 m Assume that S E m then DPQRS is a rectangle By Theorem 105 if one angle exists all triangles have defect 0 We have a contradiction to Lemma 121 Thus PS 7 m and we are done I 122 ANGLE SUMS AGAIN 77 122 Angle Sums again We have just proven the following theorem Theorem 122 In H2 rectangles do not exist and all triangles have angle sum less than 0 This tells us that in hyperbolic geometry the defect of any triangle is a positive real number We shall see that it is a very important quantity in hyperbolic geometry Corollary 1 In H2 all convex quadrilaterals have angle sum less than 360 PROOF Given any quadrilateral DABCD Take the diagonal AC and consider triangles AABC and AACD By Theorem 122 both of these triangles have angle sum less than 180 The assumption that DABCD is convex implies that ZBACo CADo BADo and ZACBo ZAC Do ZBCDquot By adding all six angles we have that the angle sum of DABCD is less that 360 I 12 3 Similar Triangles ln euclidean geometry we are used to having two triangles similar if their angles are con gruent It is obvious that we can construct two non congruent yet similar triangles In fact John Wallis attempted to prove the Parallel Postulate of Euclid by adding another postulate WALLIS POSTULATE Given any triangle AABC and given any segment DE There exists a triangle ADEF having DE as one of its sides that is similar to AABC However on page 124 of the text it is proven that Wallis7 Postulate is equivalent to Euclid s Parallel Postulate Thus we know that the negation of Wallis7 Postulate must hold in hyperbolic geometry That is under certain circumstances similar triangles do not exist We can prove a much stronger statement Theorem 123 AAA Criterion In H2 if A D 1B 1E and C 1F then AABC ADEF That is if two triangles are similar then they are congruent PROOF Assume that AABC 9 ADEF Then there can be no corresponding sides which are congruent else by ASA the triangles would be congruent Consider the triples 7 8 CHAPTER 12 H YPERB OLIC GEOMETRY AB BC AC and DEEFDF One of these triples contains at least two segments that are larger than the corresponding segments of the other Without loss of generality we can assume that AB gt DE and AC gt DF There exist points B 6 AB and C 6 AC such that AB E DE and AC E DF Then by SAS AABC g ADEF Then ABC g E g B awAC B g F g CL Thus by the Alterate Inten39or Angles Theorem we can show that B C does not intersect BC Thus it follows that DBB C C is convex Using the fact that ZAB C and BBC are supple mentary as are ZAC B and ZCC B the angle sum of DBB C C is 360 contradicting Corollary 1 to Theorem 122 I B C As a consequence of Theorem 123 we shall see that in hyperbolic geometry a segment can be determined with the aid of an angle For example an angle of an equilateral triangle determines the length of a side uniquely Thus in hyperbolic geometry there is an absolute unit of length as there is in elliptic geometry Chapter 13 Classi cation of Parallels 131 Fan Angles A line 6 subdivides an angle if it passes through the vertex of the angle and intersects the interior of the angle Let P be an arbitrary point in the hyperbolic plane The pencil of lines through P denoted PP is the set of all lines coincident with P Theorem 131 IfP 6 then there exists an angle with vertex at P whose interior contains 6 and whose bisector is perpendicular to Z m P n PROOF In the pencil PP let p be the perpendicular to Z and let in E PP be perpen dicular to p at P Thus we know that in 6 0 By the Universal Hyperbolic Theorem there is another line n E PP which does not intersect 6 Since n is distinct from m n is not perpendicular to p Thus one pair of the vertical angles formed by p and n is a pair of acute angles Thus there exists a ray 137i Q n so that ZQPAquot 04 lt 90 where Q is the foot of P in 6 Now by Congruence AxiomAhere is a ray 17B on the opposite side of p from A so that ZQPB ZQPA The line PB does not intersect Z If it did then let R be the point of intersection Choose PU on the same side ofp as A so that QR E QR Then by SAS APRQ APRQ Thus ZQPR ZQPR ZQPA Thus by Cong ence Axiom 4 PK 6 n This implies that n 6 7 0 which is a contradiction Therefore PB 6 0 Clearly ITQ bisects ZBPA We only need to show that Z is contained in the interior of ZBPA or that every point on 6 lies in the interior of ZBPA 79 80 CHAPTER 13 CLASSIFICATION OF PARALLELS Let X E 6 be on the same side of p as A and let Y be the unique point on 6 on the opposite side ofp as X so that QX E QY Now we know that the ray ITQ lies in the interior of APB and in fact using an abuse of notation ZBPA ZQPB U ZQPA Claim Q7 lies in the interior of ZQPA Likewise Q lies in the intem of ZQPB To prove this claim rst note that since X is on the same side of p PQ as A every point of Q7 lies on the same side ost A Thus we are left only to show that if R 6 Q7 then R is on the same side of n PA as Q This must be the case for if not we would have that n O l 7 0 a contradiction Thus the above claim is true We then have that loQMJTK which from above must lie in the interior of ZAPB I Corollary 1 Every line in 73P that intersects l subdiuides the angle ZAPB We have almost proven the following result Theorem 132 IfP 6 then there is a unique angle with the following properties i the angle contains 6 in its interior ii the bisector of this angle is perpendicular to 5 iii euery line passing through P that subdiuides the angle intersects l S P V U Q PROOF To prove that this angle exists we need to construct the sides of the angle We already know that there are angles with vertex at P that contain the line 6 We want to prove much more in this theorem In some sense we are proving that there is a smallest such angle Let S be a point on the line which is the perpendicular at P to the perpendicular through P to 6 Consider the line SQ Let 21 be the set of all points T on SQ h that ITT intersects 6 together with all points on the ray opposite to Let 22 SQ 21 By the Crossbar Theorem if T E SQ and T E 21 then the entire subsegment TQ CA Hence 21 22 is a De nd cut By Dedekind s Axiom there is a unique point X on SQ such that for P1P2 E SQ P1 X P2 if and only if X 7 P1 and X 7 P2 P1 6 21 and P2 6 22 By the de nition of 21 and 22 rays below W all intersect l and rays above 17X do not Now we wish to prove that PYol0 132 LIMITING PARALLEL RAYS 81 Assume on the contrary that Choose a point V on 6 so that V U Q Since V and U are on the same side of SQ Exercise 9 Chapter 3 V and P are on opposite sides so that VP meets SQ in a point Y By Proposition 67 we have