Class Note for MATH 409 with Professor Martin at KU (3)
Class Note for MATH 409 with Professor Martin at KU (3)
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CHAPTER 7 Inversions Transformations that are not rigid can be interesting too even though they are not as natural as the rigid motions of the previous chapter The inversions of this chapter are particularly appealing because they play important roles in both Euclidean and non Euclidean geometry 1 Inversions as Transformations Given a point C and a positive real number k the inversion C k is a transformation of the plane that maps any point P 5 C of the plane into the point P IC kP such that a C P are collinear with C outside the segment PP and b CPCP k2 Figure 71 illustrates the action of a typical inversion 71 71 INVERSIONS AS TRANSFORMATIONS It is clear that in general IC kP P if and only if IC kP P and hence 2 IC k Id Moreover IC kP P if and only if P is on the circle C k Otherwise the point Figure 71 The inversion C k P is inside the circle C k if and only P is outside it If g C k the inversion IC k will also be denoted by 1g Note that IC k is undefined for C and only for C The point C is called the center of the inversion C k Figure 72 displays the relation between P and P IC kP geometrically The circle of this figure has radius k and the lines SP and T P are tangent to it see Exercise 8 Figure 72 Any two points P Q inside the circle C k are transformed by the inversion C k into two points P Q such that 72 71 INVERSIONS AS TRANSFORMATIONS P Q gt PQ In fact the closer P and Q are to the center C the greater the discrepancy between PQ and P Q It follows that IC k is not a rigid motion Nevertheless it does share some of the properties of rigid motions Rigid motions transform straight lines into straight line Proposition 612 and circles into circles Exercise 612 Inversions too transform straight lines and circles into straight lines and circles although not necessarily respectively some straight lines are bent into circles and some circles are straightened out The next theorem describes these phenomena in detail THEOREM 711 The inversion IC k maps a straight lines through C onto themselves 17 straight lines not through C onto circles through C c circles through C onto straight lines not through C d circles not through C onto circles not through C PROOF a This follows directly from the definition of inversions b Let m bea straight line not through C and let M be that point of m such Figure 73 73 71 INVERSIONS AS TRANSFORMATIONS that CM J m Figure 73 whereas P is an arbitrary point of m Set M C kM and P IC kP Since CMCM k2 2 CP CP it follows that CM CP CP CM 39 Moreover since A PCM is common to both A CPM and A CM39P it follows from Proposition 359 that these triangles are similar and consequently 4 MP C 4 PMC 90 Since the points C and M are fixed whereas P is arbitrary on m it follows from Exercise 42B5 that P falls on the fixed circle with diameter CM c Let q be a circle through C let CM be a diameter of q set M IC kM39 and let m be the line through M perpendicular to CM Fig 73 It follows from part b above that IC km q and hence 2 IC kq IC km m d Let p be acircle not through C with P an arbitrary point on p Let DE be a diameter of p whose extension contains C and suppose the given inversion C k maps the points D E P onto the points D E P Figure 74 Since CPCP CDCD CECE k2 74 71 INVERSIONS AS TRANSFORMATIONS it follows that 2 OF E OF CP 2 CD and CP 2 CE Since A DCP is common to A DCP A P CD A ECP and A P CE it follows that the first two are similar to each other as are the last two Consequently 4442 and 4543 and 4142 43 244 45246290 Since D and E are fixed points it follows that P lies on the circle that has D E as its diameter see Exercise 42B5 QED The following observations are implicit in the proof of the above theorem 75 71 INVERSIONS AS TRANSFORMATIONS When the inversion IC k transforms a circle into a circle their centers are collinear with C When the inversion transform a straight line into a circle or vice versa the line through C perpendicular to the straight line contains the center of the circle EXAMPLE 712 Let I 2 106 where 0 denotes the origin and let m denote the line y 6 Fig 75 By Theorem 711b Im is a circle q that contains 0 and whose center lies on the y axis Since I fixes the point 6 0 this point must lie on q It follows that q is the circle 3 0 3 0 x j I m 6 Figure 75 EXAMPLE 713 Let I 2 106 and let q 40 1 Fig 76 It follows from Theorem 711d that 1a is also a circle with center on the x axis which of necessity contains the points 36 36 I5 0 g 0 72 0 and I3 0 g 0 2 120 