A SURVEY OF CLASSICAL AND MODERN GEOMETRIES WITH COMPUTER ACTIVITIES 1ST EDITION BY ARTHUR BARAGAR by QuentinAxel

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A SURVEY OF CLASSICAL AND MODERN GEOMETRIES WITH COMPUTER ACTIVITIES 1ST EDITION BY ARTHUR BARAGAR SOLUTION MANUAL

Contents 1 Euclidean Geometry 1 1.1 The Pythagorean Theorem ................................................................ 3 1.2 The Axioms of Euclidean Geometry ................................................. 5 1.3 SSS, SAS, and ASA ........................................................................ 7 1.4 Parallel Lines ..................................................................................... 11 1.5 Pons Asinorum .................................................................................. 12 1.6 The Star Trek Lemma ...................................................................... 12 1.7 Similar Triangles ............................................................................... 18 1.8 Power of the Point ........................................................................... 24 1.9 The Medians and Centroid .............................................................. 33 1.10 The Incircle, Excircles, and the Law of Cosines ............................ 35 1.11 The Circumcircle and Law of Sines ................................................ 42 1.12 The Euler Line .................................................................................. 48 1.13 The Nine Point Circle ..................................................................... 50 1.14 Pedal Triangles and the Simson Line ............................................. 57 1.15 Menelaus and Ceva ........................................................................... 67 2 Geometry in Greek Astronomy 75 2.1 The Relative Size of the Moon and Sun ......................................... 75 2.2 The Diameter of the Earth .............................................................. 76 3 Constructions Using a Compass and Straightedge 81 3.1 The Rules ........................................................................................... 81 3.2 Some Examples ................................................................................. 81 3.3 Basic Results ..................................................................................... 82 3.4 The Algebra of Constructible Lengths ........................................... 92 3.5 The Regular Pentagon ...................................................................... 94 3.6 Other Constructible Figures .......................................................... 102

3.7

Trisecting an Arbitrary Angle ....................................................... 105

4 Geometer’s Sketchpad 111 4.1 The Rules of Constructions ........................................................... 111 4.2 Lemmas and Theorems .................................................................. 111 4.3 Archimedes’ Trisection Algorithm ................................................ 114 v vi

CONTENTS 4.4 4.5 4.6

Veriﬁcation of Theorems ............................................................... 114 Sophisticated Results ..................................................................... 117 Parabola Paper ............................................................................... 120

5 Higher Dimensional Objects 125 5.1 The Platonic Solids ......................................................................... 125 5.2 The Duality of Platonic Solids .................................................... 127 5.3 The Euler Characteristic ................................................................ 127 5.4 Semiregular Polyhedra .................................................................. 127 5.5 A Partial Categorization of Semiregular Polyhedra ................... 130 5.6 Four-Dimensional Objects ............................................................. 138 6 Hyperbolic Geometry 143 6.1 Models ............................................................................................. 143 6.2 Results from Neutral Geometry .................................................... 143 6.3 The Congruence of Similar Triangles........................................... 145 6.4 Parallel and Ultraparallel Lines .................................................... 145 6.5 Singly Asymptotic Triangles.......................................................... 146 6.6 Doubly and Triply Asymptotic Triangles .................................... 146 6.7 The Area of Asymptotic Triangles ............................................... 147 7 The Poincaré Models of Hyperbolic Geometry 149 7.1 The Poincaré Upper Half Plane Model ........................................ 149 7.2 Vertical (Euclidean) Lines ............................................................. 149 7.3 Isometries ........................................................................................ 149 7.4 Inversion in the Circle ................................................................... 150 7.5 Inversion in Euclidean Geometry ................................................. 161 7.6 Fractional Linear Transformations .............................................. 164 7.7 The Cross Ratio .............................................................................. 169 7.8 Translations .................................................................................... 173 7.9 Rotations ......................................................................................... 177 7.10 Reﬂections ....................................................................................... 181 7.11 Lengths ............................................................................................ 185 7.12 The Axioms of Hyperbolic Geometry ........................................... 186 7.13 The Area of Triangles .................................................................... 186 7.14 The Poincaré Disc Model .............................................................. 188 7.15 Circles and Horocycles................................................................... 190 7.16 Hyperbolic Trigonometry .............................................................. 195 7.17 The Angle of Parallelism ............................................................... 207 7.18 Curvature......................................................................................... 209

8 Tilings and Lattices 211 8.1 Regular Tilings ................................................................................ 211 8.2 Semiregular Tilings ........................................................................ 211 8.3 Lattices and Fundamental Domains ............................................. 212 8.4 Tilings in Hyperbolic Space .......................................................... 212

CONTENTS 8.5

vii

Tilings in Art ................................................................................... 220

9 Foundations 221 9.1 Theories ........................................................................................... 221 9.2 The Real Line .................................................................................. 221 9.3 The Plane ......................................................................................... 221 9.4 Line Segments and Lines ............................................................... 221 9.5 Separation Axioms .......................................................................... 222 9.6 Circles ............................................................................................... 225 9.7 Isometries and Congruence ........................................................... 226 9.8 The Parallel Postulate .................................................................... 227 9.9 Similar Triangles ............................................................................. 227 10 Spherical Geometry 229 10.1 The Area of Triangles..................................................................... 229 10.2 The Geometry of Right Triangles ................................................. 231 10.3 The Geometry of Spherical Triangles ........................................... 232 10.4 Menelaus’ Theorem ........................................................................ 234 10.5 Heron’s Formula ............................................................................. 241 10.6 Tilings of the Sphere ...................................................................... 245 10.7 The Axioms ...................................................................................... 247 10.8 Elliptic Geometry ............................................................................ 247 11 Projective Geometry 249 11.1 Moving a Line to Inﬁnity ............................................................... 249 11.2 Pascal’s Theorem ............................................................................ 250 11.3 Projective Coordinates ................................................................... 250 11.4 Duality .............................................................................................. 255 11.5 Dual Conics and Brianchon’s Theorem ........................................ 257 11.6 Areal Coordinates ........................................................................... 258 12 The Pseudosphere in Lorentz Space 265 12.1 The Sphere as a Foil....................................................................... 265 12.2 The Pseudosphere .......................................................................... 272 12.3 Angles and the Lorentz Cross Product ........................................ 280 12.4 A Diﬀerent Perspective .................................................................. 284 12.5 The Beltrami-Klein Model.............................................................. 286 12.6 Menelaus’ Theorem ........................................................................ 286

Chapter 1

Euclidean Geometry Exercise 1.1. Suppose that at the spring solstice the angle between a star, the Earth, and the sun measures 79.1◦, and at the fall solstice the angle measures 100.8◦. How far away is the star from the Earth? (The Earth is 93 million miles from the sun.) The solution is in the text. Exercise 1.2. The numbers in Exercise 1.1 were cooked up. From your general knowledge, are these numbers reasonable? Explain. Solution. In Exercise 1.1, we found that the star is 100,000,000,000 miles away. Let us convert this ﬁgure to light years. A light year is how far light travels in a year. Since light travels at a speed of 186,000 miles per second, one light year is the same as 1 lightyear = 186, 000

miles 3600 sec 24 hrs 365 day yr sec hr day yr

= 5.87 × 10

miles.

Thus, the distance to the star in Exercise 1.1 is 1011 miles

1 lightyear 5.87 × 1012 miles

= .017 lightyear.

Since this is much smaller than 4.3 lightyears, the distance to the nearest star (other than the Sun), we conclude that the numbers are not reasonable.

Exercise 1.3. What order of magnitude is the diﬀerence between 180◦ and the sum of the angles for the summer and winter observations of Alpha Centauri? (See Figure 1.1.) 1

CHAPTER 1. EUCLIDEAN GEOMETRY 14hrs

15hrs

12hrs ❜ −30

13hrs

−40

−50

−60

❜

Figure 1.1. The nearest star (other than the sun) is Alpha Centauri, also known as Rigil Kentaurus, which is 4.3 light years away. This star is the α or brightest star in the Centaurus constellation, which is just south of Libra. It is visible from June to December in the nighttime sky of the southern hemisphere. It brieﬂy appears above the southern horizon in parts of Florida and southern Texas.

Solution. Let Alpha Centauri be the vertex A, let the Earth during the summer be the vertex B, and let the Earth during the winter be the vertex C, forming a long narrow triangle ∆ABC. By the Law of Sines, we have b a . = sin B sin A We are interested in the angle A. We know a = 4.3 lightyears and b = 186, 000, 000 miles. We do not know the exact measurement of B, but expect it to be close to 90◦. Thus, sin A =

b 4.3(5.87 × 1012 miles) ≈ a sin B 186, 000, 000 miles

≈ 1.357 × 10−

A ≈ .0008◦. So, the order of magnitude of the angle at A is 10−3 degrees. Exercise 1.4*. In Figure 1.1, how much does Alpha Centauri move (use units of length) between the summer and winter observations? Solution. By Exercise 1.3, the angle at the star is about 0.0008◦. We assume that the background stars are so far away that our lines of sight to

1.1 THE PYTHAGOREAN THEOREM

them for the summer and winter observations are parallel. Thus, our line of sight to Alpha Centauri changes by 0.0008◦. Let us now inspect the markings in Figure 1.1. The markings on the side are in degrees. Along the top, they are in hours, and with a little thought, we should come to the conclusion that a full circle is 24 hrs, so each hour represents 15 ◦. However, the horizontal markings (those in hours) follow lines of latitudes, so vary, while the vertical markings follow lines of longitude, which are always great circles. Using a ruler, we ﬁnd that 10 ◦ is approximately 16 mm, so 0.0008◦ is approximately 0.001 mm in this ﬁgure. Since this is so tiny, we conclude that to actually calculate this distance, we must take large pictures of a much smaller portion of the night time sky.

1.1

The Pythagorean Theorem

Exercise 1.5. The diagram in Figure 1.2(a) suggests a diﬀerent proof of the Pythagorean theorem. Fill in the details.

D c C′

A C

c B

C A

B B′

(a)

(b)

Figure 1.2. See Exercises 1.5 and 1.6.

Solution. Consider a square of side c. On each edge, place a copy of our triangle ∆ABC with the hypotenuse on each edge, and the vertex C inside the square. Since the angles A and B are complimentary, we know that the edges of these triangles match up, as in Figure 1.2(a), leaving a small square in the center whose edges have length |a − b|. Calculating the area

CHAPTER 1. EUCLIDEAN GEOMETRY

in two different ways, we find 1 2 2 c = 4( ab) + (a − b) 2 2 2 = 2ab + a − 2ab + b 2 2 2 c = a +b . Exercise 1.6. The diagram in Figure 1.2(b) suggests another proof of the Pythagorean theorem. Fill in the details. [H] Solution. Let us first look at ∆ABB′. Since the area of a triangle is half of its base times height, we know |∆ABB′| =

a2 . 2

To see this, think of BB′ as being the base. Then the altitude at A is congruent to BC. Now, note that| BB′ | = BC | ,| BA | |= BC | ′ , |and ∠ABB′ = ∠C′BC, so ′ ′ by SAS, the two triangles ∆ABB and ∆C BC are congruent. In particular, their areas are equal. We calculate the area of ∆C′BC by thinking of the base as BC′, so the altitude is congruent to CD, and hence 1 |∆C′BC| = 2 c|C′D|. Equating the areas of these two congruent triangles, we get 2

c|C′D| = a . In a similar fashion, we can show that the area of the remaining rectangle in the square with side C (the rectangle with diagonal AD) has an area of 2 b . Thus, 2 2 2 c =a +b , as desired. Exercise 1.7 (Pappus’ Variation on the Pythagorean Theorem). Let ∆ABC be a triangle (not necessarily right). Let ACDE and BCFG be parallelograms whose sides DE and FG intersect at H (see Figure 1.3). Let ABIJ be a parallelogram with sides AJ and BI parallel to and with the same length as CH. Prove that the area of ABIJ is equal to the sum of the areas of the other two parallelograms. Solution. Extend the line HC so that it intersects the parallelogram ABIJ. Label these points K and L. Also, draw a line segment MB parallel to HC with M on F G. Then, the area of F GBC is the same as the area of HMBC, since they have the same base BC and the same altitude.

1.2 THE AXIOMS OF EUCLIDEAN GEOMETRY J E

L A I

K D C B H

Figure 1.3. See Exercise 1.7 Now let us consider the parallelograms HMBC and KLIB. Note that |KL = | BI| = | HC| , by | construction, so they have equal length bases. With respect to these bases, their altitudes are also equal, so these two parallelograms have equal areas. Thus, the area of CBGF and KLIB are equal. Similarly, the areas of AECD and AJLK are equal. Thus, the area of AJIB is equal to the sum of the areas of AECD and CBGF .

1.2

The Axioms of Euclidean Geometry

Exercise 1.8. How should we measure the length of a path? Solution. This question is meant to be more of a source of discussion. One should realize that we measure paths ‘one step at a time.’ That is, consider a sequence of points along the path. Then the length of the path is approximated by the sum of the distances between these points. As we add more points, the approximation should get better. So, deﬁne the length of a path to be the limit of these sums as the length of the steps approach zero, if this limit exists. What we have described is essentially a path integral. 2

Exercise 1.9. The Cartesian plane R is a model of Euclidean geometry. In this model, explicitly describe an isometry which has no fixed points and is not a translation. Solution. Such isometries are called glide reflections, a term which might ring a bell. We create such an isometry by translating along a line and then reflecting through that line. For example, the horizontal translation 2 by one in R is given by T (x, y) = (x + 1, y). Reflection through the x-axis is given by R(x, y) = (x, −y). Thus, the composition (R ◦ T )(x, y) = R(x + 1, y) = (x + 1, −y)

CHAPTER 1. EUCLIDEAN GEOMETRY

is a glide reflection. It is clearly an isometry, since it is the composition of two isometries. It is not a translation, since it does not preserve orientation. Finally, it has no fixed points, since if it did, then we would have (x, y) = (x + 1, −y), which is not satisfied by any value of x. Footprints are common examples of objects with glide reflection symmetry. Exercise 1.10. Which of Axioms 1 – 8 are local in nature, and which are global? It may help to ask yourself if any of these properties are true on a sphere. This exercise is meant to be a source of discussion. Exercise 1.11. The triangle inequality states |P Q| + |QR|≥ |P R|. Show that we can have equality if and only if Q is a point on the line segment PR. Solution. We cannot yet answer this question with any rigor – our definition of a line segment is not yet sufficient. This exercise (as well as a few others in this section) should be thought of as motivation for defining a stronger foundation. This will be rectified in Chapter 9. At the heart of our deficiency is the question of how we should define line segments and what is meant by Axiom 1. For the moment, we are thinking of a line segment as the shortest path between two points. Thus, the existence part of Axiom 1 merely asserts the existence of a path between any two points. What we further need to know is that the length of a line segment between two points P and R is the same as the distance | PR | between the two points. If we accept this (as a refinement of our definition of a line segment), then we can answer the question. If Q is on the line segment PR, then the length of the segment PR is the same as the sum of the lengths of PQ and QR. That is, |PR| = |P Q| + |QR|. Conversely, if |PR| = |P Q| + |QR|, then the path which consists of the union of the line segments PQ and QR joins P to R and has length PR | , |which is the same length as the line segment joining P and R. Thus, if Q is not on this line segment, then we have found a path of equal length, contradicting the uniqueness part of Axiom 1. Thus, Q must be on PR.

1.3 SSS, SAS, AND ASA

Exercise 1.12. It is sometimes possible to ‘define away’ an axiom by choosing a suitable definition. For example, with our definition of a circle, Euclid’s third axiom is vacuous – all circles exist by definition, though some may be empty sets. Euclid’s first axiom can also be defined away by making the following definition for a line segment: PQ = {R : |P Q| = |PR| + |RQ|}. With this definition, between any two points P and Q there exists a unique line segment P Q. Find an example of a familiar geometry where this definition for a line segment does not correspond with our intuitive notion of what a line segment should be. Solution. Consider a sphere with North pole P and South pole Q. Then for any point R on the sphere, we have |P Q| = |PR| + |RQ|. Thus, the line segment P Q, as defined in this exercise, is the entire sphere. This does not correspond with our intuitive notion of a line segment, since we usually think of line segments as being one dimensional.

1.3

SSS, SAS, and ASA

The exercises in this section are very tough. I have never assigned them as homework. On the surface, they are indicative of the types of questions which should naturally occur to us, and that we should be able to answer. Deeper analysis might best be done in guided discussions, rather than homework, and should demonstrate a need for a better set of axioms. Exercise 1.13. Prove SAS. Solution. Let ∆ABC and ∆A′B′C′ satisfy ∠A = ∠A′, | AB | = |A′B′ ,| and |AC | = A | ′C′ . |Since ∠A = ∠A′, there exists an isometry f which sends ∠A to ∠A′ (by the deﬁnition of congruence of angles). That is, f (A) = A′, and the ray AB is sent to either the ray A′B′ or A′C′, while the ray AC is sent to the other. If AB is sent to A′B′, then f (B) = B′, since f (B) is on the ray AB and | AB| = | A′B′ |. Similarly f (C) = C′, and hence f sends ∆ABC to ∆A′B′C′, so they are congruent. If f (B) lies on the ray A′C′, then let us call B′′ = f (B) and consider the triangle ∆A′B′B′′. Since |A′B′| = |A′B′′|, we have the congruence ∆A′B′B′′ ≡ ∆A′B′′B′ by SSS, and hence there exists an isometry g such that g(A′) = A′, g(B′) = B′′, and g(B′′) = B′. Furthermore, g sends ∠A′ to ∠A′. Thus, there exists an isometry h = g◦f which sends ∠A to ∠A′ which further satisﬁes h(B) = B′. But then, as before, h(C) = C′, as desired, so ∆ABC is congruent to ∆A′B′C′.

CHAPTER 1. EUCLIDEAN GEOMETRY

Exercise 1.14. Prove ASA. Solution. Suppose there exist ∆ABC and ∆A′B′C′ with| BC| = | B′C′ ,| ∠B = ∠B′, and ∠C = ∠C′. If we also have | BA| = | B′A′ | , then by SAS, the two triangles are congruent. So let us assume, without loss of generality, that |BA |< B| ′A′ . | By the definition of congruence of angles, there exists an isometry which sends B to B′ and the rays BA and BC to the rays B′A′ and B′C′. If C is sent to a point C′′ on B′A′, then by SSS, ∆C′B′C′′ ≡ ∆C′′B′C′, so there exists an isometry g which fixes B′, sends C′′ to C′, and sends ∠C′B′C′′ to ∠C′′B′C′ (that is, it fixes the angle at B′). Thus, there exists an isometry h = g ◦ f such that h(B) = B′, h(C) = C′, and h sends the angle at B to the angle at B′. If, on the other hand, f (C) is on the ray B′C′, then f (C) = C′ already, so choose h = f . Now, A′′ = h(A) is on the ray B′A′, and since | BA| < |B′A′ ,| the point A′′ is between B′ and A′. On the ray C′A′′, find the point A′′′ such that ′ ′′′ ′ ′ ′ ′ ′′ C | A = | C|A . |Since ∆B C A is congruent to ∆ABC (since it is the image of the latter under an isometry), we know ∠A′′C′B′ = ∠ACB, so ∠A′′′C′B′ = ∠ACB = ∠A′C′B′. Thus, by SAS, ∆A′′′C′B′ ≡ ∆ACB. In particular,| B′A′′′ | = B | ′A′ . |Let D be the midpoint of A′A′′′. Note that the line C′A′′′ enters the triangle ∆DA′′′B′ at A′′′, so it must intersect the segment B′D. Let us call that point of intersection E. It is clear that E /= C′. Now, consider ∆B′A′D and ∆B′A′′′D. They are congruent by SSS. Thus, ∠A′DB′ = ∠A′′′DB′, and since these two angles sum to a straight line, they must both be right angles. Similarly, ∠C′DA′ is a right angle, so the angle ∠C′DB′ is a straight line. But this line intersects C′A′′ at C′ and E, a contradiction of Axiom 1. Thus, we could not have | BA | <| B′A′ |, as assumed, so the two triangles are congruent, as desired. Exercise 1.15. Suppose f is an isometry and suppose there exist two distinct points P and Q such that f (P ) = P and f (Q) = Q. Show that f is either the identity or a reflection. Solution. Suppose f is not the identity. Then there exists a point A such that A′ = f (A) = A. Let D be the midpoint of AA′. Since P and Q are distinct, D is not one of them, so we may assume, without loss of generality, that D = P . Suppose B /= P and f (B) = B. Then, since f is an isometry, | P A| = P | A′ |, so by SSS, ∆P AD≡ ∆P A′D. Hence, ∠ADP ≡∠A′DP , and since they sum to a straight line, they must both be right angles. Similarly, using B in place of P , we get that ∠ADB is a right angle. Thus, P , D and B are collinear. Since this is true for any B such that f (B) = B, it is in particular true for B = Q. Thus, f (B) = B only if B is on the line P Q. Now suppose B is on the line P Q. Let B′ = f (B). Then we consider three cases. Either B is between P and Q, P is between B and Q, or Q is between P and B. Let us assume B is between P and Q. Then |PB| + |BQ| = |P Q|. Since f is an isometry, we therefore also have |PB′|+|B′Q| =

1.3 SSS, SAS, AND ASA

|P Q |, which implies B′ is on the segment PQ (by Exercise 1.11). Finally, since both B and B′ are on the segment P Q, and since | PB | ′ =| PB| , the points B and B′ must be equal. The other two cases are similar, so f (B) = B if B is on the line P Q. Thus, we have shown that f is an isometry which fixes every point on the line P Q, and fixes no other points. That is, f is a reflection through the line P Q. Exercise 1.16. Suppose f is a reflection. Prove that f is not a direct isometry. Solution. Let f be the reflection through the line P Q, and let A be a point not on P Q. Then A′ = f (A) = A. Let D be the midpoint of AA′. Then D lies on PQ (see the solution to the previous exercise). But then ∆P QD and ∆P QD′ have opposite orientations, so f is not a direct isometry. Exercise 1.17. Prove that if a line l1 l is sent to itself under a reflection through l, then l1 and l intersect at right angles. Solution. Let A be a point on l1 such that A is not on l. Then A′ = f (A) = A. Let D be the midpoint of AA′. Note that D is on l1, since both A and A′ are on l1. Let P D be a point on l. Then ∆ADP ≡ ∆A′DP by SSS, so in particular, ∠ADP = ∠A′DP . Since they sum to a straight line, they are right angles. That is, l1 = AD is perpendicular to l = PD. Exercise 1.18. Suppose that f is an isometry for which there exists exactly one point P such that f (P ) = P . Prove that f is a rotation. That is, prove that f is a direct isometry. Solution. If an isometry preserves the orientation of one nondegenerate triangle, then it preserves the orientation of all nondegenerate triangles. This is shown in Exercise 9.12 and I do not think we can show this with any rigor without first coming up with a better set of axioms. So let us accept it as fact. As a corollary, if f reverses the orientation of any triangle, then it reverses the orientation of all (nondegenerate) triangles. Suppose now that f is not a direct isometry. Then it reverses the orientation of all triangles. Consider a point Q P and its image Q′ = ′ f (Q). If ∆P QQ is not degenerate, then consider the image of ∆P QQ′ under f . Note that f (P ) = P , f (Q) = Q′. Let f (Q ′ ) = Q′′. Note that Q′′ lies on the circle centered at P with radius |P Q ,| and also on the circle centered at Q′ with center |Q′Q .|There are two points of intersection, one of which is Q. If it is the other point, then ∆P QQ′ has the same orientation as ∆P Q′Q′′. Since we assumed f reverses orientation, we get f (Q′) = Q. Finally, let M be the midpoint of QQ′. Then f (M ) is a distance | M Q| from Q′, and a distance |M Q′ |from Q. Since |M Q | = M | Q′ |, these two circles intersect at exactly one point, M . Thus, f (M ) = M . But this means M = P , so ∆P QQ′ is degenerate, a contradiction. Thus, f is orientation preserving.

CHAPTER 1. EUCLIDEAN GEOMETRY

Now, suppose there does not exist a Q such that ∆P QQ′ is nondegenerate. Then P , Q, and Q′ are collinear, and since Q = Q′, P must be the midpoint of QQ′. Then it is clear that f is rotation by 180◦. Exercise 1.19 †. Suppose f and g are two isometries such that f (A) = g(A), f (B) = g(B), and f (C) = g(C) for some nondegenerate triangle ∆ABC. Show that f = g. That is, show that f (P ) = g(P ) for any point P. 1

Solution. Consider the isometry h = g− f . Note that h(A) = A, h(B) = B, and h(C) = C. By Exercise 1.15, h is either the identity, or a reflection which fixes the line AB, since h(A) = A and h(B) = B. But since h(A) = A and h(C) = C, it is also either the identity or a reflection which fixes the line AC. Since ∆ABC is nondegenerate, the lines AB and AC are distinct, so h cannot fix both of them, so it must be the identity. Thus, f = g. Exercise 1.20*. Suppose P and Q are two distinct points. Prove that there exists exactly one translation which sends P to Q. [H] Solution. There are two questions here – existence and uniqueness. I will ignore the question of existence. The only way I see of doing it is constructive, and in passing shows that Axioms 6 and 7 follow from Axiom 8. Let f be a translation which sends P to Q. Let Q′ = f (Q), and let us now show that Q′ is on the line P Q. Suppose it is not. Let M and M ′ be the midpoints of the segments PQ and QQ′, respectively. Consider the perpendicular l to PQ through M , and the perpendicular l′ to QQ′ through M ′. These lines intersect (and this is where we use the parallel postulate). To see this, consider the line l′′ which is perpendicular to PQ at Q. If l and l′′ intersect on one side of P Q, then the must intersect on the other side too, contradicting Axiom 1 (actually, we have just assumed that there are two sides to a line, a concept not covered by our current set of axioms). Thus, l and l′′ cannot intersect. Similarly, consider the line l′′′ perpendicular to QQ′ and through Q. This line is parallel to l′. Since PQ and QQ′ are distinct lines, the lines l′′ and l′′′ are distinct. Thus, l′′ must intersect l′, since by Axiom 5, there is only one line (namely l′′′) through Q which is parallel to l′. Let this point of intersection be R. Then l′′ is the unique line, through R, which is parallel to l, so since l′ = l′′ and l′ goes through R, it cannot be parallel to l. Thus, l and l′ intersect. Let the point of intersection of l and l′ be S. Let us now show that MS =| M| ′Q (the | =| M| ′S . |Note that M | Q |= MP | | first equality follows since M is the midpoint of P Q, and the second equality follows since M ′ and Q are the images of M and P under the isometry f ). Though we know the triangles ∆M QS and ∆M ′QS are congruent by the Pythagorean theorem, we really shouldn’t use that result, since we haven’t proved it (recall that the proof we gave used results from Euclidean geometry which we had not yet proved, and motivated our discussion of axioms). So, let

1.4 PARALLEL LINES

us instead reflect ∆M QS through l. Since the angle at M is a right angle, the point Q is sent to P , and the angle ∠QM P is a straight line, giving us the triangle ∆P QS with M on the side P Q. Further,| PS| = |QS .| Do the same for ∆M ′Q′S to get ∆QQ′S, which is congruent to ∆P QS and ∆QP S by SSS. Now we know ∠M QS = ∠P QS = ∠QP S = ∠Q′QS = ∠M ′QS, and we get ∆M QS≡∆M ′QS by SAS. Thus, MS | =M | ′S| . | We will now show that f (S) = S. Since S is on l, and l′ is the image of l under f , we know f (S) lies on l′. Since f (M ) = M ′, there are only two possibilities for f (S), one of which is S, the other is the reflection of S through the line QQ′. Since f is a translation, it preserves orientation, so the image of S must be S. Thus, we have shown that f has a fixed point, a contradiction. Thus, the assumption that f (Q) is not on PQ is false. Now, suppose that there exists another translation g such that g(P ) = 1 Q. Let h = g− f . Then h(P ) = P and h(Q) = Q. By Exercise 1.15, h is either the identity or a reflection. Since h is the composition of two translations, it preserves orientations, so it cannot be a reflection. Thus, it is the identity. Hence g = f , and the translation which sends P to Q is therefore unique.

1.4

Parallel Lines

Exercise 1.21*. Show that Euclid’s version of Axiom 5 implies our version of Axiom 5. Solution. Let l be a line and P a point not on l. We wish to show that there exists a unique line l′ through P which does not intersect l. Let us ﬁrst show that l′ exists. By Lemma 1.4.2, there exists a point Q on l such that l and PQ intersect at right angles. Let l′ be the line through P which is perpendicular to P Q. If l and l′ intersect, say at R, then by Axiom 8, there exists an isometry which ﬁxes the line PQ and sends R to a point R′ = R. But since the angles at P and Q in ∆P QR are both right angles, the angles at P and Q in ∆P QR′ are also right angles, so the angles ∠RPR′ and ∠RQR′ are both straight lines. That is, l and l′ intersect at two points R and R′, a contradiction of Axiom 1. Thus, R cannot exist, so l and l′ are parallel. Now, suppose l′′ is any other line through P . Then the angles l′′ makes with PQ are not right angles, so one of those angles is less than a right angle. The line PQ therefore meets l and l′′ so that the sum of the angles on one side is less than a straight line, so l and l′′ intersect. Thus, the line l′ is unique. Exercise 1.22. Prove that the angles in a quadrilateral sum up to 360 ◦. Generalize this result to an n-sided polygon. Solution. For a quadrilateral ABCD, draw the diagonal AC, creating two triangles ∆ABC and ∆BCD. The sum of the angles in these two triangles

CHAPTER 1. EUCLIDEAN GEOMETRY

are each 180◦, giving a total of 360◦. The sum of these angles is also equal to the sum of the angles in the quadrilateral. In general, if we label the vertices of an n-gon P1, P2, ... Pn, then we can ‘triangulate’ it by drawing the segments P1P3, P1P4,...,P1Pn−1. By doing so, we have drawn n − 3 segments, and created n − 2 triangles. Thus, the sum of the angles in an n-gon is (n − 2)180◦. Exercise 1.23. What is the sum of the exterior angles of a triangle? What is the sum of the exterior angles of a quadrilateral? What is the sum of the exterior angles of an n-gon? Solution. The exterior angle for any angle is the diﬀerence of 180◦ less the interior angle. Thus, using the result in Exercise 1.22, the sum of the exterior angles of an n-gon is n(180◦) − (n − 2)180◦ = 360◦.

1.5

Pons Asinorum

Exercise 1.24†. Prove the converse of pons asinorum. That is, show that if in ∆ABC we have ∠ABC = ∠ACB, then |AB| = |AC|. Solution. By ASA, ∆ABC ≡ ∆ACB, so |AB| = |AC|. Exercise 1.25 †. Prove that if a diameter of a circle bisects a chord which itself is not a diameter, then the diameter is perpendicular to the chord. Also, prove that the perpendicular bisector of a chord goes through the center of the circle. And ﬁnally, prove that if a diameter is perpendicular to a chord, then the diameter bisects the chord. [S]

1.6

The Star Trek Lemma

Exercise 1.26. Prove the Star Trek lemma for an acute angle for which the center O is outside the angle. Solution. The proof is almost identical to the one in the book, so let us repeat it here, and point out how it diﬀers. Note that OA, OB, and OC are radii, so we have several isosceles triangles. We have continued the segment OA to intersect the circle at D. Since ∆AOB is isosceles, ∠BAO = ∠OBA. Since the sum of angles in a triangle is 180◦, ∠BOD = ∠OBA + ∠BAO = 2∠BAO. Similarly, ∠DOC = 2∠OAC.

1.6 THE STAR TREK LEMMA

A O D B C

Figure 1.4 And now for the difference. Rather than adding these two angles, it is clear we want to taker the difference, which gives ∠BOC = ∠BOD − ∠DOC = 2(∠BAO − ∠OAC) = 2∠BAC. Exercise 1.27. Prove the Star Trek lemma for an obtuse angle. The proof is identical to the one in the text. Just draw a different diagram. Exercise 1.28 †. Suppose ∠ABC is a right angle inscribed in a circle. Prove that AC is a diameter. [S] Exercise 1.29† (Bow Tie Lemma). Let A, A′, B, and C lie on a circle, and suppose ∠BAC and ∠BA′C subtend the same arc (as in Figure 1.5(a)). Show that ∠BAC = ∠BA′C. Again, because of the diagram, this lemma is sometimes known as the Bow Tie lemma. We say ‘The angles at A and A′ are equal since they subtend the same arc.’ Solution. By the Star Trek lemma, ∠BAC is half the measure of the arc BC that it subtends. Also by the Star Trek lemma, ∠BA′C is half the measure of the arc BC that it subtends. Thus, the two angles are equal. Exercise 1.30. If| AB|= AC is the angle at D? (See | = | BC | , what | Figure 1.5(b).) [A] Solution. Since ∆ABC is an equilateral triangle, ∠ABC = 60◦. Since ∠ADC subtends the same arc, it must also be 60◦.

CHAPTER 1. EUCLIDEAN GEOMETRY B

A′

A O

A D

C (a)

(c)

Figure 1.5. See Exercises 1.29, 1.30, and 1.31. Exercise 1.31. If| AB| = 12,| BD| = 9, | BC | = 16, and | AC | = 20, then what is the length of the diameter? (See Figure 1.5(c).) [A] 2

Solution. Note that ∆ABC has sides of length 12, 16, and 20, so b = 2 2 a + c . Thus, by the converse of the Pythagorean theorem, the angle at B is a right angle. It therefore subtends a diameter (see Exercise 1.28), so AD is a diameter. By the Pythagorean theorem, |AD|2 = |AB|2 + |BD|2 = 12 + 9 = 15 . 2

Thus, the length of the diameter is 15. Exercise 1.32. If| AB| = | AC | = | BC | and AD is perpendicular to BC, then what is ∠BCD? (See Figure 1.6(a).) Solution. Let AD and BC intersect at E. Since| AB| = | AC |, | AE is shared, and ∠AEC = ∠AEB = 90◦, we know | BE | =| EC |by the Pythagorean theorem, and hence, ∆AEC≡ ∆AEB by SSS. Thus, 2∠BAD = ∠BAC = 60◦, so ∠BAD = 30◦. Since ∠BAD and ∠BCD subtend the same arc, we therefore get ∠BCD = 30◦. Exercise 1.33† (The Tangential Case of the Star Trek Lemma). Suppose AT is a line segment that is tangent to a circle. Prove that ∠AT B is half the measure of the arc TB which it subtends. Do this by picking a point C on the circle such that ∠TCB subtends the arc TB (as in Figure 1.6(b)). Show that ∠AT B = ∠TCB. Solution. It doesn’t matter where we pick C, since ∠TCB is always half the arc it subtends. So, pick C so that BC is a diameter. Then ∠CTB = 90◦, and ∠TCB = ∠TCO = ∠CTO. Thus, ∠OTB = 90◦ − ∠TCB. But the radius OT and the tangent TA intersect at right angles, so ∠OTB = 90◦ − ∠BTA. Thus, 90◦ − ∠BTA = 90◦ − ∠TCB

1.6 THE STAR TREK LEMMA T A B O

D B

C (a)

(b)

Figure 1.6. See Exercises 1.32 and 1.33. ∠BTA = ∠TCB. Hence, ∠BTA is half the the measure of the arc it subtends. Exercise 1.34 †. Suppose two lines intersect at P inside a circle and meet the circle at A and A′ and at B and B′, as shown in Figure 1.7(a). Let α and β be the measures of the arcs A′B′ and AB respectively. Prove that ∠AP B =

α+β 2 .

B′ P

A′

B P

α B′

A′ (b)

(a)

Figure 1.7 Solution. Draw the segment AB′, and consider the triangle ∆AB′P . By the Star Trek lemma, ∠AB′P = β/2, and ∠B′AP = α/2. Since ∠AP B is

CHAPTER 1. EUCLIDEAN GEOMETRY

the exterior angle of ∆AB′P at P , we have ∠AP B = ∠AB′P + ∠B′AP =

α+β 2 .

Exercise 1.35 †. Suppose an angle α is deﬁned by two rays which intersect a circle at four points, as in Figure 1.8(b). Suppose the angular measure of the outside arc it subtends is β, and the angular measure of the inside arc it subtends is γ. (So, in Figure 1.8(b), ∠AOB = β and ∠A′OB′ = γ.) Show that β−γ α= . [S] 2 A

B′

A′ B

O B

α B′

A′ (a)

(b)

Figure 1.8. See Exercises 1.34 and 1.35. Exercise 1.36. Prove that the opposite angles in a convex quadrilateral (called a convex quadrangle in the ﬁrst printing) inscribed in a circle sum to 180◦. Conversely, prove that if the opposite angles in a convex quadrilateral sum to 180◦, then the quadrilateral can be inscribed in a circle. Such a quadrilateral is called a cyclic quadrilateral. Solution. Let ABCD be a convex quadrilateral inscribed in circle. Then the arc subtended by the angles at B and D, together, form the whole circle, so measure 360 ◦. Thus, the sum of the angles B and D is half that, so is 180◦. Conversely, suppose the opposite angles B and D sum to 180◦ in a convex quadrilateral ABCD. Consider the circumcircle of ∆ABC. If D lies inside this circle, then by Exercise 1.34, the angle D is larger than half the measure of the arc ABC = 360◦− 2∠ABC, so the sum of the angles at B and D is greater than 180◦. Similarly, if D lies outside the circle, then by Exercise 1.35, the sum of the angles is less than 180◦. Thus, D must lie on the circle. That is, the quadrilateral ABCD is cyclic.

1.6 THE STAR TREK LEMMA

Exercise 1.37†. Let two circles Γ and Γ′ intersect at A and B, as in Figure 1.9. Let CD be a chord on Γ. Let AC and BD intersect Γ′ again at E and F . Prove that CD and EF are parallel. [S] C Γ

Γ′ E

F B D

Figure 1.9. See Exercise 1.37. This exercise is restated and proved in Lemma 11.2.2. Exercise 1.38†. Let ABCD be a nonconvex cyclic quadrilateral. That is, ABCD is inscribed in a circle and two of its opposite sides intersect (as in Figure 1.10(a)). Prove that ∠ABC = ∠CDA and ∠DAB = ∠BCD. Conversely, suppose ∠ABC = ∠DAB and ∠BCD = ∠DAB in a quadrilateral with intersecting opposite sides. Show that ABCD is cyclic. Solution. The solution is essentially the same as the solution of Exercise 1.36. The angles are equal since they subtend the same arcs. Conversely, consider the circumcircle of ∆ABC. If D is inside this circle, then by Exercise 1.34, the angle at D must be larger than the angle at B; and if D is outside this circle, then by Exercise 1.35, the angle at D is smaller than the angle at B. Thus, D must lie on the circle. Exercise 1.39. Suppose ABCDEF is a hexagon inscribed in a circle. Show that ∠ABC + ∠CDE + ∠EFA = 360◦. Prove that the converse is not true. That is, ﬁnd an example of a hexagon ABCDEF whose angles B, D, and F sum to 360◦ but which cannot be inscribed in a circle. Solution. The sum of the arcs subtended by the angles ∠ABC, ∠CDE, and ∠EFA is twice the entire circle, so 720 ◦. Thus, the sum of these angles is half that, or 360◦. The converse, though, is not true. To see this, let us ﬁrst consider a regular hexagon ABCDEF inscribed in a circle. Note that the sides AB and DE are parallel. Thus, we may stretch these sides without changing the angles. By doing so, we will have created a hexagon which is not inscribed in a circle, but whose angles satisfy the condition.

CHAPTER 1. EUCLIDEAN GEOMETRY

Exercise 1.40. Let E be a point inside a square ABCD such that ∆BCE is an equilateral triangle, as in Figure 1.10(b). Show that ∠EAD = 15◦. A

(a)

Figure 1.10 Solution 1. Note that |AB| = |BE|, so ∆ABE is isosceles. Let ∠BAE = α = ∠AEB. Note also that, ∠ABE = 90◦ − 60◦ = 30◦. Thus, 2α + 30◦ = 180◦, so α = 75◦. Finally, ∠DAE = 90◦ − α = 15◦. Solution 2. Since |BA |= BE exists a circle centered at | =| BC | , there | B through A, E, and C. Since ∠BAD = 90◦, we know AD is tangent to the circle. The angle ∠EAD subtends the arc AE, so is half of the angle ∠ABE, which as noted above, is 30◦. Thus ∠EAD = 15◦.

1.7

Similar Triangles

Exercise 1.41. Suppose ∆ABC is similar to ∆A′B′C′. Show that |∆A′B′C′| =

|A′B′| |AB|

|∆ABC|.

That is, the ratio of the areas of the two triangles is the same as the square of the ratios of the sides. Here, we have used the notation| ∆ABC | to represent the area of ∆ABC. Solution. Let the altitude at A in ∆ABC intersect the (possibly extended) side BC at D. Similarly, let the altitude at A′ intersect B′C′ and D′. Then 1 1 ′C′||A′D′|. It is clear that |∆ABC| = 2|BC||AD|, and |∆A′B′C′| = |B 2 ′ ′ ′ ∆ABD ∆A ∼ B D, since the angles at B and B are equal (since we were given that ∆ABC ∼ ∆A′B′C′) and the angles at D and D′ are both right angles. Thus, |A′D′| |A′B′| = . |AD| |AB|

1.7 SIMILAR TRIANGLES Also, since ∆ABC ∼ ∆A′B′C′, we have |B′C′| |BC| A'B'|

Let α = |

|AB|

|A′B′| |AB|

. Then,

1 1 |∆A′B′C′| = 2 |B′C′||A′D′| = 2 ( α|BC|)α|AD| = α2|∆ABC|. Exercise 1.42. Let ∆ABC be an arbitrary triangle. Let A′, B′, and C′ be the midpoints of the opposite sides. Draw lines through A′, B′, and C′ that make an angle of 60◦ with each side, as in Figure 1.11(a). These lines intersect at A′′, B′′, and C′′ as shown. Prove that ∆A′′B′′C′′ is similar to ∆ABC. Solution. Consider the quadrilateral CA′C′′B′. It is cyclic, since the angles at A′ and B′ sum to 180◦. Thus, the angles at C and C′′ sum to 180◦. Hence, ∠A′′C′′B′′ = ∠ACB. Though the quadrilateral CA′C′′B′ is convex, it doesn’t have to be. For example, the quadrilateral AC′A′′B′ is a nonconvex quadrilateral, but it is still cyclic since the angles at B′ and C′ are equal. Thus, the angles at A and A′′ are equal. In any case, by looking at the appropriate quadrilaterals at each vertex, we ﬁnd that the respective angles in ∆ABC and ∆A′′B′′C′′ are equal. That is, these triangles are similar. Note that there is nothing special about 60◦. It could have been any ﬁxed angle. C

A′′ B′

❜

C′′

❜

A′

B′′ 60 A

❜

C′ (a)

(b)

Figure 1.11. See Exercises 1.42 and 1.43. Exercise 1.43 (The Angle Bisector Theorem). In an arbitrary tri† angle ∆ABC, let the interior angle bisector at A intersect the side BC at D. Show that |BD| |AB| = . |DC| |AC|

CHAPTER 1. EUCLIDEAN GEOMETRY

Hint: Construct the line parallel to AD and through B, as shown in Figure 1.11(b). Let this intersect AC at E. Show |AB| = |AE|. Solution. Let α = ∠BAD. Since AB is the transversal of parallel lines, ∠EBA = ∠BAD = α. Since AD is the angle bisector of ∠BAD, we have ∠CAD = ∠DAB = α. Since BE and DA are parallel, α = ∠CAD = ∠CEB = ∠AEB. Thus, ∆AEB is isosceles with equal angles at B and E. Thus, |AE| = |AB|. Now, since AD is parallel to EB, we know |CE| |CA| |CA| + |AE|

|CB| |CD| |CD| + |DB|

= |CA |CD| |AB| |DB| 1+ =1+ . |CA| |CD|

Exercise 1.44. The pentagon in Figure 1.12(a) is regular and each side has length one. Show that √ |AF | 1+ 5 = . |FD| 2 B

A C

F E

(b)

(a)

Figure 1.12 Solution. Note the typo, which was corrected in the second printing. Let AC | =| x. Since the pentagon is regular, the pentagon can be inscribed in a circle with equal length arcs between consecutive vertices. Thus, ∠DAC = ∠CAB = ∠ECA = ∠ACB, so ∆ACF ≡ ∆ACB, by ASA. In particular, |AF | = |AB| = 1. Also, ∠CAD = ∠CED, since they subtend the same arc. Thus, ∆ACF ∼ ∆DEF , so |AF | |DF |

|AC| |DE|

1.7 SIMILAR TRIANGLES 1 x = x−1 1 2 0 = x −√ x −1 1+ 5 x= 2 From the ﬁrst two lines above, the ratio we are asked to ﬁnd is x.

