Class Note for MATH 409 with Professor Martin at KU
Class Note for MATH 409 with Professor Martin at KU
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CHAPTER 1 Other Geometries A Computational Introduction In order to provide a better perspective on Euclidean geometry three alternative geometries are described These are the geometry of the surface of the sphere hyperbolic geometry and taxicab geometry 1 Spherical Geometry Due to its relationship with geography and astronomy spherical geometry was studied extensively by the Greeks as early as 300 BC Menelaus circa 100 wrote the book Spherica on spherical trigonometry which was greatly extended by Ptolemy 100 178 in his Almagest Many later mathematicians including Leonhard Euler 1707 1783 and Carl Friedrich Gauss 1777 1855 made substantial contributions to this topic Here it is proposed only to compare and contrast this geometry with that of the plane Because the time to develop spherical geometry in the same manner as will be done with Euclidean geometry is not available this discussion is necessarily informal and frequent appeals will be made to the readers visual intuition 11 Strictly speaking there are no straight lines on the surface of a sphere Instead it is both customary and useful to focus on curves that share the quotshortest distancequot property with the Euclidean straight lines The following thought experiment will prove instructive for this purpose Imagine that two pins have been stuck in a smooth sphere in points that are not diametrically opposite and that a frictionless rubber band is held by the pins in a stretched state Rotate this sphere until one of the two pins is directly above the other right in front of your mind s eye It is then hard to avoid the conclusion that the rubber band will be stretched out along the sphere in the plane formed by the two pins and the eye the plane of the book39s page in Figure 11 The inherent symmetry of the sphere dictates that this plane should cut the sphere into two identical hemispheres in other words that this plane should pass through the center of the sphere It is also clear Rubber Observer Figure 11 A geodesic on the sphere that the tension of the stretched rubber band forces it to describe the shortest curve on the surface of the sphere that connects the two pins The following may therefore be concluded PROPOSITION 111 Spherical geodesics If A and B are two points on a sphere that are not diametrically opposite then the shortest curve joining A and B on 12 11 SPHERICAL GEOMETRY the sphere is an arc of the circle that constitutes the intersection of the sphere with a plane that contains the sphere s center Such circles are called great circles and these arcs are called great arcs or geodesic segments They are the spherical analogs of the Euclidean line segments Diametrically opposite points on the sphere present a dilemma A stretched rubber band joining them will again lie along a great circle but this circle is no longer uniquely determined since these points can clearly be joined by an infinite number of great semicircles For example assuming for the sake of argument that the earth is an exact sphere each meridian is a great semicircle that joins the north and south poles Hence the aforementioned analogy between the geodesic segments on the sphere and Euclidean line segments is not perfect It is necessary to either exclude such meridians from the class of geodesic segments or else to accept that some points can be joined by many such segments The first alternative is the one chosen in this text Thus by definition the endpoints of geodesic segments on the sphere are never diametrically opposite Next the spherical analog of the angle is defined Any two great semicircles that join two diametrically opposite points A and B but are not contained in the same great circle divide the sphere into two portions each of which is called a lune or a spherical 13 11 SPHERICAL GEOMETRY Figure 12 The lune at angle Fig 12 The measure of the spherical angle is defined to be the measure of the angle between their tangent lines at A or at B Alternately this equals the measure of the angle formed by the radii from the center of the sphere to the midpoints of the bounding great semicircles For example each meridian forms a 900 angle with the equator at their point of intersection In the