Consequently Iq 96 0 24 76 71 INVERSIONS AS TRANSFORMATIONS 1q Figure 76 EXAMPLE 714 Identify the inversion IC k that maps the circle 0 2 onto the straight line 6 6 By Theorem 711c C must be either 2 0 or 2 0 Fig 77 Since C Figure 77 cannot lie between a point and its image it follows that C 2 0 Finally since C k maps 20 onto 60 it follows that k 2 26 24 E Another way in which inversions resemble rigid motions is that they too preserve the measures of angles see Exercise 618 Of course since inversions bend straight lines it is necessary to allow for non rectilineal angles As is customary in calculus the measure of the angle determined by two intersecting curves is defined to be the measure of the angle between their respective tangents at that intersection 77 71 INVERSIONS AS TRANSFORMATIONS PROPOSITION 715 Inversions preserve angles but reverse their senses PROOF Figure 78 describes how an inversion centered at C transforms the angle a formed by the curves h and j into the angle a formed by the image curves h IC k h and j IC k j It is clear that the sense of the angle is reversed by the inversion Figure 78 Since 141 42 and 143 44 it will suffice to prove the special case that A 1 A 3 or that a a in Figure 79 However the equation CPCP k2 CQCQ implies that A CPQ A CQ P 78 71 INVERSIONS AS TRANSFORMATIONS and hence 4 CQ P 4 CPQ Figure 79 Since the limiting values as 6 gt 0 of A CQ P and A CPQ are 1800 a and 1800 a respectively it follows that a a QED Two intersecting circles are said to be orthogonal if their respective tangents at the point of intersection are perpendicular to each other However by Proposition 414 the tangent line is perpendicular to the radius through the point of contact and hence it follows that two intersecting circles are orthogonal if and only if their tangents at the point of intersection pass through each other s centers see Fig 710 In fact orthogonality is guaranteed by one of the tangents passing through the center of the other Exercise 3 relates orthogonality to inversions 79 71 INVERSIONS AS TRANSFORMATIONS Figure 710 Two orthogonal circles with respective centers C and D A straight line is said to be orthogonal to a circle if it is perpendicular to the tangents at the intersection points This is equivalent to saying that the straight line contains a diameter of the circle EXERCISES 7 1 1 If 0 denotes the origin to What point or curve does the inversion 104 transform the sets below a The point 3 0 b The point 0 72 c The point 2 2 d The point 71 1 e The line y 72x f The line x y 4 g The line x 4 h The line y 74 i The line x 2 j The line y 78 k The line y x 8 l The line y 7x 4 m The line y x 7 2 n The line y 7x 7 8 o The circle 0 3 p The circle 0 8 q The circle 3 0 1 r The circle 3 0 6 s The circle 0 8 2 t The circle 0 8 4 u The circle 0 8 6 v The circle 0 8 8 W The circle 0 8 10 X The circle 4 4 4 y The circle 5 5 5 z The circle 5 5 2 For each of the following pairs of curves decide Whether there exists an inversion that transforms one onto the other Identify the inversion if it exists 710 9C 10C 71 INVERSIONS AS TRANSFORMATIONS a The x axis and the line y 2 b The circle 0 5 and the line x 3 c The circle 0 5 and the line x 5 d The circle 0 5 and the line x 10 e The circle 0 5 and the circle 0 10 f The circle 0 5 and the circle 5 0 5 g The circle 0 5 and the circle 35 0 30 Let p be a circle C a point and k a positive real number Prove that IC kp 17 if and only if the circles p and C k are orthogonal Let I be an inversion and let p be a circle such that Ip is also a circle When do p and Ip have different radii Let p and q be two circles with different radii Show that there is an inversion I such that Ip q Let m be a straight line Characterize all the circles p such that there exists an inversion I for which Im p Let p be a circle Characterize all the straight lines In such that there exists an inversion I for which Ip m Prove that if the radius of the circle of Figure 72 is k then IC kP P Write a script that will take a circle C k and a point P as input and yield IC kP as output Use a computer application to verify the following parts of Theorem 711 d d aa bb cc 72 Inversions to the Rescue Inversions can be very useful in transforming problems about circles into simpler problems about straight lines Two examples of this procedure are offered EXAMPLE 721 Let two circles p and q intersect in A and B and let the extensions of the diameters of p and q through B intersect q and p in the points C lt gt and D respectively Show that the line AB contains a diameter of the circle that circumscribes A BCD 711 72 INVERSIONS TO THE RESCUE Figure 