that Y X Q Thus Y e 22 This is impossible for if Y e 22 then W m z 0 Thus I does not meet 6 lt Doing the same construction on the opposite side of PQ at P gives us a similar ray PX with similar properties We need to show that ZQPX E ZQPX Assume then we M assume that ZQPX lt ZQPX By Congruence Axiom 4 Lae ray PP between PX and ITQ such that ZQPR E ZQPX Byghe properties of PX PR must intersect 6 Let R be the point on 6 on the opposite of PQ from PU so that QR E QPJ Then by SAS AQPP E AQPR This implies that ZQPR E ZQPR E ZQPX By Congruence Axiom 4 we have to have that R E W or W 6 7 Q a contradiction Thus ZQPX E ZQPX Checking the properties clearly ITQ bisects the angle the angle contains 6 and by the de nition of the ray W any line that subdivides the angle intersects 6 For P 6 this angle at P is called the fan angle of P and Z and will be denoted by P 6 132 Limiting Parallel Rays lt gt Consider a ray E and a point P AB Characterize all rays W satisfying gt gt 1 PX AB Q 2 every ray between W and 17A intersects In euclidean geometry there is onane such ray Is this the case in H2 There is a angle of P and AB call it ZXPY and e si f this angle lies on the same side of AP as B LetAsay that it is Now PX AB Also if 17R lies between W and 137i then PR subdivides angleXPY Thus 17R AB 7 0 but due to its relative position with respect to 17A and W it must in fact intersect Thus the side of the fan angle satis es the two aforementioned conditions Do any others Let W be different from gt gt gt If W is on the line PA then PT intersects AB and the ray ITT does not have property 1 gt If P li s on the same side of PA as Y let W be between 17A and Then the line PA separates W and W from E which means that W O E 0 so that W does not satisfy property gt If ITT lies on the same side of PA as X then one of ITT and W is between the other and P lt gt If P Tl is between W and In then W must intersect AB by the properties of the fan angle Thus ITT would not satisfy property If W is between 137i and W then W cannot satisfy property 2 for P XZ O E 82 CHAPTER 13 CLASSIFICATION OF PARALLELS Thus7 IT is the only ray at P with these two properties This gives rise to the following de nition De nition 34 A ray ITQ is parallel to a ray E if gt gt i PQ AB 0 ii every ray between 172 and 17A intersects We have just shown that this ray is unique In this case the rays 172 and E are called limiting parallel rays and we say that 1762 is limiting parallelto QUESTION lf 1762 is limiting parallel to E is A is limiting parallel to 1762 We need to following lemma for the proof of the main theorem 1 leave the proof to you as an exercise lt gt Lemma 131 If ZPAB is acute then the foot ofP in the line AB lies in the ray 1 and is di erent from A I offer the following theorems without proof Theorem 133 If is limiting parallel to B then 1 lies in the interior of the angle ZAP Theorem 134 Let P l and let A e B e C in l Then 1 is limiting parallel to E if and only if 1762 is limiting parallel to Theorem 135 Let A e B e C on l E is limiting parallel to E if and only if B is limiting parallel to Theorem 136 If is limiting parallel to B then B is limiting to P Q M X H G AB F C D PROOF Let F be the foot of P in E l Let C e l so that Fe is like directed to E This means that either E C F or F C lt folloMom Theorem 1lt gtthm c2gt is limiting parallel to FY so that C lies on the same side of PF as does Q and FC PQ 0 Let F lie in the interior of ZPFC CLAIM W 11762 7 Q 132 LIMITING PARALLEL RAYS 83 From this it follows that F is limitin parallel to Applying Theorem 135 again we have that E is limiting parallel to Thus we need to establish this claim to nish the proof lt ice ZPFXo lt PFCo 90 ZPFX is acute Thus by Lemma 131 the foot of P in amp must lie in the ray Label this point G It follows that G lies on the same side of PF as If G E PQ then it must lie in alt follows that F7 P Qgt 7 Q and we are done If G lies on the opposite side of PQ from F then again F7 intersects 172 and we are done lt Assume then as the nal case that G lies on the same side of PQ as F Then this puts G in the interior of ZQPF Let 04 ZQOPG We have that 04quot lt ZQPFquot By Congruence Axiom 4 there is a unique ray ITZ in the interior of ZQPF so that ZFPZo 04 ZQPGquot Since ITQ is limiting parallel to W ITZ must intersect F in some point D From Lemma 131 G 7 F so that PC lt PF by an Exercise the hypotenuse of a right tri angle is ger than either leg Tm there is a poiM 6 PF so that PH ltPG LMY perpendiculMo PF at H Since AB is perpendicular to PF it follows that AB HY 0 Now HY does intersect one side of the triangle APFD By Pasch s Theorem it must intersect a second side That side must be PD Let the point ofintersection beE Now Ey PandEy D On ITQ there is a unique point M so that PM E PE Recalling that PH E PG and ZHPE ZFPZ E ZQPG we then haMhatAPHE E APGM by SAS Thus MM ZPHE and right angle Thus GM FX since both are perp icular to PC at G Therefore FX 1762 Since M lies on the same side of FC as Q it follows nally that M 6 F7 and F 1762 7 Q I A ray Y is parallel to a line 6 if B is a limiting parallel ray to some ray in l A line k is limiting parallel or asymptotically parallel or even horoparallel to a line 6 if some ray in k is a limiting parallel ray to some ray in Z We have just proven that these parallelisms are symmetric and we may denote them by Yul all lHk kHl gt gt If k AB 6 CD and E W then E is said to be parallel to l in the direction of W on 6 Furthermore k and l are said to be parallel in the direction of E on k and in the direction of W on 6 Theorem 137 IF P 6 then there are exactly two lines through P that are limiting parallel to 5 Each contains an arm of the fan angle P l and they are limiting parallel to l in opposite directions As we have mentioned several times already there is no simple transitiVity of parallelism in hyperbolic geometry There is a weak form of transitiVity Theorem 138 Weak Transitivity of Parallels Two lines parallel to a third in the same direction on the third are parallel to each other 84 CHAPTER 13 CLASSIFICATION OF PARALLELS 133 Hyperparallel Lines Let k and 6 be lines in H2 and let An n 1 2 be points on k Let A be the foot of An in Z We say that the points An are equidistant from 6 if AiA E AjA j for all choices of i and j Theorem 139 If 6 and 6 are non intersecting lines in H2 then any set of points on 6 equidistant from 6 has at most two elements PROOF Assume not ie assume that A BC E Z are equidistant from 6 Thus the quadrilaterals DAA B B DAA C C and DBB C C are all