Exercise 1.45. In the quadrilateral ABCD in Figure 1.12(b), AD is parallel to BC, ∠C = 2∠A, |CD| = 3, and |BC| = 2. What is |AD|? [A] Solution 1. Continue the lines AB and CD until they intersect. Label this point of intersection E. Since BC and AD are parallel, ∠EBC = ∠A. Since the angle called ∠C is the exterior angle of ∆EBC at C, it is the sum of the angles at E and B. Thus, 2∠A = ∠BCD = ∠EBC + ∠BEC = ∠A + ∠BEC, so ∠BEC = ∠A. That is, both ∆EBC and ∆EAD are isosceles, with |EC| = |BC| = 2, and |AD| = |ED| = |EC| + |CD| = 2 + 3 = 5. Solution 2. Since ∠C = 2∠A, let us bisect it. Let this bisector intersect AD at the point C′. Then ABCC′ is a parallelogram, so |AC′| = |BC| = 2. Also, ∠CC′D = ∠C′CB = ∠A, so ∆CDC′ is isosceles with |C′D| = |CD| = 3. Thus |AD| = |AC′| + |C′D| = 2 + 3 = 5. Exercise 1.46†. Let ∆ABC and ∆A′B′C′ be two triangles such that ∠BAC = ∠B′A′C′

and

|A′B′| |AB|

|A′C′| |AC|

Prove that ∆ABC ∼ ∆A′B′C′. Solution. Construct the points B′′ and C′′ on A′B′ and A′C′, respectively, such that| A′B|′′ =| AB | and | A′C|′′ =| AC | . Then ∆A′B′′C′′≡ ∆ABC by SAS. But by Theorem 1.7.1, B′C′ is parallel to B′′C′′, so ∆A′B′′C′′ ∼ ∆A′B′C′. Thus, ∆ABC ∼ ∆A′B′C′. Exercise 1.47†. Suppose ∆ABC and ∆A′B′C′ are two similar but noncongruent triangles such that AB is parallel to A′B′, AC is parallel to A′C′, and BC is parallel to B′C′, as in Figure 1.13. Prove that the lines AA′, BB′, and CC ′ are concurrent. Solution. Consider the lines AA′ and BB′. If these two lines are parallel, then AA′B′B forms a parallelogram, and hence |AB| = |A′B′|. But then, ∆ABC ≡ ∆A′B′C′, which contradicts the given conditions. Hence, AA′

CHAPTER 1. EUCLIDEAN GEOMETRY A

C B′

C′

A′

Figure 1.13. See Exercise 1.47. and BB′ are not parallel, so they must intersect at a point we will call P . By Theorem 1.7.1, |AP | = |BP| . |A′P | |B′P | Construct a point C′′ such that |CP| |C′′P |

|AP | |A′P |

and such that C′′ and C are on opposite sides of P if A′ and A are opposite sides of P (as in Figure 1.13), and on the same side if A′ and A are on the same side of P . Then, by Theorem 1.7.1, C′′A′ is parallel to CA. We also have |CP| |BP| = , ′′ |C P | |B′P | so C′′B′ is parallel to CB too. Now, consider the lines through A′ and B′ which are, respectively, parallel to AC and BC. These two lines intersect in a unique point, which must be C′ and must also be C′′. Thus, C′′ = C′. Finally, CC ′ = CC′′, and the latter, by construction, goes through P . Thus, CC ′ goes through P . Hence, AA′, BB′, and CC ′ all go through P , so are concurrent. Exercise 1.48. Let Γ be a circle. Suppose Γ′ is another circle whose center lies on Γ. Let these two circles intersect at A and B, as in Figure 1.14(a). Let P be a point on Γ, and let PB intersect Γ′ again at Q. Show that ∆P QA is isosceles. [H] Solution 1. Ignore the hint. There are two cases – P can be outside Γ′, as in the figure, or P can be inside Γ′. Let us first consider the case as in the figure. Let O′ be the center of Γ′ and draw AO′ and BO′. Let α = ∠AQP . By the Star Trek lemma, ∠AO′B = 2α. Since PBO ′ A is cyclic, we have ∠AP B +∠AO′B = 180◦, so ∠AP B = 180◦ −2α. Since we have two angles of ∆AP Q, we can calculate the third, and discover ∠P AQ = α = ∠P QA, so ∆P QA is isosceles.

1.7 SIMILAR TRIANGLES A

Γ′

C Γ′

B D

P (a)

(b)

Figure 1.14 For the other case, we again let α = ∠AQP . Then ∠AO′B = 2α, so ∠AP B = 2α (they subtend the same arc). But ∠AP B is the exterior angle at P of ∆AP Q, so ∠AQP + ∠QAP = 2α. Thus ∠QAP = α = ∠AQP , so ∆AQP is isosceles. Solution 2. First, the two cases we must consider are P outside Γ′ (as in Figure 1.14) and P inside Γ′. The two cases are similar. Let us consider the ﬁrst. Following the hint, let P ′ be another point on Γ (also outside Γ′). Note that both Q and Q′ are outside Γ. Let P ′B intersect Γ′ again at Q′. Then ∠AP B and AP ′B subtend the same arc of Γ, so are equal; and ∠AQB and ∠AQ′B subtend the same arc of Γ′, so are equal. Thus, ∆P AQ and ∆P ′AQ′ are similar. Let us now choose P ′ in a convenient place. Let O and O′ be the centers of Γ and Γ′, respectively, and choose P ′ on OO′. Then ∠O′AP ′ = 90◦, since O′P ′ is a diameter. Thus Q′ = B, and by symmetry (reﬂection through OO′) ∆AP ′Q′ = ∆AP ′B is isosceles. Since ∆AP Q is similar to ∆AP ′Q′, it must also be isosceles. The case when P is inside Γ′ is similar. This time, when we pick P ′ conveniently, we let P ′ = O′, so that P ′A and P ′Q′ are both radii, so ∆AP ′Q′ is isosceles. Exercise 1.49. Let two circles Γ and Γ′ intersect at A and B (see Figure 1.15(b)). Let P be a point on the circle Γ. Let PA intersect Γ′ again at C, and let PB intersect Γ′ again at D. Show that the length| CD|is independent of the location of P . (Note that there are two cases to consider.) [H]

Solution. There are two cases to consider – P is either inside or outside Γ′. Let us ﬁrst consider the case when P is outside Γ′, as in Figure 1.15 (b). As suggested in the hint, let us ﬁrst show that ∆PCD ∼ ∆PBA. To see this, note that the quadrilateral ACDB is cyclic, so the angles at B and C are supplementary. Thus, ∠ABP = ∠BCD. Since the angle at

CHAPTER 1. EUCLIDEAN GEOMETRY A

Γ′

C Γ′

B D

P (a)

(b)

Figure 1.15. See Exercises 1.48 and 1.49. P is shared, we therefore have ∆PCD∼ ∆PBA. Note that the length AB changes, so if we can show that the ratio of this similarity does | never | not change, we will be done. Again, as suggested in the hint, let us pick another point P ′ also outside Γ′, and construct C′ and D′ in the same way. Then ∠AP B = ∠AP ′B, since they subtend the same arc of Γ, and ∠ADP = ∠ADB = ∠AD′B = ∠AD′P , since they subtend the same arc of Γ′. Thus, ∆AP D ∼ ∆AP ′D′. In particular, |PD| |AP | = . ′ |PD′| |AP | Thus,

|CD| |AB|

|AP | |PD|

|AP ′| |PD′|

|C′D′| |AB|

That is, |CD| = |C′D′|. Now, suppose P ′ is inside Γ′ (but P is still outside Γ′). Then P AP ′B is cyclic, so the angles at P and P ′ are supplementary. Thus, ∠AP ′D = ∠AP D. Also, ∠AD′P = ∠AD′B = ∠ADB, since they subtend the same arc of Γ′, so ∆AP D ∼ ∆AP ′D, as desired. Thus, the argument above is now applicable.

1.8

Power of the Point

Exercise 1.50. If in Figure 1.16(a), |AP | = 2, |AB| = 6, and |PC| = 3, then what is |PD|? [A] Solution. By power of the point, |AP ||PB| = |CP ||P D|.

1.8 POWER OF THE POINT C

B P

D (b)

(a)

Figure 1.16 Note that |PB| = |AB|− |AP | = 4. Thus, 2(4)

|PD| =

8 3

Exercise 1.51 †. Theorem 1.8.1 is also true if the point P lies outside the circle, as in Figure 1.17(a). Prove it for this case. R′ R R P P Q

Q Q′

Q′

(a)

(b)

Figure 1.17. See Exercises 1.50 and 1.52. Solution. Following the proof of Theorem 1.8.1, let us ﬁrst draw QR′ and RQ′. Then ∠RQ′Q = ∠RR′Q, since they subtend the same arc, and since the angle at P is shared, we get ∆PRQ′ ∼ ∆P QR′. Thus, |PR| |P Q|

|P Q′| |PR′|

Cross multiplying gives us power of the point.

CHAPTER 1. EUCLIDEAN GEOMETRY

Exercise 1.52. If in Figure 1.16(b), |AB| = 5, |BC| = 12, |AC| = 13, and |BD| = 3, then what is |DE|? [A] Solution. We don’t need power of the point to prove this. After all, since ∆ABC is a (5, 12, 13) triangle, the angle at B is a right angle. Thus AD is a diameter, so ∠AED is also a right angle. Hence, ∆ABC ∼ ∆DEC, so |DE| |AB|

|DC| |AC|

We know |DC| = |BC|− |BD| = 9, so 9(5) 45 = . |DE| = 13 13 This can also be solved using power of the point to calculate |CE|, and the Pythagorean theorem to get |DE|. Exercise 1.53†(The Tangential Version of Power of the Point). In Figure 1.17(a), suppose we let R and R′ move toward each other until they meet, so that the ‘chord’ RR′ is really the tangent at R, as in Figure 1.17(b). Then we get yet another version of the power of the point. Suppose a chord QQ′ intersects a tangent at R at a point P . Show that |P Q||P Q′| = |PR|2. Solution. Draw RQ′. Then, by the tangential version of the Star Trek lemma, ∠PRQ = ∠RQ′Q, so ∆PRQ ∼ ∆P Q′R. Thus |P Q| |PR| = , |PR| |P Q′| which gives us the desired result after cross multiplying. Exercise 1.54. Use the tangential version of power of the point (Exercise 1.53) to come up with yet another proof of the Pythagorean theorem. [H][S] Exercise 1.55. In Figure 1.18(a), if BC is tangent to the circle centered at A, |BC| = 3, and |CD| = 1, then what is the radius of the circle? [A] Solution. Using the tangential version of power of the point, we get |CB|2 = |CD||CD′|, where D′ is the point where the line CA again intersects the circle. Hence, if we let r be the radius, we have 2

3 = 1(1 + 2r) r = 4. We could also solve this by noting that the angle at B is a right angle, so using the Pythagorean theorem, we get 2

9 + r = (r + 1) .

1.8 POWER OF THE POINT

B D C

B′ A′

C (a)

B (b)

Figure 1.18 Exercise 1.56. In Figure 1.18(a) as described in Exercise 1.55, what is the area of ∆ABD? [A] Solution. The area of a triangle is one half the base times height. In ∆ABC, we ﬁnd the area is 6 by using |AB| = r = 4 as the base, and |BC| = 3 as the height. Using |AC| as the base, we get that the height h is 1 6=

5h,

so h = 12/5. But h is the height of ∆ABD where| AD| = r = 4 is thought of as the base. Thus, |∆ABD| =

1 12 24 (4) = . 2 5 5

Exercise 1.57. Let ∆ABC be an isosceles triangle with | AB | =| AC|(see Figure 1.18(b)). Let A′ and B′ be the midpoints of sides BC and AC, respectively. Let Γ be the circle that goes through B, A′, and B′. Let the extended side AB intersect Γ at D. Suppose| AD | = 5 and| AB′| = 4. What are the lengths of the sides of the triangle? Solution. Since this is in the section on power of the point, let’s try to use it. The only points that we can even consider finding the power of are A and C (the only points that are not on the circle). For either one, we need to first find the point E where the line AC again intersects the circle. Let x = |AE| and let a = 2|BA′|. Note that |CB′| = |AB′| = 4, so |AC| = 8 = |AB|. By power of the point A, we get |AB′||AE| = |AD||AB| 4x = 5(8) x = 10.

CHAPTER 1. EUCLIDEAN GEOMETRY

By power of the point C, we get |CA′||CB| = |CB′||CE| a2 = 4(8 + x ) 2 2 a = 8(18) a = 12. Thus, the lengths of the sides of the triangle ∆ABC are |AB| = |AC| = 8 and |BC| = 12. Exercise 1.58. In Figure 1.19, the point C is on the diameter AB of a half circle. The two smaller half circles have diameters AC and CB. The region bounded by the curved edges of the half circles was called an arbelos or butcher’s knife by Archimedes. The perpendicular at C intersects the larger circle at D. Prove that the circle with diameter CD has the same area as the arbelos. D

Figure 1.19. See Exercises 1.58 and 1.59. Solution. Let the circle with diameter AC have radius r, and let the circle with diameter BC have radius s. Thus 2r = | AC| , 2s = | BC |, and the radius of the large circle is r + s. Hence, the arbelos has an area of 1 2 2 2 (π(r + s) − πr − πs ) = πrs. 2 By power of the point C in the larger circle, we have |CD|2 = (2r)(2s) = 4rs, √ so the circle with diameter CD has radius rs, and therefore has area πrs, the same as the area of the arbelos.

1.8 POWER OF THE POINT

Exercise 1.59. Show that the two inscribed circles in the arbelos in Figure 1.19 have the same radii. [H] Solution. Let O be the center of the large circle, let P be the midpoint of AC, and let Q be the center of the inscribed circle to the left of CD in Figure 1.19. Let this inscribed circle have radius x. Let r and s be the radii of the circles with diameters AC and BC, respectively. Then r + s is the radius of the circle with diameter AB. Find the point E on AB such that QE is perpendicular to AB. Consider the right triangles ∆QEO and ∆QEP . Note that |OQ| = r + s − x, since the line OQ intersects the larger circle at its point of tangency with the small circle. Similarly, |P Q| = r + x. Note that |EC| = x. Let y = |QE|. Then, we get the two equations (r + s − 2r + x) + y = (r + s − x) 2

(r − x) + y = (r + x) . 2

Note that these equations are valid, no matter whether r > s or s > r (or r = s). We eliminate the y by subtracting, which gives (s − r + x) − (r − x) = (r + s − x) − (r + x) s(s − 2r + 2x) = (2r + s)(s − 2x) 2

s − 2rs + 2sx = 2rs + s − 4rx − 2sx 4(r + s)x = 4rs rs x= . r+s 2

Since it doesn’t matter whether r is larger than s or not, the formula for the radius of the other circle is the same, except that the roles of r and s are exchanged. Since it is symmetric in r and s, the two are equal. Exercise 1.60. Let two circles Γ and Γ′ have distinct centers O and O′. Prove that the radical axis of Γ and Γ′ is a line perpendicular to OO′. When talking about radical axis, we are talking about powers of points with respect to diﬀerent circles. So let us deﬁne ΠΓ(P ) to be the power of the point P with respect to the circle Γ. Solution. Let P be the point on OO′ such that the power of the point P ′ with respect to each circle is the same. Let a|= PO | , b =| PO | , let r be ′ the radius of Γ, and let s be the radius of Γ . Then ΠΓ(P ) = ΠΓ' (P ) a −r = b −s . 2

Now, suppose Q is a point on the perpendicular to OO′. Let h = |QP |, x = |QO|, and y = |QO′|. Then, by the Pythagorean theorem, 2

a +h =x

CHAPTER 1. EUCLIDEAN GEOMETRY 2

b +h =y . Subtracting, we get a −b =x −y 2

(a − r ) − (b − s ) = (x − r ) − (y − s ). 2

Since the left hand side is zero, we get x −r =y −s . 2

That is, ΠΓ(Q) = ΠΓ' (Q) so Q is on the radical axis. Now, suppose Q′ is a point such that ΠΓ(Q′) = ΠΓ(Q) (where Q is as before, a point on the perpendicular). Then Q′ lies on the circle centered at O which goes through Q. Similarly, if ΠΓ' (Q′) = ΠΓ' (Q) = ΠΓ(Q), then Q′ lies on a circle centered at O′ which goes through Q. These two circles intersect at two points, both of which are on the perpendicular. Thus, the radical axis is exactly the perpendicular to OO′ which goes through P . Exercise 1.61. Suppose two circles intersect at A and B. Prove that the radical axis of the two circles is the line AB. Solution. By Exercise 1.60, the radical axis is a line. Since the power of the points A and B are zero with respect to both circles, they both lie on the radical axis. Thus, the radical axis is the line AB. Exercise 1.62. Let the circle Γ intersect the circle Γ′ at A and B, and the circle Γ′′ at C and D. Let P be the point of intersection of AB and CD. Prove that P lies on the radical axis for Γ′ and Γ′′. Solution. Since P is on AB, ΠΓ(P ) = ΠΓ' (P ). Since P lies on CD, ΠΓ(P ) = ΠΓ'' (P ). Thus, ΠΓ' (P ) = ΠΓ'' (P ). That is, P lies on the radical axis of Γ′ and Γ′′. Exercise 1.63 (The Radical Center). Let Γ, Γ′, and Γ′′ be three circles. Show that the three radical axes of these circles intersect at a common point. This point is called the radical center of the three circles.

1.8 POWER OF THE POINT

Solution. This exercise is essentially the same as the previous exercise. Let P be the intersection of the radical axis of Γ and Γ′, and the radical axis of Γ and Γ′′. Then ΠΓ(P ) = ΠΓ' (P ) = ΠΓ'' (P ), so P lies on the radical axis of Γ′ and Γ′′. Thus, the three radical axes intersect at a common point. Exercise 1.64. Let P be the radical center of three circles Γ, Γ′, and Γ′′. Suppose P is outside circle Γ. Prove that P is also outside circles Γ′ and Γ′′. Solution. Since P is outside Γ, we know ΠΓ(P ) > 0. Thus, ΠΓ' (P ) > 0 (since it has the same value), so P lies outside Γ′. Similarly, P lies outside Γ′′. Exercise 1.65. Let P be the radical center of three circles Γ, Γ′, and Γ′′, and suppose P lies outside of Γ. Show that P is the center of a circle which intersects Γ, Γ′ ,√ and Γ′′ perpendicularly. Furthermore, show that the radius of this circle is Π(P ). Solution. Let Q be a point on Γ such that PQ is tangent to Γ. Then, by the tangential version of power of the point, |P Q|2 = ΠΓ(P ). Similarly, if Q′ is a point on Γ′ such that P Q′ is tangent to Γ′, then |P Q′|2 = ΠΓ' (P ). Since ΠΓ (P ) = √ ΠΓ' (P ), the distances to these points of tangency are all equal (equal to (Π(P )), where there is no confusion caused by omitting the subscript, sin√ ce P is the radical center). Thus, the circle centered at P and with radius Π(P ) intersects the circles Γ, Γ′, and Γ′′ at the points where the tangents to these circles through P intersect the circle. Since the radius of a circle is perpendicular to its tangents, this circle therefore intersects the three circles perpendicularly. Exercise 1.66. Let Γ1, Γ2, Γ3, and Γ4 be four circles. Let Γ1 and Γ2 intersect at P1 and P2. Let Γ3 and Γ4 intersect at P3 and P4. Let Γ1 and Γ4 intersect at Q1 and Q2, and let Γ2 and Γ3 intersect at Q3 and Q4. Show that P1, P2, P3, and P4 are collinear or cyclic if and only if Q1, Q2, Q3, and Q4 are collinear or cyclic (not necessarily respectively). This exercise deserves a star, if not two. I do not remember where I ﬁrst saw it, but I think it was meant to be solved using inversion in the circle (Sections 7.4 and 7.5). I saw that it could be solved using the ideas of the radical center, so placed it here. When I returned to it to write up solutions, I was quite aggravated that I knew there was a solution using the ideas of this section, but could not see (for quite some time) just how this section was relevant.

CHAPTER 1. EUCLIDEAN GEOMETRY Γ Γ1 Γ0

Γ2

P2 P3

Γ4

P4 Γ3

Figure 1.20

Solution. Let Γ be the circle through P1, P2, P3, and P4, as in Figure 1.20. We’ll deal with the case where Γ is a line later. Let Γ 0 be the circumcircle of ∆Q1Q2Q3. Again, we’ll deal with the possibility that Q1, Q2, and Q3 are collinear later. We wish to show that Q4 lies on Γ0. Let O be the radical center of Γ1, Γ2, and Γ4. Then, O is the point of intersection of P1P2 and Q1Q2. Since Γ also goes through P1 and P2, O is the radical center of Γ, Γ1, and Γ4. Thus, the radical axis P3P4 of Γ and Γ4 goes through O. The radical center of Γ, Γ2, and Γ3 is at the point of intersection of P1P2 and P3P4, so is O. Thus, the radical axis Q3Q4 of Γ2 and Γ3 goes through O. Now consider the radical center of Γ 0, Γ1, and Γ2. It is at the intersection of Q1Q2 and P1P2, so is O. Thus, the radical axis of Γ0 and Γ2 goes through O and Q3, so is the line Q3Q4. Thus, Γ0 and Γ2 intersect at Q3 and Q4. That is, Q1, Q2, Q3, and Q4 are cyclic. Now, suppose Q1, Q2, and Q3 are collinear. If Q4 is on this line, then we are done. Otherwise, Q1, Q2, and Q4 are not collinear, so let switch the labels Q3 and Q4, and proceed as before. Suppose P1, P2, P3, and P4 are collinear. Let Q1Q2 intersect P1P2 at O. Then ΠΓ1 (O) = ΠΓ4 (O). But ΠΓ1 (O) = ΠΓ2 (O) (since P1 and P2 lie on both circles), and ΠΓ4 (O) = ΠΓ3 (O), so O is on the radical axis Q3Q4 of Γ2 and Γ3. Now, proceed as before, starting with the third paragraph.

1.9 THE MEDIANS AND CENTROID

1.9

The Medians and Centroid

Exercise 1.67. In triangle ∆ABC pictured in Figure 1.21, let A′ and B′ be the midpoints of the sides BC and AC, respectively. Let AA′ and BB′ intersect at G. Suppose |AG| = 2 and suppose also that the circle through A, A′, and B bisects the segment B′G. What is the length |BG|? [A] A

B′ G

A′

Figure 1.21. See Exercise 1.67. Solution. Since A′ and B′ are midpoints, AA′ and BB′ are medians, so G is the centroid. Thus, |BG| = 2|B′G|. Let the midpoint of B′G be D and let x = |B′D| = |DG|. Then |BG| = 4x. Also, since |AG| = 2, we know |GA′| = 1. Thus, taking power of the point G, we ﬁnd

√ Thus, |BG| = 4x = 2 2.

|AG||GA′| = |BG||GD| 2(1) = (4x)x √ 2 x= . 2

Exercise 1.68. Cut a triangle out of card stock. Locate its centroid G. Now take a needle and hang the triangle by thread through G. The triangle should be balanced, which is why we call G the center of mass or centroid of the triangle. Exercise 1.69. Suppose the medians AA′ and BB′ of ∆ABC intersect at right angles, and suppose a = 3 and b = 4. What is c? [A] Solution. It always helps to have a picture – see Figure 1.22. Let x = GA | ′ | and y = |GB′ .| Then, AG = 2x and BG = 2y. By the Pythagorean | | | | theorem, 2

x + 4y =

4 2 2 4x + y = 4

CHAPTER 1. EUCLIDEAN GEOMETRY B A′ G B′

Figure 1.22 2

4x + 4y = c . Adding the ﬁrst two equations, we get 2

5x + 5y = so 2

c = and hence, c =

√

4 5

25 =5 4

Exercise 1.70. Prove that the sum of the medians of a triangle lies between 3 p and p, where p is the length of the perimeter of the triangle. 4 Solution. The key to this question is the triangle inequality (as well as facts we just learned about medians). The triangle inequality is sharper for obtuse triangles than acute triangles, so the other idea is to look for obtuse angles. For one direction, we will look at triangles like ∆AGB, and for the other direction, we will look at triangles like ∆AB′A′. Let the medians AA′, BB′, and CC ′ have lengths of 3x, 3y, and 3z, respectively. Let m be the sum of the lengths of the medians, so m = 3x + 3y + 3z. Then, by the triangle inequality (looking at ∆BGC, ∆CGA, and ∆AGB, respectively), we get a ≤ 2y + 2z b ≤ 2x + 2z c ≤ 2x + 2y. From this, we get p ≤ 4x + 4y + 4z = (4/3)m, so m ≥ (3/4)p.

1.10 THE INCIRCLE, EXCIRCLES, AND THE LAW OF COSINES

For the other direction, we ﬁrst note that | B′|C′ = a/2,| A′C′| = b/2, and ′B′ = c/2. Thus, by considering ∆AB′A′, ∆BC′B′, and ∆CA′C′, we A | | get (by the triangle inequality), b/2 + c/2 ≥ 3x c/2 + a/2 ≥ 3y a/2 + b/2 ≥ 3z. Summing, we get p ≥ m. Exercise 1.71. Prove that the diagonals of a parallelogram bisect each other. Solution. Consider a parallelogram ABCD. Note that ∠BDA = ∠DBC, since they are opposite interior angles. Similarly, ∠CAD = ∠ACB, so by ASA, ∆ABD ∆CDB. (In passing, this proves that opposite sides of a ≡ parallelogram have equal lengths.) Let P be an arbitrary point on BD. Then, comparing the two congruent triangles, we get ∠DPA = ∠CP ′B, where P ′ is the point on BD such that | BP ′ | = | DP |. In particular, if P is the midpoint of DB, then P ′ = P . But then ∠CPA = 180◦. That is, the diagonal AC goes through the midpoint P of BD. Similarly, the diagonal BD bisects AC.

1.10

The Incircle, Excircles, and the Law of Cosines

Exercise 1.72. In a triangle ∆ABC, suppose ∠ABC = 60◦, a = 5, and c = 8. What is b? Solution. Using the Law of Cosines, we have b = a + c − 2ab cos(B) 2

b = 5 + 8 − 2(5)8(1/2) = 49 2

b = 7. Exercise 1.73. In a quadrilateral ABCD,| AB| = 3,| BC| = 11,| CD| = 8, |DA =| 4, and ∠BAD is a right angle. What is the area of the quadrilateral? [A] Solution. Draw the line BD. Note that ∆BAD is a right angle triangle with legs of length 3 and 4. Thus,| ∆BAD |= 6. The quadrilateral is the union of ∆BAD and ∆BCD. The hypotenuse BD of ∆BAD has length

CHAPTER 1. EUCLIDEAN GEOMETRY

5, so ∆BCD is a triangle with sides of length 5, 8, and 11. We can ﬁnd its area using Heron’s formula. Note that s = 12, so √ √ |∆BCD| = 12(12 − 5)(12 − 8)(12 − 11) = 4 21. √ Thus, the area of the quadrilateral ABCD is 6 + 4 21. Exercise 1.74. What is the area of ∆ABC if a = 3, b = 5, and c = 6? Solution. We will use Heron’s formula. Note that s = 7. Thus, √ √ |∆ABC| = 7(7 − 3)(7 − 5)(7 − 6) = 2 14. Exercise 1.75. What is the area of the incircle of ∆ABC if a = 5, b = 6, and c = 7? [A] Solution. Using Heron’s formula, √ √ |∆ABC| = 9(9 − 5)(9 − 6)(9 − 7) = 6 6. By Theorem 1.10.2, |∆ABC| = rs, where r is the inradius. Thus,

√ 9r = 6 6,

so r = 8/3. Thus, the area of the incircle is

8π 3

Exercise 1.76. In Figure 1.23(a), suppose |AC′| = |C′B| = |CE| = 2, |CD| = 3, and |BF | = 1. What is the area of ∆ABC? [A] Solution. Note that a couple of figures in Figure 1.23 were switched in the first printing. Let us first find the lengths of the sides of this triangle. Taking power of the point B, we find |BC′||BA| = |BF ||BD| 2(4) = 1|BD| |BD| = 8. Thus, a = |BC| = 5 and |FC| = 4. By power of the point C, we have |CA||CE| = |CD||CF | b(2) = 3(4) b = 6. We are given c = 4. Thus, by Heron’s formula, |∆ABC| =

√

√ 15 7 (15/2)(5/2)(7/2)(3/2) = . 4

1.10 THE INCIRCLE, EXCIRCLES, AND THE LAW OF COSINES D E C

F A A

C′

(a) Γ

B C

E F B

A C

D (c)

(b)

Figure 1.23. See Exercises 1.76, 1.77, and 1.79. Exercise 1.77. In Figure 1.23(b), suppose AD and EF are perpendiculars which intersect at the center of the circle. Suppose also that |AB| = 2 and |BC| = 3. Then what is |AC|? [A] Solution. By the Star Trek lemma, ∠ABC = 45◦. Cosines, we have

Using the Law of

b = a + c − 2ac cos(B) √ 2 b = 4 + 9 − 12 2/2 2

√ b = 13 − 6 2. Exercise 1.78. Explain why there is an absolute value around a −b cos C in Equation 1.1. [H] Solution. In our ﬁgure (Figure 1.33 of the text), we have drawn D between B and C, so a is larger than b cos C, so it appears that this quantity is always positive. However, the angle at B could be obtuse, in which case the quantity a—b cos C is negative. However, |DB is| a length, so is always positive. Hence, the need for absolute values.

CHAPTER 1. EUCLIDEAN GEOMETRY

Exercise 1.79. For a circle Γ and a point A outside the circle, let B and C be the points on Γ such that AB and AC are tangent to the circle Γ, as in Figure 1.23(c). Prove that the incenter I and the excenter Ia of ∆ABC both lie on Γ. [S] Exercise 1.80†. Prove that |∆ABC| = (s − a)ra. Solution. Consult Figure 1.24. Note that |∆ABC| = |∆ABIa | + |∆ACIa | − |∆BCIa | 1 = (cra + bra − ara) 2 = (s − a)ra. Exercise 1.81 †. Show that the distance from the vertex A to the tangent on the side AB of a. the incircle is s − a; b. the excircle centered at Ia is s; c. the excircle centered at Ic is s − b. Solution. Note that the distances from A to the points of intersection of the incircle on the sides AB and AC are the same. Let us call this length x. Similarly, let us call the distance from B and C to their appropriate tangent points y and z, respectively. Note that a = y + z, b = x + z, and c = x + y. Thus, 2s = a + b + c = 2x + 2y + 2z, so x = s − (y + z) = s − a. For part b, we are interested in where the excircle centered at Ia intersects the sides (or extended sides). This time, let x be the distance from A to where this excircle intersects the extended side AB (or equivalently, AC). Let y be the distance from B to the intersection on the side BC, and let z be the distance from C to the intersection on BC. Then we are interested in the length x. Note that a = y + z, b = x— z, and c = x − y, so summing, we get 2s = a + b + c = y + z + x − z + x − y = 2x. Hence x = s. For part c, note that C is a distance s away from the point of intersection of the circle centered at Ic and the side AC (by part b). Thus, since |AC| = b, the distance from A to this point is s − b.

1.10 THE INCIRCLE, EXCIRCLES, AND THE LAW OF COSINES

Ib A Ic I

Figure 1.24. A triangle together with its incircle and three excircles.

CHAPTER 1. EUCLIDEAN GEOMETRY

Exercise 1.82. Show that rrarbrc = |∆ABC|2.

[H][S]

Exercise 1.83. Show that 1 1 1 = 1. + + ra rb rc r Solution. For simplicity, let ∆ = |∆ABC|. By Exercise 1.80, ∆ = (s − a)ra, so ∆ ra

∆

= (s − a) + (s − b) + (s − c) = 3s − 2s = s =

∆

. r

Dividing through by ∆, we get the desired result. Exercise 1.84 (Theorem of Apollonius). In a triangle ∆ABC, let A′ be the midpoint of BC and let l = |AA′|. Prove that 2

2 2 b + c = 2l +

a2 2

Solution. Let θ = ∠AA′B, and let us apply the Law of Cosines to ∆AA′B and ∆AA′C: a2 − al cos θ 4 2 2 a2 ◦ c =l + − al cos(180 — θ) 4 a2 2 =l + + al cos θ. 4 2

b =l +

Adding, we get 2

2 2 b + c = 2l +

a2 2

Exercise 1.85. In this exercise, we are asked to ﬁnd a diﬀerent proof of the Law of Cosines. In ∆ABC, suppose without loss of generality that a ≥b. Draw the circle centered at C that goes through B. Extend the sides AC and AB. Find the power of the point A with respect to this circle. Conclude the Law of Cosines. Solution. Let the extended side AB intersect the circle at D; let BC intersect the circle again at E; and let AC intersect the circle at F and G, as in Figure 1.25. Let x = |BD|. By the power of the point A, |AF ||AG| = |AB||AD|

1.10 THE INCIRCLE, EXCIRCLES, AND THE LAW OF COSINES

D F A

Figure 1.25 (a − b)(a + b) = c(x − c) a − b = cx − c . 2

Since BE is a diameter, the angle at D is a right angle. Thus, x cos B = , 2a so a − b = c2a cos B − c 2

b = a + c − 2ac cos B. 2

Exercise 1.86*. Before we started using trigonometric functions, there was no need to define the measure of an angle. Come up with a suitable definition of the measure of an angle. Solution. This is a good question for discussion. Suppose two angles are defined by ∠ABC and ∠A′B′C′. It is clear that there exists an isometry which sends A′ to A and B′ onto the ray AC. We can either reflect this new angle through AC, or not, so that C′ is sent to a point on the opposite side of B. In this way, we have created an angle which is the union of the former two. If this sum is greater than a straight line, then let us ‘subtract’ a straight line in a similar fashion. Now, given an angle ∠ABC, we can define n∠ABC for any positive integer n. By subtracting straight lines, we can decide whether n∠ABC is greater than, less than, or equal to m straight lines. Thus, we can decide whether ∠ABC is greater than, less than, or equal to n/m of a straight line. In particular, if we have equality, then we will say the measure of ∠ABC is m/n of a straight line (or equivalently, mπ/n radians or (m/n)180◦). If n∠ABC is

CHAPTER 1. EUCLIDEAN GEOMETRY

never equal to m straight lines, then we instead look at the set {m/n : n∠ABC > m straight lines}. This set has a supremum, α. We will then say the measure of ∠ABC is α straight lines (or απ radians or α180◦). There are a couple of points which should probably be addressed, though it shouldn’t be too hard to convince ourselves that they are true. First, if m/n < p/q, and n∠ABC is greater than m straight lines, then q∠ABC is also greater than p straight lines. Similarly, if q∠ABC is less than p straight lines, then n∠ABC is less than m straight lines. In particular, ∠ABC is less than a straight line, so the set above has an upper bound, and hence, a supremum. (Note that we have deﬁned the measure of an angle to be no more than a straight line.)

1.11

The Circumcircle and Law of Sines

Exercise 1.87. Though we have deﬁned O to be the circumcenter, we have not shown that the circumcircle always exists. Prove that the perpendicular bisectors of two sides of a nondegenerate triangle ∆ABC intersect, and that this point is equidistant from all three vertices. Conclude that ∆ABC has a circumcircle and that the perpendicular bisectors are coincident. Solution. This question is not without merit, since it relies on the parallel axiom (and is not true in hyperbolic geometry). Since ∆ABC is nondegenerate, the lines AB and AC are distinct, so the perpendiculars at C′ and B′ are not parallel. Thus, they intersect at a point, say O. Note that ∆ABO is isosceles, with| OA| = |OB .| To see this, note that ∆OC′A ≡ ∆OC′B by SAS. Similarly, |OA | = | OC |. Thus, OB | |= | OC |. Now ∆OA′B ≡ ∆OA′C by SSS, so ∠OA′B = ∠OA′C, and must therefore be a right angle. That is, OA′ is the perpendicular bisector of BC. Hence, the three perpendicular bisectors intersect at O. Exercise 1.88. What are the areas of the incircle, circumcircle, and excircles of an equilateral triangle with sides of length 1? Solution. Let ∆ABC be an equilateral triangle. By SSS, ∆AA′B ≡ ∆AA′C, so the medians are the perpendicular bisectors. Hence O = G. Thus, √ √ 3 ′ R = |OA| = |GA| = (2/3)|AA | = (2/3)( 3/2) = . 3 Hence, the area of the circumcir√ cle is π/3. The area of the triangle is 3/4, and s = 3/2, so (using |∆ABC| = rs = ra(s — a) = rb(s − b) = rc(s − c) from Theorem 1.10.2 and Exercise 1.80) √ 3 r= 6

1.11 THE CIRCUMCIRCLE AND LAW OF SINES √ ra = rb = rc =

3 .

Thus, the area of the incircle is π/12, and the area of the excircles are 3π/4. Exercise 1.89. The Law of Sines (not the extended Law of Sines) merely states a b = c . = sin C sin A sin B There is a simpler proof of this result. Find it. [H] Solution. Let AD be the altitude of ∆ABC at A, and let h = |AD|. Then sin B =

opposite h = hypotenuse c

and

sin C =

h . b

Thus, h = c sin B = b sin C b c . = sin B sin C Similarly, by looking at the altitude at B, we get c a . = sin C sin A Combining these two results, we get the Law of Sines. Exercise 1.90†. Let D be the base of the altitude at A of the triangle ∆ABC. Let O be the circumcenter for ∆ABC, and let AO intersect the circumcircle at E, as in Figure 1.26(a). As before, let R be the circumradius. Show that bc |AD| = . 2R Conclude that abc |∆ ABC| = . 4R Hint: Show that ∆ABD is similar to ∆AEC. Or, express AD | in | terms of sin B and use the extended Law of Sines. [S] Exercise 1.91. Prove that 4R = ra + rb + rc − r. Solution. Let’s use Exercise 1.80 to expand the right hand side. As before, let ∆ = |∆ABC|. Then ∆ ∆ ∆ ∆ ra + rb + rc − r = + + − s−a s−b s−c s

CHAPTER 1. EUCLIDEAN GEOMETRY

A C

A P

O B

C D E B (a)

(b)

Figure 1.26. See Exercises 1.90 and 1.93. ∆ = ∆2 (s(s − b)(s − c) + s(s − a)(s − c) + s(s − a)(s − b) − (s − a)(s − b)(s − c)) 1 3 2 2 3 2 2 3 2 2 = (s − bs − cs + bcs + s − as − cs + acs + s − as − bs ∆ 3 2 2 2 + abs − s + as + bs + cs − abs − acs − bcs + abc) 1 3 2 = (2s − (a + b + c)s + abc) ∆ abc = ∆ = 4R. In the ﬁrst step above, we used Heron’s formula, and in the last step, we used Exercise 1.90. Exercise 1.92. Where is the circumcenter of a right angle triangle, and why? Solution. Since a right angle subtends a diameter, the diameter of the circumcircle is the hypotenuse. Thus, the circumcenter is the midpoint of the hypotenuse. Exercise 1.93 (The Four Coin Problem). Suppose three congruent circles meet at a common point P and meet in pairs at the points A, B, and C, as in Figure 1.26(b). Show that the circumcircle of ∆ABC has the same radius as the original three circles. Use a coin to draw three such circles, and note that the coin can be used as the circumcircle for the resulting triangle ∆ABC.

1.11 THE CIRCUMCIRCLE AND LAW OF SINES OB A

O P

Figure 1.27 Solution. Let’s get rid of the circles, and leave only the centers OA, OB, OC, and the points of intersection A, B, C, and P , as in Figure 1.27. Note that all the solid line segments in this ﬁgure are radii, so they are all equal (say, to r). Thus, we have three rhombuses. Now construct the line through A which is parallel to BOC, and the line through B which is parallel to AOC (two of the dotted lines in Figure 1.27). These lines intersect at O, and form a parallelogram. Thus,| BO | =| AOC| = r and | AO | = | BOC | = r. Note that OB is parallel to COA, since it is parallel to AOC, which is parallel to OBP , which is in turn parallel to COA. Since | OB | = r =| COA |, the segment OC is also parallel to BOA. Thus,| OC = | | BOA| = r. That is, the circle centered at O and with radius r passes through A, B, and C, as desired. Exercise 1.94. In a triangle ∆ABC, let O be the circumcenter, and let I be the incenter. Let the line AI intersect the circumcircle at D. Prove that ∆BID is isosceles. Solution. Please see Figure 1.28. Since I is the incenter, ∠CAI = A/2 ∠CBI = B/2. Note that ∠DBC = ∠DAC = A/2, since they subtend the same arc. Thus A+B ∠IBD = . Also, ∠BDI = ∠BDA = C, since they subtend the same 2 arc. Thus, A+B A+B A+B ∠BID = 180◦ − 2 −C =A+B+C− 2 −C = 2 . That is ∠BID = ∠DBI, so ∆BDI is isosceles, with |BD| = |ID|. Exercise 1.95*. Prove that a triangle is a right angle triangle if and only if r + 2R = s.

CHAPTER 1. EUCLIDEAN GEOMETRY A

Figure 1.28 B

Figure 1.29 Solution. Suppose ∆ABC is a right angle triangle with right angle at C, as in Figure 1.29. Let the incircle intersect the sides BC, AC, and AB at X, Y , and Z, respectively. Let x =| AY| = | AZ| , y = | BX | = | BZ |, and z = CX that c = x + y. Also, since C is a right angle, note | =| CY| . Note | that z = r, and that c = 2R. Thus, s = x + y + z = c + r = 2R + r. I suspect there must be a nicer solution than the one below which establishes the other direction. For the other direction, suppose ∆ABC has circumradius R. Let ∆A′BC be another triangle with A′ on the circumcircle of ∆ABC, as in Figure 1.30. Note that the triangles ∆A′BC and ∆ABC have the same side | BC | = a, ′ circumradius R, and angle ∠BAC = ∠BA C (since they subtend the same

1.11 THE CIRCUMCIRCLE AND LAW OF SINES A

A′

Figure 1.30 arc). For the quantities that differ, let b′, c′, s′, r′, x′, y′, and z′ be the appropriate quantities for ∆A′BC. Let θ = ∠A′BC∈ (0,π − A), and consider the function f (θ) = s′ − 2R − r′. By the Law of Sines, b′ = 2R sin B′ = 2R sin θ and c′ = 2R sin C′ = 2R sin(π − A − θ). Since a = y′ + z′ = y + z, we know x′ = s′ − a, so r′ r′ tan(A/2) = = x′ s′ − a r′ = (s′ − a) tan(A/2). Thus f (θ) = s′ − 2R − (s′ − a) tan(A/2) = (s′ − a)(1 − tan(A/2)) + a/2 − 2R = (R sin θ + R sin(π − A − θ) − a/2)(1 − tan(A/2)) + a/2 − 2R. Thus, f (θ) = kg(θ) + l, where k and l are constants, and g(θ) = sin θ + sin(π − A − θ). Differentiating g(t), we find dg = cos θ − cos(π − A − θ). dt This is zero if cos θ = cos(π − A − θ). Since the cosine function is one-to-one and onto on the domain of θ, which is (0,π − A), this equality implies θ = π − A − θ.

CHAPTER 1. EUCLIDEAN GEOMETRY

Thus, f (θ) has only one critical point. If there are three values of θ such that f (θ) = 0, then by Rolle’s theorem, there would be two critical points. Thus, f (θ) = 0 in at most two places. By the ﬁrst part of this solution, f (θ) = 0 for θ = π/2, and for π — A −θ = π/2. These two are never equal, so these are the only two values of θ such that f (θ) = 0. That is, s′ = 2R + r′ if and only if ∆A′BC is a right angle triangle.