Euclidean plane the relationships between lengths of straight line segments and measures of angles are given by well known trigonometric identities Some fundamental theorems of spherical trigonometry are now stated without proof Any three points A B C on the sphere no two of which are diametrically opposite constitute the vertices of a spherical triangle denoted by A ABC The three sides of this triangle are the geodesic segments that join each pair The sides opposite the vertices A B C and their lengths are denoted a b c respectively The interior angle a at the vertex A is the lune between AB and AC The interior angles and y at B and C are defined in a similar manner PROPOSITION 112 Spherical trigonometry On a sphere of radius R I let A ABC be a spherical triangle with sides a b c and interior angles a y Then cosa cosbcosc 1 cosa sin 1 sin c i39 cos a cos 17 cos c cos a sin 1 sin c cosa cos cosy 11 COS a sin sin y ii cos a cos a sin sin y cos cos y sin a sin sin y 1 sin a sin b sin c 39 14 11 SPHERICAL GEOMETRY These are known as the first spherical law of cosines the second spherical law of cosines and the spherical law of sines It should be noted that i and i are really the same equation as are ii and ii although as will be demonstrated by the examples below their uses are different The solution of a triangle consists of the lengths of its sides and the measures of its interior angles EXAMPLE 113 Solve the spherical triangle with sides a 1 b 2 and c 712 It follows from the first spherical law of cosines that cos 1 cos 2 cos 712 cos 1 COS a sin 2 sin 712 2 sin 2 so that 1 cos1 0 a cos Sinz 5354 The angles and y are similarly shown to have measures 119640 and 72910 EXAMPLE 114 On a sphere of radius 4000 miles solve the triangle in which an interior angle of 500 lies between sides of lengths 7000 miles and 9000 miles respectively Since the radius is the unit it follows that we may set 7000 9000 4000 175 c 4000 225 Hence from the first law of cosines 15 11 SPHERICAL GEOMETRY a cos71cos a sin 1 sin c cos 17 cos c z 9221 w 3688 miles Now that all three sides of the triangle are known the method of the previous example yields 1cosb cosacosc 0 COS sinasinc z 713905 1cosc cosacosb 0 y cos Sinasinb w 13158 Note that in both of the above examples the sum of the angles of the spherical triangle exceeds 1800 That is in fact true for all spherical triangles PROPOSITION 115 The sum of the angles of every spherical triangle lies strictly between 1800 and 5400 A spherical triangle the sum of whose angles is close to 1800 is formed by the equator together with two close meridians Thus the sum of the angles of the spherical A ABC of Figure 13 is 900 900 a 1800 a A spherical triangle AB C with C I Figure 13 A thin spherical triangle 16 11 SPHERICAL GEOMETRY angle sum near 5400 is described in Figure 14 where A B C are points that are equally spaced along a great circle As A B C approach A B C respectively the angles they form are attened out and come arbitrarily close to 1800 each For example since the spherical distance between any two of the points A B C is 2713 it might be assumed that the spherical distance between any two of the points A39 B C 39 is a 2713 000001 in which case each of the angles of A A39B39C39 is Figure 14 A nearly maximal spherical triangle 1 cos a cos a cos a COS sin a sin a z 179520 and their sum is 538560 Since by definition each of the interior angles of the spherical triangle is less than 1800 it follows that the sum of these angles can never equal 5400 Similarly as will be shown momentarily the sum of these angle cannot equal the lower bound of 1800 either The area of the spherical triangle is also of interest An elegant proof of this formula is offered in Section 32 17 11 SPHERICAL GEOMETRY PROPOSITION 116 Ifa triangle on a sphere afradius R has angles with radian measures a y then it has area a f y 71R2 For example the spherical triangle formed by the equator the Greenwich meridian and the 900 East meridian has all of its angles equal to 712 and hence its area is 71 71 71 2 71R2 rawml T This answer is consistent with the fact that the said triangle constitutes one fourth of a 2 hem1sphere Slnce the surface area of the sphere 1s 471R th1s tr1angle has area 471R2 2 2 l 4 2 which agrees with the previous calculation The quantity a f y 71 is called the excess of the spherical A