Saccheri quadrilaterals Thus ZA AB E ZB BA ZA AC E ZO CA and BBC E C CB Stringing these together we have ZBBA E ZAAB E ZCCA E ZBBC Since ZB BA and BBC are supplementary they are right angles Thus all three quadri laterals are rectangles a contradiction I This theorem states that at most two points at a time on 6 can be equidistant from 6 It does not put a limit on the number of pairs of such points though We may have pairs AB and C D so that AA E BB and CCquot E DD but it is not possible that AA E CCquot or any of the other possibilities It may also occur that the number of pairs of such points will be zero There is no guarantee that such a pair will exist This will help us distinguish types of parallel lines Theorem 1310 Let Z and 6 be non intersecting lines for which there is a pair of points A B E 6 which are equidistant from 5 Then 6 and 6 have a common perpendicular segment which is the shortest segment between 6 and 6 PROOF DAA B B is a Saccheri quadrilateral lts altitude is perpendicular to both 6 and 6 by Theorem 111 Also the quadrilateral formed by this altitude and any other segment from a point X E Z and its foot X E 6 is a Lambert quadrilateral By Theorem 116 the altitude is less than XX and hence is the shortest segment between 6 and 6 Theorem 1311 In H2 if lines 6 and 6 have a common perpendicular segment MMC then they are non intersecting and MM is unique Moreover if AB E 6 so that M is the midpoint of AB then A and B are equidistant from 6 PROOF We already know that the lines do not intersect from Corollary 1 to the Alternate Interior Angles Theorem If PP were another common perpendicular then DMM P P would be a rectangle which cannot exist Thus M M is unique 134 CLASSIFICATION OF PARALLELS 85 Let AB E t be so that M is the midpoint of AB We have that AM E BM and ZAMM E ZBMM so by SAS AAMM E ABMM Thus AM E BM and AMM E BMM By Angle Subtraction ZA M A E ZB M B Then by AAS AAA M E ABB M and AA E BB I Lemma 132 Let MM be the common perpendicular to t and 6 Let A B E 6 so that M A B thenAA lt BB If k and t are non intersecting lines which admit a common perpendicular then they are said to be hyperparallel This is denoted by k 6 They are sometimes called diuergently parallel lines Recall that for P 6 there is a unique fan angle P t for P and 6 Let F be the foot of P in t and let X and Y be on opposite arms of the fan angle The angle ZFPX E ZFPY is called the angle of parallelism at P with respect to t and its angle measure is denoted by HPF We know that its measure must be less than 90 Theorem 1312 ft is any line in H2 and 0 lt 04 lt 90 then there is a point P 6 so that HPF 04 134 Classi cation of Parallels We have de ned two types of parallel linesihyperparallel and horoparallel Are these the only types of parallel lines in hyperbolic geometry Are there lines that are both hyperparallel and limiting parallel These are the questions that we wish to answer in this section Theorem 1313 Let W be the limiting parallel rag to 6 through P and let Q and X be the feet ofP and X respectively in 5 Then PQ gt XX P Q X PROOF If P X Y then W t by Theorem 135 Thus ZX XY is the angle of parallelism for X and 6 Thus it is acute Therefore ZX XP is obtuse This implies that ZQPX lt ZX XP and therefore PQ gt XX I This shows that the distance form X to 6 decreases as the distance from P to X in creases along the limiting parallel ray In fact one can show that the distance from X to t approaches 0 Note that we have seen that for hyperparallel lines the further one gets from the common perpendicular the greater the distance between the lines This gives one the impression that these types of parallel lines should be distinct Corollary 1 ft k in a given direction then 6 and k have no common perpendicular 86 CHAPTER 13 CLASSIFICATION OF PARALLELS PROOF Let P E 6 so that W k and assume that there is a point M E 6 so that PMXand MM is a common perpendicular to l and k From Theorem 1313 PQ gt MM gt XX By Theorem 1310 MM lt XX a contradic tion Thus 6 and k have no common perpendicular I This tells us that if two lines are limiting parallel then they are not hyperparallel We now need to show that if two lines are non intersecting and are not limiting parallel then they must be hyperparallel Having done this we will then know that there are only two types of parallel lines Theorem 1314 Given lines in and n so that i in O n 0 and ii in does not contain a limiting parallel ray to n in either direction Then in n PROOF We need to nd two points HK E n that are equidistant from m ie HH E KK If this is true then by Theorem thmsq1 the perpendicular bisector of HK is the common perpendicular to m and n By Theorem 1311 this common perpendicular is unique It follows then that m n i The search for H and K Let AB E n with feet A B E m If AA BB we are done Assume that these segments are not congruent Then we may assume that AA gt BB since one must be greater than the other There is a point E 6 AA what A E E BB There is a unique ray W on the same side of AA as B so that ZA EF ZB BG with A B G CLAIM W m E 7g 0 Proof will follow 135 PROOF OF CLAIM AND ASYMPTOTIC TRIANGLES 87 A B H K Call this point of intersection H The there is a unique point K 6 H3 so that BK E EH Then by SAS AEHA E ABKB Thus A H E B K We also have that EAH E BBK so that by Angle Subtraction it follows that ZHA H E ZKB K Since HH and KK are perpendicular to m we have by Hypotenuse Angle that AHA H E AKB K Therefore HH E KK and H and K are equidistant from m Thus we are nished once we have proved the claim I It turns out that the claim is the hardest item to prove in the whole theorem It takes us through some strange territory 135 Proof of Claim and Asymptotic Triangles lf 17B and Q are limiting parallel rays the convention is to pretend that they pass through an ideal point 9 We then denote the rays by P9 and Q9 The gure consisting of these rays and the segment PO is called a singly asymptotic triangle You can have doubly asymptotic triangles with 2 ideal points and trebly asymptotic triangles with 3 ideal points Lemma 133 Exterior Angle Theorem If APQQ is a singly asymptotic triangle the exterior angles at P and Q are greater than their respective opposite interior angles PROOF Choose R 6 176 so that P Q R We need whow that ZRQQ gt ZQPQ There is a unique ray QT lying on the same side of PO as 9 so that ZRQD E ZQPQ If U 6 OD is so that U Q D then ZUQP E ZRQD E ZQPQ Thus we have that 88 CHAPTER 13 CLASSIFICATION OF PARALLELS AP 9 M Q U B D R the alternate interior angles are cgggruent