1.12

The Euler Line

Exercise 1.96. Draw an arbitrary triangle ∆ABC. Through each vertex, draw a line parallel to the opposite side. These three lines form a new triangle. Where are the perpendicular bisectors of this new triangle? Use this to give a diﬀerent proof that the altitudes of ∆ABC intersect at a common point. C′

B′

A′

Figure 1.31 Solution. Let the lines through B and C intersect again at A′. Similarly, let the lines through A and B intersect at C′, and let the lines through A and C intersect at B′, as in Figure 1.31. Since the line at B is parallel to AC, the opposite interior angles ∠CBA′ and ∠BCA are congruent. Similarly, ∠BCA′ = ∠CBA, so by ASA, ∆ABC ≡ A′CB. In particular, A B′C| = BA | ′C =| AB| . Similarly, | | . Thus, | | C is the midpoint of A′B′. ′ ′ Similarly, A is the midpoint of B C and B is the midpoint of A′C′. Since A′B′ is parallel to AB, the perpendicular bisector of A′B′ is perpendicular to AB, and goes through C. That is, it is the altitude at C. Hence, since the perpendicular bisectors of ∆A′B′C′ are coincident, the altitudes of ∆ABC are coincident. Exercise 1.97. Suppose we are told that a triangle ∆ABC is isosceles with AB | =| AC| but| are given only the length of the median BB′ and the length of the altitude BE. Use this information to construct ∆ABC. Solution. First, suppose |BE| = |BB′|. Then the median at B is also the altitude at B, so |BA| = |BC|. That is, ∆ABC is equilateral.

1.12 THE EULER LINE

Otherwise,|BE |< BB E is the point on AC which is nearest | ′ , since | to B. But now, there isn’t quite enough information. There are actually two triangles with these properties – one with the angle at A greater than 60◦, and the other with the angle at A less than 60◦. We will describe how to construct both. Observe that the altitude is perpendicular to the base, and that a right angle inscribed in a circle subtends a diameter. Thus, we begin by drawing BB′ and the circle with diameter BB′. We draw a circle centered at B and with radius| BE |. This intersects our first circle at the point E (it doesn’t matter which of the two points of intersection we call E). Divide the segment BB′ into thirds, so that we can find the centroid G, which is on BB′ and satisfies |BG |= (2/3) BB | ′ . |Draw the circle centered at G which goes through B. Since AB = | | | AC |, we must also have | GB | =| GC|. Thus, this circle intersects the line B′E at C. Unfortunately, it intersects the line at two points, and both choices are valid. Once we choose C, we use the fact that B′ is the midpoint of AC – we construct the circle centered at B′ which goes through C, which intersects the line again at A. If the point we chose to be C is on the same side of B′ as E, we get a triangle with an angle at A which is greater than 60◦. Otherwise, we get a triangle whose angle at A is less than 60◦. Exercise 1.98. Let AD be the altitude of an arbitrary triangle ∆ABC. Prove that |AD| = 2R sin B sin C. Conclude that |∆ABC| = 2R sin A sin B sin C. 2

Solution. First, note the typo, corrected above and in the second printing. Let h = |AD|. Then, h sin B = , c so h = c sin B. By the extended Law of Sines, c = 2R sin C. Thus, h = |AD| = 2R sin C sin B. We now use

1 ah, 2 and the extended Law of Sines again to get |∆ABC| =

1 (2R sin A)(2R sin C sin B) 2 2 = 2R sin A sin B sin C.

|∆ABC| =

Exercise 1.99. Suppose the Euler line of a triangle ∆ABC passes through the vertex A. Prove that ∆ABC is either a right angle triangle or an isosceles triangle (or both).

CHAPTER 1. EUCLIDEAN GEOMETRY

Solution. Suppose H = A. Then the altitudes at B and C go through A. That is, A is a right angle, and hence, ∆ABC is a right angle triangle. Otherwise, H A, in which case we can talk about the line AH. Since AH is the Euler line, it goes through G, and hence is the median at A. Thus, the altitude at A goes through A′. In particular, AA′ is perpendicular to BC. Thus, ∆AA′B ∆AA ≡ ′C, by SAS. Hence, AB| =| AC | , | so ∆ABC is isosceles. Exercise 1.100. Let ∆ABC be a triangle with orthocenter H and circumcenter O. Let A′ be the midpoint of BC and let D be the base of the altitude from A. Suppose that HOA′D is a rectangle with sides | HO | = 11 and OA | ′ |= 5. What is the length |BC ?| This question (without Figure 1.32) was question A-1 on the 1997 Putnam Mathematical Competition. [H] A

A′

Figure 1.32 Solution. Draw the line AA′. Since HO is the Euler line, we know AA′ intersects HO at G. Note that ∆AHG ∼ ∆A′OG, with | HG| = 2| GO|. Thus, | AH |= 2 OA | ′ |= 10. Now, let R be the circumradius. Then, since OA = | |= OC | | R, and both ∆AHO and ∆OA′C are right angle triangles, we get R = |AH|2 + |HO|2 = 10 + 11 2

R = |OA′|2 + |A′C|2 = 5 + |A′C|2. 2

Thus, |A′C|2 = 100 + 121 − 25 = 196 = 14 . 2

Hence, |A′C| = 14, so |BC| = 28.

1.13

The Nine Point Circle

Exercise 1.101** (Feuerbach’s Theorem). Prove that the nine point circle of ∆ABC is tangent to the incircle and excircles of ∆ABC. The nine point circle is sometimes known as the Feuerbach circle.

1.13 THE NINE POINT CIRCLE

Solution. This exercise is misplaced. My intent was to inform the reader of Feuerbach’s result without having to prove it. I don’t really expect many to solve it but I do expect that it will peak some people’s curiosity. Since the result has a name, it is not be too difficult to find a proof in the literature. The proof I’ll give here I found in Smart’s book (see the text for an exact reference). Given my intent, the placement is ok. However, in our proof, we will use Exercise 1.103 and inversion in the circle, the subject of Sections 7.4 and 7.5. I probably should have repeated this exercise in Section 7.5, perhaps with a little bit of guidance. The first guiding exercise would be to prove the following lemma: Lemma 1. Let a circle Γ go through the vertices B and C of ∆ABC with A outside Γ. Let Γ intersect AB and AC at F and E respectively. Let the angle bisector of ∠BAC intersect AB at P and EF at Q. Then ∠AQE = ∠AP B. A

αγαγ βγ E Q F

βγ B

Figure 1.33 Proof. In Figure 1.33, let 2α = ∠BAC and let β = ∠ABC. Then ∠CEF = π−β since it is the opposite angle of an inscribed quadrilateral, so ∠AEF = β. Thus ∆AEQ ∼ ∆ABP . In particular, ∠AP B = ∠AQE.

CHAPTER 1. EUCLIDEAN GEOMETRY The next exercise would be to prove the following:

Lemma 2. Let the feet of the altitudes at B and C of ∆ABC be E and F respectively. Let the angle bisector at A intersect the opposite side at P. Then the line through P that is parallel to EF is tangent to both the incircle and the excircle opposite A. A

B′

Figure 1.34 Proof. Follow along in Figure 1.34. Let the line through P parallel to EF intersect AC at B′. Note that the circle with diameter BC goes through E and F since ∠BEC and ∠CFB are right angles. In particular, we have a circle that passes through B, C, E, and F , so by Lemma 1, ∠AP B′ = ∠AP B. Thus the line PB ′ is the reﬂection of the line BC through the angle bisector of A. Since the angle bisector goes through the centers I and Ia of both the incircle and the excircle opposite A, the circles are sent to themselves under this reﬂection. Thus, the points of tangency of these

1.13 THE NINE POINT CIRCLE

circles with the line AB are sent to points of tangency under the reﬂection. That is, the line PB ′ is tangent to both circles. And another exercise: Lemma 3. Let the incircle be tangent to BC at R and let the excenter opposite A be tangent to BC at S. Let A′ be the midpoint of BC. Then |A′R| = |A′S| =

|b − c| . 2

Proof. By Exercise 1.81, |BR| = s − b, |CS| = s − b, so |A′R| = ||A′B|− |BR|| = ||A ′ C|− |CS|| = |A′S| and

.. a .. |a − a − b − c + 2b| |b − c| − (s − b) = = . . . 2 2 2 Now for some inversion. |A′R| =

Lemma 4. Let D be the base of the altitude at A in ∆ABC. Let the incircle be tangent to BC at R, and let A′ be the midpoint of BC. Let the angle bisector at A intersect BC at P. Then P is the image of D under inversion in the circle with radius |A′R| centered at A′. Proof. There is probably a nicer way of proving this, but you’ll have to suﬀer with my solution. Let’s follow along in Figure 1.35. Let the altitude at A intersect BC at D. Let the incircle be tangent to BC at R, and let the excircle opposite A be tangent to BC at S. As is usual, let the incircle be centered at I with radius r, and let the excircle opposite A be centered at Ia with radius ra. Without loss of generality, we may assume b ≥ c so that b c b c |A′R| = |A′S| = − 2. For simplicity, let d = −2 . Finally, let x = |PR|. Then |PS| = 2d − x. Now for some math. Since ∆IRP ∼ ∆IaSP , we have |RP | |P S|

|IR| = r .

|Ia S|

Since |∆ABC| = rs = ra(s − a), we get |PS| =

sx s−a

Since |PS| + |PR| = 2d, we have sx s−a

+ x = 2d

CHAPTER 1. EUCLIDEAN GEOMETRY A

P A′ B

S C

Figure 1.35 (2s − a)x = 2d s−a 2d(s − a) x= . 2s − a Since ∆ADP ∼ ∆IRP ,

|AD| |IR|

|DP | |RP |

By again comparing diﬀerent area formulas, 1 2

|AD|a = rs |AD| 2s = , r a

so 2s|RP | 2sx = . a a We wish to show that P is the inverse image of D. Since they are on the |DP | =

1.13 THE NINE POINT CIRCLE

same ray, we need only show that the ratio |A′D||A′P | |A′R|2 is one. So, let us manipulate this ... |A′D||A′P | |A′R|2

(|PD| + |PS|− d)(d − x) d2 1 2sx = 2 + 2d − x − d (d − x) d a 1 = ad2 ((2s − a)x + ad)(d − x) 1 2 2 = + ((2s − a)d − ad)x − (2s − a)x ) ad2 (ad 1 2 2d(s − a) 4d (s − a)2 =1+ 2( ) 2s − a − (2 ) d s−a s − a (2s − a)2 ad2 2 2 2 1 4d (s − a) 4d (s − a)2 =1+ 2 − ad 2s − a 2s − a = 1.

Thus, P is indeed the image of D under inversion in the circle centered at A′ and with radius d = |A′R|. We are now ready to prove Feuerbach’s theorem. We invert in the circle centered at A′ with radius| A′R |. Since the incircle is tangent at R, it intersects the circle of inversion perpendicularly. Thus its image is itself. Since |A′S =| A′|R , the | excircle is also perpendicular to the circle of inversion, so its image is itself too. Since the nine point circle goes through A′, the image of it is a line l. Since the nine point circle goes through D, the line l goes through the image of D, which by Lemma 4 is P . Since inversions preserve angles, the angle l makes with BC is the same as the angle the nine point circle makes with BC at the point D. This is the same as the angle it makes at C′. That is, it is the angle the tangent at C′ makes with BC. By Exercise 1.103, the tangent at C′ is parallel to EF . Thus, l is a line through P that is parallel to EF . By Lemma 2, that line is tangent to both the incircle and the excircle centered at Ia. Thus, the nine point circle is tangent to the incircle and the excircle centered at Ia. Since there is nothing special about a, the nine point circle is also tangent to the other excircles. Exercise 1.102**. Prove that the center of the nine point circle lies on the Euler line. Solution. The diﬃculty of this exercise is perhaps a bit overrated. Let us begin by recalling some facts we learned in the proof that O, G, and H are collinear. We noted that AH is parallel to A′O. We saw that ∆AHG ∼

CHAPTER 1. EUCLIDEAN GEOMETRY A A′′ H P O B

A′

Figure 1.36 ∆A′OG with HG | =| 2 GO | . |Thus, AH | = | 2 OA | ′ . |Since A′′ is the ′′ ′ midpoint of AH, we get that| A H| = OA let P be the intersection | . Now, | of A′′A′ with the Euler line HO, as in Figure 1.36. Since AH is parallel to A′O, we know ∠P A′′H = ∠P A′O. Thus ∆P A′′H ∼ ∆P A′O (the vertical angles at P are equal too). Since |A′′H| = |A′O|, these triangles are in fact congruent. In particular, |A′′P | = |P A′|, so P is the midpoint of A′A′′. Now, let us recall something we learned in the proof of Theorem 1.13.1 (the nine point circle). We saw that C′C′′ is a diameter, so similarly, A′A′′ is a diameter. Thus, P is the center of the nine point circle. We have therefore shown that the center of the nine point circle lies on the Euler line, and further, that it is the midpoint of OH. Exercise 1.103**. Prove that the tangent to the nine point circle at the point C′ is parallel to the line DE. Solution. The diﬃculty of this exercise is perhaps a bit overrated too. Let us ﬁrst look for an equivalent statement. Since DE is a chord of the nine point circle, its perpendicular bisector goes through the center. The tangent to the nine point circle at C′ is perpendicular to the radius of the circle at C′. Thus, the tangent at C′ is parallel to DE if and only if the perpendicular bisector to DE goes through C′. That is, the tangent at C′ is parallel to DE if and only if ∆C ′ DE is isosceles with equal sides |C ′ D| and |C ′ E|. Now, consider the circle centered at C′ that goes through A, as in Figure 1.37. Since C′ is the midpoint of AB, this circle goes through B too. Since AB is a diameter and angles ∠AEB and ∠ADB are right angles, the points D and E must be on this circle. But then both C′D and C′E are radii, so have equal lengths.

1.14 PEDAL TRIANGLES AND THE SIMSON LINE A E

C′

Figure 1.37

1.14

Pedal Triangles and the Simson Line

Exercise 1.104. The proof of Lemma 1.14.2 works for any diagram. Prove Simson’s result directly by adapting the proof of Lemma 1.14.2 for a point P on the circumcircle.

Z A P Y

Figure 1.38 Solution. Without loss of generality, we may assume P is on the arc between A and C, as in Figure 1.38. Following the proof given in the text, we wish to use Lemma 1.14.2 to establish that ∠XZY = 0. Since we are not using

CHAPTER 1. EUCLIDEAN GEOMETRY

Lemma 1.14.2 directly, let us follow its proof. First, we draw the line CP , and let it intersect the (extended) line AB at C′. Then ∠AP B = ∠C ′ PB − ∠C′P A. Note that ∠C ′ PB is an exterior angle to ∆BPC, so ∠C ′ PB = ∠PBC + ∠PCB. Note that AXP Z is cyclic, since the angles at X and Z are right angles. Thus, ∠PBC = ∠PBX = ∠PZX, since they subtend the same arc. Thus ∠C ′ PB = ∠PZX + ∠PCB.

(1.1)

Similarly, ∠C ′ PA is an exterior angle of ∆AP C, so ∠C ′ PA = ∠P AC + ∠PCA. Since AZP Y is cyclic, ∠P AC = ∠P AY = ∠PZY .Thus ∠C ′ PA = ∠PZY + ∠PCA.

(1.2)

Combining Equations 1.1 and 1.2, we get ∠AP B = ∠PZX + ∠PCB − ∠PZY − ∠PCA = (∠PZX − ∠PZY ) + (∠PCB − ∠PCA) = ±∠XZY + ∠BCA. The sign of ∠ZXY depends on whether ∠PZX is greater than ∠PZY or not. But ∠AP B = ∠ACB, since they subtend the same arc. Thus, 0 = ∠XZY. That is, ∆XZY is a degenerate triangle, so X, Y , and Z are collinear. Exercise 1.105. Let D be the center of the square constructed on the hypotenuse of a right angle triangle ∆ABC with right angle at C. Prove that ∠ACD = 45◦. [H] Solution. Note that the diagonals of a square meet perpendicularly, so ∠ADB = 90◦. Thus, ADBC is a cyclic quadrilateral, since the angles at C and D sum to 180◦. But then, ∠ACD = ∠ABD, since they subtend the same arc, and clearly ∠ABD = 45◦. Exercise 1.106. Let H′ be the reﬂection of the orthocenter H of ∆ABC through the line BC. Prove that ABCH′ is a cyclic quadrilateral. Solution. Let the altitudes at B and C intersect AC and AB at E and F , respectively. Then AEHF is a cyclic quadrilateral, since the angles at E and F are both right angles. Thus, ∠EHF + ∠CAB = 180◦. Since ∠EHF = ∠CHB = ∠CH′B, the angles ∠CAB and ∠CH′B are also supplementary, so ABH′C is a cyclic quadrilateral.

1.14 PEDAL TRIANGLES AND THE SIMSON LINE

Exercise 1.107. The Star Trek lemma is worded in terms of arcs instead of interior angles since we normally consider an angle to measure no more than 180◦. Reword the Star Trek lemma using oriented interior angles. Check that the proof given in Section 1.6 works for any diagram if we understand the angles to be oriented. Solution. Let A, B, and C be points on a circle centered at O. Then 1 ∠ABC = 2 ∠AOC, where ∠ABC is measured modulo 180◦, while ∠AOC is measured modulo 360◦. Exercise 1.108. Reword the results in Exercises 1.34 and 1.35 using oriented angles and in such a way that the wording for each is identical. Solution. Suppose A, A′, B, and B′ lie on a circle centered at O. Suppose AA′ and BB′ intersect at P . Then ∠AOB + ∠A′OB′ , 2 where the angle at P is measured modulo 180◦, while the angles at O are measured modulo 360◦. ∠AP B =

Exercise 1.109†(Ptolemy’s Theorem). Suppose that ABCD is a cyclic quadrilateral. Prove that |AC||BD| = |AB||CD| + |BC||DA|. Hint: Draw AE so that E is on BD and ∠BAE = ∠CAD, as in Figure 1.39(a). Solution. Note that ∠ABD = ∠ACD, since they subtend the same arc. Thus, ∆ABE ∼ ∆ACD, so |AB|

|BE|

|AC| |CD| |AB||CD| = |AC||BE|. Note that ∠BEA and ∠DEA are supplementary. Since ABCD is cyclic, ∠ABC and ∠ADC are also supplementary. Since ∠BEA = ∠ADC, we therefore have ∠DEA = ∠ABC. Note that ∠BCA = ∠BDA, since they subtend the same angle. Thus, ∆ABC ∼ ∆AED, so |AC|

|BC|

|AD| |ED| |AC||ED| = |AD||BC|. Since |BE| + |ED| = |BD|, summing these two results, we get |AB||CD| + |AD||BC| = |AC||BE| + |AC||ED| = |AC||BD|.

CHAPTER 1. EUCLIDEAN GEOMETRY B A E D

A C

β α

D C (a)

(b)

Figure 1.39 Exercise 1.110. Suppose triangle ∆ABC has sides of length a = 6, b = 5, and circumradius R = 4. What is the length c? [A] Solution. Consider ∆ABC, inscribed in its circumcircle. Let the diameter at C intersect the circle again at D, giving a cyclic quadrilateral ADBC. Since CD is a diameter, it has length 8 and the angles ∠CAD and ∠CBD are both right. Thus, |AD|2 + |AC|2 = |CD|2 2 2 |AD|2 + 5 = 8 |AD| =

√

|BD| + |BC| = |CD|2 2 2 |BD|2 + 6 = 8 2

|BD| =

√

√ 28 = 2 7.

Thus, by Ptolemy’s theorem, |AB||CD| = |AC||BD| + |BC||AD| √ √ 8c = 5 ·√2 7 +√6 39 5 7 + 3 39 c= . 4 Exercise 1.111. Use Ptolemy’s theorem and Figure 1.39(b) to prove the angle sum formula for sines (for acute angles α and β.) That is, show sin(α + β) = sin α cos β + cos α sin β.

1.14 PEDAL TRIANGLES AND THE SIMSON LINE

Solution. In Figure 1.39(b), let 2R = 1. Then, applying the extended Law of Sines to ∆ACD, we have 2R =

|AC| sin(α + β)

|AC| = sin(α + β) Since DB is a diameter (which we set equal to one), the angles at A and C are right angles. Thus, sin α = |BC| cos α = |CD| sin β = |AB| cos β = |AD|. Finally, by Ptolemy’s theorem, we get |AC||BD| = |AB||CD| + |BC||AD| sin(α + β) = sin β cos α + sin α cos β. Exercise 1.112. Extend the result in Exercise 1.111 to any angles. Make π the substitution α′ = 2 − α and β′ = −β to derive the angle sum formula for cosines. Solution. By Exercise 1.111, we know the angle sum formula for α ∈ [0, π/2) and β ∈ [0, π/2). We also know several properties of sines and cosines – in particular, sin(−α) = − sin α sin(π − α) = sin α sin(2π + α) = sin α sin(π/2 − α) = cos α cos(−α) = cos α cos(π − α) = − cos α cos(2π + α) = cos α cos(π/2 − α) = sin α. We begin by adapting the argument in Exercise 1.111 to expand sin(α − β) for 0 ≤ β ≤ aa < π/2. In Figure 1.39(b), reﬂect A through BD. Then, by the extended Law of Sines, 2R =

|AC| sin(α − β)

Thus, setting 2R = 1, we get | AC| = sin(α − β). By Ptolemy’s theorem, we have |AB||CD| = |BD||AC| + |AD||BC|

CHAPTER 1. EUCLIDEAN GEOMETRY |AC||BD| = |AB||CD|− |AD||BC| |AC| = cos β sin α − sin β cos α.

Thus, for α ∈ [0, π/2) and β ∈ (−π/2, 0], we have sin(α + β) = sin(α − (−β)) = sin α cos β − sin(−β) cos α = sin α cos β + sin β cos α, if α ≥ −β; and if α < −β, then sin(α + β) = − sin(−β − α) = −(sin(−β) cos(−α) + sin(−α) cos(−β) = −(− sin β cos α − sin α cos β) = sin α cos β + sin β cos α. Thus, we have extended the result to the domain α ∈ [0, π/2) and β ∈ (−π/2, π/2). Note that, if α = π/2, then sin(π/2 + β) = sin(π − π/2 − β) = cos β = sin(π/2) cos β + sin β cos(π/2). Similarly, if β = ±π/2, the result holds. Now, suppose α ∈ [π/2, π] and β ∈ [−π/2, π/2]. Then sin(α + β) = sin(π − α − β). Since π —α ∈ [0, π/2] and − β ∈ [π/2, π/2], we can use the result we already have, so sin(α + β) = sin(π − α) cos(−β) + sin(−β) cos(π − α) = sin α cos β + (− sin β)(− cos α). This extends the result to α ∈ [0, π] and β ∈ [−π/2, π/2]. Suppose now that α ∈ [−π, 0] and β ∈ [−π/2, π/2]. Then 2π − α ∈ [0, π] and −β ∈ [−π/2, π/2], so sin(α + β) = − sin(2π − α − β) = −(sin(2π − α) cos(−β) + sin(−β) cos(2π − α)) = sin α cos β + sin β cos α. If α ∈ [−π, π] and β ∈ [π/2, 3π/2], then sin(α + β) = sin(π − α − β) = sin((−α) + (π − β)). Note that — α ∈ [− π, π] and π− β ∈ [− π/2, π/2], so we can apply the result we have so far, and get sin(α + β) = sin(−α) cos(π − β) + sin(π − β) cos(−α) = sin α cos β + sin β cos α.

1.14 PEDAL TRIANGLES AND THE SIMSON LINE

Finally, for any α and β, there exist integers n and m such that α + 2m and β + 2n are in the domains [−π, π] and [−π/2, 3π/2], respectively. Then sin(α + β) = sin(α + 2mπ + β + 2nπ) = sin(α + 2mπ) cos(β + 2nπ) + sin(β + 2nπ) cos(α + 2mπ) = sin α cos β + sin β cos α. Let us now turn our attention to the angle sum formula for cosines: cos(α + β) = sin(π/2 − α − β) = sin(π − α) cos(−β) + sin(−β) cos(π/2 − α) = cos α cos β − sin β sin α. Exercise 1.113. Prove the converse of Exercise 1.111. That is, use Figure 1.39 and the angle sum formula for sines to prove Ptolemy’s theorem. Solution. Let us ﬁrst remark that the angle sum formula for cosines follows from the angle sum formula for sines (see the last paragraph of the solution to the previous exercise). In Figure 1.39(a), let α = ∠BAC, β = ∠CAD, γ = ∠DBA, and δ = ∠BCA. Note that ∠CBD = ∠CAD = β, since they subtend the same arc. Also, δ = π — α β−γ. − By the extended Law of Sines, looking at ∆ABC, |AC| = 2R sin(∠ABC) = 2R sin(β + γ). Similarly, |BD| = 2R sin(α + β) |AB| = 2R sin δ = sin(π − α − β − γ) = sin(α + β + γ) |CD| = 2R sin β |AD| = 2R sin γ |BC| = 2R sin α. So, let us consider the quantity |AC||BD|− |AD||BC|− |AB||CD| = 4R (sin(β + γ) sin(α + β) − sin γ sin α − sin δ sin β) 2

= 4R ((sin β cos γ + sin γ cos β) sin(α + β) − sin γ sin α 2

—

(sin(α + β) cos γ + sin γ cos(α + β)) sin β) = 4R (sin γ cos β sin(α + β) − sin γ sin α − sin γ sin β cos(α + β) 2 = 4R (sin γ cos β(sin α cos β + sin β cos α) 2

—

sin γ sin β(cos α cos β − sin α sin β) = 4R (sin α sin γ cos β − sin γ sin α + sin γ sin α sin β) = 0. 2

CHAPTER 1. EUCLIDEAN GEOMETRY

Exercise 1.114. There are other proofs of the angle sum formulas. Use Figure 1.40 to come up with one of them. Though it is not necessary to notice any cyclic quadrilaterals, the argument might be made neater by doing so. For which angles is this proof valid? [S] B

A β O

E α DP

Figure 1.40. See Exercise 1.114. Exercise 1.115 (Brahmagupta’s Formula). Heron’s formula generalizes to cyclic quadrilaterals in the following way. Let a, b, c, and d be the 1 lengths of the four sides of the cyclic quadrilateral. Let s = 2(a + b + c + d). Show that the area of the quadrilateral is √ (s − a)(s − b)(s − c)(s − d). Solution. It is not immediately clear how to approach this problem. Though it is clearly a generalization of Heron’s formula, it is not obvious that there is a generalization of the proof of Heron’s formula. There is, in fact, such a generalization. Let us ﬁrst set some notation: Let a| = |AB , b |= BC | , c =| CD |, and d = |DA .| Then 2s = a + b + c + d. Note that a + b + c− d = 2(s − d), and so on. Finally, let ∆ be the area of ABCD. Perhaps the key observation is that opposite angles of an inscribed quadrilateral are supplementary. We think of the area ∆ as the sum of the areas of ∆ABD and ∆BCD, and start following the proof of Heron’s formula: ∆ = |∆ABD| + |∆BCD| 1 1 = ad sin A + bc sin(π − A) 2 2 1 = (ad + bc) sin A. 2 We now use the Law of Cosines to ﬁnd sin A: |BD|2 = a + d − 2ad cos A 2

1.14 PEDAL TRIANGLES AND THE SIMSON LINE |BD|2 = b + c − 2bc cos(π − A) 2 2 = b + c + 2bc cos A 2

a + d − 2ad cos A = b + c + 2bc cos A 2 2 2 2 a +d −b −c cos A = . 2(ad + bc) 2

Thus, 1

a2 + d2 − b2 − c 2 2(ad + bc)

(ad + bc) 1 − 2 1√ = 4(ad + bc)2 − (a2 + d2 − b2 − c2)2 4 1√ = (2ad + 2bc + a2 + d2 − b2 − c2)(2ad + 2bc − a2 − d2 + b2 + c2) 4 1√ = ((a + d)2 − (b − c)2)((b + c)2 − (a − d)2) 4 1√ = (a + d + b − c)(a + d − b + c)(b + c + a − d)(b + c − a + d) 4 √ = (s − c)(s − b)(s − d)(s − a).

∆=

Exercise 1.116†. Let ∆XY Z be the pedal triangle with respect to P and a triangle ∆ABC. Show that |Y Z| = |AP | sin A.

[H]

C′

Z P

Figure 1.41

CHAPTER 1. EUCLIDEAN GEOMETRY

Solution. Consider ∆AY Z whose circumcircle is shown in Figure 1.41. By the extended Law of Sines, 2R =

|Y Z| |Y Z| = . sin(∠Y AZ) sin A

Since ∠AY P = 90◦, AP is the diameter of the circumcircle, so| AP |= 2R. Thus, |Y Z| = 2R sin A = |AP | sin A. Exercise 1.117. Let ∆ABC be an equilateral triangle. Suppose D is a point outside the triangle such that ∠ADB = 120◦. Show that |CD| = |AD| + |BD|. Solution. Let the sides of the equilateral triangle have length s. Suppose D is on the same side of AB as C. Then ∠DAB < 180◦−120◦ = 60◦. Similarly, ∠DBA < 60◦, so D is inside ∆ABC. Thus, C and D must be on opposite sides of AB, so CADB is a convex quadrilateral. Since ∠ACB + ∠ADB = 60◦ + 120◦ = 180◦, the quadrilateral is cyclic. Thus, we can apply Ptolemy’s theorem: |CD||AB| = |AC||BD| + |AD||BC| |CD|s = s|BD| + |AD|s |CD| = |BD| + |AD|. Exercise 1.118**. Let ABCDEF be a convex hexagon with | AB| = |BC | = |CD ,| DE | | = EF | |= F|A , | and ∠BCD = ∠EFA = 60◦. Let G and H be two points in the interior of the hexagon such that ∠AGB = ∠DHE = 120◦. Prove that |AG| + |GB| + |DH| + |HE| + |GH|≥ |CF |. This was Question 5 on the 1995 International Mathematical Olympiad exam. Solution. Since ∠BCD = 60◦ and| BC| = CD | ,| ∆BCD is an equilateral triangle. For similar reasons, ∆EFA is an equilateral triangle. Thus, the hexagon is composed of a ‘kite’ ABDE, which has equal sides AB | | = | BD| and |DE = EA , and on the sides BD and AE, there are placed two | | | equilateral triangles, as in Figure 1.42. Let G′ and H′ be the reﬂection of the points G and H through the line BE, the axis of symmetry of the kite. Then we must equivalently show that |BG′| + |G′D| + |AH′| + |EH′| + |G ′ H ′ |≥ |CF |. By Exercise 1.117, | BG | ′ |+ G′|D =| G′C| , and| AH′| + |H′E |= H|′F . | (Note that, since G is inside the hexagon, G′ is inside the reﬂection of the

1.15 MENELAUS AND CEVA B C

A G

F H

D E

Figure 1.42 hexagon, and hence outside ∆BCD, so Exercise 1.117 is applicable.) Thus, we must equivalently show that |CG′| + |G′H′| + |H′F | ≥ |CF |. But the left hand side is just the length of the segmented path CG′H′F joining C and F , which must be at least the length| CF |of the line segment between C and F .

1.15

Menelaus and Ceva

Exercise 1.119. Prove the trigonometric version of Ceva’s theorem. A α α′ F E

P β′

γ′

γ C

Figure 1.43 Solution. Let θ = ∠ADB in Figure 1.43. Then, applying the Law of Sines to ∆ADB, we get |BD| c = . sin α sin θ Applying the Law of Sines to ∆ADC, we get |DC| sin α′

b sin(π − θ)

CHAPTER 1. EUCLIDEAN GEOMETRY

Combining these two results, and noting that sin θ = sin(π − θ), we get |BD| |DC|

c sin α . b sin α′

Similarly, |AF | |FB| |CE|

b sin γ

a sin γ′ a sin β = . |EA| c sin β′

Thus, AD, BE, and CF are coincident if and only if 1=

|AF | |BD| |CE| |FB| |DC| |EA|

b sin γ c sin α a sin β . a sin γ′ b sin α′ c sin β′

That is, they are coincident if and only if sin γ sin α sin β = sin γ′ sin α′ sin β′. Exercise 1.120. In Figure 1.44, show that |∆ABP | |BD| = . |∆AP C| |DC| Use this to come up with a proof of Ceva’s theorem which does not use Menelaus’ theorem. A

F E

Figure 1.44 Solution. Note the slight modiﬁcation to this exercise (corrected in the second printing). Let h be the altitude at A of ∆ABC (see Figure 1.44). Then |∆ABD| =

1 2

|BD|h

1.15 MENELAUS AND CEVA 1

|∆ADC| =

|DC|h.

Let h′ be the altitude at P of ∆PBC. Then 1 |BD|h′ 2 1 |∆PDC| = |DC|h′. 2

|∆PBD| =

Thus, |∆ABP | |∆AP C|

|∆ABD|− |∆PBD| |∆ADC|− |∆P DC|

|BD|(h − h′) |DC|(h − h′)

|BD|

. |DC|

Furthermore, if we adopt the convention that the ratio of the areas of two triangles is positive if they have the same orientation, and negative otherwise, then this equality is valid for signed ratios as well (hence the slight modiﬁcation mentioned above). Similarly, |∆BCP | |∆BPA|

|CE|

|∆CAP |

and

|EA|

|∆CPB|

|AF | |FB|

Now, suppose D, E, and F are points on the sides AB, BC, and CA, respectively, of ∆ABC, and suppose AD, BE, and CF are coincident. Then |AF | |BD| |CE| |∆CAP | |∆ABP | |∆BCP | = = 1. |FB| |DC| |EA| |∆CPB| |∆AP C| |∆BPA| To prove the converse, use the argument in the text (it doesn’t use Menelaus’ theorem). Exercise 1.121. What is the analogue of Menelaus’ theorem for the case when D (in Figure 1.45) is at inﬁnity? A

B′ F A′

C′ D

Figure 1.45

CHAPTER 1. EUCLIDEAN GEOMETRY

Solution. By saying D is at infinity, we mean we consider the limiting case as D goes to infinity. As D goes to infinity, the lines FE and BC become parallel. Also as D goes to infinity, the ratio |BD| |BC| + |CD| =− |DC| |CD| goes to − 1 (where we have taken D to go to infinity on the other side of C as B). Thus, Menelaus’ theorem says |AF | |CE| (−1) = −1 |EA| |FB| |AF | |AE| = . |FB| |EC| That is, we get Theorem 1.7.1. Exercise 1.122. Use Ceva’s theorem to show that the medians intersect at a common point. Solution. The medians are Cevians which go to the midpoints of the opposite sides. But clearly, |AC′| |BA′| |CB′| |C′B| |A′C| |B′A|

= 1,

so by Ceva’s theorem, the medians are coincident. Exercise 1.123. Use Ceva’s theorem to show that the altitudes intersect at a common point. Solution. Note that |AF | = b cos A and |FB| = a cos B, and so on, so |AF | |BD| |CE| |FB| |DC| |EA|

b cos A c cos B a cos C a cos B b cos C c cos A

= 1.

Exercise 1.124. Use Ceva’s theorem to show that the angle bisectors intersect at a common point. [H] Solution. By the angle bisector theorem (Exercise 1.43), the ratio |BD| |DC|

|AB| |AC|

and so on. Thus, |AF | |BD| |CE| |FB| |DC| |EA|

|CA| |AB| |BC| |CB| |AC| |BA|

= 1,

so by Ceva’s theorem, the angle bisectors are coincident. (There may be some question about the signs, but angle bisectors always intersect the opposite sides between the vertices, so the ratios are all positive.)

1.15 MENELAUS AND CEVA

Exercise 1.125 (The Gergonne Point). Let the incircle of a triangle ∆ABC be tangent to the sides BC, AC, and AB at, respectively, the points D, E, and F . Show that AD, BE, and CF are coincident. This point is known as the Gergonne point. Solution. Since AE and AF are tangents to the incircle at E and F , we know |AE| = |AF | (a length we have called x in the past). Similarly, |BD| = |BF | = y and |CD| = |CE| = z. Thus, |AF | |BD| |CE| |FB| |DC| |EA|

xy z yzx

= 1,

so by Ceva’s theorem, the Cevians AD, BE, and CF are coincident. Exercise 1.126 (The Nagel Point). Let the excircle opposite A be tangent to BC at D. Define E and F similarly. Show that AD, BE, and CF are coincident. This point is known as the Nagel point. Solution. By Exercise 1.81, |AE| = s − c, |AF | = s − b, and so on. Thus, |AF | |BD| |CE| (s − b) (s − c) (s − a) = = 1, |FB| |DC| |EA| (s − a) (s − b) (s − c) so by Ceva’s theorem, the lines AD, BE, and CF are coincident. Exercise 1.127. Let ∆ABC be a triangle. Suppose D, E, and F are points on the sides BC, AC, and AB, respectively, and such that AD, BE, and CF are coincident. Let the reflection of the ray AD through the angle bisector of ∠BAC intersect BC at D′. Similarly define E′ and F ′, as in Figure 1.46. Show that AD′, BE′, and CF ′ are coincident. A F E

E′

F′

D′

Figure 1.46. See Exercise 1.127.

CHAPTER 1. EUCLIDEAN GEOMETRY

Solution. We will use the trigonometric version of Ceva’s theorem. Let α, α′, and so on, be the appropriate angles in ∆ABC, as labeled in Figure 1.43. Since ∠BAD = α, its reﬂection through the angle bisector, ∠CAD′, is also equal to α. Thus, AD′, BE′, and CF ′ are coincident if and only if sin α sin β sin γ = sin α′ sin β′ sin γ′. But this equality holds by trig Ceva applied to the Cevians AD, BE, and CF . Exercise 1.128. Let AD be an altitude to ∆ABC. Prove |AB|2 − |BD|2 = |AC|2 − |CD|2. Solution. By the Pythagorean theorem applied to ∆ABD, |AB|2 = |BD|2 + |AD|2. Applied to ∆ACD, |AC|2 = |CD|2 + |AD|2. Subtracting, we get |AB|2 − |AC|2 = |BD|2 − |CD|2 |AB|2 − |BD|2 = |AC|2 − |CD|2. Exercise 1.129. Let X, Y , and Z be points on the sides BC, AC, and AB, respectively, of a triangle ∆ABC. Prove that the perpendiculars to the sides of ∆ABC through the points X, Y , and Z are concurrent if and only if |AZ|2 − |ZB|2 + |BX|2 − |XC|2 + |CY |2 − |Y A|2 = 0.

[H]

Solution. Let P be a point, and let X, Y , and Z be the bases of the perpendiculars from P to the sides BC, AC, and AB, respectively. Applying Exercise 1.128 to ∆ABP (with the altitude at P ) we get |AP |2 − |AZ|2 = |BP |2 − |ZB|2 |AZ|2 − |ZB|2 = |AP |2 − |BP |2. Similarly,

|BX|2 − |XC| 2 = |BP | 2 − |CP | 2 |CY | 2 − |Y A| 2 = |CP | 2− |AP | 2.

Adding these three together, we get |AZ|2 − |ZB|2 + |BX|2 − |XC|2 + |CY |2 − |Y A|2 = |AP |2 − |BP |2 + |BP |2 − |CP |2 + |CP |2 − |AP |2 = 0.

1.15 MENELAUS AND CEVA

For the other direction, suppose the perpendiculars at X and Y intersect at P , and that the perpendicular to BC through P intersects BC at Z′. Then, since we have both |AZ|2 − |ZB|2 + |BX|2 − |XC|2 + |CY |2 − |Y A|2 = 0 |AZ ′ |2 − |Z ′ B|2 + |BX|2 − |XC|2 + |CY |2 − |Y A|2 = 0, we get |AZ|2 − |ZB|2 − |AZ′|2 + |Z′B|2 = 0. This can happen only if Z = Z′. Exercise 1.130. Generalize the previous exercise: Let X, Y , and Z be any points in the plane. Prove that the perpendiculars to the sides BC, AC, and AB through the points X, Y , and Z, respectively, are concurrent if and only if |AZ|2 − |ZB|2 + |BX|2 − |XC|2 + |CY |2 − |Y A|2 = 0. Solution. Let the perpendicular to BC through X intersect BC at X′. Deﬁne Y ′ and Z′ in a similar fashion. Applying Exercise 1.128 to ∆BCX with the altitude at X, we have |BX|2 − |BX ′ |2 = |CX|2 − |CX ′ |2 |BX|2 − |XC|2 = |BX ′ |2 − |CX ′ |2 . Similarly, |AZ| 2 − |ZB| 2 = |AZ |′ 2− |Z B| ′ 2 |CY | 2 − |Y A| 2 = |CY |′ 2− |Y A| ′ 2. Thus, the perpendiculars at X, Y , and Z are concurrent if and only if the perpendiculars at X′, Y ′, and Z′ are concurrent, which by the previous exercise, are concurrent if and only if |AZ ′ |2 − |Z ′ B|2 + |BX ′ |2 − |X ′ C|2 + |CY ′ |2 − |Y ′ A|2 = 0. But by our relations above, this is equivalent to |AZ|2 − |ZB|2 + |BX|2 − |XC|2 + |CY |2 − |Y A|2 = 0.

Chapter 2

Geometry in Greek Astronomy 2.1

The Relative Size of the Moon and Sun

Exercise 2.1. Express RM , RS, and DS in terms of θ and DM . Solution. Since

DM D 2π = S = , RM RS 720

and

RM = sin θ, RS

we have 720DM 2π 720DM RM RS = = sin θ 2π sin θ 2πRS DM DS = = . 720 sin θ

RM =

Exercise 2.2. Express RS, RM , DS, and DM in terms of θ and DE. Solution. Since

(1 + sin θ)DE , 3.5 and using the results from Exercise 2.1, we have DM =

RM =

720(1 + sin θ)DE

7π 720(1 + sin θ)DE RS = 7π sin θ 75

CHAPTER 2. GEOMETRY IN GREEK ASTRONOMY DS =

(1 + sin θ)DE 3.5 sin θ

Exercise 2.3. Explain why the ﬁgure DE− 2.5DM is an approximation. That is, what facts have we trivialized? Solution. We trivialized a couple of facts. First, we assumed that the tangent to the Earth and Sun is parallel to the line joining their centers, which is clearly not the case, though the lines are almost parallel. We also assumed that the distance from the moon to the line perpendicular to the line through the centers of the Earth and Sun is constant. Since the moon travels in a circular (or elliptical) path around the Earth, this too is not the case. Exercise 2.4. It is fairly easy to measure the angle the moon subtends. On a clear night, use a stopwatch to time the moonrise. That is, measure the amount of time between when a sliver of the moon is ﬁrst visible on the horizon and when the moon is completely above the horizon. Explain how this gives the angular measure of the moon. We know that the sun has approximately the same angular measure since solar eclipses occur and when they do, the shadow of the moon is so small on the Earth. Solution. Let’s turn this around. Given that the moon subtends 0.5◦, we can calculate how long it takes the moon to rise. We note that 24 hrs = 360◦. Thus, the time it takes the moon to rise is 0.5◦24 hrs 2 hrs = = 2 min . 360◦ 60

2.2

The Diameter of the Earth

Exercise 2.5. Use the above measurements and assumptions to ﬁnd the circumference of the Earth.