ABC The above theorem in effect states that the area of a spherical triangle is proportional to its excess This assertion is supported by the triangle below which has excess 712 712 a 71 a and whose area is clearly proportional to a as long as A and B vary along the equator and C remains at the north pole see Fig 15 C 4 18 11 SPHERICAL GEOMETRY Figure 15 A spherical triangle This area formula can be used to close a gap in the above discussion Since every triangle has positive area it follows that the sum of the angles of a spherical triangle never equals 71 or 1800 although as was seen above it can come arbitrarily close to this lower bound EXERCISES 11 1 Let ABC be a spherical triangle with a right angle at C Use the formulas of spherical trigonometry to prove that a sina sin or sin c b tan a tan or sin b c tana cos3 tanc d cosc cosacosb e cos or sin cos a f sin b sin sin c g tanb tan sina h tanb cosatanc i cos c cot acot 3 j cos 3 sin or cos b 2 On a sphere of radius R solve the spherical triangle with angles 0 0 0 0 0 0 a 60 70 80 b 70 70 70 0 0 0 O O c 120 130 140 d 6 6 6 where 60 lt 6 lt 120 3 On a sphere of radius R solve the spherical triangle with sides a R R R b R 15R 2R c nR2 HIRZ J39ERZ d d d d where 0 ltdlt 2nR e 2R 3R 4R f 02R 03R 04R g 2R 3R 4R 4 On a sphere of radius R solve the spherical triangle with a a5R 3600 y800 b bRa400y1000 c a2R y100 d a2R yl700 5 On a sphere of radius R solve the spherical triangle with a a2RbR y1000 b b5Rc12Ra100 c a2RbR y1200 d b5Rc12Ra120 6 On a sphere of radius 75 cm solve the spherical triangle with o o o a a100cmb125cmc140cm b 1100 125 y140 c a1000b 125cmc 125cm d a100cm 1250y 125 7 Evaluate the limits of the angles of the spherical triangles below both as x gt 0 and as x gt J17 a 121220 b abxc2x c a x b c 2x 8 Which of the following congruence theorems hold for spherical triangles Justify your answer 19 11 SPHERICAL GEOMETRY aSSS bSAS 0 ASA dSAA a AAA 9 Prove that the three angles a 5 y are the three interior angles of a spherical triangle if and only if they satisfy all of the following conditions a ygtm angt3y ngtay yngta 2 Hyperbolic Geometry Imagine a two dimensional universe with a superimposed Cartesian coordinate system in which the x axis is infinitely cold Imagine further that as the objects of this universe approach the x axis the drop in temperature causes them to contract see Fig 16 Thus 1 Figure 16 The shrinkage that defines the hyperbolic plane the inhabitants of this fictitious land will find that it takes them less time to walk along a horizontal line from A0 1 to B1 1 Fig 17 than it takes to walk along a horizontal line from CO 5 to D1 5 Since their rulers contract just as much as they do this observation will not seem at all paradoxical to them If it is assumed that the contraction 110 12 HYPERBOLIC GEOMETRY is such that the outside observer sees the length of any object as being proportional to its distance from the x axis then the inhabitants will find that walking from CO 05 to D1 5 takes twice as long as walking from A0 1 to B1 1 and one fifth of the time y A0 1 31 1 CO 5 D1 5 E01 F11 I Figure 17 Paths of unequal hyperbolic lengths of walking from E0 1 to F1 1 To differentiate between the Euclidean length of such a segment and its length as experienced by these fictitious beings it is customary to refer to the latter as the hyperbolic length of the segment Accordingly the hyperbolic lengths of the segments AB CD and EF of Figure 17 are 1 2 and 10 respectively In general the hyperbolic length of a horizontal line segment at distance y from the x axis is given by the formula Euclidean length y 1 hyperbolic length Other curves also have a hyperbolic length and a method for computing this is given in Exercise 16 below Not surprisingly perhaps the Euclidean straight line segment joining two points does not constitute the curve of shortest hyperbolic length between them When setting out from A0 1 to B1 1 the inhabitants of this strange land may find that if they bear a little to the north their journey will be somewhat shorter because unbeknownst to them 111 12 HYPERBOLIC GEOMETRY A0 1 BO 1 Figure 18 Which path has shorter hyperbolic length their legs are longer on this route Fig 18 However if they stray too far north the length of the detour will offset any advantages gained by the elongation of their stride and they will