Let M be the midpoint of PQ and let A and B be the feet of M in P9 and QD respectively By AAS AAPM E ABQM that M E BM and angleAMP E BMQ Th HANMband B are collinear Thus P9 and QD have a common perpendicular Thus P9 QD Thus QT lies between 2 and w from which it follows that ZRQQ gt ZQPQ I Lemma 134 If in the singly asymptotic triangles AABQ and ACDG we have BAD E ZDCG then ZABQ E CD9 if and only if AB E CD A X B PROOF First let us assume that AB E CD but ZABQ E ZODG We may suppose that ZABQ gt CDQ Then there is a unique ray W between E and so that ZABW E ZCDQ Since m and m are limiting parallel W must intersect m in a point X There is a point Y E so that AX E CY Then by SAS AABX E ACDY Therefore CDY E ZABX E CDQ Then and E are not limiting parallel a contradiction Now assume that ZABQ E CD9 and AB E CD We may then assume that AB gt CD There is a unique point P 6 AB so that BP E CD Let 179 be the limiting parallel ray to m from P Thus m is also parallel to The rst part of this proof now implies ZBPQ E DOG E ZBAQ But ZjBPQ is exterior to APAQ by the Exterior Angle Theorem ZBPQ gt ZBAQ a contradiction I Corollary 1 PQ g RS if and only if HPQ HRS We are now in a position to show that W O E 7 Q a a a Let A M H W A N H E and BF H K3 Recall that EA g BB and A EF g ZB BG By Lemma 134 EAM E BBP Choose a point L E m so that A B L a a a gt Since m and n were not limiting parallel lines we have that BL 2E B P and AL f A N Now MAL g PBL by Angle Subtraction Since A N H E A N M if and BF H E3 we must have that AN BP Thus PBL is exterior to the singly asymptotic triage AA B Q Therefore NAL lt PBL This means that A N lies between AM and AL gt a a A which implies that A M lies between AN and AA But since A N E we have that MOEJ 135 PROOF OF CLAIM AND ASYMPTOTIC TRIANGLES 89 A39 E39 H39 K39 Since J Mm it lies on thisame side of Equot as does A Thus7 A and J are on opposite sides of EF Thus7 AJ EF 7 Q But7 H E W since H lies on the same side of AA as does J It is possible to construct the limiting parallel ray to a line through a given point with a cows and straightedge Let P 6 and let Q be the foot ofP in 6 Let m be perpendicular to PO at P so that m 6 Let R E Z R 7 Q and let S be the foot of R in m Then DPQRS is a Lambert quadrilateral and PS lt QR QR lt PR since PR is a hypotenuse Take a circle centered at P of radius QR S is inside this circle and R is outside this circle There is a unique point X 6 RS where RS intersects the circle One can show that the ray W is limiting parallel to Z Chapter 14 Strange New Triangles lf 17B and Q are limiting parallel to each other7 the convention is to pretend that they go through an ideal point 9 and denote them simply by the segments P9 and Q9 The gure consisting of these rays and the segment PQ is then called a trilateral Note that here we do not assume that PQ is necessarily perpendicular to Q9 Trilaterals are also called singly asymptotic triangles The angles of a trilateral are the angles ZQPQ and ZQQP7 and these are called its interior angles Just as we have de ned a trilateralisingly asymptotic triangleiwe can de ne a DATi doubly asymptotic triangle Given a point P and two non opposite rays through P7 and Wilet t be the line that is limiting parallel to W and limiting parallel to 171 in the opposite direction Let 9 denote the ideal point in the direction of W and let 2 denote the ideal point in the direction of Then we can identify the line 6 with the segment 92 and the segments PS27 PE and 92 form a doubly asymptotic triangle The angle UPS is the fan angle of l and P Many of the properties of trilaterals are analogous to those of triangles Lemma 141 Crossbar Theorem for Trilaterals A line which svbdivides an angle of a trilateral intersects the opposite side Lemma 142 Pasch7s Theorem for Trilaterals A line which intersects a side of the trilateral AABQ but does not pass through a vertex will intersect another side provided the line is not limiting parallel to either AD or B9 at 9 Lemma 143 The sum of the interior angles of a trilateral is less than 180 Lemma 144 Exterior Angle Theorem If APQQ is a trilateral the exterior angles at P and Q are greater than their respective opposite interior angles PROOF Choose R E ITQ so that P Q R We need whow that ZRQQ gt ZQPQ Them a unique ray QT lying on the same side of PQ as 9 so that ZRQD E ZQPQ If U E QD is so that U Q D7 then ZUQP ZRQD ZQPQ Thus7 we have that the alternate interior angles are cggruent Let M be the midpoint of PQ and let A and B be the feet of M in P9 and QD7 respectively By AAS AAPM E ABQM that M BM and angleAMP BMQ Th HAm gand B are collinear Thus7 P9 and QD have a common perpendicular Thus7 PQ QD Thus QT lies between 2 and W from which it follows that ZRQQ gt ZQPQ I 90 AP 9 M Q U B D R Two trilaterals are congruent if the angles and segment of one are congruent respectively to the angles and segment of the other Lemma 145 If in the trilateml AABQ and ACDG we have ZBAQ E ZDCG then ZABQ E CD9 if and only if AB E CD A X C B D PROOF First let us assume that AB E CD but ZABQ E ZODG We may suppose that ZABQ gt CDQ Then there is a unique ray W between E and El so that ZABW E ZCDQ Since m and m are limiting parallel W must intersect m in a point X There is a point Y E so that AX E CY Then by SAS AABX E ACDY Therefore CDY E ZABX E CDQ Then and E are not limiting parallel a contradiction Now assume that ZABQ E CD9 and AB E CD We may then assume that AB gt CD There is a unique point P 6 AB so that BP E CD Let 179 be the limiting parallel ray to m from P Thus m is also parallel to The rst part of this proof now implies ZBPQ E DOG E ZBAQ But ZjBPQ is exterior to APAQ by the Exterior Angle Theorem ZBPQ gt ZBAQ a contradiction I Chapter 15 Inversion in Euclidean Circles In order to de ne congruence in the Poincare model and verify the axioms of congruence7 we must study the operation of inversion in a Euclidean circle This operation is the process of re ecting across a line in the Poincare disk model of the hyperbolic plane Let 39y be a circle of radius r and center 0 For any point P 7 O the inverse P of P with respect to the circle 39y is the unique point P E W so that OP OP r2 In this case 39y is called the circle of inversion These are Euclidean geometry theorems about circles7 so we may use the facts that we know about Euclidean geometry Proposition 151 For P7 P and 39y as above 239 P P if and only ifP E 39y it If P E int39y then P int 39y and vice versa in P P gt Proposition 152 Let P E int39y and let TU be the chord through P pemendicular