A 7.2 O

Figure 2.1

2.2 THE DIAMETER OF THE EARTH

Solution. Let us model this as in Figure 2.1, where the circle represents the Earth, O is its center, A is Alexandria, and B is Syene. The rays of light are parallel. Since the sun shines directly down the well in Syene, the points B and O are parallel to the rays of light. Since the sun shines at an angle of 7.2◦ down the well in Alexandria, the rays make an angle of 7.2◦ with the radius AO. Thus, ∠AOB = 7.2◦. The distance between Syene and Alexandria is the length of the arc AB. Thus, the ratio of the length of this arc to the angle 7.2◦ is the same as the ratio of the circumference of the Earth, C, and a whole circle, 360◦. That is, 787 km C = ◦ 7.2◦ 360 C = 39, 350 km. Exercise 2.6. Was Eratosthenes’ assumption that the sun is a point at infinity a safe assumption? Find an upper bound on the error in the above calculation of the circumference of the Earth, given that the sun is 150,000,000 km from the Earth. [S] Exercise 2.7. Note that Eratosthenes’ measurements are by no means a testimony that the Earth is round. This time, assume the Earth is flat, and use Eratosthenes’ measurements to calculate the distance from the Earth to the sun. [S] Exercise 2.8. Alexandria is not due north of Syene (Aswan). Consult a map of Egypt and measure the distance between Alexandria and Aswan, and also the distance between the lines of latitude through Aswan and through Alexandria. Use both measurements, together with the figure 7.2◦, to calculate the circumference of the Earth. How much of an error is introduced by assuming that Alexandria and Syene are on the same line of longitude? Solution. The atlas I looked at (Microsoft Bookshelf, ’95) does not have much detail. It looked, to me, like Alexandria is in fact almost 787 km north of Aswan, but also about 270 km west. The distance between them looked like 833 km. Using the figure 787 km, we get a circumference of 39, 350 km. Using the figure 833 km, and assuming that Alexandria is due north of Aswan, we get a circumference of 41, 650 km, or about 6% more. Exercise 2.9. Use Eratosthenes’ measurements (5000 stadia and 7.2◦), Aristarchus’ values (moon and sun span .5◦, the angle called θ in Figure 2.2 is 3◦), and Hipparchus’ measurement (the shadow of the Earth is 2.5 times the diameter of the moon), to find DE, DM , DS, RM , and RS in stadia. Solution. Using Eratosthenes’ measurements, we find that the circumference of the Earth is 250, 000 stadia. The diameter of the Earth is therefore 250000 DE = = 79, 577 stadia. π

CHAPTER 2. GEOMETRY IN GREEK ASTRONOMY

moon

sun

Earth

Figure 2.2 Using θ = 3◦, and the results from Exercise 2.2, we get DM =

(1 + sin θ)DE

= 23, 926 stadia 3.5 720(1 + sin θ)DE = 2, 741, 758 stadia RM = 7π 720(1 + sin θ)DE = 52, 387, 655 stadia RS = 7π sin θ (1 + sin θ)DE = 457, 168 stadia DS = 3.5 sin θ

Exercise 2.10. Of the Greek measurements in Exercise 2.9, on ly the angle 1 θ is not very accurate. This angle is approximately 10′ or 6 ◦. Use this value, together with Smart’s values (787 km, 7.2◦), Aristarchus’ values (the moon and sun span .5◦), and Hipparchus’ value (the shadow of the moon is 2.5 times the diameter of the moon) to ﬁnd DE, DM , DS, RM , and RS. Compare these values with today’s accepted values given in Table 2.1. What is the percent error in each? Solution. As we saw in Exercise 2.4, the circumference of the Earth is 39, 350 km, so DE = 12, 525 km. Using θ = 10′, and the results from Exercise 2.2, we get DM =

(1 + sin θ)DE

= 3589 km 3.5 720(1 + sin θ)DE = 411, 283 km RM = 7π 720(1 + sin θ)DE = 141, 388, 933 km RS = 7π sin θ (1 + sin θ)DE = 1, 233, 851 km DS = 3.5 sin θ

2.2 THE DIAMETER OF THE EARTH

Polar diameter of the Earth: DE = 12, 720 km Equatorial diameter of the Earth: 12, 760 km Diameter of the moon: DM = 3475 km Diameter of the sun: DS = 1, 392, 000 km From the Earth to the moon at perigee (closest): 356, 000 km From the Earth to the moon at apogee (farthest): 407, 000 km From the Earth to the sun: RS = 149, 000, 000 km Table 2.1

Chapter 3

Constructions Using a Compass and Straightedge 3.1

The Rules

3.2

Some Examples

Exercise 3.1. If the length |OP | is one, th√en the square constructed in Figure 3.1(b) (of the text) has sides of length 2. Describe how to construct a square with sides of length one. R

Figure 3.1

CHAPTER 3. CONSTRUCTIONS

Solution. Proceed as in Theorem 3.2.2, to create points P , Q, and S, as in Figure 3.1. Draw the circles centered at S and Q which both go through P . These two circles intersect at P and another point, U . Then, P QU S is a square with sides of length one.

3.3

Basic Results

Exercise 3.2. Given a constructed line segment AB, describe how to find the perpendicular at A. Solution. Draw the circle centered at A and through B. This intersects the line AB at B and another point C. Find the perpendicular bisector of BC, as described in Lemma 3.3.2. Exercise 3.3. Finish the proof of Lemma 3.3.3. That is, prove that we can construct CA (|BC|) even if |AB| ≤ 2|BC|. Solution. The proof is identical. We find the midpoint of AB. The circle the line AB at D. The circle CM ( |MD )| intersects the B C( BC | ) intersects | line AB again at D′. The circleCA( |AD′ )| has radius |BC ,| as desired. It’s only the diagram that looks different. Exercise 3.4. Prove Lemma 3.3.4. Solution. Construct the circlesC A(| BC |) and C C ( AB | ).| These two circles intersect in two points D and D′. One of the quadrilaterals ABCD or ABCD′ is a parallelogram (the other is not convex), say ABCD. Then CD is parallel to AB and goes through C, as desired. Exercise 3.5. Prove Lemma 3.3.5 and its corollary. Solution. Given ∆ABC, and two points A′ and D, we can construct the circle CA' ( |AB ). | This intersects A′D at two points. Call one of them B′. Now, constructC A' (| AC|) and C B' (| BC |). These two circles intersect at two points. Call one of them C′. Then ∆ABC is congruent to ∆A′B′C′ by SSS. There are four such triangles. To reproduce an angle ∠CAB at A′ on the line A′D, just construct ∆A′B′C as described above. Then ∠C′A′B′ = ∠CAB, and is on the line A′D, as desired. Exercise 3.6. Show how to construct a regular octagon (8-gon) using a straightedge and compass. Solution. After constructing a square SOTQ inscribed in a circle with center P , as was done in Theorem 3.2.2, find the angle bisectors of ∠SPQ and ∠SPO. The bisectors, when extended in both directions, intersect the circle at four points. The eight points now on the circle form a regular octagon.

3.3 BASIC RESULTS

Exercise 3.7** (Mohr Constructions). Show that any constructible point can be constructed using only a compass. This result is due to Mohr (1672). My grandfather tried to show me the solution to this problem when I was in graduate school. At the time, his mind was still sharp, but his short term memory was failing him. Over a Christmas holiday, he came to me several times to show me the solution. Twice, he could not remember what he had done, and returned to work out the details. The third time, he showed me his solution. The fourth time, he again could not remember the details, so I had to show him what he had done. Unfortunately, I no longer remember what he did, though I seem to remember it was shorter than the proof I’ll give here. Solution. Constructible points are created by intersecting constructible circles and lines. Let us begin with something easy, just to convince ourselves that we are getting somewhere. Theorem 5. Suppose A and B are constructible points and Γ is a constructible circle with center O. Suppose O does not lie on the line AB. Let P and Q be the points of intersection of the line AB and the circle Γ. Then P and Q can be constructed using only a compass. Γ′ O' C P

Q C'

A O

Figure 3.2 Proof. Construct the circlesCA( | AO |) and CB( BO ). | | They intersect at O and O′, as in Figure 3.2. Let the circlesCA( | AO |) and Γ intersect at C (it doesn’t matter which of the two points we call C). Let C B(| BC )| intersect ′ = ′. Then, O′C′ ( AO ) again at C = OC , so Γ A C | | | | | | CO' ( |O′C′ |) is the reﬂection of Γ through the line AB. Hence, Γ and Γ′ intersect at P and Q, the points where the line AB intersects Γ.

CHAPTER 3. CONSTRUCTIONS

We must still show that, if O is on AB, then it is possible to ﬁnd the intersection of the line AB and the circle Γ. This is, surprisingly, a more diﬃcult result to show. We will do it in a series of lemmas. Lemma 6. Let Γ be a circle with radius r and centered at O. Let A be a point outside Γ. Then we can construct a point A′ such that A′ is on the ray OA and r2 ′ |OA | = . |OA| (Throughout this solution, when we say ‘construct,’ we mean using only a compass. When we say ‘constructible,’ we mean constructible using a straightedge and compass.)

A′

B′

Figure 3.3 Proof. Construct the circle CA( |AO |), and let it intersect Γ at B and B′, as in Figure 3.3. Construct the circlesC B(| BO| ) andC B' |( B′O| ), and let them intersect at A′ . Since B and B′ are symmetric about the line OA, the point A′ lies on AB. Note that ∆AOB and ∆BOA′ are similar, since they are both isosceles and share the base angle at O. Thus, |OA′| |OB|

′

|OA | =

|OB| |OA| r2

. |OA| Lemma 7. Let O and A be constructible points. Then it is possible to construct points A1 = A, A2,..., such that Ak is on the ray OA and |OAk| = k|OA|.

3.3 BASIC RESULTS

Figure 3.4 Proof. Remember how we construct hexagons? Let’s use that idea here. Construct the circlesC O( |OA |) and CA( OA | ).| Let these intersect at B (and somewhere else), as in Figure 3.4. Construct CB( |AB ),| which intersects A( OA C | )| at A and again at (say) C. Finally, construct CC ( |AC |), which intersectsCA( |OA )| at B and again at A2. Repeat this process to construct A3, A4, and so on. Corollary 8. Suppose A and B are constructible points. Then we can construct the midpoint of AB. Proof. By Lemma 7, we can construct a point C on the line AB such that |CA =| 2 BA | . By | Lemma 6, we can construct a point C′ on the ray AC such that |AB|2 |AB|2 |AB| ′ = = . |AC | = |AC| 2|AB| 2 That is, C′ is the midpoint of AB. Lemma 9. Let Γ be a constructible circle, centered at O and with radius r. Let A be a constructible point, not equal to O. Then we can construct points B and B′ on Γ such that OB is perpendicular to OA. Proof. By Lemma 7, we may assume, without loss of generality, that A is outside Γ. Construct the point A′, as described in Lemma 6. Use the trick in the proof of Lemma 7 to construct a point A′′ on the line OA such that ′′ ′ ′ ′′ |OA = | OA | and | such that O is between A and A , as in Figure 3.5. By ′′ Corollary 8, we can ﬁnd the midpoint of AA , so we can construct the circle Γ′ with diameter AA′′. Note that the power of the point O with respect to Γ′ is r2 ′′

|OA||OA | = |OA|

|OA|

=r .

CHAPTER 3. CONSTRUCTIONS B Γ′ Γ A′′

A′

B′

Figure 3.5 Let B and B′ be the points on Γ′ such that BB′ is the perpendicular to OA through O. Then |OB = by power of the point, | OB| = r. | OB | ′ , and | ′ Thus B and B also lies on Γ. That is, they are the points of intersection of Γ and Γ′, as desired. Corollary 10. Suppose Γ is a constructible circle with center O, and A is a point not equal to O. Then we can construct the points P and Q where the circle Γ and the line OA intersect. Proof. By Lemma 9, we can construct the points B and B′ o n Γ which form the perpendicular to OA and through O. Applying the lemma again to the point B and the circle Γ, we can construct the points C and C′ on Γ which form the perpendicular to OB through O. But then, C and C′ are on the line OA, and so must be the points P and Q. Theorem 11. Let l and l′ be two constructible lines which intersect at P. Then it is possible to construct P using only a compass. Proof. Since l is constructible, there exists a constructible point A on l. Similarly, l′ must also be defined by two constructible points, one of which is not the base of the perpendicular to l through A. Let us call this point B. Let the circleCA( AB | ) |intersect l′ again at B′, a point we can construct by Corollary 10. Note that the circles C B| ( AB| ) andC B' |( AB′|) intersect at A and again at A1, the reflection of A in l′. Construct the circleC A1 |( AA1| ). If P lies inside this circle, as in Figure 3.6, then let us call it Γ. Otherwise, construct Ak (using Lemma 7) for sufficiently large k such that P lies inside A Ck ( AA | k ).| Call this circle Γ. Let Γ intersect l′ at C and C′, and let it intersect l at A and D. Finally, draw Γ′ = CA( AC | ),| which intersects l at C and C′. By power of the point P in Γ, |P A||P D| = |PC||PC′|, and by power of the point P in Γ′, |PC||PC ′ | = |PC|2 − |P A|2.

3.3 BASIC RESULTS l′

P Γ′

Γ A

l C′

Figure 3.6 Thus, |P A||P D| = |AC|2 − |P A|2 |P A|(|P D| + |P A|) = |AC|2 |P A||AD| = |AC|2. Thus, by Lemma 6, we can construct P using only a compass (using Γ′ and D). By Theorem 5, Corollary 10, and Theorem 11, every constructible point can be constructed using only a compass. Exercise 3.8**. Prove that any constructible point can be constructed using only a straightedge and a compass with a fixed radius. Is there any difference between such a compass and the lid of a jar? Solution. This is quite a project. The way points are constructed is in one of several ways. Either a point is at the intersection of two lines, the intersection of a line and circle, or at the intersection of two circles. Using a straightedge and compass with fixed radius, it is clear that we can find the point of intersection of two lines. What is not clear is whether we can find the points of intersection of two circles (without the benefit of actually drawing the circles) or the points of intersection of a line and circle. We will show that we can, in fact, find these points. Let us begin with some preliminary results. We will first show that we can construct parallel lines. Let r be the fixed radius of the compass. Lemma 1. Suppose l is a constructed line, and P is a constructed point not on l. Then we can construct a line l′ through P which is parallel to l.

CHAPTER 3. CONSTRUCTIONS

Figure 3.7 Proof. Let us ﬁrst assume that P is less than r away from l. Construct Γ = CP (r), and let it intersect l at A (and another point), as in Figure 3.7. Construct CA(r), and let this intersect the line l at B (and another point). Construct C B(r), which intersects Γ at A and again at C. Then P ABC is a rhombus, so PC is parallel to l, as desired. E

E′

Figure 3.8 Let us now extend this result. Given a line l and a constructible point A on l, we can construct CA (r), as in Figure 3.8. This intersects the line again at B. Construct CB (r), which intersects the line at A and again at C. Let it intersect our ﬁrst circle at D. Finally, construct CC (r), and let it intersect the second c√ ircle at E and E ′ . Then, either DE or DE ′ is parallel to l and a distance r 3/2 away from l. By repeat√ing this process, we can rule the plane with lines parallel to l and spaced r 3/2 apart. Thus, given any P not on l, it is possible to construct a line l′ parallel to l such that the distance from P to l′ is less than r, and by the ﬁrst part of this proof, a line through P which is parallel to l.

3.3 BASIC RESULTS

We will use this result to show that we can ﬁnd the points of intersection of a constructible line and a constructible circle. Theorem 2. Suppose A, B, O, and C are constructible points. Then it is possible to construct P, the intersection of the line AB with the circle CO(|OC|) using only a straightedge and compass with fixed radius r.

A′ B P′

C′ B′ O

Figure 3.9 Proof. The idea of this proof is to ‘scale’ the circle to one with radius r, which we can draw. Let us follow along in Figure 3.9, where we assume that AB does not go through O. Note that we cannot construct the circle O( OC C | ). | We begin by constructing the rays OA, OB, and OC, and the circle Γ = CO(r). The ray OC intersects Γ at C′. By our lemma, we can draw the lines through C′ which are parallel to CA and CB. These lines intersect the rays OA and OB at A′ and B′, respectively. Then, the line A′B′ is parallel to AB. It intersects Γ at P ′. Finally, let P be the intersection of OP ′ and AB. We claim that P is a point of intersection of the circleCO( OC | ) |and the line AB. This is clear, since the ‘prime’ picture (those points labeled with primed letters) is just a dilation of the original by the appropriate factor.

CHAPTER 3. CONSTRUCTIONS

Now, suppose the line AB goes through O. Let us first suppose that C is not on AB. Let C′ be the intersection of OC with Γ =C O(r). Let Γ intersect AB at P . We can draw the line through C which is parallel to C′P ′. This intersects AB at P , one of the points of intersection. The other point of intersection can be found in a similar way. Now, suppose C is on AB. Then C is one of the points we want, but we still have to find the other. To do so, we note that we can construct a line l through O which makes a 60◦ angle with AB. Since C is not on this line, we can follow the above procedure to find a point C′′ on the intersection of l with C O|( OC| ). Now, proceed as before, using C′′ in place of C. We can also find perpendicular bisectors. Lemma 3. Suppose AB is a constructed line segment. Then we can find the perpendicular bisector of AB. Proof. Let us first suppose that| AB| < 2r. Then CA(r) and CB(r) intersect at two points, and the line through these two points is the perpendicular bisector of AB. If AB | |= 2r, there is no problem finding its midpoint M . Then, we can find the midpoints M1 and M2 of AM and of MB. The perpendicular bisector of M1M2 is the perpendicular bisector of AB. Finally, if |AB| > 2r, then we can measure off r from both ends, to get points A′ and B′ with |A′B′| = |AB|− 2r. We continue until we find points A′′ and B′′ with |A′′B′′| < 2r, and proceed as before. Corollary 4. Suppose A, B, C, D, and E are five constructible points. Then it is possible to construct the points of intersection of the line DE and the circle CA (|BC|). Proof. Note that the proof of Lemma 3.3.3 only requires being able to find a midpoint, and the intersection of circles with lines. Theorem 5. Suppose A, B, C, and D are constructible points. Then it is possible to construct the points of intersection of the pair of circles CA (|AB|) and CC (|CD|), using only a straightedge and compass with fixed radius r. D P A B

Figure 3.10

3.3 BASIC RESULTS

Proof. Let M be the midpoint of AC. Let the circles intersect at P and Q, and let l = PQ intersect AB at E (see Figure 3.10). Our approach is to ﬁrst construct E, then the line l perpendicular to AC through E. If we can do this, then by our previous theorem, we can ﬁnd the points P and Q since they are the points of intersection of a line l and the circle CA (|AB|). Let x = |ME| and m = |M A| = |MC|. Let the radii of the circles be s = |AB| and t = |CD|. By power of the point E, |EA|2 − s = −|EP ||EQ| = |EB|2 − t 2

(m − x) − s = (m + x) − t 2

t − s = (m + x) − (m − x) (t + s)(t − s) x= . 4m 2

By the corollary above, we can move lengths, so we can construct segments of lengths (t+s), (t −s), and 4m. Since we can ﬁnd perpendicular bisectors, we can construct perpendiculars, and hence, we can create a ﬁgure such as that in Figure 3.11, where the triangles are both right angle triangles, and |RT | = t − s, |OR| = 4m, and |OS| = t + s. Note that S may be between U T

Figure 3.11 O and R, depending on which is larger, 4m or t + s. But then, by similar triangles, (t + s)(t − s) |US| = = x. 4m Thus, we can construct the length x, so we can construct the point E, and the perpendicular l at E. Thus, any point which is constructible using a straightedge and compass, is also constructible using only a straightedge and compass with ﬁxed radius r. Note that the lid of a jar is not like a compass with a ﬁxed radius, since we do not know the location of the center of a circle drawn with the lid of a jar.

3.4

CHAPTER 3. CONSTRUCTIONS

The Algebra of Constructible Lengths

√ Exercise 3.9. Con√struct 5. (For some numbers, like 5, there are easier ways to construct 5 than the above construction.)

A O

Figure 3.12 Solution. Note that a right an√ gle triangle with short sides of length 1 and 2 has a hypotenuse of length 5. Thus, the construction in Figure 3.12, where O and P are th√e initial points a distance one apart, produces a segment AB of length 5. √ Exercise 3.10. Construct 3. B

A O

Figure 3.13 Solution. Note that a triangle with h√ ypotenuse of length 2 and short side of length 1 has a third side of length 3. √ Such a triangle is constructed in Figure 3.13, where| OP| = 1 and | AB| = 3. Note that the angle at C is a right angle since it subtends a diameter.

3.4 THE ALGEBRA OF CONSTRUCTIBLE LENGTHS

Exercise 3.11. The geometric mean of two positive numbers a and b is √ ab. Find a simple construction to find the geometric mean of two lengths a and b. Solution. We use the algorithm for finding square roots (Lemma 3.4.5), only we let the segment with length one have length b instead. Then, by power of the point, |AC|2 = ab, √ so |AC| = ab, as desired. Exercise 3.12. Recall the Argand plane model of the complex numbers C. Suppose the two points we begin our constructions with are the points 0 and 1. Then, any point which is constructible represents a complex number. We call such a number a constructible number. Hence, we can talk about the sum, difference, product, and quotient of constructible numbers. Prove that if P an√d Q are constructible numbers, then so are P + Q, P − Q, P · Q, 1/P , and P . For iθthe last, recall that any complex number P can be expressed as √ P = re , where (r, θ) is the polar coordinate representation √ iθ/2 of P . Then, P = re . This is done in Section 14.1, notably in Lemmas 14.1.1, 14.1.2, and 14.1.5. Exercise 3.13. Let C ⊂ C be the set of constructible numbers. Prove that C is a field. Solution. By Exercise 3.12, the set C is closed under addition and multiplication, and that both additive and multiplicative inverses are inC . Since C ⊂C, which we know is a field, Cmust be a subfield of C. In particular, it is a field. Exercise 3.14. Suppose we have constructed a set of points S, and that K = Q[S] is the smallest field which contains S. There are only three ways of constructing new points – finding the intersection of two lines defined by points in S; the intersection of a line and circle defined by points in S; and the intersection of two circles defined by points in S. In any of these three cases, show that there exists a D ∈ K √ so that S together with the newly constructed points all lie in the field K[ D]. Conclude that [Q[S] : Q] is a power of two for any set of constructible points S. Exercise 3.15. Prove that if it is possible to double the cube, then it is √ possible to construct 3 2. Let S be the set of points required to construct √ √ 3 3 2. Then Q[ 2] ∈ Q[S]. Explain how this shows that it is impossible to double the cube. Solution. Suppose we are given a cube with sides of length s. If we can double the cube, then we can find a cube with sides of length x where 3

x = 2s

CHAPTER 3. CONSTRUCTIONS x=

√ 3

2s.

Thus, since s is given, and we can double the cube, there is a construc√ √ tion S′ which gives 3 2s. From this, we can isolate 3 2, so there exists a √3 √ √ 3 construction S that gives 2. But then, 2 ∈ Q[S], so Q[ 3 2] ⊂ Q[S]. √3 Hence, the index [Q[ 2 : Q] = 3 divides [Q[S] : Q]. By Exercise 3.14, [Q[S] : Q] = 2r for some integer r. Thus, we have 3 divides 2r , which is clearly false, so no such construction can exist. Exercise 3.16. Show that to square the circle, one must be able to construct π. In the latter part of the nineteenth century, it was proved that π is transcendental. Conclude that it is impossible to square the circle. Solution. Suppose we are given a circle with radius r. If we can square the circle, then we can ﬁnd a square of side s such that 2

πr = s . √ Thus, we can construct a length s = r π. Since r is given, we can divide √ through by it to get π. We can square this to get π. (Division and squaring are constructible operations.) Hence, if we can square the circle, then there exists a construction S of π, so again, Q[π]⊂Q[S]. But Q[π] is an infinite dimensional vector space over Q (since π is transcendental), while Q[S] is a finite dimensional vector space over Q. This is a contradiction, so there cannot be any construction of π, so it is not possible to square the circle. Let us elaborate a little on the statement that Q[π] is an infinite dimensional vector space over Q. To see this, let us first recall that Q[π] is the smallest field that contains Q and π. In particular, since fields are closed 2 3 under multiplication, the elements 1, π, π , π , ... are all in Q[π]. Suppose 2 n Q[π] has dimension n. Then the set{1, π, π , ..., π }must be linearly dependent, since it contains n+1 elements. Thus, there exist rational numbers a0, a1,..., an such that 2

a0 + a1π + a2π + ... + anπ = 0 and not all of the ai are zero. That is, there exists a polynomial f (x) = a0 + a1x + ... + anx

in Q[x] such that f (π) = 0. Finally, to say π is transcendental over Q means that there does not exist any polynomial with coeﬃcients in Q that has π as a root. That is, the polynomial f (x) cannot exist, so Q[π] is an inﬁnite dimensional vector space over Q.

3.5

The Regular Pentagon

Exercise 3.17. As mentioned, this is not the only w√ay of constructing a regular pentagon. Construct a segment of length 5, and from this,

3.5 THE REGULAR PENTAGON

√ construct a segment of length 5− 1. Rather than try to ﬁnd a quarter of this, use it as is to construct a regular pentagon inscribed in a circle of radius four. E

P-3

P-2

P-1 C

F′

E′

Figure 3.14 Solution. Here is one solution. Begin by constructing a right angle triangle ∆ABO with short sides of length |OA √| = 2 and |OB| = 1, as in Figure 3.12. Then the hypotenuse AB has length 5. Draw CA (|AB|) and let it intersect the line √ OA at C and D, with O between A and C, as in Figure 3.14. Then |PC| = 5 − 1, where P is the midpoint of OA. We find points P−1, P−2, and P−3 on the line AB by drawing circles with radius of length 1, to give a point P−3 a distance 4 away from P . We draw the perpendicular at C and let it intersect the circle CP (|P P−3 |) at E and E ′ . Then P−3 , E, and E ′ are three vertices of a regular pentagon. We can draw a circle C E|( EP−3|) to find a fourth vertex F , and a similar circle at E′ to find F ′, giving a regular pentagon P−3EFF ′E′. Alternatively, we can use Exercise 3.19 (note the correction) to observe that the perpendicular at D goes through F and F ′. √

Exercise 3.18. There is a rather nice proof that cos 72◦ = − 4 5 . Consider the isosceles triangle ∆ABC with |AB| = |AC| = 1 and ∠BAC = 36◦. Find the length of |BC|. (Hint: Construct the angle bisector of ∠ABC, 1+

CHAPTER 3. CONSTRUCTIONS

as in Figure 3.15(a).) The disadvantage of this derivation is that it works only for the regular pentagon. The derivation in the text, though, can be modiﬁed to derive the lengths one must construct to construct the regular 7-gon and 9-gon. A

A E

C (a)

(b)

Figure 3.15. See Exercises 3.18 and 3.20. Solution. Since ∆ABC is isosceles with |AB | = |AC ,| the base angles at B and C are equal. Since A = 36◦, we know B + C = 2B = 144◦, so B = C = 72◦. Let the angle bisector at B intersect the opposite side at E. Then ∠EBC = 36◦. Thus, ∆ABC ∼ ∆BCE, so |AB| |BC|

|BC| |CE|

For simplicity, let x = |BC|. Note that ∆ABE is also isosceles, so |AE| = |EB| = |BC| = x. Thus, |CE| = 1 − x, so 1

x (1 − x) 2 1−x = x 0 =x + x −1 √ −1 ± 5 x= . 2 2

Since x > 0, the sign must be positive. Finally, we note that x = 2 cos 72◦, so √ −1 + 5 . cos 72◦ = 4 Exercise 3.19. Prove that cos(4π/5) =

−1 − 4

√

3.5 THE REGULAR PENTAGON

Solution. First, note the error in the ﬁrst printing, corrected in later versions. There are a couple of ways of doing this. One is to recall how the 2 polynomial f (x) = x + x 1− was derived. It came from ﬁrst looking at the 4 3 2 2πn/5 polynomial g(ω) = ω + ω + ω + ω + 1, which has the roots ω = e . 1 We arrived at f (x) by letting x = ω + ω− . Thus, the roots of f (x) are e

2π/5

+ e−

2π/5

4π/5

4π/5 + e− = 2 cos(4π/5).

= 2 cos(2π/5)

On the other hand, the two roots of f (x) are √ −1 ± 5 x= . 2 We have already seen that 2 cos(2π/5) =

−1 +

√

2 so the other roots must be equal. That is, 2 cos(4π/5) =

−1 −

√

2 Another solution is to use the angle sum formula: cos(4π/5) = 2 cos (2π/5) − 1 = 2 2

−1+ 4

√5

−1

√ √ 3− 5 −1 − 5 −1 = . 4 4

Exercise 3.20. Let ABCDE be a regular pentagon with sides of length 1. The sides AB and CD are extended so that they intersect at F , as in Figure 3.15(b). What is the length |BF |? ◦ Solution. Let x = FB| . The | angles in a regular pentagon all measure 108 , ◦ so the angle ∠FBC = 72 . Thus, (1/2) 1 cos(72◦) = = . x 2x

We already know the value of cos(72◦), so using that ﬁgure, we get √ 1 −1 + 5 = 2x 4 √ √ 2 √ = 2(−1 − 5) = 1 + 5 . x= −4 2 −1 + 5

CHAPTER 3. CONSTRUCTIONS

But that wasn’t the intended solution. Let x = FB | . |A regular pentagon can be inscribed in a circle, and the vertices divide the circle into ﬁve equal parts. Thus, by the Star Trek lemma, ∠BDC = 36◦. Hence, ∆BDF is isosceles, with | BD | =| BF | = x. Note that AD | =| BD|= x,| and AD is parallel to BC, so ∆BFC ∼ ∆AF D. Thus, |BF| |BC| = |AF | |AD| 1 x = x+1 x 2 x = x+1 2 x −x−1=0 √ 1± 5 x= . 2 √ 1+ 5 Since x > 0, we get x = . 2 Exercise 3.21. DeMoivre’s theorem actually states n

cos nθ + i sin nθ = (cos θ + i sin θ) . Use induction to prove DeMoivre’s theorem. Explain why Euler’s formula implies DeMoivre’s theorem and so can be thought of as a generalization of it. Solution. Whoops. I should have said, “for n a positive integer.” Using induction on n, we ﬁrst note that the equality is trivially true for n = 1. Suppose it is true for n = k. Then the left hand side of the equation for n = k + 1 expands as follows: cos(k + 1)θ + i sin(k + 1)θ = cos kθ cos θ − sin kθ sin θ + i sin kθ cos θ + i cos kθ sin θ. The right hand side can be written as (cos θ + i sin θ)

k+1

= (cos θ + i sin θ) (cos θ + i sin θ),

and using the induction hypothesis, this expands as (cos θ + i sin θ)

k+1

= (cos kθ + i sin kθ)(cos θ + i sin θ) = cos kθ cos θ + i sin kθ cos θ + i cos kθ sin θ − sin kθ sin θ.

This is the same as the left hand side, so by induction, the equation is true for all positive integers n. Euler’s formula states iθ

e = cos θ + i sin θ,

3.5 THE REGULAR PENTAGON so in particular, inθ

(cos nθ + i sin nθ) = e

iθ n

= (e )

= (cos θ + i sin θ) . Thus, DeMoivre’s theorem follows from Euler’s formula. iθ

Exercise 3.22. Show that both y = Ae and y = B(cos θ + i sin θ) are solutions to the diﬀerential equation dy = iy. dθ Solve for A and B if y(0) = 1. Conclude Euler’s formula iθ

e = cos θ + i sin θ. Solution. Let’s check them: d iθ iθ (Ae ) = i(Ae ), dθ d (B(cos θ + i sin θ)) = B( − sin θ + i cos θ) = i(B(cos θ + i sin θ)). dθ Thus, both are solutions to the diﬀerential equation dy = iy. dθ If y(0) = 1, then

1 = Ae = A, and 1 = B(cos 0 + i sin 0) = B. Thus, A = B = 1. Since the solution to a continuous ﬁrst order diﬀerential equation with a single initial condition is unique, we must have iθ

e = cos θ + i sin θ. Exercise 3.23. Recall the Taylor series for sin x, cos x, and e

∞ Σ sin = (−1)k x2k+1 x3 x5 = + x x− − ... (2k + 1)! 3! 5! k=0 ∞ 2k Σ (−1)k x cos = 1 x2 x4 = + x − − ... (2k)! 2! 4! k=0 ∞ Σ xk x2 x3 ex = 1 + x + + + ... = k! . 2 3! k=0

Use this to prove Euler’s formula. You may assume that these formulas are valid for complex values of x.

10 0

CHAPTER 3. CONSTRUCTIONS x

Solution. Using the Taylor series for e , we get eiθ = =

∞ Σ (iθ)k ∞ kk!k k=0 Σ

i θ

k=0

∞ Σ

j=0

= 1 + iθ +

(iθ)2 (iθ)3 + + ... 2 3!

=1+

θ2

iθ −

−i

θ3 3!

+ ...

(—1)j θ2j + i (−1)j θ 2j+1 (2j)! (2j + 1)!

= cos θ + i sin θ. Exercise 3.24. Use Euler’s formula to prove the angle sum formulas: sin(α + β) = sin α cos β + cos α sin β cos(α + β) = cos α cos β − sin α sin β. Hint: ei(α+β) = eiαeiβ . Solution. Using Euler’s formula, we note that i(α+β)

= cos(α + β) + i sin(α + β).

On the other hand, using the hint, we note that i(α+β)

iα iβ

= e e = (cos α + i sin α)(cos β + i sin β) = cos α cos β + is ∈ α cos β + i cos α sin β − sin α sin β.

Comparing the real and imaginary parts, we get the angle sum formulas. Exercise 3.25. Prove

tan 2θ =

2 tan θ 1 − tan θ

Solution. Note that tan 2θ =

sin 2θ 2 cos θ sin θ 2 tan θ = = . 2 2 2 cos 2θ 1 − tan θ cos θ − sin θ

Exercise 3.26. Let w = 2 cos(2π/7). Show that the regular 7-gon is constructible if and only if w is constructible. Find a polynomial f (x) with rational coefficients such that f (w) = 0 and f (x) is irreducible over the rationals. [A] Solution. Suppose we have a regular 7-gon. Since the perpendicular bisectors of the sides go through the center of the circumscribed circle, we can find that point. Thus, we can construct the angle 2π/7. Let us reproduce this angle at 0 with one ray along the x-axis. Then, the other ray i2π/7 intersects the unit circle at ±e . By reflecting, we can get the other.

101

3.5 THE REGULAR PENTAGON

The line joining these two points intersects the x-axis at cos(2π/7). Thus, if the regular 7-gon is constructible, then so is 2 cos(2π/7). Conversely, if w = 2 cos(2π/7) is constructible, then so is cos(2π/7). The perpendicular at this point intersects the unit circle at two of the vertices of the regular 7gon, and these points can be used to ﬁnd the rest of the vertices. Thus, if w is constructible, then so is the regular 7-gon. i2π/7 Now, let z = e . Then 7

z = 1. Since 1 is a root of x − 1, and z = / 1, z must be a root of h(x) where 7

x − 1 = (x − 1)h(x) = (x − 1)(x + x + x + x + x + x + 1). 7

But w = 2 cos(2π/7) = z + z− , so 3

w = z + 3z + 3z− + z− 2

w = z + 2 + z− , and hence w + w − 2w − 1 = z + z + z + 1 + z− + z− + z− = 0. 3

Thus, w is the root of the polynomial f (x) = x + x − 2x − 1. 3

If f (x) is reducible, then it must have a rational root, which by the rational root theorem, must be ±1. Since neither ±1 is a root, f (x) is irreducible. Exercise 3.27. Let w = 2 cos(2π/9). Show that the regular 9-gon is constructible if and only if w is constructible. Find a polynomial f (x) with rational coeﬃcients such that f (w) = 0 and f (x) is irreducible over the rationals. Solution. The argument that w is constructible if and only if the regular 9gon is constructible is similar to the argument in Exercise 3.26. i2π/9 Let z = e . Then 9 z = 1. Note that x − 1 is a factor of x − 1, with the factorization 3

x − 1 = (x − 1)(x + x + 1) 9

Since z /= 1, z must be a root of the latter polynomial 3

g(x) = x + x + 1.

10 2

CHAPTER 3. CONSTRUCTIONS 1

Now note that w = 2 cos(2π/9) = z + z− . Thus, 3

w = z + 3z + 3z− + z− , so w − 3w + 1 = z + 1 + z− = 0. 3

Thus, w is a root of f (x) = x −3x + 1. Again, since f (x) has degree three, it factors over Q if and only if it has a rational root. By the rational root theorem, the only possible rational roots are± 1. Since neither is a root, f (x) is irreducible over Q. The next exercise is for those who have done Exercises 3.13 and 3.14 in the last section: Exercise 3.28. Conclude, from the previous two exercises, that the regular heptagon and nonagon are not constructible. Why does this imply that it is impossible to trisect an arbitrary angle? Solution. We saw, in Exercise 3.14, that if a number w is constructible, then w lies in a ﬁeld K where [K : Q] is a power of 2. That is, if w is constructible, then [Q[w] : Q] = 2r for some nonnegative integer r. But if w = 2 cos(2π/7), then [Q[w] : Q] = 3, which is not a power of two. Thus, the regular 7-gon is not constructible. Similarly, since [Q[w] : Q] = 3 for w = 2 cos(2π/9) too, the regular 9-gon is also not constructible. In particular, we cannot trisect 60 ◦, since if we could, then we could construct the regular 9-gon. Hence, it is not possible to trisect an arbitrary angle, using only a straightedge and compass.

3.6

Other Constructible Figures

Exercise 3.29. Use Theorem 3.6.4 (but not Theorem 3.6.5) to prove that the regular 7-gon is not constructible. Solution. By Exercise 3.26, the regular 7-gon is constructible if and only if w = 2 cos(2π/7) is constructible. Further, w is a root of the irreducible 3 2 polynomial f (x) = x + x — 2x 1, − which has degree three, so by Theorem 3.6.4, w is not constructible. Thus, the regular 7-gon is not constructible. Exercise 3.30. Use Theorem 3.6.4 (but not Theorem 3.6.5) to prove that the regular 9-gon is not constructible. [S] Solution. By Exercise 3.27, the regular 9-gon is constructible if and only if w = 2 cos(2π/9) is constructible. Further, w is a root of the irreducible 3 polynomial f (x) = x −3x + 1, which has degree 3, so by Theorem 3.6.4, w is not constructible. Thus, the regular 9-gon is not constructible. Exercise 3.31. Is cos(2π/15) constructible? Explain.

103

3.6 OTHER CONSTRUCTIBLE FIGURES

Solution. Yes. Since the regular 15-gon is constructible, we can construct the angle 2π/15, so we can construct cos(2π/15). Exercise 3.32. Suppose x is a root of x − 5x + 7x − 2 = 0. 3

Is the length x constructible? Explain. Solution. This is a trick question. It tests whether the reader is checking all the conditions of Theorem 3.6.4. On the surface, one notes that f (x) = 3 2 x — 5x +7x 2 is − a degree three polynomial, so it looks like x should not be constructible. But, upon further inspection, we note that f (x) is reducible, since f (2) = 8—20 + 14 −2 = 0, so it doesn’t satisfy the conditions of the theorem. Factoring, we get f (x) = (x − 2)(x − 3x + 1), 2

so the other roots of f (x) are √ 3± 5 x= . 2 It is clear that these two are constructible (be sure to note that they are positive), so all the roots of f (x) are constructible. Exercise 3.33. Is it possible to √ construct a triangle ∆ABC with ∠BAC = √ √ 1+ 24◦, |AB| = 2 5 , and |AC| = 360◦1 + 2 ? ◦ Solution. Yes. Note that 24 = , so it is constructible, since the 15-gon √ 15 √ √ 1+ 5 is constructible. Since both 2 and 1 + 2 are constructible (they only involve the usual operations and square roots), such a triangle is constructible by SAS.

Exercise 3.34. Is it possible√to construct a triangle ∆ABC with ∠BAC = 20◦, |AB| = 4, and |AC| = 2 3 ? Solution. No. We cannot construct 20◦ = be able to construct a regular 9-gon.

360◦

, since if we could, we would

Exercise 3.35. Is it possibl√ e to construct a triangle ∆ABC with |AB| = 1, BC = 2, and AC = 1 + 5 ? (Caution: You may want to think twice | | | | about this question.) Solution. No. Note that AC lengths do not | >| AB| + BC | , | so these | satisfy the triangle inequality. Thus, no such triangle exists. Exercise 3.36. Prove Theorem 3.6.3.

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CHAPTER 3. CONSTRUCTIONS

Solution. That is, prove the following: If m and n are relatively prime, and we can construct the regular m-gon and n-gon, then we can construct the regular mn-gon. Since the regular m-gon and n-gon are constructible, we can construct the angles 2π 2π and . m n Since m and n are relatively prime, there exist integers a and b such that am + bn = 1. Consider the angle

2π m

θ=b

2π . n

We can construct the angle θ by ﬁrst reproducing the angle 2π/m counterclockwise b times if b > 0 or clockwise| b| times if b < 0, and then reproducing 2π/n a total of |a |times, either counterclockwise or clockwise, depending on whether a > 0 or not. But θ=

b2πn + a2πm 2π = . mn mn

Thus, we can construct an angle of 2π/mn, so we can construct the regular mn-gon. Exercise 3.37. A computer can factor F5 very quickly, but there exists a nice proof that F5 is composite. Notice that 641 = 5 + 2 = 5 · 2 + 1. 4

Use this to show that 641 divides both 54 ·228 + 232 and 54 ·228 − 1, and hence divides the diﬀerence, which is F5. Solution. Let’s check this: 4

5 + 2 = 625 + 16 = 641 5 · 2 + 1 = 5(128) + 1 = 640 + 1 = 641. 7

Now, note that 5

F5 = 22 + 1 = 232 − 1 = (2 + 5 · 2 − 5 · 2 + 1) 32

= 2 (2 + 5 ) − (5 2 28

2 14

+ 1)(5 2

− 1)

= 2 641 − (5 2 + 1)(5 · 2 + 1)(5 · 2 − 1) 28

2 14

= 641(2 − (5 2 + 1)(5 · 2 − 1)). 28

2 14

Thus, F5 is divisible by 641, so is composite.

3.7 TRISECTING AN ARBITRARY ANGLE

3.7

105

Trisecting an Arbitrary Angle

Exercise 3.38. Describe how to trisect an arbitrary obtuse angle using a compass and twice notched straightedge. Solution. Note that we can construct 30◦. Thus, we can trisect a right angle. But every obtuse angle can be thought of as the sum of a right angle and an acute angle. So, given an obtuse angle, ﬁrst remove a right angle, trisect the resulting acute angle using Archimedes’ algorithm, and add 30 ◦, to get a third of the obtuse angle. Exercise 3.39**. Suppose we are given a piece of paper on which there is 2 drawn the parabola y = x , the points (0, 0) and (1, 0), and nothing else. Let us call this a piece of parabola paper (see Figure 3.16). Come up with a construction of the regular 7-gon using only a straightedge, compass, and the information on the parabola paper. [H][S]

Figure 3.16. Parabola paper (see Exercises 3.39 and 3.40). Exercise 3.40*. Suppose we are given a piece of parabola paper, as described in the previous exercise. Let f (x) be a polynomial of degree three with rational coeﬃcients and a real root c. Prove that it is possible to construct a length | c| using only a compass, straightedge, and the information on this paper. Solution. This is almost completely done for us in the solution to the previous exercise. We construct the circle centered at (a, b) that goes through

106

CHAPTER 3. CONSTRUCTIONS

(0, 0), which has the equation (x − a) + (y − b) = a + b . 2

We solve for the intersection of this circle and the parabola y = x by 2 substituting y = x in the above equation, which gives (x − a) + (x − b) = a + b 2

x − 2ax + x − 2b x = 0 2

2 2

x − (2b − 1)x − 2a = 0. 3

Thus, if a b|are | and | | constructible lengths, then we can construct the real 3 roots of f (x) = x (2b — 1)x−2a. − Now, suppose we are given a polynomial with rational coeﬃcients. Then, we can divide through by the leading coeﬃcient to get a polynomial 3 2 g(x) = x + Ax + Bx + C. To make this look like our polynomial f (x), we will ‘complete the cube,’ an idea that is similar to completing the square. We note that the expansion 3 2 (y —A/3)3 begins y −Ay , so we make the substitution x = y− A/3, to get g(y − A/3) = (y − A/3) + A(y − A/3) + B(y − A/3) + C3 2 3 2 A y A 2A y A AB 3 2 2 = y − Ay + − + Ay − + + By − +C 3 27 3 9 3 3 A2 A AB 3 =y + B− y+ C− − . 3 27 3 3

Thus, let us set A2 2 b − 1 = 32 − B A − 3B + 1 b= 3 6AB A 2 a= + −C 27 3 3 A + 9AB − 27C a= . 54 Since A, B, and C are rational, the lengths| a| and |b|are constructible, so we can ﬁnd the circle with center (a, b) that goes through (0, 0). The points where this circle intersects the parabola will have x components which are equal to the real roots of f (x). Thus, if r is a real root of f (x), then we can construct the length |r|.

