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he development of the axiomatic method of reasoning was one of themost profound events in the history of mathematics. In this chapterwe explore axiomatic systems and their properties.One strand running through the chapter is the search for the“ideal.” The golden ratio is the ideal in concrete form, realized throughnatural and man-made constructions. Deductive reasoning from a baseset of axioms is the ideal in abstract form, realized in the crafting ofclear, concise, and functional definitions, and in the reasoning employedin well-constructed proofs.Another strand in the chapter, and which runs through the entiretext, is that of the interplay between the concrete and the abstract. Asyou work through this text, you are encouraged to play with concreteideas, such as how the Golden Ratio appears in nature, but you are alsoencouraged to play (experiment) when doing proofs and mor

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Download Exploring Geometry 2nd Edition Hvidsten Solutions Manual and more Exams Geometry in PDF only on Docsity! Michael Hvidsten Gustavus Adolphus College Student’s Guide for Exploring Geometry, Second Edition Contents v 7.4 Project 10 - The Saccheri Quadrilateral 41 7.5 Lambert Quadrilaterals and Triangles 41 7.6 Area in Hyperbolic Geometry 43 7.7 Project 11 - Tiling the Hyperbolic Plane 44 Chapter 8 Elliptic Geometry 45 SOLUTIONS TO EXERCISES IN CHAPTER 8 45 8.2 Perpendiculars and Poles in Elliptic Geometry 45 8.3 Project 12 - Models of Elliptic Geometry 46 8.4 Basic Results in Elliptic Geometry 46 8.5 Triangles and Area in Elliptic Geometry 47 8.6 Project 13 - Elliptic Tiling 48 Chapter 9 Projective Geometry 49 SOLUTIONS TO EXERCISES IN CHAPTER 9 49 9.2 Project 14 - Perspective and Projection 49 9.3 Foundations of Projective Geometry 49 9.4 Transformations in Projective Geometry and Pap- pus’s Theorem 51 9.5 Models of Projective Geometry 52 9.6 Project 15 - Ratios and Harmonics 54 9.7 Harmonic Sets 54 9.8 Conics and Coordinates 56 Chapter 10 Fractal Geometry 59 SOLUTIONS TO EXERCISES IN CHAPTER 10 59 10.3 Similarity Dimension 59 10.4 Project 16 - An Endlessly Beautiful Snowflake 60 10.6 Fractal Dimension 60 10.7 Project 17 - IFS Ferns 61 10.9 Grammars and Productions 61 10.10 Project 18 - Words Into Plants 62 vi @ Contents
APPENDIX A# Sample Lab Report
63
C H A P T E R 1 Geometry and the Axiomatic Method The development of the axiomatic method of reasoning was one of the most profound events in the history of mathematics. In this chapter we explore axiomatic systems and their properties. One strand running through the chapter is the search for the “ideal.” The golden ratio is the ideal in concrete form, realized through natural and man-made constructions. Deductive reasoning from a base set of axioms is the ideal in abstract form, realized in the crafting of clear, concise, and functional definitions, and in the reasoning employed in well-constructed proofs. Another strand in the chapter, and which runs through the entire text, is that of the interplay between the concrete and the abstract. As you work through this text, you are encouraged to play with concrete ideas, such as how the Golden Ratio appears in nature, but you are also encouraged to play (experiment) when doing proofs and more abstract thinking. The experimentation in the latter is of the mind, but it can utilize many of the same principles of exploration as you would use in a computer lab. When trying to come up with a proof you should consider lots of examples and ask “What if ...?” questions. Most importantly, you should interact with the ideas, just as you interact with the software environment. Interaction with ideas and discovery of concepts is a primary orga- nizing principle for the text. Interaction is encouraged in three ways. First, topics are introduced and developed in the text. Next, lab projects reinforce concepts, or introduce related ideas. Lastly, project results are discussed, and conclusions drawn, in written lab reports. 1 4 Student’s Guide for Exploring Geometry, Second Edition If this topic interests you, you may want to further research the area of information theory and computability in computer science. A good reference here is Gregory Chaitin’s book The Limits of Mathematics (Springer, 1998.) Additionally, much more could be investigated as to the various philosophies of mathematics, in particular the debates between platonists and constructionists, or between intuitionists and formalists. A good reference here is Edna E. Kramer’s The Nature and Growth of Modern Mathematics (Princeton, 1981), in particular Chapter 29 on Logic and Foundations. 1.5.1 Let S be the set of all sets which are not elements of them- selves. Let P be the proposition that “S is an element of itself”. And consider the two propositions P and the negation of P , which we de- note as ¬P . Assume P is true. Then, S is an element of itself. So, S is a set which by definition is not an element of itself. So, ¬P is true. Likewise, if ¬P is true then P is true. In any event we get P and ¬P both true, and the system cannot be consistent. 1.5.3 Good research books for this question are books on the his- tory of mathematics. This could be a good final project idea. 1.5.5 Let P be a point. Each pairing of a point with P is associated to a unique line. There are exactly three such pairings. 1.5.7 Yes. The lines and points satisfy all of the axioms. 1.5.9 If (x, y) is in P , then x < y. Clearly, y < x is impossible and the first axiom is satisfied. Also, inequality is transitive on numbers so the second axiom holds and this is a model. 1.5.11 A quick listing of all points and lines and incidence relations shows that Axioms 1 and 2 are satisfied. For Axiom 3, points A, B, C, and D have the property that no subset of three of the points are collinear. For Axiom 4, line ←→ AB suffices. 1.5.13 The dual to Axiom 1 is “Given two distinct lines, there is exactly one point incident with them both.” Proof: Suppose there were two points A and B incident on both lines. This would contradict Axiom A1. 1.5.15 By Axiom 4 there is a line l with n+ 1 points, say P1, . . . , Pn+1. By Axiom 3, there must be a point Q that is not on l. Let l1 be the line incident on P1 and Q, l2 be the line incident on P2 and Q, etc. The n + 1 lines l1, . . . , ln+1 through Q satisfy the dual statement of Axiom 4. Axiomatic Method 5 1.6 Euclid’s Axiomatic Geometry In this section we take a careful look at Euclid’s original axiomatic system. We observe some of its inadequacies in light of our modern “meta” understanding of such systems, and discuss the one axiom that has been the creative source of much of modern geometry – the Parallel Postulate. 1.6.1 Good research books for this question are books on the his- tory of mathematics. 1.6.3 An explanation should be given along with a figure like the following: a a b b FIGURE 1.1: 1.6.5 123 = 3 · 36 + 15 36 = 2 · 15 + 6 15 = 2 · 6 + 3 6 = 2 · 3 + 0 Thus, gcd(123, 36) = 3. 