to OP Then P PTU the pole of TU ie the intersection of the tangents to 39y at T and U PROOF Suppose the tangent to 39y at T cuts W at the point P The right triangle AOPT is similar to right triangle OTP since they have TOP in common and the angle sum is 180 Hence corresponding sides are proportional Since OT r we get that OP r 93 gt which shows that P is the inverse to P Re ect across the line OP and we see that the tangent to 39y at U also passes through P so that P is the pole of TU Proposition 153 UP is outside of 39y let Q be the midpoint of OP Let a be the circle of radius QP centered at Then i a 39y TU gt gt ii PT and PU are tangent to 39y m P TU m OP PROOF By the circular continuity principle a and 39y do intersect in two points T and U we ZOEand ZOUP are inscribed in semicircles of a they are right angles Therefore PT and PU are tangent to 39y lf TU intersects OP in a point P then P is the inverse of P by the previous proposition Thus P is the inverse of P in 39y I The next proposition shows how to construct the Poincare line joining two ideal pointsi the line of enclosure lts proof shows that OPT in the previous gure is indeed a right angle as we needed Proposition 154 Let T and U be points on 39y that are not contained on a the same diameter and let P be the pole of TU Then i PT E PU ii ZPTU E ZPUT gt gt iii OP L TU and iu the circle 6 with center P and radius PT cuts 39y orthogonally at T and U PROOF By de nition of pole ZOTP and ZOUP are right angles so by the Hypotenuse Leg criterion AOTP E AOUP Therefore PT E PU and OPT E ZOPU The base angles ZPTU and PUT of the isosceles triangle ATPU are thus congruent and the angle bisector ITO is perpendicular to the base TU The circle 6 is tmi well ned because PU E PT and 6 intersects 39y orthogonally by the hypothesis that PT and PU are tangent to 39y I 94 CHAPTER 15 INVERSION IN EUCLIDEAN CIRCLES lk V Let P be a point in the plane and 39y a circle with center 0 The power of P with respect to 39y is de ned to be PwP l0P2 7 r2 Proposition 155 Assume P 39y and assume that two lines through P intersect 39y in points R1R2 and 5152 Then 239 PR1 PR2 P51 P52 PwP ii If one of these lines through P is tangent to 39y at T then PwP PTZ Proposition 156 Let P 39y and P 7 O the center of 39y Let 6 be a circle through P 6 is orthogonal to 39y if and only if6 passes through P the inuerse ofP with respect to 39y Proposition 157 Let 39y haue radius r 6 have radiust and let P be the center of 6 6 is orthogonal to 39y if and only if PwP t2 where the power is computed with respect to 39y Let 0 be a point and k gt O The dilation with center 0 and ratio k is a mapping of the Euclidean plane that xes O and maps every point P 7 O to a unique P E W such that OP kOP Call this map Do Proposition 158 Let 6 be a circle with center C 7 O and radius 3 D09 maps 6 to a circle 6 with center 0 D0aC and radius as If Q E 6 the tangent to 6 at Q is parallel to the tangent to 6 at PROOF Choose rectangular coordinates so that O is the origin Then the dilation is given by the mapping z y 7 om op The image ofthe line have equation azby c is them haVing equatio mby ac so the image is parallel to the original line In particular CQ is parallel to CQ and their perpendiculars at Q and 62 respectively are also parallel lf 6 has equation m 7 h2 y 7 k2 32 then 6 has equation m 7 04h2 y 7 ak2 as2 I Proposition 159 Let 39y be the circle of radius r centered at O and let 6 be the circle of radius 3 centered at C Assume 0 lies outside 6 let p PwO with respect to 6 and let oz r p The image 6 of6 under inuersion in 39y is the circle of radius as whose center is 0 D0aC UP 6 6 and P is the inuerse ofP with respect to 39y then the tangent t to 6 at P is the reflection across the perpendicular bisector of PP of the tangent to 6 at P Corollary 1 6 is orthogonal to 39y if and only if6 is mapped to itself by inuersion in 39y Proposition 1510 Let t be a line so that 0 6 The image oft under inuersion by 39y is a punctured circle with missing point 0 The diameter through 0 of the completed circle 6 is perpendicular to 5 Proposition 1511 Let 6 be a circle passing through the center 0 of 39y The image of6 minus 0 under inuersion in 39y is a line 6 so that 0 6 and l is parallel to the tangent to 6 at 0 Proposition 1512 A directed angle of intersection of two circles is preserved in magnitude by an inuersion The same applies to the angle of intersection of a circle and a line or the intersection of two lines 96 CHAPTER 15 INVERSION IN EUCLIDEAN CIRCLES Proposition 1513 Let 6 be orthogonal to P Inversion in 6 maps P onto P and the interior ofP onto itself Inversion in 6 preserves inr irtenr e L t and Lung wenee in the sense of the Poincare disk model Chapter 16 Models of Hyperbolic Geometry 161 Consistency of Hyperbolic Geometry ls this system of hyperbolic geometry a consistent mathematical system Does it lead to a contradiction This is a question of metamathematics ie a question outside of a mathematical system about the system itself The question is not about lines and points but about the whole system itself If it is inconsistent then an ordinary mathematical argument would derive a contradiction This is what Saccheri and Lambert attempted to do and failed Theorem 161 If Euclidean geometry is consistent so is hyperbolic geometry Corollary 1 If Euclidean geometry is consistent then no proof or disproof of the parallel postulate from the rest of Hilbert s postulates will euer be found ie the parallel postulate is independent of the other postulates PROOF Assume not then there is a proof of the parallel postulate This implies that hyperbolic geometry is inconsistent since the Hyperbolic Axiom contradicts a proven result Our above theorem asserts that hyperbolic geometry is as consistent as Euclidean geometry Thus Euclidean geometry is inconsistent This contradiction implies that no proof 0 the parallel postulate exists Note also that the hypothesis that euclidean geometry is consistent implies that do disproof exists either You should recognize that had mathematicians succeeded in proving Euclid s Fifth Pos tulate from the other axioms with the intention of making Euclidean geometry more secure and elegant they would have completely destroyed Euclidean geometry as a consistent body of thought What we will do now in order to see that hyperbolic geometry is as consistent as Euclidean geometry is to construct a model of hyperbolic geometry We will look at three different models for H2 162 The Beltrami Klein Model This model is often