107

3.7 TRISECTING AN ARBITRARY ANGLE

Exercise 3.41**. It is clear that it is possible to create a tool which draws ellipses. Using this tool, a straightedge, and compass, come up with a construction of some figure which is not constructible using only a straightedge and compass. Exercise 3.42. Suppose we place one notch of a twice notched straightedge on the line x = b and align the straightedge so that it passes through the origin (0, 0). Prove that the other notch lies on the conchoid r =±a+b sec θ where a is the distance between the two notches. Solution. Recall, x = r cos θ and y = r sin θ. Thus, the line x = b has the polar equation r cos θ = b r = b sec θ. If one notch is on this line at b sec θ, and the straightedge goes through the origin, then the other notch is either at b sec θ + a if that notch is on the other side of the line as the origin, or at b sec θ− a if it is on the same side as the origin. √ Exercise 3.43**. Prove that it is possible to construct 3 2 using a twice notched straightedge. [H] Solution. This solution is due to Nicomedes (ca. 250 B.C.). We first describe the procedure assuming that the notches are a unit distance apart. √ Note that it is enough to find 3 a for a < 1 (since we can construct inverses). We begin by constructing a rectangle OBCD with dimensions 2a and 2, as in Figure 3.17, where| OP| = 1 and| OA| = a. The point E is constructed so that EA is perpendicular to OA and| EB| = 1. The point F is constructed so that it is a distance 2a away from O. We then construct the line l through B that is parallel to EF . All these points are constructible using a compass and straightedge. We now use our twice-notched straightedge: we place one notch on the line l, the other notch on the line OA, and adjust these points so that the straightedge passes through the point E, giving the points G and H. Nicomedes worded this last step in terms of curves. He constructed the conchoid, described in Exercise 3.42, which has the polar equation r = sec θ + 1 using the coordinate system centered at E with x-axis along the line EB. This conchoid intersects OA at H. √ We claim that the distance x = |BH | is 2 3 a. To analyze the algorithm, let us also draw the line HC that intersects OP at J, and let y = EG | . We | begin by applying the Pythagorean theorem to triangles ∆AEB and ∆AEH: a + |AE|2 = 1, 2

(x + a) + |AE|2 = (1 + y) , 2

108

CHAPTER 3. CONSTRUCTIONS

H l

1 x

G y

1 E

A a O

2a F

Figure 3.17 x(x + 2a) = y(y + 2), x y+2 = . y x + 2a Since l is parallel to the side EF in ∆HEF , we know that y 4a = . 1 x Since ∆HBC is similar to ∆CDJ, we also know that x 2a

2 |DJ|

so |DJ| = 4a/x = y. Because ∆HOJ is similar to ∆CDJ, we get x + 2a y + 2 = , 2a y y+2 y = , x + 2a 2a which combined with (3.1), gives y x = . 2a y

(3.1)

3.7 TRISECTING AN ARBITRARY ANGLE

109

Substituting y = 4a/x, we obtain x2 4a = , 4a 2ax √ 3 which implies that x = 8a, so 3 a = x/2, as claimed. Finally, if the notches are a distance d apart, scale Figure 3.17 by d. √ √ Then x = 2d 3 a, from which we can isolate 3 a using a straightedge and compass.

Chapter 4

Geometer’s Sketchpad Most of the exercises in this chapter ask for scripts. Solutions are available at www.prenhall.com/baragar Scripts and sketches are zipped and password protected. The zipped ﬁle is about 250kB. The password is Pi∼22/7 The scripts were written on and should work on a Windows based PC. I doubt that they will work on other platforms but I plan to work on converting them. The Scripts for this chapter are labeled Ex04XXX, where XXX is the exercise number. If an exercise requires more than one script, the names of those scripts will be appended with lowercase letters – Ex04XXXa, Ex04XXXb, etc. Note that comments are included with these scripts. Those comments are repeated here as a convenience to the reader.

4.1

The Rules of Constructions

4.2

Lemmas and Theorems

Exercise 4.1. Write a script to construct an equilateral triangle with side AB. When you are done, your script should say ‘Given: Point A, Point B.’ If it asks for more points, try again. Test your script. Script Ex04001. This script creates an equilateral triangle. To run it, select two points (in an open sketch) and play the script. Exercise 4.2. Write a script to construct a square with side AB. Script Ex04002. This constructs a square with side AB. Note that it does not mimic the construction in the text, since that constructs a square with center A and vertex B. 111

112

CHAPTER 4. GEOMETER’S SKETCHPAD

Exercise 4.3. Write a script to construct a regular hexagon inscribed in the circle with center A and through B. Script Ex04003. A regular hexagon inscribed in a circle centered at A and going through B. There are, of course, other constructions. Exercise 4.4. Write a script to construct a perpendicular bisector. Write two more scripts – one to find the line through A and perpendicular to AB; and the second to find the midpoint of segment AB. Script Ex04004a. This script constructs the perpendicular bisector of a segment AB. Script Ex04004b. This script constructs the perpendicular to AB at A. It was created by doing steps 1 – 4, and then running the script Ex04004a on the point B and the hidden point C. Script Ex04004c. This finds the midpoint of the segment AB. It was created by running the script Ex04004a, and then hiding the perpendicular line. Part of the intent of this exercise is to point out the value of hiding work. Exercise 4.5. Write a script to bisect an arbitrary angle. Move around one of the points that defines an arm of the angle. Does the construction always work? Script Ex04005a. Solution 1. This solution works except when angle ∠ABC is oriented counterclockwise and is between 60 and 180 degrees. Note the curious first step of this script. Its purpose is to make sure the points appear in the right order. Without it, the resulting script would find the bisector of angle ∠BAC. An alternate solution is given in Ex04005b. Script Ex04005b. Solution 2. The (hidden) circles used in this solution always go through B. The other point of intersection of these circles is always on the side of the angle that is less than 180 degrees. This solution does not work if A, B, and C are collinear. After constructing the angle bisector, I moved A so that the angle grew past 180 degrees. When this is done, sketchpad chooses the wrong point of intersection of the circles, so the bisector disappears. This is remedied by drawing a new ray. Only one of these rays should appear at any time. Unfortunately, sometimes sketchpad doesn’t always recognize that two of the points are identical, so a random ray through B sometimes appears. Exercise 4.6. Note that the buttons of Sketchpad mimic a ‘collapsible compass.’ Write a script which mimics a ‘noncollapsing compass’ (i.e., write a script which models Lemma 3.3.3.) Test it. Does it always work? Rewrite it so that it always works, or explain why we cannot write a script which always works. Script Ex04006. This script constructs the circle centered at A with radius BC. The midpoint of AB was found using Ex04004c. Note the trick of

4.2 LEMMAS AND THEOREMS

113

drawing the line AB, and finding point beyond the segment where the circle centered at B and through C intersects the line AB. This insures that the script works even when the radius is larger than half the length AB. Finding angle bisectors, segment bisectors, and perpendiculars is very common in constructions, and after a while, the process becomes tedious. For all the remaining exercises, we allow the use of the operations under the pull-down menu ‘construct.’ Note that all of these (except ‘point on object’) can be done using the rules of construction. In particular, the operation ‘Circle by center and radius’ is the ‘noncollapsing compass.’ Exercise 4.7. Write scripts to add, subtract, multiply, and invert constructible lengths. Script Ex04007a. – addition. The script constructs a segment of length j + k. The point E and line l are for reference. Script Ex04007b. – subtraction. It constructs a segment of length n− m. The point M and line o are for reference. Script Ex04007c. – multiplication. Given the points A and B a distance one apart and a segments of length j and k, this script constructs a segment of length jk. Script Ex04007d. – inversion. Given points A and B a distance one apart and a segment j of length j, this script constructs a segment of length 1/j. Exercise 4.8. Write a script to find square roots. Script Ex04008. – square roots. Given points A and B a distance one apart and a segment j, this script constructs a segment of length the square root of j. Exercise 4.9. Use the above script to come up with your own construction of a regular pentagon. Script Ex04009. A different construction of the regular 5-gon. Scripts Ex04007a and Ex04007b aren’t all that useful, so we construct 5 in the usual way. We then use Ex04008 to find the square root of 5. We subtract one. We construct 4 and use Ex04007d to find 1/4. We use Ex04007c to find the product of 1/4 and root 5 less one, giving cos(726◦). We use that length to construct the regular 5-gon. Exercise 4.10. Write a script to construct a regular pentagon inscribed in CA (|AB|). Script Ex04010. This script constructs the regular pentagon inscribed in the circle centered at A that goes through B. The construction follows Richmond’s construction. Exercise 4.11. Write a script to construct the regular 17-gon.

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CHAPTER 4. GEOMETER’S SKETCHPAD

Script Ex04011. Richmond’s construction of the regular 17-gon. Exercise 4.12. The point F and the midpoint of DP0 are almost coincident, but not quite. If O is zero and P0 is one, what are the values of F and the midpoint of DP0?

4.3

Archimedes’ Trisection Algorithm

Exercise 4.13. Write a script which does every step of Archimedes’ construction, except the one which is not permitted in our rules. Script Ex04013. This script trisects angles, using Archimedes’ trisection algorithm. It needs a little help. Move Q until the line at Q goes through the point A (with Q not equal to A). Let P be the point that is not Q, A, B, or C. Then the angle ∠QP C is one third of the angle ∠ABC. Exercise 4.14. Write a script which satisfies the following: ‘Apply the script to points A and B. Move the point C along the circle until the line l intersects D. The resulting figure is a regular 9-gon.’ Script Ex04014. As required, all we need to do is the following: ”Apply this script to points A and B (in a fresh sketch). Move the point C along the circle until the line l intersects D (with C not equal to D). The resulting figure is a regular 9-gon.”

4.4

Veriﬁcation of Theorems

Exercise 4.15 (The Centroid). Write a script to demonstrate that the medians intersect at a common point. Once you have your sketch, use the select button to move the vertices around. Script Ex04015. – the centroid. This script draws the medians of triangle ∆ABC. Note that these medians are coincident. This ’demonstrates’ that the medians intersect at a common point. Exercise 4.16 (The Incircle). Write a script which constructs the incircle of a triangle with vertices A, B, and C. (The ‘Given’ information should be only three points.) Script Ex04016. – the incircle. This script constructs the incircle. Note, in particular, step 7, where a perpendicular to the incenter (labeled D below) to ﬁnd the point of tangency of the incircle, thereby allowing us to draw it. Exercise 4.17 (The Excircles). Write a script which constructs the excircles. Script Ex04017. – the excircles. Note that the exterior angle bisectors are perpendicular to the angle bisectors. This fact was used in this script.

4.4 VERIFICATION OF THEOREMS

115

Exercise 4.18 (The Circumcircle). Write a script which constructs the circumcircle. Script Ex04018. – the circumcircle. Exercise 4.19 (The Orthocenter). Write a script to construct the orthocenter H of ∆ABC. Does it work if the triangle is obtuse? Correct it, if not. Select a vertex in your sketch and move it around. Do you notice a symmetry between the vertices and H? Formulate a conjecture concerning this observation and prove it. Script Ex04019. – the orthocenter. This script ﬁnds the orthocenter H of a triangle ∆ABC. It works even when the angle is obtuse. The conjecture one is expected to make is that the orthocenter of triangle ∆ABH is C. This is easily demonstrated using this script. To do so, open a sketch. Apply this script to three points A, B, and C. It gives back H. Now apply it to A, B, and H. The new point, this time, is exactly at C. The proof that this always happens is in the solution manual. Solution. The altitude at A in ∆ABH is perpendicular to BH. But, by construction, BH is perpendicular to AC. Thus, the altitude at A in ∆ABH is AC. Similarly, the altitude at B in ∆ABH is BC, so the orthocenter to ∆ABH is C. Exercise 4.20 (The Euler Line). Write a script which demonstrates that the circumcenter O, the centroid G, and the orthocenter H are collinear. Script Ex04020. – the Euler line. This script constructs the centroid G, the orthocenter H, and the circumcenter O. It then constructs triangle ∆GHO. Note that this triangle is degenerate, demonstrating that G, H, and O are collinear. Exercise 4.21 (The Nine Point Circle). Write a script which constructs the nine point circle and the nine points which it goes through. Check your script. Does it work if the triangle is obtuse? Are all nine points there? Let H be the orthocenter of ∆ABC. What can you say about the nine point circle for ∆HBC? Script Ex04021. – the nine point circle. This script constructs the nine point circle of triangle ∆ABC. Note that all nine points are there even when the triangle is obtuse. Apply this script to three points ∆ABC, and let H be the orthocenter of this triangle. Now, apply it to the points A, B and H. Note that we get the same nine points. Exercise 4.22 (Feuerbach’s Theorem). On a triangle ∆ABC, run the script which constructs the nine point circle. On the same triangle, run the script which constructs the incircle and excircles. Formulate a conjecture.

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Script Ex04022. – Feuerbach’s theorem. The following script constructs the nine point circle, the incircle, and the excircles of triangle ∆ABC. It was created by opening a sketch, starting a recording, selecting three points, and running the scripts Ex04016, Ex04017, and Ex04021 on these three points. Note that the incircle and excircles are all tangent to the nine point circle. (Move a vertex around – it is still the case.) This fact is known as Feuerbach’s theorem. Exercise 4.23. On a triangle ∆ABC, find the center of the nine point circle and the Euler line. Formulate a conjecture. Script Ex04023. This script was created by opening a sketch, recording in a script, and applying scripts Ex04020 and Ex04021 to three points. The appropriate conjecture is that the center of the nine point circle lies on the Euler line, and is the midpoint of OH. Exercise 4.24 (The Simson Line). Write a script which constructs the Simson line. Script Ex04024. – the Simson line. For this script, we first ran script Ex04018 to find the circumcircle. We pick a random point P on this circle, and then find the feet of the perpendiculars from P to the (extended) sides of triangle ∆ABC. We draw the line through two of these feet. That it goes through the third foot is Simson’s result. Exercise 4.25 (The Radical Axis). Write a script which finds the radical axis of two circles. Script Ex04025a. A script to construct the center of a circle. Exercise 4.25 asks for a script which finds the radical center of two circles. Consequently, the preferred input is two circles. Since we’ll need the centers of these circles, let’s find them rather than rely on the user to include them as input. Script Ex04025b. We will employ Exercises 1.61 and 1.62 to do this problem. We begin by using Ex04025a to find the centers of the two circles. The line through these centers intersects the two given circles at several points. We pick one on each circle, and find the circumcircle of an equilateral triangle that has these two points as two of its vertices. By construction, this circle intersects both given circles; its center is not on the line joining the centers of the two given circles; and it is not tangent to either given circle, since its center, the center of a given circle, and the point of intersection are not collinear. Thus, it can be used to find a point on the radical axis (employing the result in Exercise 1.62). We then find the line through this point that is perpendicular to the line through the centers of the two given circles. By Exercise 1.61, this is the radical axis. Exercise 4.26 (The Radical Center). Write a script which finds the radical center of three circles. Have this script find the circle which is orthogonal to each of the original three circles.

4.5 SOPHISTICATED RESULTS

117

Script Ex04026. We apply the script Ex04025b to two pairs of circles to get two of the three radical axes. Where these axes intersect is the radical center P of the three circles. To find the circle that is perpendicular to all three given circles, we apply Ex04025a to find the center (say O) of one of the given circles, find the midpoint M of OP , and draw the circle centered at M and through P . It intersects the given circle at, say Q, and triangle ∆OQP is a right angle triangle with right angle at Q, since this angle subtends the diameter OP . Thus, OQ is perpendicular to QP , so QP is a tangent to the given circle. Thus, the circle centered at P and through Q is perpendicular to the given circle, and hence to all three of the given circles. Exercise 4.27. Write two scripts which demonstrate each direction of the theorem of Menelaus. Script Ex04027a. This script selects random points D and E on the appropriate sides. The point F is the intersection of DE with AB. It then calculates the product of ratios (without signs). That this product is 1 is one direction of Menelaus’ theorem. Script Ex04027b. This script calculates Menelaus’ product of ratios (without the sign). It also draws triangle ∆DEF . If the quantity is one, and either zero or two of the points D, E, and F are on the proper sides (that is, between the vertices), then triangle ∆DEF is degenerate. That is, D, E, and F are collinear. Exercise 4.28. Write a script that demonstrates both directions of Ceva’s theorem. Script Ex04028a. This script selects random points D and E on the appropriate sides of triangle ∆ABC. The lines AD and BE intersect at a point P . The line CP intersects AB at F . The script then calculates the appropriate product of ratios. That this product is 1 is one direction of Ceva’s theorem. Script Ex04028b. This script puts random points D, E, and F on the appropriate extended sides of triangle ∆ABC. It then calculates Ceva’s product of ratios. By moving one of these points so that the ratio becomes one (and either one or three of the points are between the vertices), we find that the Cevians are coincident. This demonstrates one of the directions of Ceva’s theorem. (I argue that it does not show the other direction since, with this script, the best we can say is that the product is close to one, though one might argue that this script has the same failing for the direction it is meant to show.)

4.5

Sophisticated Results

Exercise 4.29. Let ∆ABC be an arbitrary triangle. Let P be an arbitrary point, and let the perpendiculars from P to the extended sides BC, AC,

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and AB be labeled X, Y , and Z, respectively. The triangle ∆XY Z is called the pedal triangle for P . Write a script which constructs the pedal triangle for P with respect to a triangle ∆ABC. Move P about. When is ∆XY Z degenerate? Script Ex04029. This script constructs the pedal triangle ∆XY Z for the point P with respect to triangle ∆ABC. The pedal triangle is degenerate when P is on the circumcircle. Exercise 4.30. For ∆ABC and a point P , write a script which creates the pedal triangle (see Exercise 4.29) for P , finds the area of the pedal triangle, and finds the area of ∆ABC. Make this script find the circumcenter O of ∆ABC too. Finally, let R be the circumradius, r = |OP |and have this script calculate 2

R −r |∆ABC|. 4R2 Make a conjecture. Script Ex04030. In this script, we first apply Ex04018 to construct the circumcircle of triangle ∆ABC, and then Ex04029 to construct the pedal triangle for P with respect to triangle ∆ABC. We then calculate the area of the pedal triangle, and the quantity in the text. After moving P around, we note that these quantities are equal if P is inside the circumcircle, and differ by a sign when P is outside the circumcircle. I once saw this proved – I was convinced, but not terribly enlightened. Let’s look at this result though. Note that the pedal triangle is zero if R = r. This happens when P is on the circumcircle. But if the area is zero, then the triangle is degenerate. That is, we get Simson’s result. Exercise 4.31. The pedal triangle ∆XY Z is sometimes called the derived triangle or first derived triangle of ∆ABC with respect to the point P . The second derived triangle of ∆ABC with respect to P is the pedal triangle of ∆XY Z with respect to P . The third derived triangle is the derived triangle of the second derived triangle. Use the script found in Exercise 4.29 to find the third derived triangle of ∆ABC with respect to a point P . Investigate and formulate a conjecture concerning this triangle. Script Ex04031. After finding the third derived triangle, I decided it isn’t very obvious what conclusion one should come to. The result I was looking for is the observation that the third derived triangle is similar to the original. The calculations show this. Note that the first derived triangle was created by choosing A, B, C, and P , in that order. I was careful to find the second and third derived triangles using the same order. That is, I applied Ex04029 to X, Y , Z, and P , in that order; and again to the new X, Y , and Z. Then, by comparing the ratio of sides, we discover that triangle ∆ABC is similar to this third triangle ∆XY Z.

4.5 SOPHISTICATED RESULTS

119

Exercise 4.32. Write a script which finds the point of tangency of the excircles to the sides of the triangle ∆ABC. Label these points D, E, and F on sides BC, AC, and AB, respectively. Verify that the three segments AD, BE, and CF are coincident. This point of coincidence is called the Nagel point N . Find the incenter I and centroid G of the triangle ∆ABC too. Formulate a conjecture about the three points N , I, and G. Script Ex04032. – the Nagel point. We run script Ex04017 to find the excircles, from which we find D, E, and F . We note that AD, BE, and CF are coincident, giving the Nagel point. We construct the centroid G the incenter I, and the segment NI. We notice that G is on this segment, giving us a result similar to the Euler line. Given the similarity, we are inclined to check whether NG is twice GI, and we find that it is. Exercise 4.33. Write a script which demonstrates Pappus’ theorem. Script Ex04033. – Pappus’ theorem. We select four points, which describe two lines AB and CD. The script selects a random points E on AB and F on CD. It constructs the points of intersection of AF and CE, AD and BC, ED and BF . These points are labeled G, H, and I. The script draws the line GH. This line goes through I, which is what Pappus’ theorem tells us to expect. Exercise 4.34. Write a script which demonstrates Pascal’s theorem for a circle. Script Ex04034. This script chooses six points (C, D, E, F , G, and H) on a given circle. It then finds the points I, J, and K, which are the points of intersection of DH and EG, CH and EF , and CG and DF . It then constructs the line IJ. This line goes through K, which demonstrates Pascal’s theorem for a circle. Exercise 4.35. Write a script which demonstrates Desargues’ theorem. Script Ex04035. – Desargues’ theorem. Given a triangle ABC and a point P, this script chooses random points a, b, and c on PA, PB, and PC, respectively. The lines BC and bc intersect at R; the lines AC and ac intersect at S; and the lines AB and ab intersect at T. The script then constructs the line RS. Notice that this line goes through T, which is Desargues’ theorem. Pappus’ theorem is really a special case of Pascal’s theorem, since two lines may be thought of as a degenerate conic. Exercise 4.36. Let ABCDEF be a hexagon inscribed in a circle. Let AB and CD intersect at G, and let DE and AF intersect at H. Write a script which shows that BE, CF , and GH are coincident. Script Ex04036. This script chooses a random hexagon ABCDEF inscribed in a circle. It finds G, the intersection of AB and CD; and H, the intersection of AF and DE. It then draws BE, CF , and GH, which we note are coincident.

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CHAPTER 4. GEOMETER’S SKETCHPAD P

A′ R

C′ B′ A

S C B T

Figure 4.1 Exercise 4.37. State and prove the converse of Desargues’ theorem. [H] Solution. The converse of Desargues’ theorem is this: Consider two triangles ∆ABC and ∆A′B′C′. Let R be the intersection of BC and B′C′; let S be the intersection of AC and A′C′; and let T be the intersection of AB and A′B′. If R, S, and T are collinear, then AA′, BB′, and CC ′ are coincident. More succinctly, if ∆ABC and ∆A′B′C′ are perspective from a line, then they are perspective from a point. To see this, let P be the intersection of BB′ and CC ′ . Note that ∆BB′T and CC ′ S are perspective from a point. Thus, by Desargues’ theorem, the points A, A′, and P are collinear. That is, ∆ABC and ∆A′B′C′ are perspective from a point, as desired.

4.6

Parabola Paper

Exercise 4.38*. Use the parabola paper to construct a regular 7-gon (see Exercise 3.39). Script Ex04038a. – parabola paper. This was constructed as described in the text. There is another way (see Ex04038p), but it isn’t useful. Script Ex04038b. – a trisection script. From the solution of Exercise 3.39, we must ﬁnd the point (7/54, 5/3). Thus, we will have the need to trisect segments. That is what this script accomplishes. Script Ex04038c. This script constructs the point (7/54, 5/3), given the points A = (0, 0) and B = (1, 0).

4.6 PARABOLA PAPER

121

Script Ex04038d. – the regular 7-gon. In this script, we construct parabola paper using Ex04038a and the point (7/54, 5/3) using Ex04038c. We draw the circle centered at this point and going through A. This circle intersects the parabola at three points. We use D, the one furthest away from the origin. The other points of intersection will give us different vertices of the regular 7-gon, but since our parabola is really a piecewise segmented approximation, the points of intersection that are close to a ”fencepost” and far away from the origin are the best approximations. Continuing, we drop the perpendicular to the x-axis, subtract 1/3, and divide by two to get F . The perpendicular at F intersects the unit circle at two of the vertices of the regular 7-gon. We use these points to find the rest of the vertices. You’ll recognize this as the picture on the title page of the textbook. Script Ex04038p. Before we can do Exercises 4.38 – 4.40, we must produce parabola paper. The construction given here is esthetically superior to the one described in the text and the one given in Ex04038a, but it is useless. It uses the locus feature of Sketchpad, but unfortunately, Sketchpad won’t let us find points of intersection with loci. 2 Recall that the focus and directrix for the parabola y = x are at (0, 1/4) and y = −1/4. Given points O and P , which represent O and 1 in the complex plane, one can easily construct the focus (E below) and directrix (line m below). To use the locus feature of Sketchpad, we must find an arbitrary point on the parabola, given an arbitrary point on the directrix. We choose the latter (point G below). A point on the parabola is equidistant from the focus and the directrix. So, we seek the point (I) on the perpendicular from G such that | IE |= IG triangle ∆IEG is isosceles, so the altitude | . Thus, | at I is the perpendicular bisector of EG. Hence, to find I, we construct the perpendicular bisector of EG. Now, we select I and G, and apply the locus 2 feature under the construct menu. This produces the parabola y = x . Exercise 4.39*. Use the parabola paper to construct a regular 9-gon. Script Ex04039. To construct a regular 9-gon, we must construct the length cos(40◦). By Exercise 3.30, c = 2 cos(406 ◦) is the root of the poly3 nomial x −3x+1. Thus, by Exercise 3.39, we must find the point P = (a, b) where 2a = —1 and 2b− 1 = 3. That is, we must find P = ( − 1/2, 2). So, we begin by applying the script Ex04038a to construct parabola paper. We then, without much difficulty, construct P . The circle centered at P that goes through A intersects the parabola several times, one of which is at Q. We drop a perpendicular to the x-axis to get R. We bisect the segment ◦ AR to get S. The segment AR | = | 2 cos(406 ◦), so |AS |= cos(40 ). Thus, the perpendicular through S intersects the unit circle at two of the nine vertices of the regular 9-gon. We use those points to find the other vertices. Exercise 4.40**. Use the parabola paper to construct a regular 13-gon. Solution. This is quite a project. It might be appropriate to assign it to a talented student. Direct them to Chapter 14 and, in particular, Sec-

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tion 14.5. We must begin with some theory, before we can pull out our sketchpad. 2πi/13 We can construct the regular 13-gon if we can construct ω = e . Note that 13 2π ω = e = 1, so ω −1=0 13

(ω − 1)(ω + ω + ω + ... + ω + ω + 1) = 0. 12

Since ω = 1, the ﬁrst factor is not zero, so the second must be zero. That is, 12

ω + ω + ω + ω + ω + ω + ω + ω + ω + ω + ω + ω +1 = 0, (4.1) or equivalently (an expression that will be useful later), 6

ω + ω + ω + ω + ω + ω + 1 + ω− + ω− + ω− + ω− + ω− + ω− = 0. It is well known that the polynomial in 4.1 is irreducible, and that its roots are solvable by radicals. Since its degree is 12, we further know that its roots are solvable using square and cube roots. Thus, we know already that we can construct a regular 13-gon using parabola paper, but we’re still a long way from coming up with a construction. To come up with a construction, we investigate the isomorphism σ of Q[ω] defined by 2 σ(ω) = ω . Refer to Section 14.5 for the definition of a field isomorphism. This isomor0 1 phism generates a group G = σ { , σ , Note} that the images of ω under k the action of σ for k = 0, 1, 2, are, respectively, 2

ω, ω , ω , ω− , ω , ω , ω− , ω− , ω− , ω , ω− , ω− , ω, ω , .... 12

Thus, σ is the identity, so G = { σ , σ , ..., σ }. Hence, the order of G is 12. We wish to first find a subfield of Q[ω] of degree 3, which means we 0 3 6 9 want to find a subgroup of G of order 4. That group is H{= σ , σ , σ , σ } , 3 generated by σ . An element that is fixed by H is 0

x = σ (ω) + σ (ω) + σ (ω) + σ (ω) 5

= ω + ω− + ω− + ω 5

5 5

= ω + ω + ω− + ω− . Let us now ﬁnd the degree three polynomial that has x as a root. Note that 2

x = 2ω + 2ω + ω + ω + 4 + ω− + ω− + 2ω− + 2ω−

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4.6 PARABOLA PAPER 3

x = 3ω + 9ω + 3ω + 4ω + 4ω + 9ω 1

+ 9ω− + 4ω− + 4ω− + 3ω− + 9ω− + 3ω−

so x + x − 4x + 1 = 0. 3

Our subfield of degree three over Q is Q[x]. We now want to find the subgroup of H with order two. This is H′ = 0 6 {σ , σ }. An element fixed by H′ is the element 0

y = σ (ω) + σ (ω) = ω + ω−

= 2 cos(2π/13). Thus, to construct the regular 13-gon, we wish to construct y. This element y must satisfy a degree two polynomial with coeﬃcients in Q[x]. To ﬁnd that polynomial, we look at 2

y = ω + 2 + ω− 6

2 2

xy = ω + ω + ω + 2 + ω− + ω− + ω− 2

x y = 2ω + 4ω + ω + 3ω + ω + 5ω 1

+ 5ω− + ω− + 3ω− + ω− + 4ω− + 2ω− 2

6 6

xy = ω + 2ω + 2ω + 3ω + 3ω− + 2ω− + 2ω− + ω− . From this, we note that x y − xy − x + 1 = 0 2

xy − x y + (x − 1) = 0. 2

Hence, if we can construct x, then we can construct y since √ 2 x4 − 4x(x − 1) x ± y = . 2x To construct x, we ﬁrst complete the cube by substituting x = z − 1/3, to get 13 65 3 z − 3 z + 27 = 0. Thus, using Exercise 3.39, we wish to construct the circle centered at −

65 8 , . 54 3

We now know our mission – let’s get out our Sketchpad ....

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Script Ex04040a. A little bit of thought must be put into this exercise before we invoke sketchpad. This is done in the solution manual. From that work, we see that our first task is to construct the point P = (− 65/54, 8/3). We first construct the point (13, 0), and divide it in thirds twice (using Ex04038b) to get C = (13/9, 0), and a segment c of length 13/9. We then construct D = (5/3, 0) and a segment d of length 5/3. We use Ex04007c to multiply these lengths, and get a segment cd of length cd = 65/27. Bisecting this segment gives a length of 65/54. We then construct the point E = (8/3, 0). We use these two to construct the point P . Script Ex04040b. – a squaring script. We want to construct y, which involves several squares, so here is a squaring script. Points A and B are a distance one apart. The script squares the segment j. Script Ex04040c. We have to construct the root of a quadratic polynomial 2 with coefficients x,—x , and x −1. That’s what this script does. The points A and B are a distance one apart, and the segment j is to be the segment with length x. We use Ex04040b to find squares (and fourth powers). The 2 4 2 point C has a height of x— x, and D has a height of −x 4(x− x). We use Ex04008 to find E, which has a height of the square root of the previous 2 value. We add this segment to the segment with length x , to get the point E. We use Ex04007d to find 1/x, and multiply this with the previous value to get the point G. The point H is the midpoint of this segment, and GH has the desired length. Script Ex04040d. – the regular 13-gon. We construct parabola paper using Ex04038a. We construct the point P using Ex04040a. We draw the circle centered at P that goes through A. This intersects the parabola in three points. We are not keen on using the intersection close to A, since our approximate parabola is not very good near A. Our parabola doesn’t go far enough for other points of intersection, so we plot more points – the points (1.6, 2.56), (1.7, 2.89), and (1.8, 3.24). We now get the point Q of intersection of the circle and the parabola. We drop a perpendicular to the x-axis, and subtract 1/3 from this point to get R. The value y we seek is the length AR. We apply Ex04040c to the segment AR to get the segment GH with length x. The segment AS is half the length of GH, so the perpendicular at S intersects the unit circle at a vertex of the regular 13-gon. We use it to get the rest of the vertices. The particular vertex we get depends on our choices – which intersection with the parabola we choose, and whether we use + or − in the quadratic formula.

Chapter 5

Higher Dimensional Objects 5.1

The Platonic Solids

Exercise 5.1. Prove that the angles in a regular n-gon measure

n−2 n

180◦.

Solution. We know that the sum of angles in a polygon with n sides is (n − 2)180◦. Thus, if the n-gon is regular, then each angle measures (n − 2)180◦ . n Exercise 5.2. Construct a regular pentagon inscribed in a circle with radius two inches. Make a cutout of this pentagon on card stock (to use as a template). Construct templates of a square, equilateral triangle, and hexagon, each with sides equal in length to the side of the pentagon you constructed. Exercise 5.3. Construct models of the Platonic solids using the templates constructed in the previous exercise. Exercise 5.4*. When we talk about regular polygons, we usually mean convex regular polygons. There are, though, polygons whose sides cross, but whose angles and edges are all identical. For example, the five-pointed star in Figure 5.1 is one such regular star polygon. Note that this polygon has only five vertices and five sides. In a similar fashion, we can construct the regular star polyhedra (see, for example, Figure 5.1). Identify the set of all regular star polyhedra. How many faces does the regular star polyhedron in Figure 5.1 have? Exercise 5.5. We consider the faces of the star polyhedron in Figure 5.1 to be regular pentagons which intersect. If we instead think of the triangles 125

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CHAPTER 5. HIGHER DIMENSIONAL OBJECTS

Figure 5.1. A star polygon and a star polyhedron. The vertices of the ﬁve pointed star are the vertices of a regular pentagon. The vertices of this star polyhedron are the vertices of a regular icosahedron.

as the faces, then how many triangular faces are there? Find the lengths of the sides of each of these triangles. These triangular faces are constructible using a straightedge and compass. Construct a template, and use it to construct one of these star polyhedra.

Solution. If we look carefully at Figure 5.1, we notice the underlying structure of an icosahedron. Associated to each edge of the icosahedron, there are two triangular faces of the star polyhedron. Thus, this solid has 60 triangular faces. Now, consider a pentagonal face of the star polyhedron. The other faces cut a star out of it. Now, reﬂect on Exercise 1.44. In Figure 1.12(a), the triangle ∆EFD is what we are looking for. Notice that |AF | = |AE|, and hence, √ 2 −1 + 5 √ |AF | = |AF |. |FD| = 2 1+ 5 (Note the correction in Exercise 1.44.) Thus, this polyhedron is constructed with isosceles triangle whose base is s and equal sides has length √ (−1 + 5)s . To construct such triangles, ﬁrst construct a regular pen2 tagon, and then join every other vertex to get the interior star. The triangles on the edges are the ones we are interested in.

5.2 THE DUALITY OF PLATONIC SOLIDS

5.2

The Duality of Platonic Solids

5.3

The Euler Characteristic

127

Exercise 5.6. What is the Euler characteristic of a polyhedron which is topologically equivalent to a torus? Prove your assertion. Solution. The Euler characteristic of such a polyhedron is zero. Let us think of our toroidal polyhedron as looking like the inner tube of a bicycle. Choose a path of edges on the polyhedron which goes around the tube (going through the donut hole). Suppose that this path has n edges. Then it also goes through n vertices. Cut the polyhedron along this path, and on each open end, glue a face. By doing so, we have added two faces, added n edges, and added n vertices. That is, we have increased the Euler characteristic by 2. But now the resulting polyhedron is topologically equivalent to a sphere, which has Euler characteristic equal to 2. Thus, the Euler characteristic of a polyhedron which is topologically equivalent to a torus is zero.

5.4

Semiregular Polyhedra

Exercise 5.7. What is the truncated tetrahedron better known as? Solution. An octahedron. If we truncate a tetrahedron, then each triangular face becomes a new triangle (with one fourth the area), and the slice at each vertex is now a triangle. Thus, our new polyhedron has eight triangular faces. Exercise 5.8*. In Exercise 5.4, we introduced the notion of a regular star polyhedron. There exist, also, semiregular star polyhedra. Find and classify all of them. Exercise 5.9*. A geodesic dome can be created by making shallow pentagonal and hexagonal based pyramids and gluing these pyramids on the pentagonal and hexagonal faces of a soccer ball. What lengths should we choose for the sides of the faces of the pyramids if we want to make the geodesic dome as spherical as possible? Note that the triangular faces used for the pentagonal pyramids will be different from the triangular faces used for the hexagonal pyramids. Solution. The vertices of a semiregular solid all lie on a sphere. The proposed geodesic dome will most resemble a sphere if the new vertices are also on the sphere. Thus, our first objective is to find out the radius of the sphere that circumscribes the soccer ball. I didn’t see how to do this problem until I pulled out a soccer ball, so let us make that the first step. Let us imagine a pole running through the center of one of the pentagons. Then the equator with respect to this pole runs through ten hexagons, as shown in Figure 5.2. Thus, there is a regular

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CHAPTER 5. HIGHER DIMENSIONAL OBJECTS

A B

Figure 5.2 10-gon inscribed in the equator. If we normalize the sides of the pentagons and hexagons to be one, then the sides of the 10-gon have length 3/2. A pictoral proof of this appears in Figure 5.3. This gives us the relation r sin(2π/20) =

4 3 r = csc(π/10) √ = 4 (1 + 5). 3 4

I’ll leave it to the reader to verify the last equality.

A B

Figure 5.3 Let us now calculate the lengths of the edges of the triangles of the pyramids that are stacked on the hexagons. Let A be the center of a hexagon and let B be one of its vertices, as in Figure 5.2. Let O be the center of the sphere and let A′ be the point on the sphere such that A is on the radius OA′. We wish to calculate |A′B|, as in Figure 5.4.

129

5.4 SEMIREGULAR POLYHEDRA B r

A A′

Figure 5.4 Let θ = ∠BOA. It is clear that |AB| = 1, so we have r sin θ = 1 and by the Pythagorean theorem, |A′B|2 = 1 + (r − r cos θ) 2 2 = 1 + r 1 − 2 cos θ + cos θ √ 2 2 2 = 1 + r2(1 − 2 1 − sin θ + 1 − sin θ) =1+r 1 1 2−2 1 — 2− 2 2 r r = 1 + 2r − 2r √ r2 − 1 − 1 √ 2 = 2r − 2r r2 − 1 |A ′ B|≈ 1.022974785. 2

Though |A′B |is a constructible length, its exact algebraic representation is rather messy. It’s straightedge and compass construction, though, is not too bad, though we’ll leave that as an exercise. Let C′ be the point on the sphere such that C is on the radius OC′. We calculate |BC′ |in a similar way. We must ﬁrst ﬁnd the length| BC |, which is the radius of the circle that circumscribes the pentagon. We note that the sides of a regular pentagon inscribed in a circle of radius R is s = 2R sin(2π/10), so |BC′| = R =

1 csc(π/5), 2

since s = 1. Let θ′ = ∠BOC. Then r sin θ′ = R and |BC′|2 = R + (r − r cos θ′) 2

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CHAPTER 5. HIGHER DIMENSIONAL OBJECTS 2

= R + r (1 − 2 cos θ′ + cos θ′) √ 2 2 2 = R + r (1 − 2 1 − sin2θ′ + 1 − sin θ′) 2 2 2 2 = R R R +r 2− 2 1− − 2 r2 r √ 2 2 = R + 2r − 2r r2 − R2 − R2 √ 2 = 2r − 2r r2 − R2 |BC ′ |≈ .8644701515. Again, we leave it as an exercise to come up with a straightedge and compass construction of |BC′|.

5.5

A Partial Categorization of Semiregular Polyhedra

Exercise 5.10. How many pentagonal faces are on a snub dodecahedron? The solution is in the text. Exercise 5.11. How many square faces are on the snub cube? Solution. One solution is to just make a snub cube, but there is also an algebraic solution. The snub cube is represented by (3, 3, 3, 3, 4). Let F3 and F4 be, respectively, the number of triangular faces and the number of square faces on the snub cube. Let V be the number of vertices and let E be the number of edges. Since there is only one square at each vertex, we know V = 4F4. Every corner of a triangle is at a vertex, and there are four corners from triangles at each vertex, so 3F3 V = . 4 Hence, 16F4 F3 = . 3 We count edges by noting that every triangular face joins three edges, every square face joins four edges, and every edge adjoins two faces, so E=

3F3 + 4F4 2

Since the Euler characteristic is two, we get 2=F +V −E

5.5 A PARTIAL CATEGORIZATION OF SEMIREGULAR POLYHEDRA = F3 + F4 + 4F4 —

131

3F3 + 4F4

2 16F4 + 4F4 16F4 = + F4 + 4F4 — 3 2 F4 = (16 + 3 + 12 30) − 3 F4 = . 3 Thus, there are six square faces on a snub cube. Exercise 5.12. How many square faces are on a semiregular polyhedron represented by (3, 4, 4, 4)? The answer might surprise you. Solution. The answer is not surprising if you’ve already made a model. If not, let’s go through the algebraic solution. Let F3 and F4 be, respectively, the number of triangular and square faces. The corner of every triangle is at a vertex and there is only one corner from a triangle at each vertex, so V = 3F3. Similarly, since there are three corners from squares at each vertex, V =

4F4 3

Thus,

4F4 . 9 Triangular and square faces have three and four edges, respectively. Every edge adjoins two faces, so F3 =

3F3 + 4F4 2

4F4

+ 4F4 2

4F4 + 12F4 6

8F4 3

Thus, using the Euler characteristic, we get 2=F +V −E = F3 + F4 + 4F4 − 8F4 3 3 4F4 = + F4 + 4F4 − 8F4 3 3 9 F4 = (4 + 9 + 12 24) − 9 F4 = . 9 Thus, there are 18 square faces. Why is this surprising? Since this ﬁgure has square faces, we sort of expect the underlying structure of a cube, so

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we might expect 6 square faces. The ﬁgure does, in fact, have such an underlying structure, but it also has squares where the edges of the cube would be. Since a cube has 12 edges and 6 sides, it, after all, isn’t surprising that there are 18 square faces. Exercise 5.13. Find formulas for F , V , and E in terms of m and n for a Platonic solid with m n-gons at each vertex. Solution. The corner of each face is at a vertex, and there are m corners at each vertex. Thus, mV = nF. Each edge adjoins two faces, so E=

nF 2

The Euler characteristic is two, so 2=F +V −E nF nF =F + − 2 m F = (2m + 2n − nm). 2m Thus,

4m F =

2m + 2n − nm

From this, we get nF 4n = m 2m + 2n − nm nF 2mn E= = . 2 2m + 2n − nm V=

Exercise 5.14. Using the templates you made earlier, construct a model of your favorite semiregular polyhedron. If you can, involve a young child in the project. Exercise 5.15. Answer the ﬁrst ‘Why’ in this section. That is, if there are four faces to a vertex and only one of them is a triangle, explain why two of the other faces must be identical. Solution. Let the semiregular solid be represented by (3, k, m, n) where n not equal to 3 or k. Look at a triangular side with vertices labeled A, B, and C, as in Figure 5.5. Without loss of generality, we may assume that the face on the side AC is a k-gon. Then, by looking at the vertex A, the face on the side AB must be an n-gon. Hence, by looking at the vertex B, the face on the side BC must be a k-gon. Finally, looking at the vertex

5.5 A PARTIAL CATEGORIZATION OF SEMIREGULAR POLYHEDRA

133

m A n

k 3

Figure 5.5 C, we have that k = n, which is a contradiction. If n = 3, then a similar argument shows that three of the faces at the vertex C are triangles, so two of k, m, and n are three. In either case, two of the other faces are identical. Let us further observe that the two identical sides are either triangles, or are adjacent to the triangle. This observation that will help in the following exercise. Exercise 5.16. Answer the second and third ‘Whys’ in this section. That is, if there are four faces to a vertex and two of them are triangles, explain why these triangles cannot be adjacent, and explain why the other two faces must be identical. Solution. Suppose four faces are at each vertex and that (exactly) two of those faces are triangles. Then, the polyhedron can be represented by (3, k, m, n), where one of k, m, and n is three. By the observation made at the end of the proof of Exercise 5.15, k = n. Thus, the polyhedron is represented by (3, k, m, k). If k = 3, then there are three triangles at a vertex, and we have the m-gonal drum. Since there are only two triangles at each vertex, k /= 3, so the triangles are not adjacent. Exercise 5.17. Identify the two diﬀerent semiregular polyhedra represented by (3, 4, 4, 4).