1.6.7 This exercise is a good starting off point for discussing the importance of definitions in mathematics. One possible definition for a circle is: Definition 1.1. A circle with center O and radius length r is the set of points P on the sphere such that the distance along the great circle from O to P is r. 6 Student’s Guide for Exploring Geometry, Second Edition Note that this definition is itself not entirely well-defined, as we have not specified what we mean by distance. One workable definition is for distance to be net cumulative arc length along a great circle as we move from a point O to a point P . An angle ABC can be most easily defined as the Euclidean angle made by the tangent lines at B to the circles defining ←→ AB and ←→ CB. Then, Postulate 1 is satisfied most of the time, as we can construct a unique great circle passing through two points on the sphere, if the points are not antipodal. We simply intersect the sphere with the plane through the two points and the center of the sphere. Postulate 2 is satisfied as we can always extend an arc of a great circle, although we may retrace the existing arc. Postulate 3 is satisfied if we use the cumulative distance definition as discussed above. Postulate 4 is automatically satisfied as angles are Euclidean angles. Postulate 5 is not satisfied, as every pair of lines intersects. An easy proof of this is to observe that every line is uniquely defined by a plane through the origin. Two different planes will intersect in a line, and this line must intersect the sphere at two points. 1.6.9 This is true. Given a plane through the origin, we can always find an orthogonal plane. The angle these planes make will equal the angle of the curves they define on the sphere, as the spherical angles are defined by tangent lines to the sphere, and thus lie in the planes. 1.6.11 This is true. An example is the triangle that is defined in the first octant. It has three right angles. 1.7 Project 2 - A Concrete Axiomatic System After the last few sections dealing with abstract axiomatic systems, this lab is designed so that you can explore another geometric system through concrete manipulation of the points, lines, etc of that system. The idea here is to have you explore the environment first, then make some conjectures about what is similar and what is different in this system as compared to standard Euclidean geometry. Euclidean Geometry 9 2.1.9 Assume Playfair and let lines m and n be parallel to line l. If m 6= n and m and n intersect at P , then we would have two different lines parallel to l through P , contradicting Playfair. Thus, either m and n are parallel, or are the same line. Conversely, assume that two lines parallel to the same line are equal or themselves parallel. Let l be a line and suppose m and n are parallel to l at a point P not on l. Then, n and m must be equal, as they intersect at P . 2.2 Congruent Triangles and Pasch’s Axiom This section introduces many results concerning triangles and also dis- cusses several axiomatic issues that arose from Euclid’s treatment of triangles. 2.2.1 Yes, it could pass through points A and B of ∆ABC. It does not contradict Pasch’s axiom, as the axiom stipulates that the line cannot pass through A, B, or C. 2.2.3 No. Here is a counter-example. A B D C l FIGURE 2.1: 2.2.5 If A = C we are done. If A, B, and C are collinear, then B cannot be between A and C, for then we would have two points of intersection for two lines. If A is between B and C, then l cannot intersect AC. Likewise, C cannot be between A and B. If the points are not collinear, suppose A and C are on opposite sides. Then l would intersect all three sides of ∆ABC, contradicting Pasch’s axiom. 2.2.7 Let ∠ABC ∼= ∠ACB in ∆ABC. Let −−→ AD be the angle bisector 10 Student’s Guide for Exploring Geometry, Second Edition of ∠BAC meeting side BC at D. Then, by AAS, ∆DBA and ∆DCA are congruent and AB ∼= AC. 2.2.9 Suppose that two sides of a triangle are not congruent. Then, the angles opposite those sides cannot be congruent, as if they were, then by the previous exercise, the triangle would be isosceles. Suppose in ∆ABC that AC is greater than AB. On AC we can find a point D between A and C such that AD ∼= AB. Then, ∠ADB is an exterior angle to ∆BDC and is thus greater than ∠DCB. But, ∆ABD is isosceles and so ∠ADB ∼= ∠ABD, and ∠ABD is greater than ∠DCB = ∠ACB. 2.2.11 Let ∆ABC and ∆XY Z be two right triangles with right angles at A and X, and suppose BC ∼= Y Z and AC ∼= XZ. Suppose AB is greater than XY . Then, we can find a point D between A and B such that AD ∼= XY . By SAS ∆ADC ∼= ∆XY Z. Now, ∠BDC is exterior to ∆ADC and thus must be greater than 90 degrees. But, ∆CDB is isosceles, and thus ∠DBC must also be greater than 90 degrees. This is impossible, as then ∆CDB would have angle sum greater than 180 degrees. A C B X Z Y D FIGURE 2.2: 2.2.13 We use AAS to show that ∆BFH ∼= ∆AFG and ∆CEI ∼= ∆AEG. Thus BH ∼= AG ∼= CI and BHIC is Saccheri. Also, by adding congruent angles in the left case we get that the sum of the angles in the triangle is the same as the sum of the summit angles. In the right case, we need to re-arrange congruent angles. 2.2.15 Given quadrilaterals ABCD and WXY Z we say the two quadrilaterals are congruent if there is some way to match vertices so that corresponding sides are congruent and corresponding angles are congruent. SASAS Theorem: If AB ∼= WX, ∠ABC ∼= ∠WXY , BC ∼= XY , Euclidean Geometry 11 ∠BCD ∼= ∠XY Z, and CD ∼= Y Z, then quadrilateral ABCD is con- gruent to quadrilateral WXY Z. A B D C W X Z Y FIGURE 2.3: Proof: ∆ABC and ∆WXY are congruent by SAS. This im- plies that ∆ACD and ∆WY Z are congruent. This shows that sides are correspondingly congruent, and two sets of angles are congruent (∠ABC ∼= ∠WXY and ∠CDA ∼= ∠Y ZW ). Since ∠BAC ∼= ∠XWY and ∠CAD ∼= ∠YWZ, then by angle addition ∠BAD ∼= ∠XWZ. Similarly, ∠BCD ∼= ∠XY Z. 2 2.3 Project 3 - Special Points of a Triangle You are encouraged to explore and experiment in this lab project. Are there any other sets of intersecting lines that one could construct for a given triangle? Are there interesting properties of constructed intersecting lines in other polygons? You may be taking this class to become a secondary math teacher. This project is one that could be easily transferred to the high school setting. 2.4.1 Mini-Project:Area in Euclidean Geometry This section includes the first “mini-project” for the course. These projects are designed to be done in the classroom, in groups of three or four students. Each group should elect a Recorder. The Recorder’s sole job is to outline the group’s solutions to exercises. The summary should not be a formal write-up of the project, but should give enough a brief synopsis of the group’s reasoning process. The main goal for the mini-projects is to have a discussion of geo- metric ideas. Through the group process, you can clarify your under- 14 Student’s Guide for Exploring Geometry, Second Edition ∠AGH ∼= ∠ABC and ∠AHG ∼= ∠ACB. Thus, ∆AGH and ∆ABC are similar. Therefore, AB AG = AC AH . Equivalently, AB DE = AC AH . We are given that AB DE = AC DF . Thus, AH ∼= DF . Also, ABAG = BC GH and AB AG = AB DE = BC EF . Thus, GH ∼= EF . By SSS ∆AGH and ∆DEF are congruent, and thus ∆ABC and ∆DEF are similar. A B C D E F G H FIGURE 2.6: 2.5.5 Any right triangle constructed so that one angle is congruent to ∠A must have congruent third angles, and thus the constructed triangle must be similar to ∆ABC. Since sin and cos are defined in terms of ratios of sides, then proportional sides will have the same ratio, and thus it does not matter what triangle one uses for the definition. 2.5.7 If the parallel to ←→ AC does not intersect ←→ RP , then it would be parallel to this line, and since it is already parallel to ←→ AC, then by Exercise 2.1.9 ←→ RP and ←→ AC would be parallel, which is impossible. By the properties of parallels, ∠RAP ∼= ∠RBS and ∠RPA ∼= ∠RSB. Thus, by AAA ∆RBS and ∆RAP are similar. ∆PCQ and ∆SBQ are similar by AAA using an analogous argument for two of the angles and the vertical angles at Q. Thus, CP BS = CQ BQ = PQ QS , and AP BS = AR BR = PR SR . So, CP AP BQ QC = CP AP BS CP = BS AP And, CPAP BQ QC AR RB = BS AP AR RB = BS AP AP BS = 1. 