find the length of the tour to be excessive They are therefore faced with a trade off problem Some deviation to the north will shorten the duration of the trip from A to B but too much will extend it Which path then is it that makes the trip as short as possible The answer to this question is surprisingly easy to describe though not to justify The path of shortest hyperbolic length that connects A0 1 to B1 1 is the arc of the Figure 19 A hyperbolic geodesic circle that is centered at 5 0 and contains A and B Fig 19 Its hyperbolic length turns out to be 0962 in contrast with the hyperbolic length 1 of the horizontal segment AB More drastically the arc of the semicircle centered at 50 0 and which 112 12 HYPERBOLIC GEOMETRY joins the points A0 1 and X100 1 has hyperbolic length 921 a mere 9 of the hyperbolic length of the segment AX Given any two points their hyperbolic distance is the minimum of the hyperbolic lengths of all the curves joining them As was the case for spherical geometry the geodesic segments of hyperbolic geometry are those curves that realize the hyperbolic distance between their endpoints The hyperbolic plane consists of the portion of the Cartesian coordinate system that lies above the x axis PROPOSITION 121 Hyperbolic geodesics The geodesic segments of the hyperbolic plane are arcs of circles centered on the xaxis and Euclidean line segments that are perpendicular to the xaxis P QI pP Q Q R Q Q 39 Q P i P I Figure 110 Six hyperbolic geodesics The geodesics of the first variety are called bowed geodesics whereas the vertical ones are the straight geodesics see Fig 110 This distinction is only meaningful to the outside observer The inhabitants of this geometry perceive no difference between these two kinds of geodesics It so happens that as the inhabitants of the hyperbolic plane approach the x axis they shrink at such a rate as to make the x axis unattainable Technically speaking the hyperbolic lengths of all of the geodesic segments in Figure 110 diverge to infinity as 113 12 HYPERBOLIC GEOMETRY the endpoints Q approach the x axis This claim will be given a quantitative justification at the end of this section A hyperbolic angle is the portion of the hyperbolic plane between two geodesic rays Fig 111 The measure of the angle between two geodesics is by definition the 1 Figure 111 Three hyperbolic angles measure of the angle between the tangents to the geodesics at the vertex of the angle Accordingly two geodesics are said to form a hyperbolic right angle if and only if their tangents are perpendicular to each other as Euclidean straight lines Fig 112 Given any three points that do not lie on one hyperbolic geodesic they constitute the vertices of a Figure 112 Hyperbolic right angles hyperbolic triangle formed by joining the vertices two at a time with hyperbolic geodesics Fig 113 114 12 HYPERBOLIC GEOMETRY Q A Figure 113 Three hyperbolic triangles The geometry of the hyperbolic plane has been studied extensively Its trigonometric laws are surprisingly not to say mysteriously similar to those of spherical geometry PROPOSITION 122 Hyperbolic trigonometry Let A ABC be a hyperbolic triangle with sides a b c and interior angles a y Then cosh b cosh c cosh a 1 COS a sinh b sinh c i39 cosh a cosh b cosh c cos a sinh b sinh c cosa cos cosy 11 cosh a Sin Sin Y ii cos a cosh a sin sin y cos it cos y sin a sin sin y 1 sinh a sinh b sinh 0 EXAMPLE 123 Solve the hyperbolic triangle with sides a I b 2 and c 7M2 It follows from Formula i of hyperbolic trigonometry that cosh 2 cosh 712 cosh 1 COS 0 sinh 2 sinh 712 9461 115 12 HYPERBOLIC GEOMETRY and so a z cos719461 z 18890 Similarly f 87670 and y z 3934quot EXAMPLE 124 Solve the hyperbolic triangle with two sides of lengths 2 3 respectively if they are to include an angle of 300 Set a 300 b 2 c 3 It follows from Formulai of hyperbolic trigonometry that a cosh71cosh b cosh c cos a sinh b sinh c 2545 Now that all three sides of the triangle are known the method of the previous example yields 1 cosh a cosh c cosh b 0 COS sinh a sinh c z 163964 1 cosh a cosh b cosh c 0 sinh a sinh b 5228 y cos As for the sum of the angles of a hyperbolic triangle the situation is diametrically opposite to that on the sphere PROPOSITION 125 The sum of the angles of every hyperbolic triangle is less than 180 2 x V x Figure 114 A hyperbolic triangle with