Figure 5.6

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Solution. See Figure 5.6. The second figure is obtained from the first by rotating the top third (or top two thirds) by one edge. Though it does not seem symmetric, it does satisfy the definition of a semiregular solid. Exercise 5.18*. Classify the semiregular polyhedra with three faces to each vertex. Solution. Let the representation of the polyhedron be (k, m, n). Without loss of generality, we may assume k≤ m ≤ n. Suppose k = 3. Then, using an argument similar to that given in Exercise 5.15, we know m = n. If m = 3, we have a tetrahedron. Otherwise, if m is odd, then using a similar argument again, we get that n = 3 (the argument in Exercise 5.15 works because 3 is odd). Thus, m = n and m is even. Note that (3, 12, 12) represents a tiling, and hence that m < 12. Thus, we get the possible representations (3, 4, 4), (3, 6, 6), (3, 8, 8), and (3, 10, 10). Now, suppose k = 4. Note that the representation (4, 4, n) gives the infinite set of n-gonal prisms. If m is odd, then, as noted before, n = 4, which contradicts our order ≤ k m ≤ n. Thus, m is even. Note that both (4, 6, 12) and (4, 8, 8) give tilings, so if m = 6, then n < 12, and if m = 8, then n < 8, contradicting our order. Thus, we get the possible representations (4, 6, 6), (4, 6, 8), and (4, 6, 10). Now suppose k = 5. As noted before, this means m = n. If m = n = 5, then we get the dodecahedron. If m/= 5, then it must be even. Note that (5, 8, 8) gives a set of faces whose angles at a vertex sum to more than 360◦, so m < 8. Thus, we have the one possible representation (5, 6, 6). Finally, if k = 6, note that the representation (6, 6, 6) gives a tiling, so no such polyhedron can exist. Finally, we must show that each of these representations gives a polyhedron. For this, please see Table 5.1. I’ll leave it to the reader to convince themselves that each representation gives rise to no more than one polyhedron. Let me explain a little of what is meant by the description of (4, 6, 8) and (4, 6, 10). The polyhedron represented by (4, 6, 8) can be arrived at by first considering a truncated cube. We can shave off each vertex just enough so that the square faces become octagons, the triangular faces become hexagons, and where the vertex was is now a square. Thus, we are hypo-truncating the truncated cube. Exercise 5.19. Note that (3, 3, 3, 3, 6) represents a tiling. Identify the complete set of tilings (both regular and semiregular) with five faces at each vertex. Are there any tilings with six faces to a vertex? Solution. Suppose there are four triangles at each vertex. Since the representation is supposed to give a tiling, the sum of the angles at each vertex is 360◦. Thus, the angle of the fifth face is 120 ◦, so it must be a hexagon, giving us (3, 3, 3, 3, 6). If there are only three triangles, then the other two must be at least squares. Since three triangles and two squares is already flat, this is the only possibility. However, there is still some choice of order.

5.5 A PARTIAL CATEGORIZATION OF SEMIREGULAR POLYHEDRA Representation (3,4,4) (3,6,6) (3,8,8) (3,10,10) (4,4,n) (4,6,6) (4,6,8) (4,6,10) (5,6,6)

135

Description Triangular prism Hypo-truncated tetrahedron Hypo-truncated cube Hypo-truncated dodecahedron The n-gonal prisms Hypo-truncated octahedron Hypo-truncated truncated cube Hypo-truncated truncated dodecahedron Hypo-truncated icosahedron Table 5.1

Either the squares are adjacent, giving the representation (3, 3, 3, 4, 4), or they are not, which gives the representation (3, 3, 4, 3, 4). If there are only two triangles, then the others must be at least squares, giving us a sum of angles greater than 360◦. Thus, we have all the possibilities. Diagrams of these tilings can be found in the textbook (Figures 8.2 and 8.3). Now, let us consider the possibility that there are six faces to a vertex. If all of them are triangles, then the angles sum to 360◦, so we get a tiling (3, 3, 3, 3, 3, 3), which is shown in Figure 8.1 of the text. If one of the faces is not a triangle, then we have too much. Thus, these are all the tilings with five or six faces to a vertex. Exercise 5.20*. Consider the set of convex polyhedra whose faces are all congruent equilateral triangles (but with possibly different numbers of triangles at each vertex). Show that this set is finite. How many such polyhedra are there? Solution. Let’s start by constructing some and then worry about whether we have them all. We have two basic building blocks. We can create nbased pyramids and n-gonal drums for n = 3, 4, or 5. We can glue the n-gons together. We have the three Platonic solids, the tetrahedron, octahedron, and icosahedron. We can glue two n-pyramids together for n = 3 or 5 (n = 4 gives the octahedron). On the 4-gonal drum, we can glue two 4-based pyramids. If we do the same with the 5-gonal drum, we get the icosahedron. The 3-gonal drum is better known as the octahedron and presents a more special case. Since all the faces are triangles, we can start by gluing only one pyramid to some face. But recall that the ‘truncated tetrahedron’ is the octahedron, so this object can also be thought of as a tetrahedron (with sides twice as long) from which we have cut off three of the corners. This makes it clear that there are pairs of faces which meet at 180◦, so such a pair should be thought of as a single rhombic face (a face in the shape of a rhombus). This makes this object questionable. After all, if we accept such an object, then we should have to consider vertices with six triangles at a vertex (to make either a flat surface or a folded surface).

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For similar reasons, we do not include the octahedron with pyramids glued on opposite sides, since this gives an object with six rhombic faces and no triangular faces. There is one more object that we have not considered. Take a triangular prism and glue square pyramids on each of the three square faces. Such a process cannot be done with a square prism (a cube) or pentagonal prism since then we would have to stack pyramids on the square or pentagonal faces too, giving six triangles at a number of vertices. So far, we have seven such polyhedra: the tetrahedron with 4 faces; two triangular pyramids glued together to give a polyhedron with 6 faces; the octahedron with 8 faces; two pentagonal pyramids glued together to give a polyhedron with 10 faces; a triangular prism with square based pyramids glued to the sides, giving a polyhedron with 14 faces; the 4-gonal drum with square based pyramids glued to the square faces, giving a polyhedron with 16 faces; and the icosahedron with 20 faces. Are these are all of them? Let us ﬁrst use the Euler characteristic. Let F be the number of faces. Then there are exactly 3F/2 edges, so F must be even. There must be at least three faces to each vertex and there can be at most 5 faces at each vertex, so 3F ≤ V ≤ F. 5 Using the Euler characteristic, 2=F +V−E = F +V −

3F 2

4 + F = 2V. Thus 6F 5

≤ 4 + F ≤ 2F

F ≤4 ≤F 5 F ≤ 20 ≤ 5F 4 ≤ F ≤ 20. We now investigate each case. In Table 5.2 we list the possible number of faces together with the number of vertices with 5, 4, or 3 faces going to it. Let’s look at these cases. In Case 2, there is a vertex with five faces. The other corners of these faces give five more vertices and since there cannot be a pentagonal face, there must be at least one more vertex, for a total of seven vertices. Thus, Case 2 is not possible. This takes care of Cases 4, 5, and 6 too. Though we have an example that satisfies Case 3, we should show that it is the only object. It includes a vertex with three faces. If one of the adjacent vertices also has only three, then we get the tetrahedron. If an adjacent vertex has four faces, then after placing those faces, the remaining open edges form an equilateral triangle. It is clear, in Case 3, that this must be filled with a triangle, since there are only six

5.5 A PARTIAL CATEGORIZATION OF SEMIREGULAR POLYHEDRA

Case 1: Case 2: Case 3: Case 4: Case 5: Case 6: Case 7: Case 8: Case 9: Case 10: Case 11: Case 12: Case 13: Case 14: Case 15: Case 16: Case 17: Case 18: Case 19:

F 4 6

V 4 5

18 20

11 12

V5 0 1 0 3 2 1 0 4 3 2 6 5 4 7 6 9 8 10 12

V4 0 1 3 0 2 4 6 1 3 5 0 2 4 1 3 0 2 1 0

V3 4 3 2 3 2 1 0 2 1 0 2 1 0 1 0 1 0 0 0

137

Comments The tetrahedron – Two triangular based pyramids – – – The octahedron – Has rhombic faces Two pentagonal based pyramids Has rhombic faces – Build it ... – Triangular prism with pyramids – Square drum with pyramids – Icosahedron

Table 5.2 faces. However, further analysis is useful for other cases, so let us note that the other two adjacent vertices have at least four faces each. If one of them has five, then so must the other one, and we have built a pyramid on this open gap. Such an object is not convex. To see that, note that the angle between the faces of a tetrahedron is greater than 60◦, so gluing three along an edge (as we have done) gives two faces that meet at an angle greater than 180◦. Thus, we have shown that the vertices adjacent to a vertex with three faces must all have the same number of faces. Hence, in Case 8, the vertices adjacent to the vertex with three faces must all have five faces, giving us a drum stacked on a pyramid. Such a structure is rigid. If we fill the gap with a triangle, then the resulting structure is the only possibility for Case 9. It is a tetrahedron with three corners sliced off, which has rhombic faces. This rules out all cases with V3 ≥1 and V4 < 3, which are cases 8, 11, 12, 14, and 16. The case where a pyramid is stacked on this gap is Case 11. Suppose now that V3 = 0, as in our remaining cases. Suppose there is a face whose three vertices all have four faces. The object we have so far built is an octahedron with a face removed. The empty face must be filled with a triangle, since a pyramid gives rhombic faces (and anything else gives vertices with more than five faces). Thus, in our remaining cases, we cannot have a face whose vertices all have four faces. Suppose two adjacent vertices have four faces. Then the two remaining vertices of the

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faces with this edge must have five to them. The object we have so far is two pentagonal pyramids glued together at three adjacent edges. If the other two set of edges are also joined, we get Case 10. If not, then the most we can do is insert two triangles to give an object with 12 faces and four vertices with four faces. This object, if it exists, is included in Case 13. The only way I can think of showing that it exists is to build it, which I have and it does exist. It really helps to have appropriate toys, like Zaks. We have now exhausted all possibilities with adjacent vertices with four faces. Suppose now that we have a vertex with four faces and that all adjacent vertices have five faces. This gives us a 4-gonal drum glued to one pyramid. The remaining gap has four edges. It cannot be glued together, since this gives a vertex with six triangles. If one of these vertices has four faces, then we are gluing a triangle to two adjacent edges. The remaining gap is a triangle and since two of the vertices already have four faces, all we can do is fill it with a triangle. The resulting figure is the triangular prism with square pyramids glued to the square faces. Rather than fill the gap with two triangles, we can instead fill it with a square pyramid. This gives the object in Case 17. This exhausts all possible objects with V≥ 4 1, so all that remains is the case when V3 = V4 = 0, which gives the icosahedron, Case 19. No other cases are possible, including objects different from those described above but with the same number of faces and vertices of each type. Thus, there are eight such polyhedra.

5.6

Four-Dimensional Objects

Exercise 5.21. We can unfold the faces of a cube to get the region in Figure 5.7. What do we get if we unfold the hypercube?

Figure 5.7. An unfolded cube. See Exercise 5.21. Solution. Let’s look at the lower dimensional analogue ﬁrst – the square. The square is composed of four edges, joined at the vertices. Unfolding happens at the vertices. The edges are segments which join the vertices in the lines x = 0, x = 1, y = 0, and y = 1.

5.6 FOUR-DIMENSIONAL OBJECTS

139

The cube consists of six squares, the regions joining the vertices in each of the planes x = 0, x = 1, y = 0, y = 1, z = 0, and z = 1. These squares are joined at the edges, and when unfolding, the folds are at the edges. The hypercube consists of eight cubes, the regions joining the vertices in each of the planes w = 0, w = 1, x = 0, x = 1, y = 0, y = 1, z = 0, and z = 1. We will abuse notation a bit, and refer to ‘the cube in the plane w = 0’ as just ‘w = 0.’ These cubes should be joined at the faces, and unfolding should happen at the faces. Let’s take one of these cubes, say the cube given by w = 0. When unfolded (into three dimensions), the adjacent cubes are joined at the faces of this cube. Faces are given by pairs of equations, like w = 0 and z = 0. Thus, the cube z = 0 shares a face with the cube w = 0. Similarly, each of the six cubes x = 0, x = 1, y = 0, y = 1, z = 0, and z = 1 all share a face with the cube w = 0. Thus, unfolded, we have a cube with six cubes attached to each face. The one cube which does not share a face with the cube w = 0 is the cube w = 1. This cube shares a face with each of the other six cubes. Furthermore, since none of the vertices of the cube w = 0 are vertices of the cube w = 1, the cube w = 1 joins each of the other six cubes on the face opposite from the face joining the cube w = 0. Thus, we get a ﬁgure like the one in Figure 5.8 .

Figure 5.8 This ﬁgure is also featured in Dal´ı’s painting shown in the text. One should not be disturbed if it is diﬃcult to imagine these cubes being folded along the faces. After all, in two dimensions, there is no way of folding unfolded cube along the edges. Folding requires us to think in four dimensions. Exercise 5.22. We saw that the dual of the cube is the octahedron, and it can be thought of as having vertices which are the centers of the faces of

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CHAPTER 5. HIGHER DIMENSIONAL OBJECTS

the cube. Find the centers of the faces of the hyper-cube. Check that this gives a four-dimensional regular hyper-polyhedron. Solution. Let us look at the situation in 3-D ﬁrst. Rather than describe the vertices as we have, let us describe the vertices as the points (x, y, z) where x, y, and z ± are 1. This cube is now centered at the origin (0, 0, 0). The faces are described by the planes x =± 1, y = ± 1, and z = ± 1. The centers of these faces are the points ± ( 1, 0, 0), (0,± 1, 0), and (0, 0,± 1). The faces of the octahedron formed by these vertices lie in the planes ± x± y± z = 3. Ok, so what happens in 4-D? The vertices of the hyper-cube are the points (w, x, y, z) where w, x, y, and z are ± 1. The hyper-faces are the cubes described by the vertices in the planes w = ± 1, x = ± 1, y = ± 1, and z = ± 1. The center of the cube w = 1 (where we have abused notation as we did in the previous exercise) is the point (1, 0, 0, 0). Similarly, the centers of the hyper-faces of the hyper-cube are the sixteen points ± ( 1, 0, 0, 0), 0, 1). I claim that this forms a regular ± (0, 1, 0, 0), (0, ± 0, 1, 0), and (0, 0,± hyper-polyhedron. To show this, we must show that every hyper-face is a regular polyhedron. So the next question should be – what are the hyperfaces? They lie in the planes ±w ± x ± y ± z = 4. The vertices on the hyper-plane w + x + y + z = 4 are the points (1, 0, 0, 0), (0, 1, 0, 0)√ , (0, 0, 1, 0), and (0, 0, 0, 1). The distance between any two of these points is 2. Thus, since these four points lie in a three dimensional space, they must form a regular tetrahedron. Thus, the dual of the hypercube is a regular hyperpolyhedron with 16 tetrahedral faces. Exercise 5.23. What does the dual of the hyper-cube look like when it is unfolded? Exercise 5.24. What is the analogue of a cube (a hypo-cube?) in one dimension? What is a hypo-sphere in one dimension? Solution. We can think of the cube as the region whose vertices are at (x, y, z), where x, y, and z are all either 0 or 1. A square has the vertices (x, y) where x and y are either 0 or 1. That is, it has the vertices (0, { 0), (1, 0), (0, 1), (1, 1) }. So, by analogy, the ‘hypo-cube’ has the vertices x where x is 0 or 1. That is, a square is just two points. A sphere is the set of points a distance r from a point P . In one dimensions, this is just two points. Thus, the ‘hypo-sphere’ and ‘hypocube’ are the same thing in one dimension. Exercise 5.25. What is the hyper-surface area (a volume) of the hypersphere in four dimensions? [H]

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5.6 FOUR-DIMENSIONAL OBJECTS

Solution. As is pointed in the hint, the surface area of the sphere is the derivative of its volume. This makes sense – think of it this way. The change in volume ∆V of the sphere as r changes by ∆r is a skin that has thickness ∆r. The volume of the skin is its surface area times its thickness. Thus, ∆V = S∆r, or

. dr This argument works in any dimension. Thus, the surface hyper-volume of the hyper-sphere is 2 4 d πr =22 3 π r . dr 2 Exercise 5.26. Find the hyper-volume of the interior of a hyper-sphere in ﬁve dimensions. [A] Solution. The ﬁve dimensional hyper-sphere has the equation 2

x +x +x +x +z =r . Like we did for the sphere and 4-D hyper-sphere, we take a slices by holding z constant. Such a cross section has the equation x1 + x 2 + x 3 + x 4 = r − z , √ which is a 4-D hyper-sphere of radius r2 − z2. It therefore has a hyperπ2(r2 − z2)2 volume of . We sum up these slices to get 2 ∫ r 2 2 2 π (r z )2 − V = dz 2 −r ∫ π2 r 4 2 2 4 = r − 2r z + z dz 2 −r 2

2 3

2r z z π2 4 rz− 3 + 5 2 −r 5 5 2(−r)5 (−r)5 π2 5 2r + r ( )5 + + = r − − −r 2 3 5 3 5 2 5 2 1 =π r 1− + 3 5 2 5 8π r = . 15 =

Chapter 6

Hyperbolic Geometry 6.1

Models

6.2

Results from Neutral Geometry

Exercise 6.1. Prove that the sum of angles in a quadrilateral is at most 360◦. Conclude that the angles A and D in the Saccheri quadrilateral in Figure 6.3 [of the text] are at most right angles. [S] Exercise 6.2. Suppose ABCD is a rectangle. That is, suppose ABCD is a quadrilateral whose angles are all right angles. Prove that | AB | = | CD| and |BC = that the line AB does not intersect the line CD. | DA | . Prove | Let a line l intersect AB perpendicularly at E. Show that l intersects CD at a point F and that AEF D is a rectangle. Solution. Suppose |AB =| /CD| . Without loss of generality, we may assume | ′ ′ |AB |< CD | . | Let D be the point on the segment CD such that | CD | = ′ ′ |AB . |Then, since D is between C and D, the angle ∠D BA is smaller than 90◦. Note that ABCD′ is symmetric with respect the reﬂection through the perpendicular bisector of BC. Thus, ∠BD′C = ∠D′BA < 90◦. Hence ∠DD′A > 90◦, so the sum of the angles in ∆ADD′ is greater than 180◦. This contradicts the Saccheri-Legendre theorem (Theorem 6.2.2). Thus, we must have | AB | =| CD| . Similarly,| BC| = | AD|. Suppose AD intersects BC at P . Then we have a triangle ∆ABP with angles at A and B that sum to 180 ◦. This contradicts Lemma 6.2.3, so no such P can exits. Now, suppose l intersects AB perpendicularly at E. Then l cannot intersect BC or AD, since that would again give a contradiction to Lemma 6.2.3, so l must leave the rectangle through the side CD. Let it intersect CD at F . By Exercise 6.1, the sum of the angles in any quadrilateral is at most 360◦. Thus, by looking at AEF D, we have ∠EFD ≤ 90◦, and by 143

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CHAPTER 6. HYPERBOLIC GEOMETRY

looking at BEFC, we have ∠EFC ≤ 90◦. Since these two angles sum to 180◦, they must both equal 90◦. In particular, AEF D is a rectangle. Exercise 6.3. Prove that if there exists a triangle whose angles sum to 180◦, then there exists a right angle triangle whose angles sum to 180 ◦. Conclude that if there exists a triangle whose angles sum to 180 ◦, then there exists a rectangle. [H] Solution. Since the sum of the angles in a triangle is no more than 180 ◦, there can be at most one obtuse angle in the triangle. In an arbitrary triangle ∆ABC, we may assume, without loss of generality, that the angles at B and C are acute. Let the perpendicular to BC through A intersect the line BC at D. Since the sum of the angles in ∆ADB is at most 180◦, and the angle at D is 90◦, the other two angles must be acute. Thus, if D is not between B and C, then either the angle at B or the angle at C in ∆ABC supplements an acute angle, so must be obtuse. This is a contradiction, so D must be between B and C. If the sum of the angles in ∆ABD is 180◦, then we are done. Otherwise, the sum of the angles must be less than 180 0, so ∠ABD + ∠BAD < 90◦. But 180◦ = ∠CAB + ∠ABC + ∠BCA = ∠CAD + ∠DAB + ∠ABC + ∠BCA, so 90◦ < ∠CAD + ∠BCA = ∠CAD + ∠DCA (since ∠ABD = ∠ABC). Thus the sum of the angles in ∆ACD is more than 180◦, which is a contradiction. Hence, the sum of the angles in ∆ABD must be 180◦, so there exists a right angle triangle whose angles sum to 180◦. Finally, we can glue two copies of this triangle together to get a rectangle. Exercise 6.4. Suppose ABCD is a rectangle, and that l is a line through A which is not the line AD. Prove that l intersects the line BC. Conclude that there are no rectangles in hyperbolic geometry. [S] Exercise 6.5. Prove that the existence of a rectangle (or a triangle whose angles sum to 180◦) implies the Euclidean parallel postulate. [H] Solution. We must show that if l is a line and P is a point not on l, then there exists a unique line through P which does not intersect l. We know there exists a point Q on l such that the line PQ is perpendicular to l. Since there exists a rectangle ABCD, we can translate, rotate it until A is at Q and D is on l. Then B is on P Q. We can further have B on the same side of l as P by reﬂecting too, if necessary.

6.3 THE CONGRUENCE OF SIMILAR TRIANGLES

145

If P is between A and B, then let l′ be the line through P that is perpendicular to AB. By Exercise 6.2, l′ intersects CD at (say) F ; AP FD is a rectangle; and l′ does not intersect the line AD, which is l. By Exercise 6.4, there is only one line through P which does not intersect l. That is, we have the parallel postulate, provided P is between A and B. If P is not between A and B, then all we have to do is create a longer rectangle. We stack a copy A1B1C1D1 of ABCD on top of ABCD such that A1 is at B and D1 is on BC and on the same side of AB as C. Since |BC |= AD | = | A|1D1, the point D1 ends up at C. Since all angles are right angles, the point B1 is on the line AB, and C1 is on the line BC. Thus, we get a new rectangle AB1C1D with | AB1| = 2| AB |. If P is on the segment AB1, then we proceed as before. Otherwise, we double its length again. Repeating, we will eventually have a rectangle AB′C′D with P on the line segment AB′. Thus, the existence of a rectangle (or by Exercise 6.4, the existence of a right angle triangle), implies the Euclidean parallel postulate.

6.3

The Congruence of Similar Triangles

Exercise 6.6. Show that, in the proof of Theorem 6.3.3, the edges BC and B′′C′′ cannot intersect. [H] Solution. We can go one further – the lines BC and B′′C′′ do not intersect, for suppose they do, at (say) D. Then we have a triangle ∆BB′′D in which the sum of two of the angles (at B and B′′) is 180◦. That contradicts Lemma 6.2.3.

6.4

Parallel and Ultraparallel Lines

Exercise 6.7. Let M be a point at inﬁnity and let ∠P QM = 90◦. Given ϵ > 0, show that there exists a point R on QM such that ∠PRQ < ϵ. Figure 6.1 is a hint. P

M Q

Figure 6.1. See Exercise 6.7. Solution. Let R0 be a point on the ray QM such that QR | 0 =| P Q |. | Generate Rk inductively by deﬁning Rk+1 to be the point on the ray RkM

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CHAPTER 6. HYPERBOLIC GEOMETRY

such that| RkRk+1 |= PR | k .| Since ∆PRkRk+1 is isosceles, ∠Rk+1PRk = ∠PRk+1Rk. Since the sum of the angles is less than 180◦, we have 2∠PRk+1Rk < ∠PRkQ. Thus,

∠PR

k+1

R < k

1 2k

∠PR Q. 0

Thus, given any ϵ > 0, there exists a k such that ∠PRk+1Q < ϵ. We set R = Rk+1. Exercise 6.8. Prove that if alternate interior angles of a transversal l to two lines l1 and l2 are equal, then l1 and l2 are ultraparallel. Solution. Since opposite interior angles are equal, l1 and l2 have a common perpendicular (see the argument in the proof of Theorem 6.4.2). Let this common perpendicular intersect l1 and l2 at P and Q, respectively. If l1 and l2 intersect at, say, R, then the sum of the angles in ∆P QR is greater than 180◦. Thus, no such R can exist. If l1 and l2 are parallel, then Π( |P Q|) = π/2. But Π( P |Q ) |< π/2 for any length P| Q , | so l1 and l2 cannot be parallel. They are therefore ultraparallel. Exercise 6.9. Two parallel lines do not have a common perpendicular (see Theorem 6.5.1 below). Where does the proof of Theorem 6.4.2 fail in this case? That is, where was it important that the two lines in Theorem 6.4.2 be ultraparallel? Solution. In the proof of Theorem 6.4.2, we are using the intermediate value theorem. If R is close to Q, ∠PRQ is close to 90◦. Since ∠NPR < ∠NPQ < 90◦, there exists an R such that ∠NPR < ∠PRQ. At the other end, ∠NPR > ∠NPM > 0, but we can make ∠PRQ as small as we like, so there exists an R such that ∠PRQ < ∠NPR. Hence, somewhere in between, they are equal. But if the two lines are parallel, then ∠NPM = 0, so we do not know whether we can ﬁnd an R such that ∠PRQ < ∠NPR, and the argument falls apart.

6.5

Singly Asymptotic Triangles

6.6

Doubly and Triply Asymptotic Triangles

Exercise 6.10. Prove that any two triply asymptotic triangles are congruent.

6.7 THE AREA OF ASYMPTOTIC TRIANGLES

147

Solution. Let ∆LM N be a triply asymptotic triangle. Reﬂect N through LM to get P . Then the line NP is sent to itself, but is not the line LM , so must be perpendicular to LM . Let NP intersect LM at A. Then both ∆LAN and ∆M AN are doubly asymptotic triangles with right angles at A. Now, suppose ∆L′M ′N ′ is another triply asymptotic triangle with point P ′ on L′M ′ such that N ′P ′ is perpendicular to L′M ′. Then ∆LAN and ∆L′A′N ′ are congruent by Theorem 6.6.2. Thus, there exists an isometry f such that f (L) = L′, f (A) = A′, and f (N ) = N ′. Since f (L) = L′ and f (A) = A′, the isometry f sends the line LA to the line L′A′. In particular, this means it sends the M to M ′. That is, f (M ) = M ′, so f is an isometry that sends ∆LM N to ∆L′M ′N ′. That is, the two triangles are congruent.

6.7

The Area of Asymptotic Triangles

Exercise 6.11. Prove that the area of any doubly asymptotic triangle is finite. Solution. Let ∆AM N be a doubly asymptotic triangle with vertices M and N at infinity. Pick any point B on MN . Then the area of ∆AM N is the sum of the areas of ∆ABN and ∆ABM . Since these two triangles are singly asymptotic triangles, their areas are finite, so their sum is finite. Exercise 6.12. Prove that the area of all triply asymptotic triangles are finite and equal. Solution. Let ∆LM N be a triply asymptotic triangle. Let P be any point inside this triangle. Then ∆LM N is the union of ∆P LM , ∆PMN , and ∆PNL. Since these there triangles are doubly asymptotic, their areas are all finite (by Exercise 6.11), so their sum is finite. Since all triply asymptotic triangles are congruent (see Exercise 6.10), their areas must all be equal.

Chapter 7

The Poincaré Models of Hyperbolic Geometry 7.1

The Poincaré Upper Half Plane Model

7.2

Vertical (Euclidean) Lines

Exercise 7.1. What is the distance between the points 3 + i and 3 + 5i in the Poincaré upper half plane H ? Remember, the Poincaré upper half plane H comes equipped with a non-Euclidean arclength element, so the answer is not 4. Solution. Since the x-components of the two points are the same (equal to 3), the distance between the points is the absolute value of the logarithm of the ratio of the y-components. Thus, for P = 3 + i and Q = 3 + 5i, |P Q| = | ln(5/1)| = | ln(1/5)| = ln 5. Exercise 7.2. What is the distance between the points −2+2i and −2 + i in the Poincaré upper half plane H?

1 7

1 Solution. Since the points P = −2 + 2i and Q = −2 + i 7 have the same x-component, we have |P Q| = | ln(2/(1/7))| = ln 14.

7.3

Isometries

Exercise 7.3. Prove that the dilation δλ(x, y) = (λx, λy) preserves the Poincaré arclength element. 149

150 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Solution. Let (u, v) = (λx, λy). Then, 2

du + dv (λdx)2 + (λdy)2 = 2 v (λy)2 2 2 dx + dy = . y2

7.4

Inversion in the Circle

Exercise 7.4. Let P = 4+4i and Q = 5+3i. Find M and N , the endpoints of the Poincaré line through P and Q. Solution. The Poincaré line through P and Q is the Euclidean half circle through P and Q that is perpendicular to the x-axis. Such a circle has a center on the x-axis, so the center is a real value. Let x be this center. Then, the distance from x to P and to Q are equal (the radius of the circle), so using the Pythagorean theorem, (x − 4) + 4 = (x − 5) + 3 2

x − 8x + 32 = x − 10x + 34 2x = 2 2

x = 1. Now that we have the center, we can ﬁnd the radius: r = (1 − 4) + 4 = 25 r = 5. 2

Finally, M and N are the points where the circle intersects the x-axis. That is, M and N are 1 − 5 = −4 and 1 + 5 = 6. Exercise 7.5. Let P = 12i and Q = 7 + 5i. Find M and N , the endpoints of the Poincaré line through P and Q. Solution. As in the solution to Exercise 7.5, we let x∈R be the center of the circle that represents the Poincaré line through P and Q. Then, x + 12 = (x − 7) + 5 2

x + 144 = x − 14x + 49 + 25 14x = 70 2

x = 5. Thus,

r = 5 + 12 = 13 , so r = 13. Finally, M = x+r = 5+13 = 18 and n = x−r = 5−13 = −8.

151

7.4 INVERSION IN THE CIRCLE

Exercise 7.6. Finish the proof of Theorem 7.4.1. That is, show that the image of a line l under inversion in the unit circle is a circle through O when l is tangent to the unit circle, or when l intersects the circle in two places, but does not go through O. What is the image of l if l goes through the origin O? Exercise 7.7. In the proof of Lemma 7.4.2, we assumed that Γ lies outside the unit circle. What are the remaining cases which must be checked? Prove them. Exercise 7.8. The proof of Lemma 7.4.3 is not quite complete. What other cases must be considered? Exercise 7.9. Show that the map 2

Φr(x, y) =

r x x2 + y

, 2

r y x2 + y 2

is inversion in the circle with radius r and centered at the origin O. Solution. Whoops, please note the correction (to appear in the second printing). Note that r2 φr(x, y) = 2 (x, y). x + y2 1 1 As we saw before, ||(x, y)|| = , where O is the origin and P = x2 + y 2 |OP | (x, y). Thus, r2 ||φr (x, y)|| = . |OP | Since φr(x, y) is in the same direction as (x, y), it is the inversion of (x, y) in the circle of radius r centered at (0, 0). Exercise 7.10. Prove Lemmas 7.4.1 and 7.4.2 analytically. That is, for Lemma 7.4.1, let l be a line with equation ax + by = c and c/= 0. Show that the image of l under the action of Φ (or Φr) gives the equation of a circle which goes through O. Do the same for Lemma 7.4.2. Solution. We ﬁrst note that 2

u +v =

2 1 x2 y = 2 . + 2 2 2 2 2 (x + y ) x + y2 (x + y )

Thus, u u2 + v 2 v y= 2 . u + v2

152 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY We could have also seen this by noting that φ is its own inverse. Thus an arbitrary line ax + by = c becomes u v a 2 +b 2 =c 2 u +v u + v2 2 2 au + bv = c(u + v ). If c 0, this is an equation of a circle. Furthermore, it includes the solution (0, 0) so this circle goes through the origin. Let us also point out that if c = 0, then we get the equation of the line we began with. Thus, the image of a line that does not go through the origin is a circle that goes through the origin, and the image of a line that goes through the origin is itself. An arbitrary circle (x − a) + (y − b) = r 2

becomes x − 2ax + a + y − 2by + b = r 2

x + y − 2ax − 2by = r − a − b 2au 2bv 2 2 2 — a − b − − =r u 2 + v2 u 2 + v2 u 2 + v 2 2 2 2 2 2 1 − 2au − 2bv = (r − a − b )(u + v ). 2

If r − a − b 0, this is an equation of a circle. If r − a − b = 0, then it is an equation of a line and the original circle includes the solution (0, 0). Thus, the image of a circle that does not go through the origin is a circle, and the image of a circle that does go through the origin is a line. 2

Exercise 7.11. Let P be a point inside a circle centered at O. Let T be a point where the perpendicular to PO intersects the circle. Let P ′ be the point where the tangent to the circle at T intersects OP . Show that P ′ is the image of P under inversion in the circle. Solution. Since T is a point of tangency, ∠P ′ TO is a right angle (see Figure 7.1). Thus, ∆OTP ′ ∼ ∆OPT , so |OT |

|OP ′| |OT | |OP | |OT |2 . |OP ′| = |OP | =

Since OT | =|r, the radius of the circle, and P ′ lies on the ray OP , P ′ is the image of P under inversion in the circle.

153

7.4 INVERSION IN THE CIRCLE T

P′

Figure 7.1 Exercise 7.12. Two circles, as shown in Figure 7.2, intersect at right angles. The circle centered at O has radius one unit. The point R is an arbitrary point on the circle centered at P . Prove that |OS| =

1 |OR|

Solution. Let us apply power of the point to the point O with respect to the circle centered at P : Π(O) = |OS||OR|. Let T be a point of intersection of the two circles. Then OT is tangent to the circle centered at P , since the circles intersect perpendicularly. Thus, Π(O) = |OT |2 = 1. Combining, we have |OS||OR| = 1 |OS| =

|OR|

Exercise 7.13. Write a Geometer’s Sketchpad script which draws the Poincaré line through two points in H . The input might be the two points and the line which deﬁnes the real axis. Script Ex07013. This script constructs the Poincaré line through two points A and B. The input is the two points and the real line. It doesn’t work if the points lie on a Euclidean line perpendicular to the real axis. Exercise 7.14. Write a Geometer’s Sketchpad script which constructs the inversion of a line in a given circle. Use only the buttons to the left and the ‘Construct’ pull-down menu. Does your script work for all lines? If not, consider the use of rays instead of lines at some places in your script.

154 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY R S O

Figure 7.2. See Exercise 7.12. Script Ex07014. In this script, the input is the circle of inversion and the line we are inverting. We begin by running the script Ex04025a to first find the center O of the circle. The point Q is where the perpendicular to the given line j through O intersects j. Note that this line is immediately replaced with a ray. This ensures that future points of intersection are on the correct side. To find the inverse of Q, we pick an arbitrary point on j and draw a similar triangle to find the appropriate length. This length is translated to the ray OQ to give the point P , the inverse of Q. The inverse of the given line j is the circle with diameter OP . Exercise 7.15. Write a Sketchpad script which demonstrates that inversion in a circle preserves angles. Script Ex07015. Inversion in the circle preserves angles. Given an angle ∠ABC and a circle c1, we run Ex07014 to find the images of the lines AB and BC under inversion. The heavy arcs are the portion of those images that correspond to the rays BA and BC. The segments B′A′ and B′C′ are tangents to these images. The angle between these tangents is angle ∠A′B′C′. The measurement shows that this angle is the same as the measure of angle ∠ABC. Unfortunately, I saw no way of convincing sketchpad to always measure the correct angle, so I have included the measure of angle A′B′C′′ too, which for some choices of A, B, and C, is the correct angle to look at. Exercise 7.16. Write a Sketchpad script which inverts a circle in a given circle. Again, use only the buttons and the ‘Construct’ pull-down menu. Script Ex07016. The inversion of a circle c1 in the circle c2. The input is just the circles, so Ex04025a was used to find the centers P and O, respectively. The real script starts after that (line 18). The line through the centers intersects the circle c1 at two points. We use a random point on the circle and similar triangles to find the inverses of these points. Note the use of rays instead of lines in some places. This insures the correct image is found. This script works for all circles except when c1 goes through the center of c2.

155

7.4 INVERSION IN THE CIRCLE

Exercise 7.17. Write a Sketchpad script which inverts a circle in a given circle. This time, include, in your script, an arbitrary point P on the original circle and its image under inversion. As you drag the point P around the circle, does its inverse move appropriately? If not, rethink your method of constructing the inverse of P . Script Ex07017. In this script, we first run Ex07016 to find c3, the image of the circle c1 under inversion in c2. The real script starts after step 39. Since that script hides the center of c2, we run Ex04025a again to get the point O (line 48). To find the image of P , a random point on c1, we can’t just find the intersection of OP with c3, since there are two points. To get the correct point P ′, we instead use similar triangles again. Exercise 7.18 (The Necklace Theorem or Steiner’s Porism). Let the circle Γ′ be inside the circle Γ. Insert a circle c1 externally tangent to Γ′ and internally tangent to Γ, as in the diagram. Insert another circle c2 tangent to c1, Γ, and Γ′. Continue around Γ′. Suppose that, for some n, we have cn tangent to c1 (an example with n = 6 is shown in Figure 7.3.) Prove that if this happens for some n and some pair of circles Γ and Γ′, Γ

c1 c2 Γ′ c3

Figure 7.3. A necklace with n = 6. See Exercise 7.18. then no matter where we ﬁrst draw c1, we will always have cn tangent to c1. Solution. First, we note that if Γ and Γ′ are concentric, then the result is obvious, since all circles that are tangent to both Γ and Γ ′ are congruent (with diameter equal to the diﬀerence of the radii of Γ and Γ′).

156 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY For the general case, our idea is to ﬁnd a inversion which sends Γ and Γ′ to concentric circles. If we can do this, then we have shown that the general case is equivalent to the trivial case where Γ and Γ′ are concentric. So let us show that such an inversion exists. Let us construct a coordinate system on the line that goes through the centers of Γ and Γ′. Let Γ intersect this line at a and b, where a and b are coordinates on the line. Let Γ′ intersect this line at c and d, and let us consider inversion in the 1 unit circle centered at x. The image of y under this inversion is + x. y−x Thus, the center of the image of Γ under this image is at 1

a−x

+x+

1 b−x

+x .

If the two circles are concentric, then we have 1

+ 2x

2 a−x b−x 2 c−x (b − x)(c − x)(d − x) + (a − x)(c − x)(d − x)

1 c−x

+ 2x

= (a − x)(b − x)(d − x) + (a − x)(b − x)(c − x). This is a cubic polynomial in x, so it has at least one real solution. Thus, there exists an inversion such that the images of Γ and Γ′ under this inversion are concentric. Exercise 7.19. Use Sketchpad to draw a picture like the one in Figure 7.3. Script Ex07019a. Let’s start with a nice symmetric version of Figure 7.7 (of the text). We’ve all placed pennies around a central one – that’s what we’ve done in this script. Script Ex07019b. To get a less symmetric version of Figure 7.7, we invert the symmetric version of Ex07019a (run on A, B) in a circle centered at O and through P , using Ex07016 repeatedly. Script Ex07019c. In this script, we create a symmetric necklace with ﬁve beads. We do this by constructing a regular pentagon using Ex04010, and then draw the circles centered at the vertices and going through the midpoints of the sides. The line through the center and a vertex intersects one of the circles at two points. The inside and outside circles are the ones centered at the center and going through each of these points. Script Ex07019d. A ﬁve bead necklace. We invert the symmetric one created with Ex07019c and the points A and B in the circle centered at Q and through R. Exercise 7.20. Let Γ′ and Γ be internally tangent and have centers on the x-axis, as in Figure 7.4. Let c0 be the circle with center on the x-axis and such that c0 is tangent to Γ and Γ′. Let c1 be the circle tangent to Γ, Γ′, and c0, and in general, let cn be the circle tangent to Γ, Γ′, and cn−1. Let

157

7.4 INVERSION IN THE CIRCLE

hn be the distance from the center of cn to the x-axis, and let dn be the diameter of cn. Prove that hn = ndn. This result is attributed to Pappus. c3

Γ c2 c1

Γ′ c0

Figure 7.4. See Exercise 7.20. Proof. Let the circles Γ and Γ′ be internally tangent at O and let c0 intersect Γ at P . Consider inversion in the circle Γ′′ centered at O with radius r where r is chosen so that cn intersects Γ′′ perpendicularly. Since Γ and Γ′ intersect tangentially at the center of inversion, their images l and l′ are parallel lines. Since cn intersects Γ′′ perpendicularly, its image is itself. Since the line OP goes through the center of inversion, its image is itself. Finally, since inversions preserve angles, the lines l and l′ are perpendicular to OP . Thus, the images of c0, ..., cn forms a string of circles between l and l′, starting with the image of c0 centered on the line OP , and ending with cn, as in Figure 7.5. Hence, hn = ndn. The following Exercises 7.21 – 7.24 are related: Exercise 7.21. Suppose three congruent circles are concurrent at H and intersect in pairs at A, B, and C, as in Figure 7.6(a). Write a Sketchpad script which demonstrates that H is the orthocenter of ∆ABC. Script Ex07021. Let the three congruent circles that intersect at H be centered at P , Q, and R. Then P , Q, and R are all equidistant from H, so lie on a circle centered at H. Let us begin our script by choosing H and P . The points Q and R are random points on the (hidden) circle centered at H and through P . The other points of intersection of pairs of circles are the vertices of the triangle ∆ABC. We ﬁnd the altitudes, and note that they go through H. Thus, H is in fact the orthocenter of triangle ∆ABC. Exercise 7.22. Three tangents to a circle centered at I intersect in pairs at A′, B′, and C′, as in the Figure 7.6(b). Write a Sketchpad script which demonstrates that the line A′I goes through the circumcenter of ∆B′C′I.

158 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY

Γ′′ c4

Γ c2 c1

Γ′

c0 O

Figure 7.5 Script Ex07022. Given I and P , this script constructs the circle centered at I and through P . It picks random points Q and R on this circle and constructs tangents to the circle at P , Q, and R. These tangents meet at A′, B′, and C′. It then constructs the circumcircle of triangle ∆IB′C′ and the ray A′I. We note that the ray A′I goes through O, the center of the circumcircle of triangle ∆IB′C′. Exercise 7.23. For A, B, C, and H as in Exercise 7.21 and A′, B′, C′, and I as in Exercise 7.22, show that H is the orthocenter of ∆ABC if and only if A′I goes through the circumcenter of ∆B′C′I. (Hint: Think inversion in some circle.) Solution. Let three congruent circles intersect at H, and in pairs at A, B, and C. Let the congruent circles have radius r. Note that the circle Γ with radius 2r and centered at H is tangent to all three smaller circles. Let us invert this situation in a circle centered at H (the radius of the circle of inversion is not important). Let A′, B′, and C′ be the images of A, B, and C, respectively. Since the congruent circles all go through H, their images are lines. Since Γ is tangent to all three circles, its image is tangent to the sides of ∆A′B′C′, so it is the incircle. Since the center of Γ is H, it’s image is centered at the image of inﬁnity. That is, the incenter I is the image of inﬁnity. Now consider the line AH. Since it goes through H, its image is a line; since it is a line, it’s image goes through I; and of course the image of A is A′. Thus, the image of the line AH is the line A′I. Let us now consider the side BC. Since it is a line, its image goes through I. Thus, the image of BC is a circle that goes through the points B′, C′, and I. That is, it is the circumcircle of ∆B′C′I.