2.5.1 Mini-Project: Finding Heights This mini-project is a very practical application of the notion of sim- ilarity. The mathematics in the first example for finding height is ex- tremely easy, but the interesting part is the data collection. You will Euclidean Geometry 15 need to determine how to get the most accurate measurements using the materials on hand. The second method of finding height is a calculation using two simi- lar triangles. The interesting part of this project is to see the connection between the mirror reflection and the calculation you made in part I. You should work in small groups with a Recorder, but make sure the Recorder position gets shifted around from project to project. 2.6 Circle Geometry This section is an introduction to the basic geometry of the circle. The properties of inscribed angles and tangents are the most important properties to focus on in this section. 2.6.1 Case 2: A is on the diameter through OP . Let α = m∠PBO and β = m∠POB. Then, β = 180−2α. Also, m∠AOB = 180−β = 2α. Case 3: A and B are on the same side of ←→ PO. We can assume that m∠OPB > m∠OPA. Let m∠OPB = α and m∠OPA = β. Then, we can argue in a similar fashion to the proof of the Theorem using α− β instead of α+ β. 2.6.3 Consider ∠AQP . This must be a right angle by Corol- lary 2.33. Similarly, ∠BQP must be a right angle. Thus, A, Q, and B are collinear. 2.6.5 Let AB be the chord, O the center, and M the midpoint of AB. Then ∆AOM ∼= ∆BOM by SSS and the result follows. 2.6.7 Consider a triangle on the diagonal of the rectangle. This has a right angle, and thus we can construct the circle on this angle. Since the other triangle in the rectangle also has a right angle on the same side (the diameter of the circle) then it is also inscribed in the same circle. 2.6.9 If point P is inside the circle c, then Theorem 2.41 applies. But, this theorem says that m∠BPA = 1 2(m∠BOA+m∠COD), where C and D are the other points of intersections of ←→ PA and ←→ PB with the circle. If P is inside c, then C and D are different points. The assumption of Theorem 2.42 says that m∠BPA = 1 2m∠BOA. But, m∠BPA = 1 2(m∠BOA+m∠COD) would then imply that m∠COD = 0, which is impossible as C and D are not collinear with O. 2.6.11 The angle made by BT and l must be a right angle by Theorem 2.36. Likewise, the angle made by AT and l is a right angle. Thus, A, T , and B are collinear. 2.6.13 Suppose one of the circles had points A and B on opposite 16 Student’s Guide for Exploring Geometry, Second Edition sides of the tangent line l. Then AB would intersect l at some point P which is interior to the circle. But, then l would pass through an interior point of the circle and by continuity must intersect the circle in two points which is impossible. Thus, either all points of one circle are on opposite sides of l from the other circle or are on the same side. 2.6.15 By Theorem 2.36, we have that ∠OAP is a right angle, as is ∠OBP . Since the hypotenuse (OP ) and leg (OA) of right triangle ∆OAP are congruent to the hypotenuse (OP ) and leg (OB) of right triangle ∆OBP , then by Exercise 2.2.10 the two triangles are congru- ent. Thus ∠OPA ∼= ∠OPB. 2.6.17 Let A and B be the centers of the two circles. Construct the two perpendiculars at A and B to ←→ AB and let C and D be the intersections with the circles on one side of ←→ AB. If ←→ CD does not intersect ←→ AB, then these lines are parallel, and the angles made by ←→ CD and the radii of the circles will be right angles. Thus, this line will be a common tangent. Otherwise, ←→ CD intersects ←→ AB at some point P . Let ←→ PE be a tangent to the circle with center A. Then, since ∆PAC and ∆PBD are similar, we have AP BP = AC BD . Let ←→ BF be parallel to ←→ AE with F the intersection of the parallel with the circle centered at B. Then, AC BD = AE BF . So, AP BP = AE BF . By SAS similarity, ∆PAE and ∆PBF are similar, and so F is on ←→ PE and ∠PFB is a right angle. Thus, ←→ PE is a tangent to the circle centered at B. A T B C D P E F FIGURE 2.7: C H A P T E R 3 Analytic Geometry This chapter is a very quick review of analytic geometry. In succeeding chapters, analytic methods will be utilized freely. SOLUTIONS TO EXERCISES IN CHAPTER 3 3.2 Vector Geometry 3.2.1 If A is on either of the axes, then so is B and the distance result holds by the definition of coordinates. Otherwise, A (and B) are not on either axis. Drop perpendiculars from A and B to the x-axis at P and Q. By SAS similarity, ∆AOP and ∆BOQ are similar, and thus ∠AOP ∼= ∠BOQ, which means that A and B are on the same line ←→ AO, and the ratio of BO to AO is k. 3.2.3 The vector from P to Q is in the same direction (or opposite direction) as the vector v. Thus, since the vector from P to Q is ~Q− ~P , we have ~Q − ~P = tv, for some real number t. In coordinates we have (x, y)− (a, b) = (tv1, tv2), or (x, y) = (a, b) + t(v1, v2). 3.2.5 By Exercise 3.2.3 the line through A andB can be represented by the set of points of the form ~A+ t( ~B− ~A). Then, M = 1 2( ~A+ ~B) = ~A + 1 2( ~B − ~A) is on the line through A and B, and is between A and B. Let A = (x1, y1) and B = (x2, y2), then the distance from A to M is √ (x1 2 − x2 2 )2 + (y12 − y2 2 )2, which is equal to the distance from B to M . 19 20 Student’s Guide for Exploring Geometry, Second Edition 3.3 Project 5 - Bézier Curves 3.4 Angles in Coordinate Geometry 3.4.1 Let ~A = (cos(α), sin(α)) and ~B = (cos(β), sin(β)). Then, from Theorem 3.11 we have cos(α − β) = ~A ◦ ~B, since ~A and ~B are unit length vectors. The result follows immediately. 3.4.3 By Exercise 3.4.1, cos( π 2 − (α+ β)) = cos( pi 2 ) cos(α+ β) + sin( pi 2 90) sin(α+ β)) = sin(α+ β). Then, use the formula from Exercise 3.4.2 with the term inside cos being (pi2 − α) + (−β). 3.5 The Complex Plane 3.5.1 eiθeiφ = (cos(θ) + i sin(θ))(cos(φ) + i sin(φ)) = (cos(θ) cos(φ)− sin(θ) sin(φ)) + i(cos(θ) sin(φ) + sin(θ) cos(φ)) = cos(θ + φ) + i sin(θ + φ) = ei(θ+φ) 3.5.3 Let z = eiθ and w = eiφ and use Exercise 3.4.1. 3.5.5 The rationalized complex numbers have the form i−12 , i, and 1 10 − i 1 5 . 3.6 Birkhoff’s Axiomatic System for Analytic Geometry 3.6.1 First, if A is associated to xA = tA √ dx2 + dy2, where A = (x, y) = (x0, y0) + tA(dx, dy), and B is associated to xB in a similar fashion, then |xA − xB| = |tA − tB| √ dx2 + dy2. On the other hand, d(A,B) = √ (tAdx− tBdx)2 + (tAdy − tBdy)2 = √ dx2 + dy2|tA − tB| 3.6.3 Given a point O as the vertex of the angle, set O as the origin of the coordinate system. Then, identify a ray −→ OA associated to the angle θ, with A = (x, y). Let a = || ~A|| = √ x2 + y2. Then, sin2(θ) + cos2(θ) = (xa )2 + (ya)2 = x2+y2 a2 = 1. 3.6.5 Discussion question. One idea is that analytic geometry al- lows one to study geometric figures by the equations that define them. Thus, geometry can be reduced to the arithmetic (algebra) of equa- tions. C H A P T E R 4 Constructions In this chapter we cover some of the basic Euclidean constructions and also have a lot of fun with lab projects. The origami project should be especially interesting, as it is an axiomatic system with which you can physically interact and explore. The third section on constructibility may be a bit heavy and ab- stract, but the relationship between geometric constructibility and al- gebra is a fascinating one, especially if you have had some exposure to abstract algebra. Also, any mathematically literate person should know what the three classical construction problems are, and how the pursuit of solutions to these problems has had a profound influence on the development of modern mathematics. SOLUTIONS TO EXERCISES IN CHAPTER 4 4.1 Euclidean Constructions 4.1.1 Use SSS triangle congruence on ∆ABF and ∆DGH. 4.1.3 Use the SSS triangle congruence theorem on ∆ADE and ∆ABE to show that ∠EAB ∼= ∠BAE. 4.1.5 Use the fact that both circles have the same radius. 