three small angles 116 12 HYPERBOLIC GEOMETRY This proposition is borne out by the above two examples Figure 114 demonstrates that this sum can be quite small In fact in the hyperbolic triangle with sides a b c 10 each angle equals 1 cosh 10 cosh 10 cosh 10 0 a COS sinh 10 sinh 10 7 397739 EXAMPLE 126 Solve the hyperbolic triangle with a 2 y 600 By formula ii39 of hyperbolic trigonometry cos a cosh 2 sin 600sin 600 cos 600 cos 600 z 257 Since the cosine of an angle cannot exceed 1 such a hyperbolic triangle does not exist Note that a Euclidean triangle with the same specifications does exist Exercises 4 5 contain some related information The area of the hyperbolic triangle is of course of interest too Its formula is quite surprising PROPOSITION 127 The area of the hyperbolic triangle whose angles have radian measures a y is n a y This formula is given some support by Figure 115 Note that the sum of the angles of the larger hyperbolic A ABC is less than the sum of the angles of the smaller hyperbolic A AB39C39 The quantity 1 a y is called by analogy with its spherical counterpart the defect of the hyperbolic triangle Thus the above theorem asserts that the area of a hyperbolic triangle is equal to its defect 117 12 HYPERBOLIC GEOMETRY Figure 115 Two hyperbolic triangles with different areas and defects A somewhat peculiar aspect of the hyperbolic notion of length is its independence of the choice of unit of length Regardless of what scale is chosen for the Cartesian coordinate system that is used to define the hyperbolic version of length the hyperbolic distance between any two points remains the same Note that the three parts of Figure 116 correspond to three different scales and yet according to Formula 1 above in Figure 116 A curve with hyperbolic length 1 each of the three cases the hyperbolic length of AB is Hp MIN II can H H 118 12 HYPERBOLIC GEOMETRY It follows from this independence of scale that in hyperbolic geometry it is not necessary to specify units of length It might be instructive to show that this independence holds in the vertical direction as well The hyperbolic length of a vertical segment of the hyperbolic plane is easily computed with the aid of calculus For if dy denotes the Euclidean length of the P b ydy dy dh y y Q a I Figure 117 Hyperbolic length along the y axis infinitesimally small vertical line segment at height y above the x axis then its hyperbolic length is dyy see Fig 117 Consequently the total hyperbolic length of the segment PQ is lt SW J lnb lna ln 2 In particular if a 1 and b e 2718 then the hyperbolic length of the y axis between P0 1 and Q0 e is e 1n 2 Inc 2 1 119 12 HYPERBOLIC GEOMETRY Moreover if the scale on the axes is changed by a factor of s then P 0 s and Q 0 es and again their hyperbolic distance is Thus if P and Q are points with coordinates 0 1 and 0 8 relative to some unit of the Cartesian coordinate system then the line segment joining P and Q has hyperbolic length 1 regardless of the scale that is actually used Moreover this Euclidean line segment PQ which has hyperbolic length 1 is also a hyperbolic geodesic in contrast with the aforementioned segment AB which also has hyperbolic length 1 but is not a hyperbolic geodesic Consequently this geodesic can be taken as the natural unit or absolute unit of length of hyperbolic geometry Both spherical and hyperbolic geometry look very different from Euclidean geometry Nevertheless it is well known that on a large sphere a small portion of the surface may be practically indistinguishable from a piece of a plane This resemblance accounts for the fact that people first thought that the world was flat and small children still do so today The same confusion could occur in the hyperbolic plane If the portion of the hyperbolic plane that is subject to the direct experience and observation of its inhabitants is sufficiently small their geometry would appear to them as practically indistinguishable from that of the Euclidean plane This affinity between the hyperbolic and Euclidean planes is a topic that will be revisited many times in the subsequent discussion The explanation of how the trigonometry of a small portion of the hyperbolic plane may be confusable with Euclidean trigonometry can be found in the references At this point it will be demonstrated that just like the Euclidean plane and in