711 Let k be any real number and suppose the inversion I 2 13k is applied to the given configuration of circles so that A IA C IC D IDp Ip q 61 r Ir Fig711 Since BC and BD are orthogonal to p and q respectively it follows from Proposition 715 that BC J p and BD J 61 The concurrence of the three altitudes of the triangle Exercise 42B11 now implies that BA J r and hence BA contains a diameter of r Proposition 723 below was first proved by Ptolemy It was an important tool in his construction of the table of chords which appears in his definitive book on Greek astronomy the Almagest The proof of the required lemma is relegated to Exercise 1 LEMMA 722 Suppose P IC kP and Q IC kQ Then kZPQ P Q CPCQ 712 72 INVERSIONS TO THE RESCUE PROPOSITION 723 In a cyclic quadrilateral the product of the diagonals equals the sum of the products of the two pairs of opposite sides A B C D39 Figure 712 PROOF Let ABCD be a cyclic quadrilateral inscribed in a circle of diameter k Fig 712 It follows from Theorem 711b that I A k inverts this circle into a tangent line that contains the points B 2 IA kB C A kC and D 2 IA kD Since BC CD 2 RD it follows from the lemma that kZBC kZCD 1831 7 7 AB AC AC AD AB AD or upon multiplying this equation by AB AC ADk2 BCAD CDAB BDAC QED In order to illustrate one more application of inversions we return to the issue of constructibility In elementary geometry classes it is customary to construct figures using rulers and compasses These two tools however are qualitatively different The 713 72 INVERSIONS TO THE RESCUE compass generates a circle because of its mechanical properties whereas the ruler simply allows us to copy a given straight line onto the paper The circular analog of a ruler would be a coin or the lid of a jar What then is the linear analog of the compass In other words what mechanical device consisting of linked rods would constrain a pencil to move so as to draw a straight line Such devices are called linkages the simplest one being a single rod AB with fixed point A It is clear that a pencil attached at B would be constrained to draw a circle The utility of linkages in drawing curves has been studied for hundreds of years but the first one capable of drawing a straight line was invented in 1864 by A Peaucellier 1832 1913 who was an engineer in the French army In his honor this device which contains 7 rods is called Peaucellier s cell In 1874 Harry Hart 1848 1920 invented a 5 rod linkage for drawing straight lines and it is unknown whether there are any such linkages with fewer than 5 rods Figure 713 Peaucellier s cell Peaucellier s cell is depicted in Figure 713 where the points C and X are fixed XP 2 XC and the solid lines XP AP 2 PB 2 BP 2 PA and CA 2 CB denote rods that are loosely linked at their endpoints The dashed lines denote auxiliary lines that serve 714 72 INVERSIONS TO THE RESCUE only for the purpose of demonstrating the properties of the linkage Since APBP is a rhombus we have CP CP CD DPCD DP CD2 DP2 CD2 DA2 DP2 DA2 CA2 APZ If k l CA2 AP2 then k is constant and we have P 10km Since the rod XP constrains P to move in a circle q that contains C it follows from Proposition 711c that P traces out a straight line m EXERCISES 72 7C Prove Lemma 722 Two circles 17 and q have a common tangent at a point T and a variable circle through T intersects p and q orthogonally in points P and Q Prove that PXQ passes through a fixed point Suppose ABCD is a cyclic quadrilateral If T is the point of contact of a circle containing A and B with another circle containing C and D show that the locus of T is a circle Prove that if ABCD is a convex quadrilateral then BC 39AD CD 39AB 2 BD AC Show that equality holds if and only if ABCD is cyclic Let p be a fixed circle and let P be a fixed point not on p Prove that there exists a point P distinct from P such that every circle through P that is orthogonal to 17 also passes through P Let p be a fixed circle and P a fixed point Show that the locus of the centers of all the circles that pass through P and are orthogonal to p is a straight line Use a computer application to verify Proposition 723 715 73 INVERSIONS AS HYPERBOLIC RIGID MOTIONS 73 Inversions as Hyperbolic Rigid Motions In addition to the role they play in Euclidean geometry inversions also provide a powerful tool for describing the rigid motions of non Euclidean geometry in the context of the upper half plane model of Section 12 As was the case in that section the exposition here is informal and no proofs are given Instead several examples are offered that are easily implemented on computers and substantiate the discussion Exercises 