4.1.7 Let the given line be l and let P be the point not on l. Construct the perpendicular m to l through P . At a point Q on m, but not on l, construct the perpendicular n to m. Theorem 2.7 implies that l and n are parallel. 4.1.9 On −−→ BA construct A′ such that BA′ = a. On −−→ BC construct C ′ such that BC ′ = b. Then, SAS congruence gives ∆AB′C ′ congruent to any other triangle with the specified data. 21 C H A P T E R 5 Transformational Geometry In this chapter we make great use of functional notation and somewhat abstract notions such as 1 − 1 and onto, inverses, composition, etc. You may wonder how such computations are related to geometry, but that is the very essence of the chapter—that we can understand and investigate geometric ideas with more than one set of foundational lenses. With that in mind, we will make use of synthetic geometric tech- niques where they are most elegant and can aid intuition, and at other times we will rely on analytical techniques. SOLUTIONS TO EXERCISES IN CHAPTER 5 5.1 Euclidean Isometries 5.1.1 Define the function f−1 by f−1(y) = x if and only if f(x) = y. Then, f−1 is well-defined, as suppose f(x1) = f(x2) = y. Then, since f is 1− 1 we have that x1 = x2. Since f is onto, we have that for every y in S there is an x such that f(x) = y. Thus, f−1 is defined on all of S. Finally, f−1(f(x)) = f−1(y) = x and f(f−1(y)) = f(x) = y. So, f ◦ f−1 = f−1 ◦ f = idS . Suppose g was another function on S such that f ◦ g = g ◦ f = idS . Then, g ◦ f ◦ f−1 = f−1, or g = f−1. 5.1.3 Since g−1 ◦ f−1 ◦ f ◦ g = g−1 ◦ g = id and f ◦ g ◦ g−1 ◦ f−1 = f ◦ f−1 = id, then g−1 ◦ f−1 = (f ◦ g)−1. 5.1.5 Let f be an isometry and let c be a circle centered at O of radius r = OA. Let O′ = f(O) and A′ = f(A). Let P be any 25 26 Student’s Guide for Exploring Geometry, Second Edition point on c. Then, O′f(P ) = f(O)f(P ) = OP = r. Thus, the image of c under f is contained in the circle centered at O′ of radius r. Let P ′ be any other point on the circle centered at O′ of radius r. Then, Of−1(P ′) = f−1(O′)f−1(P ′) = O′P ′ = r. Thus, f−1(P ′) is a point on c and every such point P ′ is the image of a point on c, under the map T . 5.1.7 Label the vertices of the triangle A, B, and C. Then, consider vertex A. Under an isometry, consider the actual position of A in the plane. After applying the isometry, A might remain or be replaced by one of the other two vertices. Thus, there are three possibilities for the position occupied by A. Once that vertex has been identified, consider position B. There are now just two remaining vertices to be placed in this position. Thus, there are a maximum of 6 isometries. We can find 6 by considering the three basic rotations by 0, 120, and 240 degrees, and the three reflections about perpendicular bisectors of the sides. 5.1.9 First, we show that f is a transformation. To show it is 1−1, suppose f(x, y) = f(x′, y′). Then, kx+a = kx′+a and ky+b = ky′+b. So, x = x′ and y = y′. To show it is onto, let (x′, y′) be a point. Then, f(x ′−a k , y ′−b k ) = (x′, y′). f is not, in general, an isometry, since if A = (x, y) and B = (x′, y′) then f(A)f(B) = kAB. 5.1.11 Let ABC be a triangle and let A′B′C ′ be its image under f . By the previous exercise, these two triangles are similar. Thus, there is a k > 0 such that A′B′ = kAB, B′C ′ = kBC, and A′C ′ = kAC. Let D be any other point not on ←→ AB. Then, using triangles ABD and A′B′D′ we get that A′D′ = kAD. Now, let DE be any segment with D not on ←→ AB. Then, using triangles ADE and A′D′E′ we get D′E′ = kDE, since we know that A′D′ = kAD. Finally, let EF be a segment entirely on ←→ AB, and let D be a point off ←→ AB. Then, using triangles DEF and D′E′F ′ we get E′F ′ = kEF , since we know that D′E′ = kDE. Thus, in all cases, we get that f(A)f(B) = kAB. 5.2.1 Mini-Project:Isometries Through Reflection In this mini-project, you will be led through a guided discovery of the amazing fact that, given any two congruent triangles, one can find a sequence of at most three reflections taking one triangle to the other. Transformational Geometry 29 is given by T−1 ◦ rx ◦T (x, y) = T−1 ◦ rx(x, y−K) = T−1(x,−y+K) = (x,−y + 2K). So, r(x, y) = (x,−y + 2K). 5.3.9 Let T be a translation with (non-zero) translation vector parallel to a line l. Let m be perpendicular to l at point P . Let n be the perpendicular bisector of PT (P ), intersecting PT (P ) at point Q. Then, rn, reflection about n maps P to T (P ). Consider rn◦T . We have rn◦T (P ) = P . Let R 6= P be another point on m. Then, PRT (R)T (P ) is a parallelogram, and thus ∠PRT (R) and ∠RT (R)T (P ) are right angles. Let S be the point where n intersects RT (R). Then, ∠RSQ is also a right angle. Also, by a congruent triangle argument, we have RS ∼= ST (R), and so n is the perpendicular bisector of RT (R) and rn ◦T (R) = R. Since rn ◦T fixes two points on m we have rn ◦T = rm, or T = rn ◦ rm. T(P)Pl m Q n R T(R)S FIGURE 5.2: 5.4 Rotations 5.4.1 First, T−1 ◦Rotφ ◦ T (C) = T−1 ◦Rotφ ◦ T (x, y) = T−1 ◦Rotφ(0, 0) = T−1(0, 0) = (x, y) = C Suppose T−1◦Rotφ◦T fixed another point P . Then, Rotφ◦T (P ) = T (P ), which implies that T (P ) = (0, 0), or P = T−1(0, 0) = (x, y) = C. Thus, T−1 ◦ Rotφ ◦ T must be a rotation. What is the angle for this rotation? Consider a line l through C that is parallel to the x-axis. Then, T will map l to the x− axis and Rotφ will map the x-axis to a 30 Student’s Guide for Exploring Geometry, Second Edition line m making an angle of φ with the x-axis. Then, T−1 will preserve this angle, mapping m to a line making an angle of φ with l. Thus, the rotation angle for T−1 ◦Rotφ ◦ T is φ. 5.4.3 A book on flowers or diatoms (algae) would be a good place to start. 5.4.5 By the preceding exercise, the invariant line must pass through the center of rotation. Let A be a point on the invariant line. Then, RO(A) lies on ←→ OA and OA ∼= ORO(A). Either A and RO(A) are on the same side of O or are on opposite sides. If they are on the same side, then A = RO(A), and the rotation is the identity, which is ruled out. If they are on opposite sides, then the rotation is 180 degrees. If the rotation is 180 degrees, then for every point A 6= O we have that A, O, and RO(A) are collinear, which means that the line ←→ OA is invariant. 5.4.7 Suppose R(m) is parallel to m. Let O be the center of rota- tion. If m passes through O, then R(m) also passes through O, which contradicts the lines being parallel. So, we assume m does not pass through O. Let A,B be two points on m and construct AR(A). If the angle ∠BAO is a right angle, then ∠R(B)R(A)R(O) is also a right angle, and the rotation angle must be 180 degrees, which is impossible. So, WLOG we assume that ∠BAO is less than a right angle. Drop a perpendicular from O to m at D. Since R is an isometry , then ∆DAO ∼= ∆R(D)R(A)O and the vertical angles at O are congruent, which means A, O, and R(A) are collinear. This would mean the rotation angle is 180 degrees, which is impossible. So, m and R(m) intersect at a point P . Transformational Geometry 31 Drop perpendiculars from O to m and R(m) at A and B. Note that A 6= P , as if A = P then m and R(m) would coincide. Thus, OBPA is a quadrilateral. Let θ be the rotation angle. Since the sum of the angles in a quadrilat- eral is 360 degrees, then ∠BPA must have measure 180− θ. But, then the vertical angle at P has measure 180−m∠BPA = θ. 5.4.9 Let R1 and R2 be two rotations about P . Let l and m be the lines of reflection for R1. By Theorem 5.15 we can choose m as a defining line of reflection for R2 and there is a unique line n such that R2 = rn ◦ rm. Then, R2 ◦R1 = rn ◦ rm ◦ rm ◦ rl = rn ◦ rl, which is again a rotation about P . 