contrast with the sphere hyperbolic geometry extends indefinitely in all directions In other words the inhabitants of the hyperbolic plane have no reason to suspect that part of their 120 12 HYPERBOLIC GEOMETRY t a Figure 118 The hyperbolic plane extends indefinitely in all directions universe is missing To see this note that by Eq n 2 the hyperbolic distance from the point a 1 to the point a t in Figure 118 is 1n lnt Hence if travel in the direction of the x axis is simulated by letting I approach 0 then this quantity diverges to 00 00 In other words for the hyperbolic people the x axis lies infinitely far away HYPERBOLIC DISTANCE The hyperbolic distance between any two points Ax1 y and Bx2 y can be determined by means of the following formulas y i If X 262 then the hyperbolic distance from A to B is Ilny I 2 ii If X 5 x2 and c o is the center of the geodesic segment that connects x1y1 and 262 y and r is its radius then the hyperbolic distance from A to x1 c ry2 x2 c ry1 B is Iln 121 12 HYPERBOLIC GEOMETRY EXAMPLE 128 The hyperbolic distance between the points 5 4 and 57 is ln 47 z 056 EXAMPLE 129 To find the hyperbolic distance between A8 4 and B0 8 8 4 1 note that the line segment AB has slope E and midpoint M4 6 see Fig 119 Hence the perpendicular bisector of AB has equation y 6 2x 4 and is easily seen to intersect the x axis in the point Cl 0 Thus 0 1 and r 1 2 2 81416 38 0 1 8 0 Hence the required d1stance1s Iln016 54 I 145 B0 8 Aamp 4 x C10 Figure 119 Computing the length of a hyperbolic geodesic EXERCISES 12 1 Let ABC be a hyperbolic triangle with a right angle at C a sinh a sin asinh c b tanha tan a sinh b c tanh a cos 3 tanh c d cosh c cosh a cosh b 6 cos a sin cosh a f sinh b sin sinh c g tanhb tan sinha h tanhb cosatanhc i cosh c cot a cot 3 j cos 3 sin a cosh b 2 Solve the hyperbolic triangle with angles 0 0 0 0 0 0 a 60 50 40 b 50 50 50 c 20 50 70 d 6 6 6 where 0 lt 6 lt 50 3 Solve the hyperbolic triangle with sides 122 10 11 12 13 14 15 16 17C 18C 12 HYPERBOLIC GEOMETRY a 111 b 234 C d d d d where 0ltd e 2 3 4 f 12 12 12 02 03 04 Solve the hyperbolic Triangle with a a5 3600 y400 b 175 a40 y500 c a2 y400 d a10 y400 e a1 y600 f a1 y600 g a2 y1000 For which values of a does there exist a hyperbolic triangle with 3 y 600 Solve the hyperbolic triangle with a a2b1 y300 b b5c12a1200 c a2b1 y450 d b5c12a1200 Evaluate the limits of the angles of the hyperbolic tiiangles below both as x gt 0 and as x gt 00 a abcx b abxc2x c axbc2x Does the SSS congruence theorern hold for hyperbolic triangles Justify your answer Does the SAS congruence theorern hold for hyperbolic triangles Justify your answer Does the ASA congruence theorern hold for hyperbolic tiiangles Justify your answer Does the SAA congruence theorern hold for hyperbolic tiiangles Justify your answer Explain why the AAA congruence theorern holds for hyperbolic triangles Find the hyperbolic distances between each pair of the three points A0 6 B10 4 C10 16 Explain why the hyperbolic length of every Euclidean line segment that is parallel to the xiaxis is independent of the unit of the underlying Cartesian coordinate system Explain why the hyperbolic length of every Euclidean line segment that is parallel to the yiaxis is independent of the unit of the underlying Cartesian coordinate system Explain why b 2 J V f dx f a is a reasonable formula for the hyperbolic length of a differentiable curve defined by y x a sx s b Write a script that takes two distinct points as input and yields the hyperbolic geodesic joining them as well as its hyperbolic length as output Write a script that takes three distinct points as input and yields a sketch of the hyperbolic triangle 71 4 l 2 they form as well as its solution as output Recall that cosh x lnx x 1 123 13 OTHER GEOMETRIES 3 Other Geometries All school children learn about the geometry of the plane As was seen in the previous two sections there are other geometries which broadly speaking can be classified into two categories One way to obtain a new geometry is to distort the plane A piece of paper can be rolled into either a cylinder or a cone A film of soap can assume many other shapes including that of a sphere The best known of this type of geometries is that of the sphere some of whose properties were presented in Section 1 Of course the distortion of surfaces may or may not result