13 17 provide an opportunity for an exploration of the properties of the hyperbolic rigid motions It was proven in Section 64 that every hyperbolic rigid motion can be expressed as the composition of hyperbolic re ections These hyperbolic re ections are now described in the context of the upper half plane geometry PROPOSITION 731 There are two kinds of re ections of the upper halfplane a Euclidean re ections whose axes are vertical 1 Inversions whose centers are on the xaxis It stands to reason that Euclidean re ections with vertical axes should also double as hyperbolic re ections After all the distortion of lengths that was used to create this geometry depends only on the distances from the x axis and since these particular re ections do not change these distances it is not surprising that they constitute hyperbolic 716 73 INVERSIONS AS HYPERBOLIC RIGID MOTIONS 3 Cu Figure 714 Hyperbolic re ections rigid motions Figure 714 illustrates the effect of the re ections of both types on a triangle Hyperbolic A A B C is both the Euclidean and the hyperbolic re ection of A ABC in the vertical geodesic m and A A B C IgA ABC is the hyperbolic re ection of A ABC in the bowed geodesic g EXAMPLE 732 Find a hyperbolic re ection that transforms the point PI I to thepoint Q3 5 Fig 715 l The line P Q has equation y 1 2 526 1 and intersects the x axis at the Figure 715 point C 1 0 Since CPCQ 1121 02 5123 02 717 73 INVERSIONS AS HYPERBOLIC RIGID MOTIONS zap 515 it follows that the inversion 7va is the required hyperbolic reflection EXAMPLE 733 Find a hyperbolic reflection that transforms the geodesic consisting of the upper half of the circle 4 0 2 onto the upper half of the straight line 6 3 Figure 716 Any inversion centered at 6 0 will transform the given semicircle into a vertical ray Since 6 2X6 3 36 it follows that the requisite hyperbolic reflection is the inversion 16Y 0m with fixed circle g 3 2 6 Figure 716 A hyperbolic reflection Inasmuch as the definition of a Euclidean rotation makes no reference to parallelism or any of its consequences it can also serve as the definition of a hyperbolic rotation Unfortunately because of the distortion of distances in the upper half plane this definition is not very helpful in trying to visualize this non Euclidean transformation However Proposition 622 which states that the composition of two reflections with 718 73 INVERSIONS AS HYPERBOLIC RIGID MOTIONS intersecting axes is a rotation about the point of intersection is neutral and so it applies here as well Hence if g denotes the inversion whose fixed circle is g Fig 717 and A Bquot A 3 Figure 717 Hyperbolic re ections C M 0 we recall that pm is also a hyperbolic re ection then pm I g is a hyperbolic rotation R about the point X Note that RAABC pm1gAABC pmAA B C AA B C If g 0 3 and m is the line 6 2 it follows from Exercise12 that 2 4 g m LXOM cos 1g 4820 Consequently R RX 9640 Figure 718 illustrates a hyperbolic rotation R RY600 719 73 INVERSIONS AS HYPERBOLIC RIGID MOTIONS m P 2 8 P1 30quot Y P 3 P 6 P5 P4 0 x M Figure 718 A hyperbolic rotation In this figure PH1 RPi for i 1 2 5 and P1 RP6 Obviously these points all lie on a hyperbolic circle centered at Y Exercise 15C examines this issue further The definition of a Euclidean translation does involve parallelism and is therefore of no use in the hyperbolic context Instead motivated by Proposition 621 a hyperbolic translation is defined as the composition of two hyperbolic reflections whose axes do not intersect If the axes of both the hyperbolic reflections are straight geodesics then their composition is a horizontal Euclidean translation This is illustrated in Figure 719 where 7 pnpm and 1Pi PH1 for i12 34 Figure 719 Both a Euclidean and a hyperbolic translation 7 20 73 INVERSIONS AS HYPERBOLIC RIGID MOTIONS Such horizontal Euclidean translations constitute hyperbolic rigid motions for the same reason that the Euclidean reflections pm and p do They do not alter the distances of points from the x axis Figure 720 illustrates the composition of hyperbolic reflections of mixed types with non intersecting axes If 7 Ig pm then P 1Pi for i 1 2 3 4 Note 11 m g I 2 P 3 Pl39 P4 0 J P5 Figure 720 A hyperbolic translation that the orbit of this translation 395 does go out to hyperbolic infinity just as Euclidean translations go out to Euclidean infinity This stands in marked contrast with the orbit of the hyperbolic rotation R of Figure 718 Figure 721 displays an orbit of the hyperbolic glide reflection y 7 105 where 7 is the horizontal shift 6 y gt x 2 y 721 73 INVERSIONS AS HYPERBOLIC RIGID MOTIONS Figure 721 A hyperbolic glide