5.4.11 If r1 ◦ R = r2, then R = r1 ◦ r2. Thus, the lines for r1 and r2 must intersect. If they intersected at a point other than the center of rotation for R, then R would fix more than one point, which is impossible. 5.4.13 Let m = ←→ AB. Then, we can choose lines l and n such that HA = rm ◦ rl and HB = rn ◦ rm. Note that l and n are both perpen- dicular to m and thus parallel. Then, HB ◦ HA = rn ◦ rl, which is a translation parallel to l. 5.5 Project 7 -Quilts and Transformations This project is another great opportunity for the future teachers in the class to develop similar projects for use in their own teaching. One idea to incorporate into a high school version of the project is to bring into the class the cultural and historical aspects of quilting. 5.6 Glide Reflections 5.6.1 As with translations, it will be hard to find a perfect example of a glide symmetry in nature. But, the are many plants whose branches alternate in a glide fashion. 5.6.3 Suppose m is invariant. Then, the glide reflection can be written as G = TAB ◦ rl = rl ◦ TAB. If G(G(m)) = m, then (TAB ◦ rl) ◦ (rl ◦ TAB)(m) = T2AB(m) = m. So, m must be parallel or equal C H A P T E R 6 Symmetry This chapter is quite algebraic in nature—focusing on the different discrete symmetry groups that arise for frieze patterns and wallpaper patterns. SOLUTIONS TO EXERCISES IN CHAPTER 6 6.1 Finite Plane Symmetry Groups 6.1.1 Flowers and diatoms make good examples. 6.1.3 The dihedral group of order 5. This has order 10, and there can be at most 10 symmetries. (Use an argument like that used in the preceding exercise) 6.1.5 By the previous exercise there are at most 2n symmetries. Also, by the work done in Section 5.4 we know there are n rotations, generated by a rotation of 360 n , that will be symmetries. Let r be a reflection across a perpendicular bisector of a side. This will be a re- flection, as will all n compositions of this reflection with the n rotations. This gives 2n different symmetries 6.1.7 The number of symmetries is 2n. The only symmetries that fix a side are the identity and a reflection across the perpendicular bisector of that side. The side can move to n different sides. Thus, the stated product is 2n as claimed. 6.2 Frieze Groups 6.2.1 Since γ2 = τ , then < τ, γ,H > is contained in < γ,H >. Also, it is clear that< γ,H > is contained in< τ, γ,H >. Thus,< τ, γ,H >=< γ,H >. 35 36 Student’s Guide for Exploring Geometry, Second Edition 6.2.3 Let ru and ru′ be two reflections across lines perpendicular to m. Then, the composition ru ◦ ru′ must be a translation, as these lines will be parallel. Thus, ru ◦ ru′ = T k for some k, and ru′ = ru ◦T k. 6.2.5 Consider g2. This must be a translation, so g2 = Tkv for some k where Tv is the fundamental translation. Then, g = Tk 2 v ◦rm, where m is the midline. Suppose k 2 is an integer, say k 2 = j. Then, since T(v−jv) is in the group, we have T(v−jv) ◦ g = T(v−jv) ◦ Tk 2 v ◦ rm = Tv ◦ rm is in the group. Otherwise, k 2 = j + 1 2 for some integer j. We can find T−jv in the group such that T−jv ◦ g = T v 2 ◦ rm is in the group. 6.2.7 The composition rv ◦ ru must be a translation. Also, if rv ◦ ru(A) = rv(A) = C, then the translation vector must be ~AC. But, the length of AC is twice that of AB. So, we get that 2 ~AB = k′v for some k′. Now, either k′ is even or it is odd. The result follows. 6.2.9 From Table 5.3 we know that τ ◦ H or H ◦ τ is either a translation or a rotation, so it must be either τk for some k or HA for A on m. Thus, any composition of products of τ and H can be reduced ultimately to a simple translation or half-turn, or to some τ j ◦HB or HB ◦τj , which are both half-turns. Thus, the subgroup generated by τ and H cannot contain rm or ru or γ and none of < τ, rm > or < τ, ru > or < τ, rm > can be subgroups of < τ,H >. 6.2.11 The compositions τk◦rm or rm◦τk generate glide reflections with glide vectors kv. The composition of τ with such glide reflections generates other glide reflections with glide vectors (k+j)v. The compo- sition of rm with a glide in the direction ofm will generate a translation. Thus, compositions of the three types of symmetries—glides, rm, and τk—will only generate symmetries within those types. Thus, < τ, γ > cannot be a subgroup of < τ, rm >, since γ has translation vector of v2 which cannot be generated in < τ, rm >. Also, neither < τ, ru > nor < τ,H > can be subgroups of < τ, rm >. 6.2.13 First Row: < τ >, < τ, γ >. Second Row: < τ, γ,H >, < τ, ru >. Third Row: < τ, rm, H >, < τ,H >. Last Row: < τ, rm >. 6.3 Wallpaper Groups 6.3.1 The first is rectangular, the second rhomboidal, and the third is square. 6.3.3 The translation determined by f2 will be in the same direc- tion as T , so we do not find two independent directions of translation. 6.3.5 The lattice for G will be invariant under rotations about C H A P T E R 7 Hyperbolic Geometry The discovery of non-Euclidean geometry is one of the most impor- tant events in the history of mathematics. Much more time could be spent on telling this story, and, in particular, the history of the col- orful figures who co-discovered this geometry. The book by Boyer and Merzbach, and the University of St. Andrews web site, both listed in the bibliography of the text, are excellent references for a deeper look at this history. Greenberg’s book is also an excellent reference. SOLUTIONS TO EXERCISES IN CHAPTER 7 In section 7.2 we see for the first time the relevance of our earlier dis- cussion of models in Chapter 1. The change of axioms in Chapter 7 (replacing Euclid’s fifth postulate with the hyperbolic parallel postu- late) requires a change of models. As you work through this section, it is important to recall that, in an axiomatic system, it not impor- tant what the terms actually mean; the only thing that matters is the relationships between the terms. We introduce two different models at this point to help you recog- nize the abstraction that lies behind the concrete expression of points and lines in theses models. 7.2.2 Mini-Project: The Klein Model It may be helpful to do the constructions (lines, etc) of the Klein model on paper as you read through the material. 7.2.1 Use the properties of Euclidean segments. 39 40 Student’s Guide for Exploring Geometry, Second Edition 7.2.3 The special case is where the lines intersect at a boundary point of the Klein disk. Otherwise, use the line connecting the poles of the two parallels to construct a common perpendicular. 7.3 Basic Results in Hyperbolic Geometry In this section it is important to note the distinction between points at infinity and regular points. Omega triangles share some properties of regular triangles, like congruence theorems and Pasch-like properties, but are not regular trianglesthus necessitating the theorems found in this section. 7.3.1 Use the interpretation of limiting parallels in the Klein model. 