159

7.4 INVERSION IN THE CIRCLE

A′ B′

A B

C C′

(a)

(b)

Figure 7.6. See Exercises 7.21 – 7.24. Now, suppose H is the orthocenter of ∆ABC. Then AH is perpendicular to BC. Thus, their images are perpendicular. That is, A′I is perpendicular to the circumcircle of ∆B′C′I. Hence, A′I goes through the circumcenter of ∆B′C′I. Conversely, suppose A′I goes through the center of ∆B′C′I for any triangle ∆A′B′C′. Then A′I is perpendicular to the circle through B′, C′, and I. Thus, their inverses, AH and BC, are perpendicular. Similarly, BH is perpendicular to AC. Thus, H is the orthocenter of ∆ABC. Exercise 7.24. For A′, B′, C′, and I as in Exercise 7.22, prove that A′I goes through the circumcenter of ∆B′C′I. (Hint: Let O be the center of the circumcircle of ∆B′C′I. What is ∠IC′B′? What is ∠IOB′? What type of triangle is ∆IOB′? What is ∠A′IO?) Conclude that for A, B, C, and H as in Exercise 7.21, H is the orthocenter of ∆ABC. Solution. Let us look at the quadrilateral A′B′OI. Since I is the incenter, 1 2 ′ (where ∠A′, ∠B′, and ∠C′ are the angles of ∆A′B′C′). ∠B′A′I = ∠A 1 ′ ′ ′ ′ ′ By the Star Trek lemma, ∠B C I = 2 ∠B OI, so ∠B OI = ∠C . Since ∆OB′I is isosceles (|OB′| and |OI| are radii), ∠IB′O = ∠B′IO. Thus, ∠C′ + 2∠IB′O = 180◦ = ∠A′ + ∠B′ + ∠C′, so ∠IB′O =

1 (∠A′ + ∠B′). 2

Since ∠A′B′I = 2∠B′, we get 1

∠A′B′O = ∠B′ +

∠A′ . 2

160 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Thus, ∠IA′B′ + ∠A′B′O + ∠B′OI =

∠A′ + ∠B′ + 2

∠A′ + ∠C′ = 180◦. 2

Hence, ∠OIA′ = 180◦, so O is on the line A′I, as claimed. As a corollary of this and Exercise 7.23, we get the following: Suppose three congruent circles are coincident at H and intersect in pairs at A, B, and C. Then H is the orthocenter of ∆ABC. Exercise 7.25*. Two circles Γ and Γ′ intersect at A and D. A common tangent intersects Γ and Γ′ at E and F , respectively. A line through D parallel to EF intersects Γ and Γ′ and C and B, respectively, as in Figure 7.7. Show that the circumcircles of ∆BDE and ∆CDF intersect again on the line AD. Hint: Exercise 1.125 might help. Γ′ D

Figure 7.7. See Exercise 7.25. Solution. This section is on inversion, so let’s invert. In what point? The usual rule of thumb is to invert in the point that has the most action – in this case D. The radius of the circle of inversion is not important. Let the points A′, B′, etc. be the images of A, B, etc., and follow along in Figure 7.8. The image of D is the point at infinity. Let P be the image of the point at infinity. The lines and circles through D go to lines, so we get a triangle ∆A′B′C with P on B′C′, E′ on A′C′, and F ′ on A′B′. The image of the line EF is a circle (since EF does not go through D). Let us call this circle Γ′′. Since EF is tangent to Γ and Γ′ at E and F , the circle Γ′′ is tangent to A′C′ and A′B′ at E′ and F ′. Since EF is parallel to BC, they are ‘tangent at infinity’, so Γ′′ is tangent to B′C′ at P . That is, Γ′′ is the incircle of ∆A′B′C′. Now, what of the circumcircles of ∆BDE and ∆CDF ? Since they go through D, their images are the lines B′E′ and C′F ′. By Exercise 1.125, these lines and A′P are coincident at (say) G′, the Gergonne point. Finally, the line A′P is the image of the line AD, so the line AD and the circumcircles of ∆BDE and ∆CDF all intersect at G, the inverse image of G′.

161

7.5 INVERSION IN EUCLIDEAN GEOMETRY A′ E′

F′

B′

C′

Figure 7.8 Exercise 7.26. Check the result in Exercise 7.25 using Sketchpad. Script Ex07026. First, we must produce a drawing like the one in Figure 7.10 (of the text)[Figure 7.7 of this document]. There are a number of choices one could make for input. I chose E, F , and A. We construct the circles through A and tangent to EF at E, and the circle through A and tangent to EF at F . We construct BC through D, parallel to EF , and with endpoints on the appropriate circles. This is Figure 7.10. We then construct the circumcircles of ∆BDE and ∆CDF , to get the point of intersection G. We draw AD, and note that it goes through G, as claimed.

7.5

Inversion in Euclidean Geometry

Exercise 7.27 (Ptolemy’s Inequality). For an arbitrary quadrilateral ABCD, show that |AC||BD|≤ |AB||CD| + |BC||DA|. When does equality hold? The solution is in the text. Exercise 7.28. Prove Theorems 7.5.1 and t7.5.2. Proof of Theorem 7.5.1. I’m sure there are a number of ways of doing this but today, I feel like using the power of the point. Follow along in Figure 7.9. Let Γ be the circumcircle of ∆ABA′ and let r be the radius of the circle of inversion. Then the power of the point O with respect to Γ is r2 ′ = Π ( )= = 2 O |OA||OA | Γ |OA| r . |OA| Let the line OB intersect Γ at B′′. Then r = ΠΓ(O) = |OB||OB′′|, 2

162 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Γ

A′ O

B′′

Figure 7.9 so |OB′′ | =

. |OB| Thus, B′′ = B′ and the quadrilateral ABB′A′ is cyclic. Thus, ∠A′AB and ∠BB′A′ are supplementary, so ∠OAB = ∠A′B′O. Similarly, ∠OBA = ∠OA′B′. Proof of Theorem 7.5.2. By Theorem 7.5.1, ∆OAB ∼ ∆OB′A′, so |A′B′|

|OA′|

|AB| |OB| |AB| A ′ B′ = OA′ | | | | |OB| 2 |AB| r = . |OB| |OA| Exercise 7.29. Let Q be the image of O under reﬂection through the line l. Prove that the image Q′ of Q under inversion in a circle centered at O is the center of the circle Γ which is the image of l under the same inversion. Solution. Let q = OQ | and | let OQ intersect l at P . Since Q is the image of O under reﬂection through O, the line OQ is perpendicular to l and P is the midpoint of OQ. Thus |OP = | q/2. The image of l under inversion is the circle with diameter OP ′ where P ′ is the image of P , so is the point on the ray OP such that |OP ′| =

2r2

The center of this circle is the midpoint OP '|

such that |OR| = | 2 = q . Since the ray OP is the same as the ray OQ, we have established that R is the r2 point on the ray OQ such that |OR| = q, which is exactly the description of Q′. of OP ′, so is the point R on the ray OP

163

7.5 INVERSION IN EUCLIDEAN GEOMETRY

Exercise 7.30. Four circles are coincident at a point O in such a way that they are externally tangent at O in pairs, as in Figure 7.10(a). These circles intersect again at points A, B, C, and D. Show that |AB||OD||OC| = |CD||OA||OB|.

A B O C D C

B D (a)

(b)

Figure 7.10. See Exercises 7.30 and 7.31. Solution. After dividing both sides with the terms that include O, we have some relations that look a lot like the expression in Theorem 7.5.2. Thus, we are inclined to invert in a circle centered at O. Let A′, B′, etc., be the images of A, B, etc. under inversion in a circle centered at O. Since all four circles go through O, the images of these circles are lines. Since the circle through A, B, and O, and the circle through C, D, and O are tangent at O, their images are parallel lines. That is, A′B′ and C′D′ are parallel lines. Similarly, B′C′ and A′D′ are parallel lines. Thus, A′B′C′D′ is a parallelogram. In particular, |A′B′| = |C′D′|, so using Theorem 7.5.2, 2 |CD|r |AB|r = |OC||OD| |OA||OB| |AB||OC||OD| = |CD||OA||OB|. 2

Exercise 7.31. Two circles intersect at A and C. The tangents at A to these circles intersect them again at B and D, as in Figure 7.10(b). Show that |AB||CD| = |AC||AD|.

164 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Solution. Since this section is on inversion, let’s try to use that. Again, using the maxim that we should invert through a circle centered at a point where there is a lot of action, let us invert in a circle centered at A. Let the image of the point at inﬁnity be P , and let B′, C′, and D′ be the images of B, C, and D, respectively. Since BA is a line through A, its image is the line B′P . Similarly, the image of DA is D′P . The image of the circle through A, B, and C, is a line through B′C′. Since DA is tangent to this circle at D, its image D′P is parallel to B′C′. Similarly, B′P is parallel to C′D′, so PB ′ C ′ D ′ is a parallelogram. In particular, |PB′| = |C′D′| 2 r2 |CD|r = |AC||AD| |AB| |AC||AD| = |AB||CD|.

7.6

Fractional Linear Transformations

Exercise 7.32. Prove Theorem 7.6.1. That is, prove that if γ1 and γ2 are two invertible 2 × 2 matrices, then Tγ1γ2 (z) = Tγ1 (Tγ2 (z)), or equivalently, that (γ1γ2)z = γ1(γ2z). Solution. Let b d

and

Tγ1γ2 (z) =

a b c d

γ1 =

a c

γ2 =

A C

B D

B . D

Then, A C

aA + bC aB + bD z cA + dC cB + dD

(aA + bC)z + (aB + bD) . (cA + dC)z + (cB + dD)

For the other side, Tγ1 (Tγ2 )z = =

a b Az + B c d Cz + D a

Az+B Cz+D

165

7.6 FRACTIONAL LINEAR TRANSFORMATIONS a(Az + B) + b(Cz + D) c(Az + B) + d(Cz + D) (aA + bC)z + (aB + bD) = . (cA + dC)z + (cB + dD)

Since these are equal, the result has been established. Exercise 7.33. Prove that if γ ≡ γ′, then γ′ = kγ for some scalar k. Solution. Let γ=

a b c d

γ′ = A C

and

B . D

Then

az + b Az + B = cz + d Cz + D for all z. In particular, we have this equality for z = 0, so b

Thus, there exists a k1 = B/b such that B = k1b and D = k1d. By setting z = ∞, we get a A = , c C so there exists a k2 such that A = k2a and C = k2c. Finally, let us combine this with an evaluation at 1, which gives a+b k a + k1b = 2 c+d k2c + k1d k1(ad + bd) + k2(ac + bc) = k1(bc + bd) + k2(ac + ad) k1(ad − bc) = k2(ad − bc). Since ad − bc /= 0, we get k1 = k2. Thus, γ′ = k1γ. Exercise 7.34. In the upper half plane model H , carefully draw the asymptotic triangle with vertices i, 1 + i, and 1. Is the map γ=

−1

an isometry of H ? In the same diagram, carefully draw the image of the asymptotic triangle under the action of γ.

166 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY i

i +1

i +1 2

Figure 7.11 Solution. We know the map γ is an isometry of H since it has real entries and positive determinant. Under the action of this map, 1 −1 i = i − 1 = 1 + i i 1 0 1+i−1 i(1 − i) 1+i 1 −1 = = γ(1 + i) = 1 0 (1 + i) = 2 2 1+i 1−1 1 −1 γ(1) = 1= = 0. 1 0 1 γ(i) =

These two triangles are shown in Figure 7.11. Exercise 7.35. In the Poincaré upper half plane H , carefully draw the triangle with vertices i, −1 + i, and 1 + i. In the same diagram, carefully draw the image of this triangle under the isometry γ=

2 1 1 1

. [S]

8+i 13 + i 2 −1 . What are ,Q= , and γ = −3 2 20 13 γP and γQ? Sketch P , Q, and their images. Is γ an isometry? Why? Use all this information to ﬁnd the distance between P and Q in H. Exercise 7.36. Let P =

Solution. We calculate 2 −1 8 + i 2 8+i − 1 γP = 13 = −3 2 8+i 13 −3 13 + 2 16 + 2i − 13 3 + 2i i(−3i + 2) = = = =i −24 − 3i + 26 2 − 3i 2 − 3i 2(13 + i) − 20 26 + 2i − 20 6 + 2i 2i(−3i + 1) γQ = = = = = 2i. −3(13 + i) + 40 −39 − 3i + 40 1 − 3i −3i + 1 A sketch of P , Q, γP , and γQ is in Figure 7.12. The point of this sketch is to emphasize that points near the real axis (in the Euclidean sense) are far apart using the Poincaré metric. We note that γ is an isometry since it has real coeﬃcients and positive determinant. Since γP and γQ have the same

167

7.6 FRACTIONAL LINEAR TRANSFORMATIONS γP

γQ

P Q 0

Figure 7.12 real part, we can calculate the distance between them. Since isometries preserve distance, we are actually calculating |P Q|. Thus, |P Q| = | ln(1/2)| = ln 2. The Poincaré distance between P and Q is ln 2. 6 + 4i Exercise 7.37. Let P = 2 + 4i and Q = be two points in the 3 Poincaré upper half plane. Let 2 γ= 1 . −1 2 What are γP and γQ? What is the Poincaré distance from P to Q?

[S]

Exercise 7.38 †. Suppose T is a fractional linear transformation such that T (1) = 1, T (0) = 0, and T ( ∞) = ∞ . Prove that T is the identity map. That is, show that T (z) = z for all z. Solution. Let az + b T (z) = . cz + d From what we know of T , we get the following: b 0 = T (0) = (7.1) d a ∞ = T (∞) = (7.2) c a+b 1 = T (1) = . (7.3) c+d From (7.1), we get b = 0. From (7.2), we get c = 0. Combining these with (7.3), we get a = d. Thus, T (z) =

az + 0 0z + a

= z.

168 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY That is, T is the identity map. Exercise 7.39. Show that the dilation δ λ (z) = λz is an isometry of H . Find an isometry which sends a + ib to i by composing a dilation δλ and a horizontal translation τr for some λ and r. Solution. We note that δλ(z) = λz =

λz + 0 0z + 1

λ 0 0

To send a + ib to i, let us ﬁrst subtract a, and then divide by b. That is, i = δ1/b τ−a (i), so we calculate δ1/b τ −a =

1/b 0 0 1

1 −a 1 0 = 0 1 0 b

1 −a 1 −a . = 0 1 0 b

Exercise 7.40. Write the dilation δλ(x, y) = (λx, λy) as a product of σ’s and τa’s. Solution. In Exercise 7.39, we saw that dilations are given by δλ =

λ 0 , 0 1

which does not have determinant equal to one, while both σ and τa do. Thus, we must ﬁrst represent δλ with an element of SL2(R). That’s not √ diﬃcult to do ... λ 0 √ . δλ = 0 1/ λ √ √ Following the proof o√f Theorem 7.6.4, we set s = − λ, r = −1/ λ = 1/2 −λ− , and t = −1/ λ. Thus, δλ = στ 1/√λ στ −

√ √ −1/ λ στ− λ.

Exercise 7.41. Suppose γ ∈ SL2(R). Show that γ is a direct isometry. Hint: Show that τa and σ are direct isometries. Solution. Since τa is just a horizontal translation by a, it is clear that it preserves orientations. Recall that inversion in the circle can be thought of as reﬂection through a circular mirror, so it is orientation reversing. Since σ is the composition of an inversion and a reﬂection (which is also orientation reversing), it is in the end an orientation preserving or direct isometry. Since the product of direct isometries is still direct, any isometry that can be expressed as a product of τa’s and σ’s is direct. Thus, every element of SL2(R) is a direct isometry.

169

7.7 THE CROSS RATIO

7.7

The Cross Ratio

Exercise 7.42. Find the fractional linear transformation which sends 1 to 1, −i to 0, and −1 to ∞. The solution is in the text. Exercise 7.43. Find the fractional linear transformation which ﬁxes i, sends ∞ to 3, and 0 to −1/3. The solution is in the text. Exercise 7.44. Suppose a, b, and c are distinct elements of C ∪ {∞} , and that T is a fractional linear transformation such that T (a) = 1, T (b) = 0, and T (c) = ∞. Prove that T is unique. [H] Solution. Suppose there are two maps T and S with this property. Then 1 1 1 ST − (1) = 1, ST − (0) = 0, and ST − ( ∞) = ∞. By Exercise 7.38, the 1 map ST − is the identity. That is, S = T . Hence, such a T is unique. Exercise 7.45 †. There is an important result hidden in the solution to Exercise 7.43. Suppose γ is a fractional linear transformation and that γa = a′, γb = b′, and γc = c′. Prove that (z, a; b, c) = (γz, a′; b′, c′). Solution. Let’s follow the solution to Exercise 7.43 using variables instead of speciﬁc numbers. We let γ1z = (z, a; b, c)

and

γ2w = (w, a′; b′, c′).

Then γ1a = 1, γ1b = 0, and γ1c = ∞. Also, γ2a′ = 1, γ2b′ = 0, and γ2c′ = ∞. If we set γ1z = γ2w

(7.4)

γ2−1 γ1 z = w, then γ ′ = γ2−1 γ1 has the property that γ ′ a = a′ , γ ′ b = b′ , and γ ′c = c′. Let us now use Exercise 7.44 to show that γ′ = γ. We compose with γ2 so that γ2γ′a = γ2γa = 0, γ2γ′b = γ2γb = 1, and γ2γ′c = γ2γc = ∞ . By Exercise 7.44, γ2γ′ = γ2γ, so γ′ = γ. Finally, since γz = γ′z = w, equation (7.4) says (z, a; b, c) = (γz, a′; b′, c′). Exercise 7.46†. Find the fractional linear transformation which sends 0 to —i, 1 to 1, and ∞ to i. What is the image of H under this map? If l is a vertical line or half circle perpendicular to the real axis, what is its image under this map? (The image of H under this map is the Poincaré disc model D of hyperbolic geometry introduced as our ‘crutch’ in Chapter 6.) [S]

170 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY iθ

Exercise 7.47. The map r(z) = e z rotates the complex plane an angle θ about the origin. Compose this map with the inverse of the map found in Exercise 7.46 to get a map of H. Is this new map an isometry? Solution. Wh at is meant by ‘compose’ ? The only thing that makes sense 1 −i is to use τ = (the map found in Exercise 7.46) to map the upper −i 1 1 half plane H to the unit disc D, apply r, and apply τ − to get back to H . Let us do this. 1

z−i − iz + 1 iθ iθ 1 i e z − ie = i 1 −iz + 1 iθ iθ e z − ie + z + i = ieiθ (z − i) − iz + 1 iθ iθ (e + 1)z − i(e − 1) = i(eiθ − 1)z + eiθ + 1 (eiθ/2 + e−iθ/2)z − i(eiθ/2 − e−iθ/2) = i(eiθ/2 − e−iθ/2)z + eiθ/2 + e−iθ/2 2 cos(θ/2)z + 2 sin(θ/2) = −2 sin(θ/2)z + 2 cos(θ/2) cos(θ/2) sin(θ/2) = z. — sin(θ/2) cos(θ/2) 1

τ − rτz = τ − r

Since this map has real entries and determinant equal to one, we have an isometry of H . It is the rotation of ρθ ofH by θ about the point i (see Section 7.9). Exercise 7.48. A circle centered at P and with radius r is the set of points a distance r away from P . Prove that a circle in the Poincaré model is a (Euclidean) circle which lies entirely within the model. [H] cos(θ/2)

sin(θ/2)

— sin(θ/2) found in Exercise 7.47. Note that

cos(θ/2)

Solution. Let ρθ = τ − rτ =

ρθi = =

be the isometry of H

cos(θ/2) sin(θ/2) i — sin(θ/2) cos(θ/2) i cos(θ/2) + sin(θ/2)

−i sin(θ/2) + cos(θ/2) (i cos(θ/2) + sin(θ/2))(i sin(θ/2) + cos(θ/2)) = 1 = i.

171

7.7 THE CROSS RATIO

Thus, ρθ fixes i for all θ. Let P be a point in H. Then ρθ(P ) lies on the Poincaré circle centered at i and through P , since ρθ is an isometry, whence the distance from i to P is preserved. Since θ is arbitrary, this gives us a lot of points on the Poincaré circle. Let us now address the question of what it looks like. Let’s first look at the image of τP under the action iθ of r, which is just multiplication by e . As θ ranges over [0, 2π] (or even R), the image of τP describes a circle centered at the origin. Thus, the image of P under the action of ρθ as θ varies is the image of this circle 1 under the action of τ − . Since τ is a fractional linear transformation, its inverse sends circles to circles or lines. Thus, a Poincaré circle centered at i in contains as a component either a Euclidean circle or line. Since H Euclidean lines include points at infinity (inH), and a circle is the set of points a fixed finite distance from its center, a Poincaré circle cannot include a Euclidean line. Thus, it includes a Euclidean circle. But thinking geometrically, it is clear that a Poincaré circle cannot include any more points. Thus, Poincaré circles centered at i are Euclidean circles. If this circle intersects the real axis, then it contains points arbitrarily far away, so cannot be a Poincaré circle. Thus, the Euclidean circle must be entirely in H . Finally, if the Poincaré circle is centered somewhere else, we can first move it to i (whoops, that we can do that is the subject of the next section). This move does not change its shape, since fractional linear transformations send circles and lines to circles and lines (and as mentioned before, neither the circle nor its image can be a line). Exercise 7.49. Suppose w = γz where γ ∈ SL2(R). Suppose also that (z, A; B, C) = (w, B; C, A) for all z ∈ H and some distinct points A, B, and ∈C H . What can be said about ∆ABC? What can be said about γ? [A] Solution. Note that γ is an isometry, and that γA = B, γB = C, and γC = A. Thus, ∆ABC is congruent to ∆BCA. That is, ∆ABC is an equilateral triangle. Furthermore, γ must be a rotation by 120◦ about the center of ∆ABC. Exercise 7.50. Let (a, b; c, d) = λ. Show that (a, b; d, c) = 1/λ and (a, c; b, d) = 1 − λ. Use this to find (a, x; y, z) for the other three permutations (x, y, z) of {b, c, d}. Solution. We first note that a − d,b − d b − c , a − c (a, b; d, c) = = = 1/(a, b; c, d) a−c b−c b−d a−d and (a, c; b, d) =

a − b,c − b a−d c−d

172 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY (a − b)(c − d) (a − d)(c − b) (a − b)(c − d) − (a − d)(c − b) =1+ (a − d)(c − b) ac − bc − ad + bd − ac + dc + ab − bd =1+ (a − d)(c − b) −bc − ad + dc + ab =1+ (a − d)(c − b) (a − c)(b − d) =1+ (a − d)(c − b) a − c,b − c =1− a−d b−d = 1 − (a, b; c, d). =

Thus, (a, b; c, d) = λ (a, b; d, c) = λ−

(a, c; b, d) = 1 − λ

(a, c; d, b) = 1/(a, c; b, d) =

1−λ 1 (a, d; b, c) = 1 − (a, b; d, c) = 1 − λ− λ (a, d; c, b) = 1/(a, d; b, c) = . λ−1 Exercise 7.51. Suppose λ = (a, b; c, d) = (a, c; d, b). What is λ? Solution. From Exercise 7.50, (a, c; d, b) =

1 1−λ

so λ=

1−λ −λ + λ = 1 2 λ −λ+1 =0 √ 1± 1−4 λ= 2 √ 1±i 3 = . 2 2

173

7.8 TRANSLATIONS

7.8

Translations

Exercise 7.52. Suppose γ =

a b

c d ∈ SL2(R) and c = 0. Under what condition(s) is γ a parabolic translation? Solution. The map γ is a parabolic translation if it has one fixed point at infinity. A point z is fixed if γz = z az + b =z d az + b = dz (a − d)z = −b. It looks like there is only one solution, but remember, the symbol ∞ is also at infinity (of H), so we must check γ∞: γ∞ = ∞. Thus, if there is only one fixed point, then it must be infinity, so a = d and b = 0. If b = 0, we have the identity. Thus, the map γ with c = 0 is a parabolic translation if and only if a = d and b = 0. Exercise 7.53. Suppose ∈ γ SL2(R) and γ∞ = . ∞ Prove that γ is a translation. a b . Then Solution. Let γ = c d γ∞ =

a , c

so c = 0. If γz = z, then

az + b = z, d which never has a complex solution. Thus, γ has no fixed points inH so it is a translation. Exercise 7.54. Is horizontal translation by a (the map τa) a parabolic or hyperbolic translation? Solution. Note that τaz = z + a, so if τaz = z, then a = 0. But that describes the identity. Thus, if τa is not the identity, then it is a parabolic translation, since it has only one fixed point, namely ∞. Exercise 7.55. Prove that a map γ∈SL2(R) is either the identity onH or has at most one fixed point in H. Thus, every element in SL 2(R) is either a rotation or translation.

174 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Solution. Suppose γz = z. Then az + b =z cz + d 2 az + b = cz + dz cz + (d − a)z − b = 0 2

If c =/ 0, then this is a quadratic equation, which has either one or two real roots, or two conjugate complex roots. Real roots are not in H . If the roots are complex, then only one is in the upper half plane, so γ has at most one fixed point in .H If c = 0, then az + b =z d az + b = dz (a − d)z + b = 0. If a d, then this equation is linear so has real roots. If a = d, then either b = 0, in which case γ is the identity, or b = / 0, in which case there are no fixed points. Thus, there is at most one fixed point. By Exercise 7.41, γ is a direct isometry, so γ must be a rotation or translation. What’s the point? How do we otherwise know that our classification of proper isometries inHis complete. That is, how do we know there are no proper isometries that do not fit the definition of either translation or rotation? Exercise 7.56. Suppose γ is a hyperbolic translation. Show that there exists exactly one line l in H such that for any P ∈ l, we have γP ∈ l. Show that no such line exists if γ is a parabolic translation. What is the corresponding statement for translations in Euclidean geometry? Solution. Since γ is hyperbolic, it fixes two real values, say M and N . Thus, γ fixes the line MN . Now suppose there exists another line l′ = M ′N ′ fixed by γ. If γM ′ = M ′ and γN ′ = N ′, then l′ = l, since γ can have at most two fixed points. Thus, we must have γM ′ = N ′ and γN ′ = M ′. Let P be an arbitrary point on l′ and let Q = γP . Let A be the midpoint of P Q. Let us consider A′ = γA, the image of A. Suppose A is on the M ′ side of P . Then A′ is on the N ′ side of Q. Further, A′ is a distance| AP| =| AQ| away from Q. Thus, A′ = A, so γ is not a translation. Thus, no such l′ exists, so γ fixes exactly one line l. Suppose γ is a parabolic translation that fixes the line l with endpoints M and N . If γM = M , then γN = N and γ is not parabolic. Thus, we must have γM = N and γN = M , which again leads to a fixed point A ∈ H. Hence, γ cannot fix any lines.

175

7.8 TRANSLATIONS

In Euclidean geometry, there is only one type of translation. Every translation fixes some line l (not pointwise), and therefore fixes all lines parallel to l. Exercise 7.57. Prove that for any two distinct points P and Q inH, there exists an infinite number of translations f such that f (P ) = Q. Contrast this with Exercise 1.20. Solution. In the text, we chose one of the fixed points to be∞. But any real point represents a point at infinity for H . So for any a∈ R, we should be able to find an isometry that sends P to Q and fixes a. Since a is fixed and is in R, such an isometry is a translation (the roots of the quadratic include a, so must be real). To find these translations, let us first find a map γ that sends a to ∞. Any such map γ will do. In the text, we establish that for any pair of points in H, there exists a translation that sends one to the other and fixes∞. Let us consider the map f that sends γP to γQ, and consider 1 τ = γ− fγ. Note that

τP = γ− fγP = γ− γQ = Q, as desired, and that τa = γ− fγa = γ− f ∞ = γ− ∞ = a. 1

Thus, τ fixes a and sends P to Q. We have therefore found an appropriate translation for any a ∈ R. Since a translation fixes at most two real points, this correspondence with the reals is at most two to one. Thus, we have found an infinite number of translations that send P to Q. Contrast this with Euclidean geometry, where there exists exactly one translation that sends P to Q. Let us now establish that such a γ exists. There are a lot of choices; let us choose that γi = i. The Poincaré line through a and i is a Euclidean half circle through a and i with center on the real axis. To find the other endpoint of the line, let us apply power of the point to the origin O. The power of this point is 1, which we see by considering the chord through i and O, which intersects the circle again at −i. Note that x and a have opposite signs, so by considering the chord through a and O, we get −ax = 1. Thus, the other endpoint is −1/a. Now, we find γ: (z, i; −1/a, a) = (w, i; 0, ∞) z + 1/a , i + 1/a w = z− a i− a i az + 1 i(i − a) =w z− a ia + 1 a 1 z = w. −1 a

176 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Exercise 7.58. Find a map γ ∈ SL2(R) which sends 1 + i to i and ∞ to 1. Does this map have a fixed point? The solution is in the text. Exercise 7.59. Find the isometry of γ of H in GL2(R) which sends 2i to 3i + 4 and 2 to −5. Does this isometry have any fixed points in H? Solution. Our first task to to find another point whose image we know. We first find the other endpoint of the Poincaré line through 2 and 2i. That point is clearly—2i. We then find the other end x of the line through−5 and 3i + 4. Let us use power of the point again (as in Exercise 7.57). The power of the point 4 is 9, and is also —(4− − ( 5))(4− x) =− 36 + 9x. Thus, x = 5. Hence, (z, 2i; 2, −2) = (w, 3i + 4; −5, 5) z − 2 , 2i − 2 w + 5 , 3i + 9 = z + 2 2i + 2 w − 5 3i − 1 z − 2 i(i + 1) w+5 i(3i − 1) = z+2 3(−1 + 3i) −1 − i w−5 3z − 6 w + 5 z+2 = w−5 3 −6 z = 1 5 1 2 1 −5 w −6 3 z=w 5 −51 2 −1 1 10 −40 4 −4 z = w 5 −20 z = w. 2 −2 Thus, γ =

5 −20 . To see if this has a ﬁxed point, we solve 2 −2 γz = z 5z − 20 =z 2z − 2 2 5z − 20 = 2z − 2z 2 2z − 7z + 20 = 0 √ 7 ± 49 − 160 z= √4 7 + i 111 z= . 4

Thus, γ is a rotation.

177

7.9 ROTATIONS

Exercise 7.60. Find an isometry γ ∈ GL2(R) of H which sends 2i to 3i and ∞ to −1. [S]

7.9

Rotations

Exercise 7.61. Fill in the algebraic manipulation that establishes the above result. Solution. From the text, we must solve (z, P ; a, ∞) = (w, P ; N, M ). (Note the correction – ﬁxed in the second printing.) Recall P = a + bi and we have set P = i, so a = 0. Also derived in the text are the following: M = − csc θ − cot θ and N = csc θ − cot θ. Thus, we must solve (z, i; 0, ∞) = (w, i; csc θ − cot θ, − csc θ − cot θ) z w − csc θ + cot θ , i − csc θ + cot θ = i w + csc θ + cot θ i + csc θ + cot θ i(i sin θ + 1 + cos θ) w sin θ − 1 + cos θ z= w sin θ + 1 + cos θ i sin θ − 1 + cos θ

iθ

Recall that e = cos θ +i sin θ. For convenience, let us also set c = cos(θ/2) and s = sin(θ/2). Then, by the angle sum formula, 2 scw − 1 + 1 − 2s

iθ

i(e + 1) eiθ − 1

2scw + 1 + 2c2 − 1 iθ/2

Let us divide top and bottom by e iθ/2

i(e

to get

iθ/2 + e− )

eiθ/2 − e−iθ/2 c s(cw − s) = s c(sw + c) c −s = w. s c

s (cw − s) c(sw + c

Finally, we isolate w: c −s

s z = w. c

Exercise 7.62. Use the above result to ﬁnd the map which rotates an angle θ about a point P . (Hint: Translate P to i, then do the rotation, and translate back.)

178 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Solution. Let us translate P to i, holding ∞ ﬁxed. This is done in Section 7.8, and the map is 1 −a γ= , 0 b where P = a + ib. We therefore seek ρ(P,θ) = γ−1ρθγ =

b a 0 1

c −s

s c

1 −a , 0 b

where, as in the previous exercise, we set c = cos(θ/2) and s = sin(θ/2). Continuing, we get c −ac + bs − s as + 2bc 2 bc − as −abc + b s + a s + abc = −s as + bc

ρ(P,θ) =

b a 0 1

bc − as (a + b )s . −s as + bc 2

Thus, the rotation through a counterclockwise angle of θ centered at P = a + ib is given by ρ(P,θ) =

b cos(θ/2) − a sin(θ/2) — sin(θ/2)

(a2 + b ) sin(θ/2) . a sin(θ/2) + b cos(θ/2)

Exercise 7.63. Find an isometry which ﬁxes 1 + i and sends 1 to 2.

[S]

Exercise 7.64. Find an isometry in GL2(R) which sends i +1 to i and ∞ to 1. Is this isometry a rotation or translation? [A] Solution. The line through ∞ and i + 1 also goes through 1. The Poincaré line through i and 1 also goes through −1. Thus, the isometry γ sends 1 to −1, so

−1 −1

(z, i + 1; 1, ∞) = (w, i; −1, 1) z−1 w + 1,i + 1 = i+1−1 w−1 i−1 i+1 w+1 (z − 1) = w− 1 i(i − 1) 1 1 i+1 ( z − 1) = 1 i) w — − 1 −1 −1 − 1 1 z = w 0 1 1 1 −2 z = w. 1 0

179

7.9 ROTATIONS Now we look for its ﬁxed points: z− 2 =z z 2 z−2 = z 2 0=z −z+2 √ 1± 1−8 z= 2 √ 1+i 7 z= . 2 Thus, the isometry is a rotation about the point

√ 1+i 7 2

Exercise 7.65. The map γ=

1 − 5 1 3

is a rotation of H. What is the center of this rotation? Solution. We look for the ﬁxed points: 1 −5 z = z 1 3 z− 5 =z z+3 z − 5 = z + 3z 2 0 = z + 2z + 5 √ −2 ± 4 − 20 z= 2 = −1 + 2i. 2

Note that we take the positive root since it gives a point in H, while the other does not. Thus, the center of rotation is the point −1 + 2i. Exercise 7.66. Find an isometry γ ∈GL2(R) which ﬁxes i and sends ∞ to 1. The isometry γ ﬁxes i and hence is a rotation about i through some angle. What is that angle? Solution. The Poincaré line through ∞ and i is the vertical line that goes through 0. The Poincaré line through 1 and i is the half circle centered at 0, so goes through −1. Thus, (z, i; 0, ∞) = (w, i; −1, 1) z w + 1,i + 1 = i w−1 i−1

180 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY i+1 −1 −1 −1 1

1 i(i − 1) −1 0 z = w 0 1

−1

1 −1 z = w. 1 1 −1 . Since the Euclidean vertical line through 0 intersects 1 1 the Euclidean unit circle centered at 0 at a right angle, this rotation is through an angle of π/2. Since ∞ is sent to 1, the direction is clockwise. Thus, γ =

Exercise 7.67. Recall that all triply asymptotic triangles are congruent. Find three distinct isometries in GL2(R) which send the triply asymptotic triangle with vertices 1, 2, and 3 to the triply asymptotic triangle with vertices 0, — 1, and .∞Remember, for an element of GL 2(R) to be an isometry, it must have a positive determinant. Identify which of these isometries are rotations and which are translations. Solution. One can do all six combinations and pick out those with positive determinants, or one can be a little more clever. The points 1, 2, and 3, in that order, are oriented counterclockwise (think of where these points are in relation to a point inH ). The points 0, −1, and ∞, in that order, are oriented clockwise. Thus, the relations that should give isometries are (z, 1; 2, 3) = (w, −1; 0, ∞) (z, 2; 3, 1) = (w, −1; 0, ∞) (z, 3; 1, 2) = (w, −1; 0, ∞). Let’s calculate the ﬁrst: (z, 1; 2, 3) = (w, −1; 0, ∞) z − 2,1 − 2 w = z−3 1−3 −1 −2z + 4 =w z−3 −2 4 z = w. 1 −3

Since this matrix has determinant equal to 2, which is positive, our analysis was correct. For the second relation, z − 3,2 − 3 = −w z−1 2−1 1 −3 z = w. 1 −1

181

7.10 REFLECTIONS And the third: z − 1,3 − 1 = −w z− 2 3 − 2 −1 −1 z = w. 2 −4 Now we’ll check which are rotations and which are translations: −2z + 4 =z z−3 2 −2z + 4 = z − 3z 2 0=z −z−4 √ 1 ± 1 − 16 z= √2 1 + i 15 z= . 2 Thus, the ﬁrst is a rotation. z− 3 =z z−1 2 z−3=z −z 2 0 = z − 2z + 3 √ 2 ± 4 − 12 z= 2 = 1 + 3i. Thus, the second is a rotation. −z − 1 =z 2z − 4 2 −z − 1 = 2z − 4z 2 0 = 2z − 3z + 1 √ 3± 9−8 z= 4 1 = 1, . 2 Thus, the third is a translation.

7.10

Reﬂections

Exercise 7.68. Find the image of i under reﬂection through the line l whose endpoints are 1 and 3.

182 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY The solution is in the text. Exercise 7.69. Find the reflection of 1+i through the line l with endpoints 2 and 5. Solution. We must first find the reflection. The reflection fixes 2 and 5, which gives the equations a c

b (−2) = 2 a −2a + b =2 −2c + a −2a + b = −4c + 2a −4a + b + 4c = 0, and a b (−5) = 5 c a −5a + b =5 −5c + a −5a + b = −25c + 5a −10a + b + 25c = 0. Taking the difference, we get 6a − 21c = 0 7 a = c. 2 Since we have choice for one variable, let us set c = 2. Then a = 7, and b = 4a − 4c = 28 − 8 = 20. Finally, we find the image of 1 + i: 13 + 7i 7(i − 1) + 20 7 20 = = (13 + 7i)(5 − 2i) 2 7 (−(1 − i)) = 2(i − 1) + 7 25 + 4 5 + 2i 65 + 35i − 26i + 14 79 + 9i = = . 29 29 Exercise 7.70. Find a formula for the reflection through the line l with endpoints −1 and 1. Solution. Since the reflection fixes ±1, we get the relations a b (−1) = 1 c a

183

7.10 REFLECTIONS −a + b =1 −c + a −a + b = −c + a −2a + b + c = 0, and b 1 = −1 a a+b = 1 c+a − a + b = −c − a 2a + b + c = 0. a c

Taking the diﬀerence of these two, we get 4a = 0, so a = 0. Thus, b + c = 0 so b = −c. Let us pick c = 1, so the map is 1 0 − Φ(z) =

1 (−z) = . 0 z This is inversion in the unit circle, which might be easier to recognize if we z write it as Φ(z) = |z| 2. Exercise 7.71. Find a formula for the reflection through the line l which goes through 3i and 1 + 4i. Solution. We could find the endpoints of the line and proceed as before, but I don’t think this is easier than proceeding in a straight forward manner. Since reflections fix points on the line of reflection, we get the equations a c

b (−(−3i)) = 3i a 3ai + b = 3i 3ci + a 3ai + b = −9c + 3ai b = −9c,

and a c

b (−(1 − 4i)) = 1 + 4i a −a + 4ai + b = 1 + 4i −c + 4ci + a −a + 4ai + b = −c + 4ci + a − 4ic − 16c + 4ai

184 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY −2a + b + 17c = 0. Combining these two equations, we get −2a − 9c + 17c = 0 2a = 8c a = 4c. So let us choose c = 1, so a = 4 and b = − 9. Thus, the formula for the reﬂection through the line l which goes through 3i and 1 + 4i is given by R(z) =

(−z). 1 4 4−9

Exercise 7.72. Prove Theorem 7.10.1. [Every isometry f of H which is not direct can be written in the form f (z) = γ(−z) for some γ ∈ SL2(R). Furthermore, if γ =

a c

b , then f (z) is a reﬂection d

if and only if a = d.] Solution. Let G be the group of isometries onH. Let H = SL2(R) be the subgroup of direct isometries in G. Note that g(z) =− z is not a direct isometry, so Hg is a coset of G. Suppose f is another isometry that is not direct. Then gf is direct, so Hgf = H. Hence H has index two in G and Hg is the full set of isometries that is not direct. That is to say, every isometry f which is not direct can be written in the form f (z) = γg(z) = γ(−z) where γ ∈ H. Now, suppose f (z) = γ(−z). Let us look for its ﬁxed points: −az + b =z −cz + d −az + b = −czz + dz czz + b = dz + az.

(7.5)

Since the left hand side is real, the right hand side must also be real. Thus, f (z) has fixed points if and only if a = d. In particular, if f is a reflection, then it has fixed points (the line of reflection), so a = d. Now, suppose a = d. Then (7.5) becomes c(x + y ) − 2ax + b = 0 2

where z = x + iy. If c /= 0, this is the equation of a circle centered on the real axis, so is a Poincaré line. If c = 0 then a 0 (since det γ = 1) so this is the equation of a vertical line, so is a Poincaré line. Thus, f fixes every point on a line and fixes no other points, so is a reflection.

185

7.11 LENGTHS

7.11

Lengths

Exercise 7.73. Show that P|Q is| independent of which endpoint we call M and which we call N . That is, show that |ln(Q, P ; M, N )| = |ln(Q, P ; N, M )|. Show also that |P Q| = |QP |. Solution. From Exercise 7.50, we know that (Q, P ; N, M ) = (Q, P ; M, N )−1. Thus, | ln(Q, P ; N, M )| = | ln((Q, P ; M, N )− )| = | − ln(Q, P ; M, N )| 1

= | ln(Q, P ; M, N )|. Similarly (though this is not shown in Exercise 7.50), (Q, P ; M, N ) (P, Q; M, N )−1. Thus,

|QP | = | ln(Q, P ; M, N )| = | ln((P, Q; M, N )− )| 1

= | − ln(P, Q; M, N )| = | ln(P, Q; M, N )| = |P Q|. Exercise 7.74. Find the Poincaré distance |P Q | between P = 4 + 4i and Q = 5 + 3i. The solution is in the text. Exercise 7.75. Find the Poincaré distance between P = 1 + 3i and Q = 8 + 4i. Write your answer in the form ln(a/b) where a and b are positive integers. Solution. Let us first find the endpoints of the line P Q. To do this, we first find the center x∈R of the Euclidean half circle through these points. By comparing radii, we get (x − 1) + 3 = (x − 8) + 4 2

x − 2x + 1 + 9 = x − 16x + 64 + 16 14x = 70 x = 5. 2

From this, we ﬁnd that r = (5 − 1) + 3 = 5 , 2

so r = 5, M = 0, and N = 10. Thus,

. . 1 + 3i , 8 + 4i . . . |P Q| = | ln(1 + 3i, 8 + 4i; 0, 10)| = ..ln −9 + 3i −2 + 4i . . . . . . . , 4(2i − 1) .. .. 2i i−3 . = .ln = ln . . . . (3i)4 3i(−3 + i) 2i(−1 + 2i) = | ln(1/6)| = ln 6.

186 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Exercise 7.76. Find the distance |P Q| between P = i and Q = in the Poincaré upper half plane H. Solution. Note . that √ .2 1+ 3 .. . .

.. . . =

√ 1+ i 3

√ 1−i 3

1+3

√ 1+i 3 2

so both P and Q are on the Euclidean unit circle centered at 0. Thus, the end points of the line PQ are M = −1 and N = 1, so . . √ √ + 1, 3 + 3 . . i i . . |P Q| = | ln(i, (1 + i 3)/2; −1, 1)| = .ln √ .. i−1 . √ −1 + i 3 . . . (−i − 3 i(i + 1) . .. −i .. . . √ .. = .ln = .ln 1 i i √3( √3 + i) −i 3 .. . —. − 1 √ = ..ln √1 . = ln( 3) = ln 3. 2 3 . . Exercise 7.77. Find the Poincaré distance between 12 + 5i and 5 + 12i. Write your answer in the form ln(a/b) where a and b are integers. [S]

7.12

The Axioms of Hyperbolic Geometry

7.13

The Area of Triangles

Exercise 7.78. Consider the doubly asymptotic triangle ∆AM N in H 8+i 5 where A = , M = , and N = 2. What is the image of ∆AM N under 5 3 the isometry 2 −3 γ= ? −1 2 Use this to ﬁnd the hyperbolic area of ∆AM N .