referred to as the Klein model because of the extensive work done by the German mathematician Felix Klein in geometry with this model Fix once and for all 97 98 CHAPTER 16 MODELS OF HYPERBOLIC GEOMETRY a circle P in E2 with center 0 and radius OR Let th X e E2 l OX lt OR Identify the points of H2 with the points in intP A chord of P is the Euclidean segment AB joining two points A B E P AB intF A B is an open chord These open cmds represent the lines of the hyperbolic plane To say that P lies on A B means P 6 AB the Euclidean line and A P B It is easy to see that this model satis es the Hyperbolic Axiom Figure 161 Parallel lines in the Klein model Here k O P A B The lines a anMPjBgt are o lines that do not intersect k in H2 as is the line 6 The difference is that PA and PB are limiting parallel to k in different directions while 6 and k are hyperparallel admitting a common perpendicular Recall that the points of P do not belong to the hyperbolic plane They are called ideal points of H2 The points outside of P are called ultraideal points By saying that the lines k and Z admit a common perpendicular raises the question about how one de nes congruence of segments and angles It is not obVious for it would seem that lines must be of nite extent no line measuring more than twice the diameter of P If this were the case we would not have a model for hyperbolic geometry as the congruence axioms Archimedes7 axiom and Dedekind s axiom de nitely would not hold We must de ne a different method for measuring the length of segments and the measure of angles in this model I will at this point only mention the method for measuring the length of segments and the de nition of right angles The remaining measurement will follow from work that we do later and will be noted then 163 THE POINCARE HALF PLANE MODEL 99 Let A B E H2 and let P Q E P denote the endpoints of the chord through A and B Let E denote the Euclidean distance from A to B or the length of the segment AB De ne the Klein distance MAB llogmapol 1 372 og 2 P Q Figure 162 Length in the Klein model Note then that as B approaches Q m approaches 0 thus 1 d A B BLIPQ k 7 00 This allows us to see then that we can nd a segment of any length on any ray Axiom C l Also it will follow that Dedekind s and Archimedes7 axioms are valid Our technique for measuring angles will be introduced at a later time It depends on the model for hyperbolic geometry due to Poincare It is not the way in which angles are measured in Euclidean geometry For this reason the Klein model is not a conformal model of geometry We can talk about right angles without too much dif culty Let Z and m be two lines in the Klein model of the hyperbolic plane or K lz39nes l l m in the Klein sense if a l or m is a diameter and l and m are perpendicular in the Euclidean sense or b if neither 1 or m is a diameter do the following Let t1 and 252 be the tangents to P at the ends of 1 Since 1 is not a diameter 251 O 252 7 0 Put Pl 251 O 252 Pl is called the pole of l m is perpendicular to l in the Klein sense if and only if the Euclidean line extending m passes through Pl All of this will be veri ed shortly 163 The Poincar HalfPlane Model In this model take H2 to be the set of all points in the coordinate Euclidean plane that satisfy H2 E E2 l y gt 0 100 CHAPTER 16 MODELS OF HYPERBOLIC GEOMETRY We interpret lines in this model to be one of the following 1 the intersection of a Euclidean vertical line with H2 or 2 the intersection of a Euclidean circle whose center lies on the z axis with H2 Once again our concept of congruence must change As before we must de ne a new measure for segments However the manner of measuring angles is exactly the method that is used in Euclidean geometry To measure the angle formed by two of these lines we simply measure the Euclidean angle formed by the tangent lines to the two Poincare lines The method of measuring segments is more easily described in the next model which is derived from this model 164 The Poincar Disk Model Let 39y denote the circle of radius 1 centered at the point 01 in the real plane Let N denote the point 0 2 Then N and each point on the circle 39y determine a line which has non zero slope It must then intersect the m axis in some point On the other hand given any point on the z axis it and the point N determine a line which must intersect the circle 39y in some point Thus there is a one to one correspondence between the points of the real line here the m axis and the points of the circle 39y less N The mappings are 747 272 fiRH YVV giVeHby fTmvm i 2 g yNHR g1venby gzy2 34 Not only can you see that the mappings are continuous but are differentiable as well What this means to us is that we may add a point at in nity to the real line and we get a circle in some sense Think of taking the Poincare half plane model and adding a point 164 THE POINCARE DISK MODEL 101 at in nity to the real axis We will then have the interior of a circle sitting in the real plane The vertical lines will correspond to diameters of the circle and the circles in the half plane model will correspond to circles which are orthogonal to the boundary circle To make the setting more precise let H2 int P A p line in this model is either 1 an open diameter of P or to an open arc of a circle orthogonal to P A circle 6 is orthogonal to P if at each point of intersection of 6 and P the radii of P and 6 through that point are perpendicular 6 th m an open arc which represents a p line in H2 A point in H2 lies on a p line if and only if it lies on it in the Euclidean sense Between ness has the same interpretation though if A B and C are on an open arc from 6 with center P then A B C if and only if 17B is between 17A and P If A B E H2 and if P and Q are the ends of the p line through A and B then we de ne the Poincare distance from A to B to be MA 3 How13 PQM log AQBP 372 As with the above model this Poincare model is conformal If two directed circular arcs meet at A then the measure of the angle they form is the measure of the angle between their tangent rays at A Question IF A B E H2 and A and I do not lie on a diameter of P how do you construct the p line through A and B We will verify these steps later a Construct the perpendicular to W at A intersecting P in P and Q b Construct the tangents to P at P and Q These tangents intersect at a point A c Construct the circle 6 containing A A and B 6 is orthogonal to P and m 6 intF is the p line through A and B Let us compare the two disk models by looking at certain familiar