7.3.3 First, if m is a limiting parallel to l through a point P , then rl(m) cannot intersect l, as if it did, then r2l (m) = m would also inter- sect l. Now, drop a perpendicular from rl(P ) to l at Q, and consider the angle made by Q, rl(P ), and the omega point of rl(m). If there were another limiting parallel (n) to l through rl(P ) that lies within this angle, then by reflecting back by rl we would get a limiting parallel rl(n) that lies within the angle made by Q, P and the omega point of l, which is impossible. Thus, rl(m) must be limiting parallel to l and reflection maps omega points to omega points, as rl maps limiting par- allels to l to other limiting parallels. Also, it must fix the omega point, as it maps limiting parallels on one side of the perpendicular dropped to l to limiting parallels on that same side. 7.3.5 Let P be the center of rotation and let l be a line through P with the given omega point Ω. (Such a line must exist as Ω must correspond to a limiting parallel line m, and there is always a limiting parallel to m through a given point P ) Then, we can write R = rn ◦ rl for another line n passing through P . But, since rl fixes Ω, and R does as well, then, rn must fix Ω. But, if n and l are not coincident, then n is not limiting parallel to l and thus cannot have the same omega points as l. By the previous exercise, rn could not fix Ω. Thus, it must be the case that n and l are coincident and r is the identity. 7.3.7 Let PQΩ be an omega triangle and let R be a point interior to the triangle. Drop a perpendicular from Q to ←→ PΩ at S. Then, either R is interior to triangle QPS, or it is on ←→ QS, or it is interior to ∠QSΩ. If it is interior to ∆QPS it intersects ←→ PΩ by Pasch’s axiom for triangles. If it is on ←→ QS it obviously intersects ←→ PΩ. If it is interior to ∠QSΩ, it intersects ←→ PΩ by the definition of limiting parallels. Hyperbolic Geometry 41 P Q R T Ω S FIGURE 7.1: 7.3.9 Let l be the line passing through R. Then, either l passes within Omega triangle PRΩ or it passes within QRΩ. In either case, we know by Theorem 7.5 that l must intersect the opposite side, i.e. it must intersect PΩ or QΩ. 7.3.11 Suppose we had another segment P ′Q′ with P ′Q′ ∼= PQ and let l′ be a perpendicular to P ′Q′ at Q′. Let ←−→ P ′R′ be a limiting parallel to l′ at P ′. Then, by Theorem 7.8, we know that ∠QPR ∼= ∠Q′P ′R′ and thus, the definition of this angle only depends on h, the length of PQ. 7.3.13 Suppose a(h) = a(h′) with h 6= h′. We can assume that h < h′. But, then the previous exercise would imply that a(h) > a(h′). Thus, if a(h) = a(h′) then h = h′. 7.4 Project 10 - The Saccheri Quadrilateral As you do the computer construction of the Saccheri Quadrilateral, you may experience a flip of orientation for your construction when moving the quad about the screen. The construction depends on the orientation of the intersections of circles and these may switch as the quad is moved. A construction of the Saccheri quad that does not have this unfortunate behavior was searched for unsuccessfully by the author. A nice challenge problem would be to see if you can come up with a better construction. If you can, the author would love to hear about it! 7.5 Lambert Quadrilaterals and Triangles 7.5.1 Referring to Figure 7.6, we know ∆ACB and ∆ACE are congru- ent by SAS. Thus, ∠ACB ∼= ∠ECA. Since ∠ACD ∼= ∠FCA, and both are right angles, then ∠BCD ∼= ∠FCE. Then, ∆BCD and ∆FCE are congruent by SAS. We conclude that BD ∼= FE and the angle at E is a right angle. 44 Student’s Guide for Exploring Geometry, Second Edition A C B E F I H E’’ A’’ K J n FIGURE 7.4: 7.6.3 This question can be argued both ways. If we could make incredibly precise measurements of a triangle, then we could possibly measure the angle sum to be less than 180. However, since the universe is so vast, we would have to have an incredibly large triangle to measure, or incredibly good instruments. Also, we could never be sure of errors in the measurement overwhelming the actual differential between the angle sum and 180. 7.7 Project 11 - Tiling the Hyperbolic Plane A nice artisitic example of hyperbolic tilings can be found in M. C. Es- chers Circle Limit figures. Consult Doris Schattschnieders book M. C. Escher, Visions of Symmetry for more information about these tilings. C H A P T E R 8 Elliptic Geometry Hyperbolic and Elliptic geometry are the fundamental examples of non-Euclidean geometry. The connection between the three geometries —Euclidean, Hyperbolic, and Elliptic —and the three possible parallel properties —1, > 1, or 0 parallels through a point to a given line —is one of the most interesting and thought-provoking ideas in geometry. SOLUTIONS TO EXERCISES IN CHAPTER 8 8.2 Perpendiculars and Poles in Elliptic Geometry 8.2.1 Referring to Figure 8.1, we know ∆ACB and ∆ACE are congru- ent by SAS. Thus, ∠ACB ∼= ∠ECA. Since ∠ACD ∼= ∠FCA, and both are right angles, then ∠BCD ∼= ∠FCE. Then, ∆BCD and ∆FCE are congruent by SAS. We conclude that BD ∼= FE and the angle at E is a right angle. 45 46 Student’s Guide for Exploring Geometry, Second Edition FIGURE 8.1: 8.2.3 By Exercise 8.2.2 we can create a Saccheri quadrilateral from the Lambert quadrilateral. Since the summit angles in the Saccheri quadrilateral will equal the fourth (non-right) angle in the Lambert quadrilateral, the result follows. 8.2.5 Since there are two perpendiculars from n that meet at O, then by the work in this section, all perpendiculars to n meet at O and O is the polar point. 8.3 Project 12 - Models of Elliptic Geometry All of the projects in Chapter 8 use the software program Geom- etry Explorer. If you are not already using this program, go to http://www.gac.edu/ hvidsten/geom-text for instructions on how to download and use this software. 8.4 Basic Results in Elliptic Geometry 8.4.1 Consider the line OR. This line must intersect l at some point Q. Also, it is perpendicular to l at Q, for if it were not perpendicular, then at Q we could construct the perpendicular which must pass through the polar point O. But, then we would have two lines through O and C H A P T E R 9 Projective Geometry Projective geometry is the natural culmination of a study of geome- try that proceeds from Euclidean to non-Euclidean geometries. While much of the material in this chapter is much more opaque than in pre- vious chapters, the beauty and elegance realized from abstraction can (hopefully) shine through. SOLUTIONS TO EXERCISES IN CHAPTER 9 9.2 Project 14 - Perspective and Projection 9.3 Foundations of Projective Geometry 9.3.1 Suppose the lines are l and m. If they both pass through points P and Q, then by axiom A1 the two lines must be the same line. 9.3.3 By the previous exercise, we know that an Affine geometry must have at least four distinct points, say P , Q, R, and S. For each pair of points, we have a unique line, by axiom A1. A quick check of every pairing of one of the six lines with a point not on that line shows that axiom A2 is satisifed. Thus, for these four points and six lines, all three axioms are satisfied. 9.3.5 Let P be a point. By axiom P3 there must be two other points Q and R such that P , Q, and R are non-collinear. Then, l = ←→ PQ and m = ←→ PR are two distinct lines. Now, Q and R define a line n by axiom P1, and this line is different than l or m. By axiom P4, n must have a third point, say S. Then, ←→ PS exists and is distinct from both ←→ PQ and←→ PR. 9.3.7 It is clear that the example satisfies P3 and P4. Checking all pairs of points and lines will suffice for P1 and P2. 