in distortions of lengths of curves Rolling a piece of paper into a cone has no such effect the straight lines on the paper are merely twisted into spirals 0r circles but their lengths are unaffected On the other hand when a flat soap film waves in the air the lengths of the curves on it are continuously altered It took mathematicians several centuries to realize sometime around 1850 that this notion of distortion of lengths and distances could be and should be studied independently of the shape distortion that induced it The geometries that are obtained by changing the way distance is measured in the plane are called Riemannian geometries and hyperbolic geometry is their best known and studied instance Riemannian geometry has found many applications in science the most spectacular of these being the theory of relativity Hyperbolic geometry is still the subject of much contemporary research and has had many surprising applications to other mathematical disciplines Every Riemannian geometry has geodesics which are defined as the shortest curves joining two points Such geodesics will form triangles and these triangles will have interior angles These angles in turn provide a means for quantifying the distortion or curvature of a geometry If ABC is a triangle of a geometry with interior angles of radian measures a y then the expression a f y n is called its total 124 13 OTHER GEOMETRIES curvature Accordingly every triangle of the Euclidean plane has total curvature 0 It is therefore reasonable to interpret this quantity as a measure of the extent to which that triangle differs from a Euclidean triangle By this definition the total curvature of a spherical triangle is always positive and so the sphere is said to be positively curved Hyperbolic geometry on the other hand is negatively curved It was Gauss who formally defined this notion and pointed out its central role in the study of geometry This chapter concludes with the discussion of yet another specific geometry which while also arising from an esoteric way of measuring distance is not for reasons that cannot be explained here a Riemannian geometry Taxicab geometry was first defined in 1973 but unlike spherical and hyperbolic geometry has not been integrated into the mathematical mainstream Nevertheless is has proven useful as a pedagogical tool that sheds a light on Euclidean geometry and also provides students with an elementarily defined mathematical territory they can explore on their own Like hyperbolic geometry taxicab geometry takes the Euclidean plane as its starting point and redefines distance The taxicab distance between the points P x1 y I and Q x2 y2 is dtP Q x1 x2y1 y2 Thus the taxicab distance between P 0 0 and Q 11 is dtPQ 0101 2 whereas the Euclidean distance dEP Q between them is 0 120 12 IE 125 13 OTHER GEOMETRIES Similarly the taxicab distance between 2 3 and 3 5 is 1 8 9 and their Euclidean distance is m l6 5 This geometry receives its name from the fact that it models the way a taxicab driver would think of distances in a city all of whose blocks are perfect squares It is clear that taxicab distances agree with Euclidean distances along both horizontal and vertical straight lines If p is any other straight line with inclination 6 from the positive x axis then the taxicab distances along p are different from but still proportional to the Euclidean distances As indicated by Figure 120 Figure 120 dtP Q x2 261 lyZ y1 cos 6 dEP Q sin 6 dEP Q cos 6 sin 6 dEP Q It follows that along any fixed straight line taxicab distances behave very much like Euclidean distances In particular the geometrical notion of betweenness can still be expressed numerically PROPOSITION 131 If the distinctpoints P Q R are collinear then Q is between P and R ifand only if 126 13 OTHER GEOMETRIES dtPQ dtQR dtPR There are many other similarities between the taxicab and Euclidean geometries and these are relegated to the exercises in the subsequent chapters Some of these differences are qualitative For example note that in Figure 120 dtPQ dtPR dtR Q In other words the line segment PQ is not the only shortest path joining P and Q In fact if y is any polygonal path joining P and Q all of whose segments have non negative slope then the taxicab length of y still equals dtP Q Exercise 6 Nevertheless no path joining P and Q has taxicab length shorter than