objects in each of them 102 CHAPTER 16 MODELS OF HYPERBOLIC GEOMETRY Klein Model Poincare Model Lines in the different models 1644 THE POINCARE DISK MODEL Figure 163 Measuring angles in the Poincare Half Plane Model 103 104 CHAPTER 16 MODELS OF HYPERBOLIC GEOMETRY Figure 164 A p line in the Poincar disk model 1644 THE POINCARE DISK MODEL Figure 165 Angle measure in the Poincare Disk Model 105 106 CHAPTER 16 MODELS OF HYPERBOLIC GEOMETRY 6 Figure 166 Construction of the p line through A and B B 1644 THE POINCARE DISK MODEL Figure 167 Limiting Parallel Rays Figure 168 Hyperparallel Lines 107 108 CHAPTER 16 MODELS OF HYPERBOLIC GEOMETRY Figure 169 Horoparallel Lines 1644 THE POINCARE DISK MODEL Figure 1610 Lambert Quadrilateral Klein Model Model Quadrilaterals in the different models 109 Poincare 110 CHAPTER 16 MODELS OF HYPERBOLIC GEOMETRY x 1 v 1 1 1 1 r 1 l I I Figure 1611 Saccheri Quadrilateral 165 ISOMORPHISM OF MODELS 111 165 Isomorphism of Models You will doubtless be surprised to learn that these two models are isomo rphtc 239e there is a one to one correspondence between the points and lines of the two models such that the relations of incidence betweenness and congruence are preserved Let 52 be a sphere of the same radius as the radius of P tangent to intF at the origin Use orthogonal projection of the Klein model in P upward onto the lower hemisphere of the sphere and then stereographic projection from the north pole back to the plane The image will be the Poincare model Let my 6 intP Orthogonal projection maps my onto the bottom hemisphere of by z2y2z7a2a any H x7y7a7xa27z27y2i Stereographic projection of 52 onto the coordinate plane is the analogous map to that described for mapping the punctured circle onto the line A point z y z E 2 is sent to the point on the zy plane at which the line through 0 0 2a and z yz intersects the plane We can parameterize the line by at lt0 0 2agt tltz 212 7 2agt tm ty 2a 1 tz 7 2a t passes through the zy plane when 2a 1 tz 7 2a 0 t 7 2a 7 2a 7 z 2am Qay Thus stereographic projection sends zy z to lt20 7 2 2a 7 Z We need a shrinking map to shrink the image back into P The equator of the sphere z ya maps to 2m 2y m 3 Thus z 2 32 472 Shrinking back into P it is only necessary to diVide by 2 Combining the three maps we have a mapping from the Klein model to the Poincare model given by am ay 1 l gt y H 1273527342011 12352342 It is easy to check that this is a continuous and differentiable map from intF to itself It sends k lines to p lines and is an isomorphism of geometries 166 Constructions in the Poincare Model To construct the Poincare line joining two ideal points or the so called line of enclosure we have the following result Lemma 161 Let T U E P not on a diameter and letP PTU the pole ofTU Then 239 PT g PU 112 CHAPTER 16 MODELS OF H YPERB OLIC GEOMETRY 2392 ZPTU g PUT gt gt m OP l TU 23911 the circle 6 with center P and radius PU PT is orthogonal to 39y at T and U The previous description for the construction of the unique line between two ordinary points follows from the fth item above We have already discussed the measure of segments as well How do you copy a given angle at a given point A on a given ray E 1 If A is the center of P then the ray E is a diameter of P and the angle is copied as in Euclidean geometry using another diameter of P to If A 7 O the center of P then we must nd the unique circle 6 through A orthogonal to P and tangent to the given Euclidean line 6 which forms the Euclidean angle with the tangent to E at A a For 6 orthogonal to P by the above facts 6 must pass through A the inverse of A with respect to P The center C of 6 lies on the perpendicular bisector of AA Let m be this bisector gt b For 6 to be tangent to 6 at A we must have CA perpendicular to 6 Thus C lies on the unique perpendicular to 6 at A say 71 C m O n 166 CONSTRUCTIONS IN THE POINCARE MODEL 113 emil e117 where r is the radius ofP and e is the number so that loge 1 PROOF If P and Q are the ends of the diameter through OB then x logOBPQ em OP BQ 062 BP 7 g 7 r OB 7 BP 7 r 7 OB 12 OB e 1 ex 7 1 which is what was to be proven I This will give us enough information to prove that all of the axioms of neutral geometry are satis ed7 except for the SAS axiom It requires a bit of work Assume that we are given 114 CHAPTER 16 MODELS OF H YPERB OLIC GEOMETRY two p triangles AABC and AXYZ inside P so that A E X AC E XZ and AB E XY We need to show that these triangles are congruent in the Poincare model First we show that we may assume that A X 0 where O is the center of P Let 6 be the circle through A and B orthogonal to P and let a be the circle through A and C orthogonal to P Then since both circles are orthogonal to P it follows that 6O 0 A A Let 52 AAA O and let 6 be the circle of radius 3 centered at A Then AA A O 7 A0 so 32 A O2 7 A OAO AO2 7 r2 by the de nition of inverse Thus r2 52 A O2 so that r is perpendicular to s by the converse of the Pythagorean theorem This means then that P and e are orthogonal circles By the con struction of this circle 6 O is the inverse of A in 6 By the last item above inversion in 6 maps AABC onto a p congruent triangle AOB C Likewise AXYZ can be mapped by inversion onto a p congruent triangle AOY Z By Lemma 162 and the SAS axiom for Euclidean geometry we have that dO B dOY dOC dO Z and B00 E ZYOZ Thus the Euclidean triangles AOBC and AOYZ are congruent by SAS Therefore there is a rotation about the center of P and possibly a re ection through a diameter which sends AOBC to AOYZ This will map F onto itself and the orthogonal circle through B and C to the orthogonal circle through Y and Z This preserves the p length and hence the triangles are p congruent Theorem 162 Two triangles in the Poincare model of hyperbolic geometry are p congruent if and only if they can be mapped onto each other by a succession of inversions in circles orthogonal to P andor reflections in diameters of P At this point I would rather think of measuring angles in radian measure not degree measure In this case we know that the sum of the angles of any triangle in hyperbolic geometry is less than 7139 Also let Hd denote the radian measure of the angle of parallelism corresponding to the hyperbolic distance d We can de ne the standard trigonometric functions not as beforeiusing right trianglesibut in a standard way De ne 00 n m2n1 71 7 161 sinz 2n1 00 m2n TL cosz 204 W 162 n tanm 163 m In this way we have avoided the problem of the lack of similarity in triangles the premise upon which all of