49 50 Student’s Guide for Exploring Geometry, Second Edition 9.3.9 Since Axioms P(n)1 and P(n)2 exactly match P1 and P2, then P1 and P2 are satisfied. P(n)3 guarantees the existence of three non-collinear points, so Axiom P3 is satisfied. Let P , Q, R, and S be the four points guaranteed by Axiom P(n)3. Then, ←→ PQ, ←→ PR , ←→ PS, ←→ QR, ←→ QS and ←→ RS must be distinct lines. Suppose there were a line m with only two points A and B. Since m intersects←→ PQ and ←→ PR, then either one of A or B is P , or A is on one of the lines and B is on the other. Suppose A = P . Now m intersects ←→ QR and ←→ QS at a point other than P , so B = Q. Then, m = ←→ PQ. But, we know that ←→ PQ must have a third point as it must intersect ←→ RS at a point other than P or Q. So, A 6= P . Suppose A is on ←→ PQ and B is on ←→ PR. Since m must intersect ←→ QR at some point, then A is on ←→ QR or B is on ←→ QR. Then, A = Q and B = R. Then, m = ←→ QR. But, we know that ←→ QR must have a third point as it must intersect ←→ PS at a point other than Q or R. So, it cannot be the case that A is on ←→ PQ and B is on ←→ PR. We conclude that there cannot be a line with only two points. A similar proof (simpler) shows that there cannot be a line with only one point. 9.3.11 Since Moulton lines are built from pieces of Euclidean lines, then P3 and P4 follow immediately. For P2, any pair of Moulton lines that are standard Euclidean lines will either intersect in the regular Euclidean plane or will intersect at a point at infinity. For other Moulton lines, if the two lines are defined by different slopes m1 and m2, then consider the regular Euclidean lines of those slopes. Either those lines will intersect in the region x ≤ 0 or x > 0. Since the Moulton “pieces” defining the line have the same slope, then the Moulton lines with slopes m1 and m2 will also intersect where the Euclidean lines did. The case left is the case where the Moulton lines have the same slope. In this case, they intersect at the point at infinity. So, P2 holds. For P1, let A and B be two point in the plane. If A and B are both in the left half-plane (x ≤ 0) we just find the line y = m′x+ b joining them and define y = 2m′x + b in the right half-plane (x > 0) to get the Moulton line incident on A and B. A similar construction works if A and B are both in the right half-plane (x > 0). If A is in the left half-plane and B in the right, but A is above or at the same level as B, then the slope of the line joining the two Projective Geometry 51 points is negative or zero, and the Moulton line is just the Euclidean line incident on the points. The only case left is where A is in the left half-plane and B in the right, and A is below B. It is clear from the hint that the fraction slope( ←→ BY ) slope( ←→ AY ) range from 0 to infinity as Y ranges from Y1 to Y0. Thus, by continuity there is a Y value where this fraction is equal to 2. Let m = slope( ←→ BY ). This will generate the unique Moulton line from A to B. 9.4 Transformations in Projective Geometry and Pappus’s Theorem 9.4.1 Closure: Let π1 and π2 be projectivities. Then, each is con- structed from a composition of perspectivities. The composition of π1 with π2 will again be a composition of perspectivities and thus is a projectivity. Associativity: This is automatic (inherited) from function com- position. Identity: A perspectivity from a pencil of points back to itself (or pencil of lines back to itself) is allowed, and thus the identity is a projectivity. Inverses: Given any basic perspectivity from, say a pencil of points on l to a pencil of points on l′, will have its inverse be the revers map- ping from the pencil at l′ to the pencil at l. Since a projectivity is the composition of perspectivities, then the inverse will be the composition of inverse perspectivities (in reverse order). 9.4.3 Dual to Theorem 9.6: Let a,b,c be three distinct lines incident on point P and a′, b′, and c′ be three distinct lines incident on point P ′, with P 6= P ′. Then, there is a projectivity taking a,b,c to a′,b′,c′. Dual to Corollary 9.7: Let a,b,c and a′, b′, and c′ be two sets of distinct lines incident on point P . Then, there is a projectivity taking a,b,c to a′,b′,c′. 9.4.5 Let π1 and π2 be projectivities having the same value on points A, B, and C on l. Then π1 ◦π−12 will leave the points A, B, and C on l invariant. By P7, π1 ◦ π−12 = id, or π1 = π2. 9.4.7 Consider a projectivity of pencils of points from line l to line l′. Theorem 9.6 gives us a construction for a projectivity as the composition of two perspectivities if l 6= l′. If l = l′, then let A, B, and C be projectively related to A′, B′, and C ′ on l′. Let m be a line 54 Student’s Guide for Exploring Geometry, Second Edition 9.6 Project 15 - Ratios and Harmonics 9.7 Harmonic Sets 9.7.1 For the quadrangle, ABCD let E = AB · CD, F = AC · BD, and G = AD ·BC. A B C D E F G FIGURE 9.2: Consider a line through two of the diagonal points, say the line EF . Suppose A was on EF . Then, F is on AE = AB. But, this implies that D is on AB, which contradicts the fact that A, B, and D must be non- collinear. A similar argument will show that EF cannot intersect any of the other three points B, C, or D. Similarly, the result holds for EG and FG. 9.7.3 Let O be the intersection of FF ′ and HH ′. Then, the per- spectivity from O maps EF to E′F ′, thus maps l to l′. (See Fig. 9.3) The perspectivity maps E,F , and H to E′, F ′, and H ′. Let I ′′ be th image of I under this perspectivity. Then, by Corollary 9.31 we have that H(E′, F ′;H ′, I ′′). But, by Theorem 9.29 we know that the fourth point in a harmonic set is uniquely defined, based on the first three points. Thus, I ′ = I ′′. Projective Geometry 55 E F H I F’ I’ O H’ FIGURE 9.3: 9.7.5 Let A,B,C, and D have parametric homogeneous coordinates (α1, α2), (β1, β2), (γ1, γ2) and (δ1, δ2). We can assume that the homo- geneous parameters for the four points are equivalently (α, 1), (β, 1), (γ, 1) and (δ, 1), where α = α1 α2 , β = β1 β2 , γ = γ1 γ2 , and δ = δ1 δ2 . From Theorem 9.33 we have R(A,B;C,D) = dd(A,C) dd(B,D) dd(B,C) dd(A,D) = (γ − α)(δ − β) (γ − β)(δ − α) If we do a replacement using α = α1 α2 , β = β1 β2 , γ = γ1 γ2 , and δ = δ1 δ2 . we get R(A,B;C,D) = (γ1γ2 − α1 α2 )( δ1δ2 − β1 β2 ) (γ1γ2 − β1 β2 )( δ1δ2 − α1 α2 ) = 1 γ2α2 (γ1α2 − γ2α1) 1 δ2β2 (δ1β2 − δ2β1) 1 γ2β2 (γ1β2 − γ2β1) 1 δ2α2 (δ1α2 − δ2α1) = (γ1α2 − γ2α1)(δ1β2 − δ2β1) (γ1β2 − γ2β1)(δ1α2 − δ2α1) 9.7.7 Use the coordinates for A, B, C, and D as set up in Theo- rem 9.33. Let D′ have coordinates (δ′, 1). Then, R(A,B;C,D) = (γ − α)(δ − β) (γ − β)(δ − α) = R(A,B;C,D′) = (γ − α)(δ′ − β) (γ − β)(δ′ − α) 56 Student’s Guide for Exploring Geometry, Second Edition Thus, δ′ − β δ′ − α = δ − β δ − α . Or, (δ′ − β)(δ − α) = (δ − β)(δ′ − α). So, −δ′α− δβ = −δ′β − δα. Thus, δ′(α− β) = δ(α− β) and δ′ = δ. 9.7.9 Let the parametric coordinates of A, B, C, D, and E be (α, 1), (β, 1), (γ, 1), (δ, 1), and (ε, 1). Then, R(A,B;C,D)R(A,B;D,E) = (γ − α)(δ − β) (γ − β)(δ − α) (δ − α)(ε− β) (δ − β)(ε− α) = (γ − α)(ε− β) (γ − β)(ε− α) = R(A,B;C,E) 9.7.11 If the coordinates for A and C are α and γ then the coor- dinates for B would be β = α+ γ 2 . Then, for a fourth point D with coordinate δ we have R(A,C;B,D) = (β − α)(δ − γ) (β − γ)(δ − α) = (γ−α2 )(δ − γ) (α−γ2 )(δ − α) = − δ − γ δ − α If D has coordinates of the point at infinity, then the cross-ratio will be equal to −1 and the four points will form a harmonic set. 9.8 Conics and Coordinates 9.8.1 According to Theorem 9.12 the projectivity defining the conic is equivalent to the composition of two perspectivities. The proof is by contradiction. Suppose that the line AB corresponded to itself under the projectivity. Interpret Lemma 9.10 as a statement about the pencils of points on l, m, and n. The dual to this lemma would be: ”Given three pencils of lines at P , R, and Q with P 6= Q, suppose there is a projectivity taking the pencil at P to the pencil at Q. If l = PQ is invariant under the projectivity, then, then the pencil at P is perspective to the pencil at Q.” In the case of the exercise, the line l would be AB. If AB it is invariant, then by the lemma the projectivity is a perspectivity. This implies that the point conic is singular. 