dtP Q Exercise 9 and so it is not unreasonable to agree to regard the Euclidean straight lines as the straight lines of taxicab geometry Not surprisingly the taxicab measure of angles agrees with their Euclidean measure There is however no consensus yet on how areas should be measured in this outlandish geometry Only one more striking difference between the taxicab and Euclidean geometries will be mentioned here The SAS congruence theorem does not hold for taxicab geometry For the two triangles in Figure 121 we have dtA B dtD E 2 dtA C dtDF and ABAC AEDF 900 and yet dtB C 2 d 4 dtE F 127 13 OTHER GEOMETRIES c E o 1 3 5 A Figure 121 Two almost congruent triangles Maxi geometry yet another variant of the Cartesian plane is defined in Exercise 4 below Some of its properties are the subject of Exercise 4 EXERCISES 13 1 Compute the total curvature of the triangles in Exercise 112 2 Compute the total curvature of the triangles in Exercise 122 3 A metric is a function fP Q of pairs of points such that for any points P Q R a fP Q 2 0 and equality holds if and only if P and Q are identical points b f0 Q QvP C fP Q QvR 2 g R Show that the taxicab distance is a metric 4 The maxi geometry is defined on the Cartesian plane by redefining the distance between its points The maxi dismnce between P x1y1 and Q x2 y2 is dmP Q Maximum of le 7 x2 y1 7 yzl In other words the maxi distance is the larger of the horizontal and vertical Euclidean distances between P and Q The straight lines of maxi geometry are by definition the Euclidean straight lines and the taxicab measures of angles are also taken to be identical with their Euclidean measures a Show that the maxi distance is a metric b Show that if P Q R are collinear then Q is between P and R if and only if dmP Q dmQ R dmP R 5 Determine the taxicab perimeter and curvature of the triangles with the following vertices a 0 0 0 3 3 4 b 71 72 2 3 0 6 6 Suppose y is a polygonal path joining P and Q such that all of its segments have nonnegative slopes Show that the taxicab length of y equals dtP Q Is this also true if all of the segments have nonpositive slopes 128 10 11 13 OTHER GEOMETRIES b Explain why f I If dx is a reasonable formula for the taxicab length of a differentiable a curve defined by y z x a sx sb Let y 2 x a sx s b be a monotone either increasing or decreasing differentiable function Show that if P afa and Q bfb then the taxicab length of the curve defined by f is dP Q Explain why no path joining P and Q has taxicab length shorter that dP Q Determine the maxi perimeter and total curvature of the triangles with the following vertices a 0 0 03 3 4 b 71 72 2 3 0 6 Find a maxi geometry analog for d Exercise 9 a Exercise 6 b Exercise 7 c Exercise 8 CHAPTER REVIEW EXERCISES QMJgtWN Compute the perimeter of the triangle formed by joining the midpoints of an equilateral triangle all of whose sides have length a l in a Euclidean geometry b spherical geometry c hyperbolic geometry Repeat Exercise 1 for a 5 Repeat Exercise 1 for a 10 Repeat Exercise 1 for a 01 Compute the areas of all the triangles in a Exercise 1 b Exercise 2 c Exercise 4 Compute the total curvature of all the triangles in a Exercise 1 b Exercise 2 c Exercise 4 Are the following statements true or false Justify your answers a There is a spherical triangle with angles 12 13 16 b There is a hyperbolic triangle with angles 12 13 16 c There is a taxicab triangle with angles 12 13 16 d If two spherical triangles have angles 12 12 12 then they are congruent e If two hyperbolic triangles have angles 14 14 14 then they are congruent f If two taxicab triangles have angles 13 13 13 then they are congruent g Euclidean spherical hyperbolic taxicab maxi are all the geometries there are h Given any two points of spherical geometry there is a unique geodesic that joins them i Given any two points of hyperbolic geometry there is a unique geodesic that joins them j Given any two points of taxicab geometry there is a unique geodesic that joins them k On a sphere of radius 1 there is a triangle of area 4 1 In hyperbolic geometry there is a triangle of area 4 129 CHAPTER REVIEW In Every proposition that is valid in spherical geometry is false in hyperbolic geometry n Every proposition that is valid in hyperbolic geometry is false in spherical geometry 130
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