reflection Since the proof of Theorem 642 is neutral it follows that every hyperbolic rigid motion is the composition of no more than three hyperbolic reflections The group of all the rigid motions of the upper half plane is one of the best studies structures of advanced mathematics It has also proved to be an indispensable tool in such diverse areas as geometry of course number theory and analysis An elementary discussion of this topic and its surprising connection with complex numbers can be found in the author s The Poincare HalfPlane A Gateway to Modern Geometry Given a point P and a transformation f the orbit of P generated by f is the sequence of points PfPf2Pf3P Thus the sequences P1 P2 P3 of Figures 718 21 are all orbits These orbits help visualize the action of the transformation that generated them It is clear that orbits of Euclidean translations are contained in straight lines whereas orbits of Euclidean rotations are circular in nature Exercise 15 offers the reader the opportunity to explore the orbits generated by hyperbolic transformations EXERCISES 73 7 22 10 11 12 13C 14C 15C 16C 73 INVERSIONS AS HYPERBOLIC RIGID MOTIONS Find a hyperbolic reflection that transforms the point 1 1 to the point 73 5 Find a hyperbolic reflection that transforms the point 1 1 to the point 73 1 Find a hyperbolic reflection that transforms the upper half of x 4 to the upper half of the circle 0 4 Find a hyperbolic reflection that transforms the upper half of x 4 to the upper half of the circle 0 2 Find a hyperbolic reflection that transforms the upper half of x 4 to the upper half of the circle 0 5 Find a hyperbolic reflection that transforms the upper half of x 4 to the upper half of x 17 Find a hyperbolic reflection that transforms the upper half of O 5 to the upper half of O 3 Find a hyperbolic reflection that transforms the upper half of O 5 to the upper half of 9 0 2 Prove that given any two points of the upper half plane there is a hyperbolic reflection that transforms one onto the other Prove that given any two intersecting geodesics of the upper half plane there is a hyperbolic reflection that transforms one onto the other Prove that given any two non intersecting geodesics of the upper half plane there is a hyperbolic re ection that transforms one onto the other Prove thatin Figure 717 A g m A XOM Write a script that will reflect any two points of the upper halfiplane in any bowed geodesic Use the script of Exercise 1217 to substantiate the claim that this transformation is indeed a hyperbolic rigid motion Write a script that will reflect any triangle of the upper halfiplane in any bowed geodesic Let P an arbitrary point of the upper halfiplane m an arbitrary vertical straight line and g an arbitrary circle centered on the xiaXis Write a script that takes P m and g as its input and yields several iterations of the action of pm 0 lg on P Use this script to eXplore the following questions a When In and g intersect whatis the geometrical nature of the orbits of pm 0 lg b What does a hyperbolic circle look like to a Euclidean observer c When In and g do not intersect what is the geometrical nature of the orbits of pm 0 lg To be specific what is the nature of each orbit of pm 0 lg and how do these orbits relate to each other Use a computer application to model noniEuclidean glideireflections 17C Use a computer application to eXplore the notion of a hyperbolic inversion 723 CHAPTER REVIEW CHAPTER REVIEW EXERCISES 7 l Prove that if fis any Euclidean rigid motion and I is any inversion then f 0 I 0 f is also an inversion 7 2 Suppose F is any Euclidean rigid motion and I is any inversion Is I 0 f 0 I necessarily a Euclidean rigid motion 3 Is the composition of two inversions ever an inversion 4 When is the composition of two inversions a Euclidean rigid motion 5 Are the following statements true or false Justify your answers a Every inversion is a rigid motion of Euclidean geometry b Some inversions are rigid motions of Euclidean geometry c Every inversion is a rigid motion of hyperbolic geometry d Some inversions are rigid motions of hyperbolic geometry e Every rigid motion of hyperbolic geometry is an inversion f Some rigid motions of hyperbolic geometry are inversions g lnversions transform circles into circles h Given a straight line there exists no inversion that will transform it into another distinct straight line i Peaucellier s cell was a notorious torture chamber in the Bastille j Given any two circles there exists either a Euclidean rigid motion or an inversion that transforms one into the other 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