real Euclidean trigonometry is based What we have done is to de ne these functions analytically in terms of a power series expansion These functions are de ned for all real numbers z and satisfy the usual properties of the trigonometric functions Theorem 163 In the Poincare model of hyperbolic geometry the angle of parallelism sat e7d tan is es the equation 166 CONSTRUCTIONS IN THE POINCARE MODEL 115 EX Figure 1612 SAS in the Poincare Model PROOF By the de nition of the angle of parallelism7 d dPP7 Q for some point P to its foot Q in some p line Z Now7 Hd is half of the radian measure of the fan angle at P7 or is the radian measure of ZQPQ7 where 179 is the limiting parallel ray to 6 through P We may choose 6 to be a diameter of P and Q 07 the center of P so that P lies on a diameter of P perpendicular to Z The limiting parallel ray through P is the arc of a circle 6 so that a 6 is orthogonal to P b 6 is tangent to 6 at Q The tangent line to 6 at P must meet 6 at a point B inside P This point B is the pole of P9 from 6 By Lemma 161 ZQPQ ZQQP B radians Let us denote Hd 04 Then in APQQ7 04 1 2B g or 7139 5 1 mils 116 CHAPTER 16 MODELS OF H YPERB OLIC GEOMETRY Figure 1613 Angle of Parallelism Now dP Q rtan rtan 7 Applying Lemma 162 we have d rdPQ 1tan e 7 7 dPQ 17tan 1 7 t 2 Using the identity tan 7 22 it follows that 1 d 7 e 7 tan 042 Simplifying this it becomes 7d ltHdgt e tan 7 2 Also we can write this as Hd 2 arctane d I 167 Return to the Klein Model Now we have that the Poincare model is a model for hyperbolic geometry Since the Klein model is isomorphic to the Poincare model it too is a model of the hyperbolic plane but that does not explain why we do the things the way that we do 167 RETURN TO THE KLEIN MODEL 117 We shall use another isomorphism to study the Klein model Let 2 denote the unit shpere in Euclidean threespace7 2 412 lm2y222 1 and let 0 denote the equatorial circle a 410 1 352 y 1 We shall consider the interior of a for both the Klein and Poincare models Let N 07 07 1 and map the interior of 0 onto the southern hemisphere by projection from N Figure 1614 Central Projection from N 2m 2y m2y271 957970 lt1z 2y 1z2y2 72y2 71 Now7 apply the projection up to the my plane7 749571172 9072470 The composition of these two maps is the map from the interior of a to itself given by 2m 2y 0gt F7y70lt1m 2y271m 2y27 118 CHAPTER 16 MODELS OF H YPERB OLIC GEOMETRY A common way of studying the circle is to recall that the Euclidean plane can represent the complex line by the point ab represents the complex number a bi The circle is then the set a z E C l 2 1 De ning the modulus of 2 by 25 we can rewrite the above function F by 22 7 1 121239 Clearly F maps a diameter of 0 onto itself but xes only the origin and the endpoints of the diameter In fact as F maps the interior of 0 onto itself it xes only the origin and the points on 0 Consider a p line in a say 67a circle orthogonal to a 6 0 P Q What is F67 Claim If A E 6 inta and PQ is the chord of 6 in int 0 then FA PQ PROOF Let C a b be the center of6 Then we found P and Q by taking the circle with CO as diameter and intersecting it with a It s center is g g and its radius is a2 1 2 Applying standard analytic geometry we nd that its equation is 2 7 am y2 7 by 0 lntersecting this with a m2 y2 1 gives am 1 by 1 for the line joining P and Q Now CQ is perpendicular to OQ since a and 6 are orthogonal circles Thus 002 OQ2 C39Q2 or r2 C39Q2 a2 b2 7 1 where r is the radius of 6 Then 6 has equation m2y22am2by71 Let A 043 6 6 imfa Put FA 1412 Solving we nd u mum and v Thus au by 1 which implies that 1412 6 PQ We can now justify the de nitions we made earlier in the Klein model 0 AB gk CD if and only if F1AB g1 F 1CD 0 ABC gk DEF if and only if F 1ZABC g1 F 1ZDEF PERPENDICULARITY Now we see that Z 1 m if and only if F 1 1 F 1m in the Poincare model 1 lf 6 and m are both diameters then F 1 and F 1m are both diameters In fact F 1 Z and F 1m m Hence the usual Euclidean meaning applies to Assume that only 6 is a diameter Then F 1 Z F 1m is a circular arc perpendicular to 6 Thus the tangent to F 1m is perpendicular to Z and 6 passes through the center of F 1m m is the chord of F 1m Since 6 passes through this center 6 is the perpendicular bisector of m in the Euclidean sense Therefore 6 is perpendicular to m in our previous de nition Conversely if 6 1k m then 6 is the perpendicular bisector of m Thus 6 passes through the center of the circle and Z is perpendicular to the circular arc F 1m OJ Assume that neither 6 nor m is a diameter F 1 and F 1m are arcs of circles and u orthogonal to a 167 RETURN TO THE KLEIN MODEL a A T V 119 Suppose A l u in the Poincare model By one of the facts7 the center of A is P and that of u is Pm We need to show that m passes through P 7 then we will have that m lk 6 Let m 0 P7 Q Since A and u are orthogonal inversion in A interchanges P and Q since A and u are also orthogonal Thus7 P Q in A and the Euclidean line through P and Qinamely mipasses through the center of A7 which is P Suppose m passes through P Then P Q in A so that A and u are orthog onal Chapter 17 Area in Hyperbolic Geometry 171 Preliminaries A polygonal region is a plane gure which can be expressed as the union of a nite number of triangular regions in such a way that if two of the triangular regions intersect their intersection is an edge or a vertex of each of them Let R be a polygonal region A triangulation of R is a nite collection K T1T2Tn of triangular regions Ti such that 1 the Ti s intersect only at edges and vertices and 2 their union is B Let R1 and R2 be polygonal regions Suppose that they have triangulations K1 T1T2 Tn K2 T12 T25 T such that for each i we have T Ti Then we say that R1 and R2 are equivalent by nite decomposition and we write R1 E R2 Let 6T defectAT for any triangular region T Theorem 171 If K1 and K2 are triangulations of the same polygonal region R then 6K1 6K2 Theorem 172 If two Saccheri quadrilaterals have the congruent summits and equal de fects then their summit angles are congruent in which case the two Saccheri quadrilaterals are congruent Given AABC with BC considered as the base Let D and E be the midpoints of AB and ACltlet F G and H be the feet of the perpendiculars from B A and C respectively to l DE As you have proven in the homework DHCBF is a Saccheri quadrilateral It is known as the quadrilateral associated with AABC It depends on the choice of the base but it should be clear which base we mean 120
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