9.8.3 Suppose the conic is defined by pencils of points at A and C H A P T E R 10 Fractal Geometry Much of the material in this chapter is at an advanced level, especially the sections on contraction mappings and fractal dimension—Sections 10.5 and 10.6. But this abstraction can be made quite concrete by the computer explorations developed in the chapter. In fact, the computer projects are the only way to really understand these geometric objects on an intuitive level. SOLUTIONS TO EXERCISES IN CHAPTER 10 10.3 Similarity Dimension The notion of dimension of a fractal is very hard to make precise. In this section we present one simple way to define dimension, but there are also other ways to define dimension as well, each useful for a particular purpose and all agreeing with integer dimension, but not necessarily with each other. 10.3.1 Theorem 2.27 guarantees that the sides of the new triangles are parallel to the original sides. Then, we can use SAS congruence to achieve the result. 10.3.3 At each successive stage of the construction, 8 new squares are created, each of area 1 9 the area of the squares at the previous stage. Thus, the pattern for the total area of each successive stage of 59 60 Student’s Guide for Exploring Geometry, Second Edition the construction is l = 1− 1 9 − 8 81 − 64 93 − . . . = 1− 1 9 ∞∑ k=0 ( 8 9 )k = 1− 1 9 1 1− 8 9 = 1− 1 = 0 Thus, the area of the final figure is 0. 10.3.5 The similarity dimension would be log(4) log(3) . 10.3.7 Split a cube into 27 sub-cubes, as in the Menger sponge construction, and then remove all cubes except the eight corner cubes and the central cube. Do this recursively. The resulting fractal will have similarity dimension log(9) log(3) , which is exactly 2. 10.4 Project 16 - An Endlessly Beautiful Snowflake If students want a challenge, you could think of other templates based on a simple segment, generalizing the Koch template and the Hat tem- plate from exercise 10.4.4. 10.6 Fractal Dimension Sections 10.5 and 10.6 are quite “thick” mathematically. To get some sense of the Hausdorff metric, you can compute it for some simple pairs of compact sets. For example, two triangles in different positions. Ample practice with examples will help you get a feel for the mini-max approach to the metric and this will also help you be successful with the homework exercises. 10.6.1 A function f is continuous if for each ε > 0 we can find δ > 0 such that |f(x) − f(y)| < ε when 0 < |x − y| < δ. Let S be a contraction mapping with contraction factor 0 ≤ c < 1. Then, given ε, let δ = ε (if c = 0) and δ = ε c (if c > 0). If c = 0 we have 0 = |S(x)− S(y)| ≤ |x− y| < δ = ε. If c > 0, we have |S(x)− S(y)| ≤ c|x− y| < c εc = ε. 10.6.3 Property (2): Since dH(A,A) = d(A,A), and since d(A,A) = max{d(x,A)|x ∈ A}, then we need to show d(x,A) = 0. Fractal Geometry 61 But, d(x,A) = min{d(x, y)|y ∈ A}, and this minimum clearly occurs when x = y; that is, when the distance is 0. Property (3): If A 6= B then we can always find a point x in A that is not in B. Then, d(x,B) = min{d(x, y)|y ∈ B} must be greater than 0. This implies that d(A,B) = max{d(x,B)|x ∈ A} is also greater than 0. 10.6.5 We know that d(A,C ∪D) = max{d(x,C ∪D)|x ∈ A} = max{min{d(x, y)|x ∈ A and y ∈ C or D}} = max{min{min{d(x, y)|x ∈ Ay ∈ C},min{d(x, y)|x ∈ A, y ∈ D}}} = max{min{d(x,C), d(x,D)}|x ∈ A} The last expression is clearly less than or equal to max{d(x,C)|x ∈ A} = d(A,C) and also less than or equal to max{d(x,D)|x ∈ A} = d(A,D). 10.6.7 There are three contraction mappings which are used to construct Sierpinski’s triangle. Each of them has contraction scale fac- tor of 1 2 . Thus, we want (12)D+(12)D+(12)D = 1, or 3(12)D = 1, Solving for D we get D = log(3) log(2) . 10.7 Project 17 - IFS Ferns Do not worry too much about getting exactly the same numbers for the scaling factor and the rotations that define the fern. The important idea is that you get the right types of transformations (in the correct order of evaluation) needed to build the fern image. For exercise 10.7.5 it may be helpful to copy out one piece of the image and then rotate and move it so it covers the other pieces, thus generating the transformations needed. 10.9 Grammars and Productions This section will be very different from anything you have done before, except for those who have had some computer science courses. The connection between re-writing and axiomatic systems is a deep one. One could view a theorem as essentially a re-writing of various symbols and terms used to initialize a set of axioms. Also, turtle geometry is a very concrete way to view re-writing and so we have a nice concrete realization of an abstract idea. 64 Student’s Guide for Exploring Geometry, Second Edition Then, we construct a square on side a and a square on side b. The purpose of doing this is to create two regions whose total area is a2+b2. Clever huh? Constructing the squares involved several rotations, but was otherwise straightforward. A CB a b D E F G FIGURE A.2: The next construction was a bit tricky. We define a translation from B to A and translate point C to get point H. Then, we connect H to D and H to G, resulting in two right triangles. In part II, we will prove that both of these right triangles are congruent to the original right triangle. A CB a b D E F G H FIGURE A.3: Next, we hide segment BC and create segments BH and HC. This is so that we have well-defined triangle sides for the next step - rotating right triangle ADH 90 degrees about its top vertex, and right triangle HGC -90 degrees about its top vertex. Sample Lab Report 65 A CB a D E F G H FIGURE A.4: Part II: We will now prove that this construction yields a square (on DH) of side length c, and thus, since the area of this square is clearly equal to the sum of the areas of the original two squares, we have a2 + b2 = c2, and our proof would be complete. By SAS, triangle HCB must be congruent to the original right triangle, and thus its hypotenuse must be c. Also, by SAS, triangle DAH is also congruent to the original triangle, and so its hypotenuse is also c. Then, angles AHD and CHG(= ADH) must sum to 90 degrees, and the angle DHG is a right angle. Thus, we have shown that the construction yields a square on DH of side length c, and our proof is complete. Conclusion: This was a very elegant proof of the Pythagorean Theorem. In re- searching the topic of proofs of the Pythagorean Theorem, we discov- ered that over 300 proofs of this theorem have been discovered. Elisha Scott Loomis, a mathematics teacher from Ohio, compiled many of these proofs into a book titled The Pythagorean Proposition, published in 1928. This tidbit of historical lore was gleaned from the Ask Dr. Math website (http://mathforum.org/library/drmath/view/62539.html). It seems that people cannot get enough of proofs of the Pythagorean Theorem.