[S]

Exercise 7.79. Find √ the area in H of the doubly asymptotic triangle with vertices i, 1, and 1 + 2. √ Solution. Let us ﬁrst characterize the line through 1 + 2 and i. Let x ∈ R be the other endpoint of this line. Using power of the point 0, we get −x(1 +

√

2) = 1

√ √ −1 + 2 − 1 = 1 − 2. √ = x= 1−2 1+ 2

187

7.13 THE AREA OF TRIANGLES

√ Thus, the Poincaré line throug√h 1 + 2 and i is the Euclidean half circle centered at 1 and with radius 2. The Poincaré line through i and 1 is, of course, the Euclidean half circle centered at 0 with radius 1. The tangent at i to the half circle centered at 0 is a (Euclidean) horizontal line. The tangent at i to the half circle centered at 1 has a (Euclidean) slope of π/4. Thus, the angle between the two Poincaré lines thr√ ough i is π/4. Thus, the area of the triangle with vertices at i, 1, and 1 + 2 is 3π π = . 4 4 Exerc√ ise 7.80. Draw the asymptotic triangle ∆ABM in H with A = i, B = i 3, and M = 1. What is the Poincaré length |AB|? What is the area of ∆ABM ? π−

√ i 3

Figure 7.13 Solution. The Poincaré line through i and 1 is the Euclid√ean unit half circle centered at 0. To ﬁnd the Poincaré line through 1 and i 3, we let x be the other end point, and use power of the point 0, which gives −1x = 3. Thus, this line is the Euclidean half circle centered at− 1 with radius 2. A picture of the asymptotic triangle ∆ABM is shown in Figure 7.13. Note that the tangent at i to the larger circle is perpendicular to the radius. This radius makes an angle of π/3 with the x-axis. Thus, the circle intersects the vertical line at 0 at an angle of π/3, so the area of the triangle is |∆ABM | = π − π/3 − π/2 = π/6. √ 1 The Poincaré length |AB| is ln 3 = ln 3. Since the angle at B is π/2, we 2 have just shown that the angle of parallelism for the length

ln 3 is π/3.

188 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY That is, 1 ln 3 2

7.14

= π/3.

The Poincaré Disc Model

Exercise 7.81. Prove that the arclength element in D is √ 2 dr2 + r2dθ2 ds = . [H][S] 1 − r2 Exercise 7.82. Prove that Γ is the group of proper isometries of D. [S] Exercise 7.83. Characterize the set of reﬂections in D.

[H][A]

Solution. Note that reﬂections are improper isometries and have the property that they are their own inverse. Let R be a reﬂection. Because R is improper, we can write a

R(z) =

(−z).

b a Because R is its own inverse, we know 2

R (z) = z, 2

so let us calculate R (z). Note that R(z) = so 2

R (z) = =

−az + b −bz + a

−a

−az+b −bz+a

−b

−az+b −bz+a

aaz − ab − b z + ab 2

. 2 baz − b − abz + aa Since the coeﬃcient of z in the denominator must be zero, we get ba − ab = 0 a(b − b) = 0 b = b. But a number’s conjugate b is equal to itself if and only if b ∈ R. Now that 2 we know b ∈ R, we simplify R (z) further: 2

R (z) =

(aa − b )z = z. −b2 + aa 2

189

7.14 THE POINCARÉ DISC MODEL

We now have a characterization, but it is not clear that all such maps are actually reflections. To show this, let us first look at the fixed points of this map: a b (−z) = z b a −az + b =z −bz + a −az + b = −bzz + az bzz − (az + az) + b = 0. Let us write z = x + iy and a = a1 + ia2. Then we have b(x + y ) − 2(a1x + a 2 y)+ b = 0 2 2 a +a a 2 a 2 2 x− 1 + y− 2 = 1 . b2 − 1 b b 2

If b 0, this is an equation of a circle. Let us ﬁnd out more. Since γ ∈ SL2(C), we know det γ = 1. That is, aa − b = 1 2 2 a + a − b = 1. 2

Thus x−

a1 b a1

+ y−

a2 b a2

2 2

1+b 1

−1

+ y− = 2. b b b 2 a 1 a2 This circle is centered at b , b and has radius 1/b . Let O be the origin. x−

Then the power of the point O with respect to this circle is 2

a +a

− 2 = 1, b2 b which is the radius of the unit disc. Thus, this circle intersects the unit disc at right angles so (the portion in the disc) is a Poicaré line. If b = 0, then the ﬁxed points satisfy 1

−2(a1x + a2y) = 0, which is the equation of a line through the origin, so again (the portion inside the unit disc) is a Poincaré line. Thus, in either case, γ( − z) fixes every point on a Poincaré line and fixes no other point. That is, it is a reflection. Exercise 7.84. What is the area element in D?

[A]

190 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Solution. Let us consider a small region described by a small change ∆r in r and a small change ∆θ in θ. Such a region approximates a Euclidean rectangle. Its sides correspond to holding r or θ constant. If we hold r constant, we get a curve of length √ 2 r2∆θ2 2r∆θ = , ∆s = 2 1−r 1 − r2 and if we hold θ constant, we get a curve of length √ 2 ∆r2 2∆r = . ∆s = 1 − r2 1 − r2 Thus, the area of this region is approximately the product of these two values, so 4r∆r∆θ ∆A = . (1 − r2)2 As ∆r and ∆θ approach zero, the approximation gets better, and we get the area element 4rdrdθ dA = . (1 − r2)2 Exercise 7.85. Let P and Q be two points inD and let M and N be the endpoints of the line P Q. What is the formula for the length | P Q| in terms of P , Q, M , and N . [A] Solution. Let us bring these points back to the upper half plane H ; let 1 1 1 1 P ′ = φ− P , Q′ = φ− Q, M ′ = φ− M , and N ′ = φ− N . Then |P Q| = |P ′Q′| = | ln(P ′, Q′; M ′ ,N ′)|. But recall, in Exercise 7.45, we showed that for any fractional linear transformation γ, (a, b; c, d) = (γa, γb; γc, γd). In particular, (P ′, Q′; M ′ ,N ′ ) = (P, Q; M, N ), so |P Q| = | ln(P, Q; M, N )|.

7.15

Circles and Horocycles

Exercise 7.86. In the upper half plane model, describe the horocycles centered at ∞. Justify your answer.

191

7.15 CIRCLES AND HOROCYCLES

Solution. The horocycles at ∞ are horizontal lines. To see this, consider any γ ∈ SL2(R) that sends to∞ a point a R.∈ Under such a map, a horocycle at ∞ will be sent to a horocycle at a, which is a (Euclidean) circle through a. Moving the point a back to ∞ sends that circle to a line. Since the circle is tangent to the real axis, the horocycle at ∞ is a line parallel to the real axis. Exercise 7.87. Use Geometer’s Sketchpad to draw a circle and several of its radii in H. Though we ask for a sketch here, it occurred to me that we can write a script that does this. Script Ex07087. Given a circle c1 and a line j that does not intersect the circle, this script constructs the center of the circle, if thought of as a Poincaré circle in the upper half plane deﬁned by the line j. It also picks a random point P on the circle, and draws the radius OP . Move P around to see how the radius changes. Move the circle around too. Exercise 7.88*. Show that the area A of a circle of radius r in H is 2

A = 4π sinh (r/2), ex

−x

where sinh(x) = −e is the hyperbolic sine function. Note that, unlike 2 triangles, the area of a hyperbolic circle can be arbitrarily large. Solution. We will ﬁnd the area of a circle centered at 0 in the Poincaré disc + model. If this circle goes through x∈R , then it is the Euclidean circle centered at 0 with radius x. The area A is given by ∫ 2π ∫ x 0

4rdrdθ , 2 2 0 (1 − r )

(7.6)

where the integrand is the area element found in Exercise 7.84. So, to find the area of a Poincaré circle with radius r, we must find out what the appropriate x is. We will do this using the distance formula found in Exercise 7.85. (A derivation from first principals is done in Section 7.16.) + Let P be the point in D represented by x ∈ R . The line through 0 and P has the endpoints ±1, so . . x + 1, 1 . 1+x . . . = ln . |0P | = | ln(x, 0; −1, 1)| = .ln . x − 1 −1 1−x We now set |OP | = r, so ln

1+x 1−x 1+x 1−x

=r = er

192 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY 1 + x = e − xe r r x + xe = e r− 1 e −1 r

er + 1 er/2 − e−r/2 = er/2 + e−r/2 = tanh(r/2).

See Section 7.16 for a brief review of the deﬁnitions of hyperbolic trigonometric functions and their relations. We put this in (7.6) to get ∫ 2π ∫ tanh(r/2) 4rdrdθ . A= (1 − r2)2 0 0 2

Let us make the substitution u = 1 — r . Then du = − 2rdr; at r = 0, u = 1; and at r = tanh(r/2), cosh (r/2) − sinh (r/2) = sech 2(r/2). u = 1 − tanh (r/2) = 2 cosh (r/2) 2

Thus, we get A = 2π

∫ sech 2(r/2) 1

2 sech (r/2) −2du = 2π u 1 u2 2

2 = 2π 2 cosh (r/2) − 2 = 4π sinh (r/2).

Let us conclude with a remark. If r is small, then sinh(r/2)≈ r/2, so 2 the formula is approximately πr , the formula for the area of a circle in Euclidean geometry. Exercise 7.89*. Use the result found in Exercise 7.88 to ﬁnd an equation for the circumference of a circle in hyperbolic geometry. [H] Solution. Consider the area of a washer constructed by removing a disc of radius r from a disc of radius r + ∆r. We can calculate this area by subtracting the two areas: A = 4π sinh ((r + ∆r)/2) − 4π sinh (r/2), 2

or approximate this area by multiplying its thickness ∆r by its circumference C. Thus, 2

C≈

4π sinh ((r + ∆r)/2) − 4π sinh (r/2)

. ∆r This approximation approaches equality as ∆r goes to zero. That is, C=

(4π sinh (r/2)) = 8π sinh(r/2) cosh(r/2)(1/2) dr

7.15 CIRCLES AND HOROCYCLES

193

= 4π sinh(r/2) cosh(r/2) = 2π sinh r. Let us conclude as we did in the previous exercise; with a remark about what happens if r is small. If r is small, then sinh(r) ≈ r, so the formula is approximately 2πr, the formula for the circumference of a circle in Euclidean geometry. Exercise 7.90. The area of a triangle is at most π. The area of a circle is unbounded. Hence, there must be a largest circle that can be inscribed in a triangle. What is the diameter of this circle? [S] Exercise 7.91. What is the diameter of the largest circle that can be inscribed in a quadrilateral? Solution. When I ﬁrst thought of this question, I placed it here because it had something to do with circles, and expected one to model the situation in H , as is done in the previous exercise. But it occurs to me that it makes a lot more sense to use the Poincaré disc model.

Figure 7.14 We ﬁrst observe that every quadrilateral can be thought of as a subset of a quadrilateral whose vertices are all at inﬁnity. Also, without loss of generality, we may assume that the center of the inscribed circle is the center 0 of the Poincaré disc. Then it is clear that the inscribed circle is largest if the vertices are uniformly distributed about the boundary of the circle. That is, without loss of generality, we may assume the vertices of the quadrilateral are at ± 1 and ± i. Since the Poincaré lines are Euclidean circles that are perpendicular to the boundary of the disc, the centers of these circles are where the tangents intersect. That is, they are centered at the vertices of the Euclidean square that circumscribes the dis√ c, as in Figure 7.14. Hence, the inscribed circle has a Euclidean radius of 2 − 1.

194 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY To ﬁnd the Poincaré radius of this circle, we√use Exercise 7.85 to ﬁnd the distance between the point O = 0 and P = 2 − 1. The endpoints of the line OP are ±1, so we get √ r = |OP | = | ln( 2 − 1, 0; 1, −1)| . . . . = .ln √2 − 2 , −1 .. . √ . 1 . 2 √ = | ln( 2 − 1)| √ = ln( 2 + 1) Thus, of the largest circle that can be inscribed in a quadrilateral the is 2diam ln(√eter 2 + 1). Exercise 7.92. What is the diameter of the largest circle that can be inscribed in a pentagon?

Figure 7.15 Solution. After solving Exercise 7.91, I think we should be able to do this in general. The Euclidean circles that describe the edges of the appropriate regular asymptotic n-gon in the Poincaré disc model have centers at the vertices of the regular Euclidean n-gon that circumscribes the unit disc. A ﬁgure with n = 5 is shown in Figure 7.15. These vertices are a Euclidean distance sec(π/n) away from the center 0. The radii of these circles are tan(π/n). Thus, the inscribed circle goes through x = sec(π/n) − tan(π/n). The Poincaré distance from 0 to x is r, where r = | ln(x, 0; 1, −1)|

195

7.16 HYPERBOLIC TRIGONOMETRY . . . = .ln x − 1 , −1 . . . x+1 1 . . . . . 1 x − = ..ln 1 + x .. = ln

1 + x 1−x

This can be simpliﬁed further. Note that 1+x 1−x

cos(π/n) + 1 − sin(π/n) cos(π/n) − 1 + sin(π/n)

For convenience, let us set c = cos(π/2n) and s = sin(π/2n), so cos(π/n) = 2 2 2 2 c − s = 2c − 1 = 1 − 2s and sin(π/n) = 2sc. Then c(c − s) c 2 1+x 2c − 2sc = = = cot( 2 ) = π/ n . 1−x −2s2 + 2sc s(c − s) s Thus, the diameter is 2r = 2 ln(cot(π/2n)). √ √ √ √ is ln(5 + 2 5). For n = 5, we get cot(π/10) = 5 + 2 5, so the√diameter I’ll leave it to the reader to verify cot(π/10) = 5 + 2 5. Exercise 7.93. In Euclidean geometry, all triangles have a circumcircle. In hyperbolic geometry, this is not the case. Draw a (nonasymptotic) triangle ∆ABC in H which does not have a circumcircle.

Figure 7.16 Solution. All we need to do is pick any three points in the upper half plane with the property that their Euclidean circumcircle intersects the real axis. One such triangle is shown in Figure 7.16.

7.16

Hyperbolic Trigonometry 2

Exercise 7.94. Prove sech x = 1 − tanh x. Compare this with the iden2 tity sec2 x = 1 + tan x.

196 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Solution. Note that sech

1 =

cosh x − sinh x 2

tanh

2 — cosh x 2 cosh x In the usual trigonometry, the corresponding relation is

sec

1 cos2 x

cos2 x + sin x

= 1 − tan x.

cos2 x Note that the relation is very similar and the derivation is almost identical. This is common in hyperbolic trigonometry. Exercise 7.95. Prove the multiple angle formulas cosh(x + y) = cosh x cosh y + sinh x sinh y sinh(x + y) = sinh x cosh y + sinh y cosh x. 2

Use the ﬁrst to ﬁnd formulas for cosh x and sinh x. Solution. Note that

e = cosh x + sinh x, so ex+y = exey = (cosh x + sinh x)(cosh y + sinh y) = cosh x cosh y + sinh x sinh y + cosh x sinh y + sinh x cosh y. Similarly, e− = cosh x − sinh x, x

so e−x−y = e−xe−y = (cosh x − sinh x)(cosh y − sinh y) = cosh x cosh y + sinh x sinh y − cosh x sinh y − sinh x cosh y. Thus, cosh(x + y) =

ex+y + e−x−y = cosh cosh +y sinh sinh x x y, 2

sinh(x + y ) =

ex+y − e−x−y = cosh x sinh y + sinh x sinh y. 2

and

Now note that cosh(2x) = cosh x + sinh x = 2 cosh x − 1 = 1 + 2 sinh x, 2

197

7.16 HYPERBOLIC TRIGONOMETRY so cosh(2x) + 1 2 cosh(2x) −1 sinh2 x = . 2

cosh2 x =

Note that, subtracting these last two, we get 1, as expected. Exercise 7.96. Let ∆ABC be a right angle triangle with right angle at C. Prove that cos A = cosh a sin B. Solution. Since sin B = we have cosh a sin B =

sinh b sinh c

cosh a sinh b sinh c

= cos A.

Exercise 7.97. Let ∆ABC be a right angle triangle with right angle at C. Prove that cot A cot B = cosh c. Solution. Let’s just do it... cos A cos B cosh a sinh b cosh b sinh a sinh2 c = 2 sin A sin B sinh c sinh a sinh b = cosh a cosh b = cosh c.

cot A cot B =

In the last step we used the Pythagorean theorem. Exercise 7.98. Prove the hyperbolic Law of Sines. b h

A c

x D a –x B

Figure 7.17

198 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Solution. Consult Figure 7.17, which is the upper half plane version of the usual diagram for the proving the Law of Sines or Law of Cosines (for example, compare this with Figure 1.33 of the text). Let h =| AD| be the altitude at A. Then, by considering the right angle triangle ∆ADC, we get sin C =

sinh h sinh b

and by considering ∆ABD, we get sin B =

sinh h sinh c

Thus, isolating sinh h in both expressions and equating, we get sin C sinh b = sin B sinh c sinh b sinh c = . sin B sin C Similarly, by considering an altitude at B, we get sinh a sinh c = , sin A sin C so combining, we get the Law of Sines: sinh a sinh b sinh c = = . sin A sin B sin C Exercise 7.99. Prove the hyperbolic Law of Cosines for sides. Solution. We again consider Figure 7.17. We let h = | AD|. It is important to get the next labeling right: We set x =| CD | so that| DB| = a− x. The proof now follows essentially the same way as it does in Euclidean geometry. By the trig rules, we get cosh h sinh x sinh b sinh b cos C = cosh h sinh x. cos C =

(7.7)

By the Pythagorean theorem applied to ∆ABD, we get cosh c = cosh h cosh(a − x) = cosh h(cosh a cosh(−x) + sinh a sinh(−x) = cosh h cosh a cosh x − cosh h sinh a sinh x.

(7.8)

By the Pythagorean theorem, cosh h cosh x = cosh b. Using this and (7.7) in (7.8), we get cosh c = cosh a cosh b − sinh a sinh b cos C.

199

7.16 HYPERBOLIC TRIGONOMETRY a h

c –x

b x A

Figure 7.18 Exercise 7.100. Prove the hyperbolic Law of Cosines for angles. Solution. I don’t know of any proof other than the following uninspired one. Recall, we want to prove cos C = − cos A cos B + sin A sin B cosh c. We expand the left hand side. In Figure 7.18, we let C1 = ∠FCA and C2 = ∠BCF . Let x = |AF | so |FB| = c − x. Let h = |CF |. Then cos C = cos(C1 + C2) = cos C1 cos C2 − sin C1 sin C2 cosh x sinh h cosh(c − x) sinh h sinh x sinh(c − x) − sinh b sinh a sinh b sinh a 2 (cosh c − sinh x sinh(c − x)) sinh h − sinh x sinh(c − x) = sinh a sinh b 2 2 cosh c sinh h − sinh x sinh(c − x)(sinh h + 1) = sinh a sinh b 2 2 cosh c sinh h − sinh x sin(c − x) cosh h = . sinh a sinh b =

Expanding the right hand side, we get — cos A cos B + sin A sin B cosh c cosh h sinh x cosh h sinh(c − x) sinh h sinh h =− + cosh c sinh sinh a 2 b 2 sinh b sinh a − cosh h sinh x sinh(c − x) + sinh h cosh c = . sinh a sinh b Since the left and right hand sides are equal, our result follows. Exercise 7.101. Verify that the Pythagorean theorem, sine and cosine rules, and the Law of Sines are consistent with the corresponding rules in Euclidean geometry if a, b, and c are very small lengths.

200 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Solution. Recall the Taylor series for the trig and hyperbolic trig functions. The conclusion of the Pythagorean theorem is cosh c = cosh a cosh b a2 b2 1+ + ... = 1 + + ... 1+ + ... 2 2 2 2 2 2 c 1+ + ... = 1 + a + b + ... 2 2 2 2 2 2 c = a + b + ..., c2

where the ellipses (the ‘...’) refer to higher order terms. If a, b, and c are very small, then these higher order terms are extremely small, so we are left with the Euclidean Pythagorean theorem. We also have sinh a sinh c a + ... = c + ... a 1 + ... = , c 1 + ... cosh a sinh b cos A = sinh c (1 + ...)(b + ...) = c + ... b + ... = . c + ... sin A =

That is, these approximate opposite over hypotenuse and adjacent over hypotenuse. Finally, the law of sines says sinh a sinh b sinh c = = sin A sin B sin C a + ... b + ... c + ... = = . sin A sin B sin C Note that we did not say A, B, or C are very small, so they remain. Thus, this approximates the Euclidean law of sines. Exercise 7.102. What does the Law of Cosines for angles say if c is very small? Solution. The law of cosines says cos C = − cos A cos B + sin A sin B cosh c. For small c, cosh c is approximately 1, so we get cos C = − cos A cos B + sin A sin B

201

7.16 HYPERBOLIC TRIGONOMETRY — cos C = cos A cos B − sin A sin B cos(π − C) = cos(A + B).

Thus, A + B + C = π. Note that A, B, and C are in [0, π), hence the choice of angles. Also, the cosine function is invertible on this domain. One might also interpret that the Law of Cosines implies the angle sum formula for cosines. Exercise 7.103* (The Extended Law of Sines in Hyperbolic Geometry). Suppose a triangle ∆ABC has a circumcircle with radius R. Prove that tanh(a/2) tanh(b/2) tanh(c/2) tanh R = = = . B+C−A A+C−B cos cos cos A+B−C 2

A γ β

D β α

γ α

B O

Figure 7.19 Solution. The thought is to adapt the proof of Theorem 1.11.1 (the Euclidean extended Law of Sines) to hyperbolic geometry. The problem is we do not have the Star Trek lemma, since the sum of angles is not constant. However, we still have isosceles triangles. Let O be the circumcenter of ∆ABC (note that we are given that ∆ABC has a circumcircle). Then ∆OBC, ∆OAC, and ∆OAB are isosceles triangles. Let their base angles be α, β, and γ, respectively, as shown in Figure 7.19. Note that some of these angles may have a negative measure, as in the diagram. Then, A= β+ γ B= α+ γ C = α + β,

202 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY so

B+C−A . 2 In light of angles like this, let us introduce the notation α=

A+B+C 2

an analogue of the semiperimiter. Then α = S − A. Let us now look at ∆BOA′, where A′ is the midpoint of BC. Since this triangle is a right angle triangle, we have cos(α ) = cos(S − A ) =

sinh(a/2) cosh(|OA′|) . sinh R

(7.9)

By the Pythagorean theorem, cosh(a/2) cosh(|OA′|) = cosh R cosh R cosh( ′ ) = |OA | . cosh(a/2) We plug this into (7.9) to get cos(S − A ) =

sinh(a/2) cosh R . sinh R cosh(a/2)

Separating the R’s, we get tanh R =

tanh(a/2) cos(S − A)

Since there is nothing special about the angle A, we get the same results for B and C. That is, tanh R =

tanh(a/2) cos(S − A)

tanh(b/2) tanh(c/2) = . cos(S − B) cos(S − C)

Note that, in Euclidean geometry, S = π/2, so cos(S − A) = sin A. Exercise 7.104*. Suppose ACBD is a quadrialteral with right angles at A, B, and C. Let a = |BC| and b = |AC|. Prove cos D = sinh a sinh b.

Solution. Let c = AB | and | think of the quadrilateral as a union of two triangles, ∆ABC and ∆ABD, as in Figure 7.20. Let A′ and B′ be the angles at A and B of ∆ABC. Let A′′ and B′′ be their complements (since the angles at A and B in the quadrilateral are right angles). Note that A′′

203

7.16 HYPERBOLIC TRIGONOMETRY C a

b c

Figure 7.20 and B′′ are the angles at A and B of ∆ABD. Let us apply the Law of Cosines for angles to ∆ABD. Then cos D = − cos A′′ cos B′′ + sin A′′ sin B′′ cosh c. Since A′ and A′′ are complementary, cos A′′ = sin A′, and so on, so we get cos D = − sin A′ sin B′ + cos A′ cos B′ cosh c. The angle at C is a right angle, so we can use the trigonometry rules to ﬁnd sin A′, etc., in terms of a, b, and c: sinh a sinh b sinh b cosh a sinh a cosh b cos D = − + cosh c sinh c sinh c sinh c sinh c sinh a sinh b = (−1 + cosh a cosh b cosh c) . 2 sinh c Since the angle at C is a right angle, we have, by the Pythagorean theorem, that cosh a cosh b = cosh c, so cos D =

sinh a sinh b 2

( −1 + cosh c)

sinh c sinh a sinh b = sinh2 c 2 sinh c = sinh a sinh b.

Exercise 7.105* (Heron’s Formula in Hyperbolic Geometry). Since the area of a triangle is determined by its angles, and since the sides of a triangle determine the angles, there must exist a formula for the area of a triangle in terms of its sides. That is, there must be a version of Heron’s formula in hyperbolic geometry. State and prove this formula. [H][A]

204 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Solution. We have Section 10.5 to guide us but it doesn’t offer much intuition, so let us first recall the proof of Heron’s formula in Euclidean geometry (Theorem 1.10.4). It begins with a formula for area that involves lengths. The only formula we have for area involves angles. Suddenly, the results of Exercises 10.21 and 10.22 make more sense. The first deals with the area of right angle triangles and the second is an analogue of the ‘one half base times height’ formula. So let us begin by evaluating sin ∆ and cos ∆ for right angles ∆ABC with area ∆ and right angle at C. Since the area formula involves angles, we do in fact want to apply regular trig functions, rather than hyperbolic trig functions. We have sin ∆ = sin(π − A − B − π/2) = sin(π/2 − A − B) = cos(A + B) = cos A cos B − sin A sin B sinh b cosh a sinh a cosh b − sinh a sinh b = 2 sinh c sinh a sinh b = (cosh a cosh b − 1) 2 cosh c − 1 sinh a sinh b = (cosh c − 1) (cosh c − 1)(cosh c + 1) sinh a sinh b = , cosh c + 1 cos ∆ = cos(π/2 − A − B) = sin(A + B) = sin A cos B + sin B cos A 2

sinh a cosh b + sinh b cosh a 2 sinh c (cos2 a − 1) cosh b + (cos2 b − 1) cosh a = 2 cosh c − 1 (cosh a + cosh b)(cosh a cosh b − 1) = 2 cosh c − 1 (cosh a + cosh b)(cosh c − 1) = (cosh c − 1)(cosh c + 1) cosh a + cosh b = . cosh c + 1 =

This gives us our ﬁrst result: Lemma 1. Let ∆ABC be a right angle triangle with right angle at C and area ∆. Then sin ∆ =

sinh a sinh b cosh c + 1

205

7.16 HYPERBOLIC TRIGONOMETRY cos ∆ =

cosh a + cosh b . cosh c + 1

Let us now use this to ﬁnd the area of an arbitrary angle. Let the altitude at A intersect the opposite side at D. Let x = ±| BD ,| where we take ‘+’ if D is on the same side of B as C and ‘−’ otherwise. Then ±| DC| = a − x. Let ∆1 = ±| ∆ABD | and ∆2 = ±| ∆ADC |, where we choose the sign depending on whether the triangle is oriented the same way as ∆ABC or not. Then, in all cases, ∆ = ∆1 + ∆2, so cos ∆ = cos ∆1 cos ∆2 − sin ∆1 sin ∆2 (cosh x + cosh h)(cosh(a − x) + cosh h) = (cosh b + 1)(cosh c + 1) (sinh x sinh h sinh(a − x) sinh h) + . (cosh b + 1)(cosh c + 1) Note that if one of ∆1 or ∆2 is negative, then one of x or a − x is negative, and sinh x sinh(a − x) is negative, as desired. Continuing, we have cosh x cosh(a − x) + cosh h cosh(a − x) + cosh x cosh h + cosh h (cosh b + 1)(cosh c + 1) 2 sinh x sinh(a − x) sinh h + (cosh b + 1)(cosh c + 1) 2 2 cosh x cosh(a − x) cosh h − cosh x cosh(a − x) sinh h = (cosh b + 1)(cosh c + 1) 2 2 cosh b + cosh c + 1 + sinh h − sinh x sinh(a − x) sinh h + (cosh b + 1)(cosh c + 1) 2 2 cosh b cosh c − cosh a sinh h + cosh b + cosh c + 1 + sinh h = (cosh b + 1)(cosh c + 1) 2 sinh h(cosh a − 1) =1 — . (cosh b + 1)(cosh c + 1) This gives us an area formula for arbitrary triangles: 2

cos ∆ =

Lemma 2. Let the altitude at A of ∆ABC intersect BC at D, let h = |AD|, and let ∆ = |∆ABC|. Then 1 − cos ∆ =

sinh h(cosh a − 1) . (cosh b + 1)(cosh c + 1) 2

Now, we solve for sinh h. We note that sin B =

sinh h sinh c

so sinh h = sinh c sin B,

206 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY and sinh c sin B(cosh a − 1) (cosh b + 1)(cosh c + 1) 2 2 2 sinh c sin B(cosh a − 1) = (cosh a + 1)(cosh b + 1)(cosh c + 1) 2

1 − cos ∆ =

sinh c sin B sinh a (cosh a + 1)(cosh b + 1)(cosh c + 1)

(7.10)

The Law of Cosines gives us cosh b = cosh a cosh c − sinh a sinh c cos B,

sin B = 1 − cos B = 1 − 2

cosh

cosh cosh b− a c sinh a sinh c

We substitute this into (7.10) to get sinh a sinh c − (cosh b − cosh a cosh c)2 (cosh a + 1)(cosh b + 1)(cosh c + 1) (sinh a sinh c − cosh b + cosh a cosh c) (sinh a sinh c + cosh b − cosh a cosh c) = (cosh a + 1)(cosh b + 1)(cosh c + 1) (cosh(a + c) − cosh b)(cosh b − cosh(a − c)) = . (7.11) (cosh a + 1)(cosh b + 1)(cosh c + 1)

1− cos ∆ =

Things have been going great. Our difference of squares factorization was reminiscent of the proof of Heron’s formula in Euclidean geometry, but now we have a difference of hyperbolic cosines. For small values of a, b, and c, we note that this does look like a difference of squares, but we need something else. We look back in Section 10.5 and find an interesting trig identity (Exercise 10.24). What is the analogue in hyperbolic trigonometry? Such a result comes from the angle sum formula. Consider cosh(r + s) = cosh r cosh s + sinh r sinh s cosh(r − s) = cosh r cosh s − sinh r sinh s. Taking the difference, cosh(r + s) − cosh(r − s) = 2 sinh r sinh s. So let us set r+s = α r − s = β,

207

7.17 THE ANGLE OF PARALLELISM which means r=

α+β

2 α−β s= . 2

Thus, we get: Lemma 3. cosh α − cosh β = 2 sinh

α+β 2

sinh

a+c−b

sinh

α−β 2

Let’s use this in (7.11), to get 1 − cos ∆ =

4 sinh

a+c+b 2

sinh

b+a−c 2

sinh

b−a+c 2

(cosh a + 1)(cosh b + 1)(cosh c + 1) 4 sinh s sinh(s − b) sinh(s − c) sinh(s − a) = . (cosh a + 1)(cosh b + 1)(cosh c + 1)

Thus, we get Theorem 4 (Heron’s formula in hyperbolic geometry). Suppose a+b+c ∆ABC has sides of length a, b, and c. Let s = be the semiperimiter 2 and let ∆ = |∆ABC|. Then 1 —cos ∆ =

4 sinh s sinh(s − a) sinh(s − b) sinh(s − c) . (cosh a + 1)(cosh b + 1)(cosh c + 1)

This points out another typo. The answer to this exercise on page 343 of the ﬁrst printing has cosh ∆ rather than cos ∆. This is corrected in the second printing.

7.17

The Angle of Parallelism

Exercise 7.106. Prove the Law of Cosines for asymptotic triangles. Solution. Whoops, I should have asked this question after the next one, since we will use that result. Let ∆ABM be an asymptotic triangle with vertex M at inﬁnity. There exists a perpendicular to the segment AB that goes through M . Let this perpendicular intersect AB at C. There are two possibilities – either C is between A and B (so both angles are acute), or not. Without loss of generality, we may assume A is acute. Let m1 =| AC| and let m2 = ±|CB ,| where we take ‘+’ if C is between A and B, and ‘−’ otherwise. Then, m = m1 + m2 ,

208 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY so cosh m = cosh(m1 + m2) = cosh m1 cosh m2 + sinh m1 sinh m2. By Theorem 7.17.2, cosh m1 = csc A and, if B is acute, cosh m2 = csc B. If B is obtuse, then cosh m2 = csc(π − B) = csc B too. By Exercise 7.107, sinh m1 = cot A and, if B is acute, sinh m2 = cot B. If B is obtuse, then m2 < 0, so − sinh m2 = cot(π − B) = − cot B. Thus, in both cases, we get cosh m = csc A csc B + cot A cot B 1 + cos A cos B = sin A sin B sin A sin B cosh m = 1 + cos A cos B 1 = − cos A cos B + sin A sin B cosh m. Exercise 7.107. Prove that cos Π(m) = tanh m. Solution. We know sin Π(m), so let’s use this: cos Π(m) = 1 − sin2 Π(m) √ = 1 − sech 2m cosh m − 1 2 cosh m 2

sinh m 2 cosh m = tanh m.

There may be some question about the sign, but since Π(m) ∈ [0, π/2) and m > 0, it is correct to choose the positive root. Exercise 7.108. Prove that ∆ABC has a circumcircle if and only if C < Π(a/2) + Π(b/2). Solution. Let A′ and B′ be the midpoints of BC and AC, respectively. Consider the perpendiculars at these points. If these lines intersect at a point O at inﬁnity, then CO is parallel to both A′O and B′O, so ∠A′CO = Π(a/2) and ∠B′CO = Π(b/2). Thus, these perpendiculars intersect at a point O in H if and only if C < Π(a/2) + Π(b/2). The point O is, of course, the circumcenter, and the circumcircle exists if and only if O exists.

209

7.18 CURVATURE

Exercise 7.109. There is a different proof of Theorem 7.17.2. Consider the triangle ∆ABC with right angle at C. Solve for sin A in terms of a and b, and take the limit as a goes to infinity. Solution. Note typo in the first printing, corrected in the second printing. Using our trig formula, sinh a sinh c sinh a = √ 2 cosh c − 1 sinh a = √ 2 2 cosh a cosh b − 1 sinh a 1 = . cosh a cosh2 b − cosh1 2 a

sin A =

We note that

lim tanh x→∞

= lim ex − e−x = 1, x→∞ ex + e−x

so if we let a go to inﬁnity, then sin A = lim tanh a a→∞

1 cosh b

1−

cosh a cosh b

= sech b. As a goes to inﬁnity, the angle A approaches Π(b), so we get sin Π(b) = sech b.

7.18

Curvature

Exercise 7.110. How does scaling the arclength element by k change the Pythagorean theorem? Solution. Suppose the arclength ds has been scaled by k. Then every distance has been scaled by k, so to apply the Pythagorean theorem as we have proved it, we must ﬁrst normalize by dividing all lengths by k. Thus, 2 if the curvature is −1/k , then the Pythagorean theorem reads cosh(c/k) = cosh(a/k) cosh(b/k). Exercise 7.111. How does scaling the arclength element by k change the Law of Cosines for sides? Solution. The Law of Cosines for sides would then read cosh(c/k) = cosh(a/k) cosh(b/k) − sinh(a/k) sinh(b/k) cos C.

210 CHAPTER 7. THE POINCARÉ MODELS OF HYPERBOLIC GEOMETRY Exercise 7.112. How does scaling the arclength element by k change the formula for the area of a circle? Solution. The area formula for circles would then be r A = 4π sinh 2 . 2k

Chapter 8

Tilings and Lattices 8.1

Regular Tilings

Exercise 8.1. Prove that these three tilings are the only regular tilings of Euclidean space. Solution. If the tiles are equilateral triangles, then there are six to a vertex. If they are squares, then there are four to a vertex. If they are pentagons, then four to a vertex give a sum of angles around the vertex that is greater than 360◦, and three to a vertex is less than 360 ◦, so there are no tilings with regular pentagons. If they are hexagons, then there are three to a vertex. Finally, if they are n-gons with n ≥7, then three at a vertex gives more than 360◦, and two to a vertex is always less than 360 ◦. Thus, there are no more tilings. Exercise 8.2. The idea of duality introduced for Platonic solids extends to regular tilings. What are the duals of each of the regular tilings? Solution. Let us represent these tilings with (3, 3, 3, 3, 3, 3), (4, 4, 4, 4), and (6, 6, 6). The tilings (3, 3, 3, 3, 3, 3) and (6, 6, 6) are duals of each other, and the dual of (4, 4, 4, 4) is itself.

8.2

Semiregular Tilings

Exercise 8.3. Show that if a semiregular tiling has three tiles to a vertex and includes an n-gon for n odd, then the other two tiles must be identical. Explain why this eliminates all possibilities for (l, m, n) above, except those which we claimed are possible. Solution. Let the vertices of the n-gon be P1, P2,...,Pn. Without loss of generality, we may assume the face along P1P2 is an l-gon. Then, by looking at the vertex P2, the face along P2P3 must be an m-gon. It is clear that the faces must alternate, so the face along the P2kP2k+1 edge 211

212

CHAPTER 8. TILINGS AND LATTICES

must be an m-gon, and the face along the P2k−1P2k edge must be an l-gon. In particular, the face along the Pn−1Pn edge must be an m-gon, since n is odd. Thus, the face along the PnP1 edge must be an l-gon, so the face along the P1P2 edge must be an m-gon. But we already decided it is an l-gon, so l = m. Thus, there are no tilings with characterization (3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (4, 5, 20), and (5, 5, 10). This leaves the possibilities (3, 12, 12), (4, 6, 12), (4, 8, 8), and (6, 6, 6). That’s one more than stated, pointing out another error in the text, corrected in the second printing. Exercise 8.4. Construct the tiling with a square, hexagon, and dodecagon, at each vertex. See Figure 8.1.

Figure 8.1 Exercise 8.5. Construct the tiling with a square and two octagons at each vertex. See Figure 8.2. Exercise 8.6. Construct the tilings (3, 3, 6, 6) and (3, 6, 4, 4). Solution. Neither of these are possible (see the solution to Exercise 5.15). This was not meant to be a trick question. I just don’t know what I was thinking. I expect I’ll reword this soon.

8.3

Lattices and Fundamental Domains

8.4

Tilings in Hyperbolic Space

Exercise 8.7. What is the area of each square tile in Figure 8.8 [of the text]?

8.4 TILINGS IN HYPERBOLIC SPACE

213

Figure 8.2 Solution. Let us ﬁrst ﬁnd the formula for the area of an n-gon. Since an n-gon can be triangulated into n − 2 triangles, as was done in Exercise 5.1. The area of each of these triangles is π less the sum of the angles of the triangle. The sum of the angles of all the triangles is the sum of the angles of the n-gon, so the area of the n-gon is (n − 2)π less the sum of the angles. There are six tiles at each vertex, so the angles of the squares are all 2π/6 = π/3. Thus, the area of each square is 2π − 4(π/3) = 2π/3. Exercise 8.8. What is the area of each hexagonal tile in Figure 8.9 [of the text]? Solution. See the solution to Exercise 8.7 for the derivation of the area of an n-gon. Since there are four hexagons at each vertex, the angles in the hexagon are all 2π/4 = π/2. Thus, the area of each hexagon is (6 − 2)π − 6π/2 = π. Exercise 8.9. What is the area of each regular triangular tile if tiled with seven of these triangles at each vertex?

H can be [S]

Exercise 8.10. In this exercise, we find the vertices of a square ABCD centered at P = i which gives a tiling of H with six squares to a vertex. Let one of the vertices be at A = ia. Where is the diagonal vertex C? Use rotation through π/2 (see Section 7.9) to find the other two vertices. Find a formula in terms of a to find the angle ∠P AB. Use this to solve for a. [A]

Solution. If the square is centered at P = i, then the vertex diagonal to A is C = i/a. This is easy to see, since | AP | = ln a = |CP .| As suggested, to ﬁnd B we rotate A = ia by π/2. The rotation by π/2 is given by √ √ 2/2 2/2 cos(π/4) sin(π/4) 1 1 √ √ ≡ . — sin(π/4) cos(π/4) = −1 1 − 2/2 2/2

214

CHAPTER 8. TILINGS AND LATTICES

Thus, we get B=

1 −1

ia = ia + 1 = (1 + ai)2 = (1 − a ) + 2ia . 1 + a2 1 + a2 1 −ia + 1 2

The point D is the reﬂection of B through the imaginary axis, so (a − 1) + 2ia 2

. 1 + a2 These are the vertices of an arbitrary square. We are interested in the square that tiles the plane with six squares at a vertex. To find such a square, we must calculate the appropriate a. Let us first find the length AB | . |I know what you’re thinking – how will we find the endpoints of the line AB? Fortunately, we do not have to. Note that ∆ABP is a right angle triangle with right angle at P , so by the Pythagorean theorem, cos |AB| = cos |P A| cos |PB| cos |AB| = cos (ln a). 2

For simplicity, let us set r = ln a. Since there are six squares at a vertex, the angle ∠ABC = 2π/6 = π/3. Thus, ∠P AB = π/6, so sin(π/6) = 1 2

sinh r sinh |AB| sinh r cosh |AB|− 1 sinh r 2

=√

cosh r − 1 2 sinh r 4

(cosh r − 1)(cosh r + 1) 2 sinh r = 2 sinh r(cos2 r + 1) 1 = 2 cosh r + 1 2 cosh r + 1 = 4 4

cosh r = 3 √ cosh r = 3 er + e−r √ = 3 2 √ 1 a+ =2 3 a √ 2 a − 2 3a + 1 = 0

8.4 TILINGS IN HYPERBOLIC SPACE

215

√ √ 2 3 ± 12 − 4 a= 2 √ √ = 3 ± 2. √ √ Since a > 1, we get a = 3 + 2. Thus, √ √ A = i( 3 + 2) 2 √ √ √ (1 − a ) + 2ia = −4 − 2 6 + 2i( 3 + 2) √ 1 + a2 6+2 6 √ √ √ (2 + 6) + i( 3 + 2) = − √ 3+ 6 √ √ C = i( 3 2) − √ √ √ (2 + 6) + i( 3 + 2) √ D= . 3+ 6 Exercise 8.11. Write a Geometer’s Sketchpad script which inverts a point in a circle. The input can vary, but one possibility is the circle of inversion and the point.

B =

Script Ex08011. This script finds the inverse of A in the circle c1. Since the input includes a circle without its center, we first run Ex04025a to find the center O. Exercise 8.12. Write a script that draws the line through two points in H. Suggested input is the two points and the real line. Script Ex08012. This script finds the Poincaré line through A and B with respect to the real axis j. It is a repeat of Exercise 7.13. Exercise 8.13. The solution for a found in Exercise 8.10 is a constructible length. Construct the point A = ia in a Geometer’s Sketchpad sketch. Construct the other three vertices. (Hint: Construct the diagonal point C and the circumcircle for the square.) Use the scripts found in Exercises 8.11 and 8.12 to tile around one of the vertices. Script Ex08013a. This script finds the four vertices of the square centered at P = i with diagonal AC along the imaginary axis and appropriately sized so that it can tile the upper half plane with six squares per vertex. The input are the points O a√ nd Q√ , which are at 0 and 1. By √ Exerc√ ise 8.10, the point A has coordinates 3+ 2 and C has coordinates 3 − 2. The points B and D are on the circle with diameter AC, and on the unit circle. Script Ex08013b. This script finds the inverse of the point P in the arc a1. Ex08011 is not quite satisfactory since our Poincaré lines are only arcs of circles. The script in Ex08011 requires the full circle before it will run. Script Ex08013c. This script runs Ex08013a to find the tile. It runs Ex08012 repeatedly to fill in Poincaré lines. Points are reflected through Poincaré lines using Ex08013b. It tiles around the vertices A and D. Some of the points were found by reflecting through the imaginary axis.

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i 1 −1 which sends −i H to the disc model to ﬁnd the vertices of the square centered at 0 which tiles the plane with six squares to a vertex. Repeat Exercises 8.11, 8.12, and 8.13 to tile a portion of the disc. Compare your sketch with Figure 8.8 [of the text]. [A] Exercise 8.14. Use Exercise 8.10 and the map φ =

Solution. We evaluate 1 −i

√ √ i( 3 + 2 − 1) − i i(√ 3 + √2) = 1 √ √ 3+ 2+1 √ √ √ √ = i( 3 + 2 − 1)( 3 − 2 − 1) √ 3 − ( 2 + 1)2 √ √ √ √ √ 6 − 3 + 6 − 2 − 2 − 3 + 2 + 1) √ 3 − 2 − 2 2 − 1 √ i(2 − 2√ 3) = −2 2 =

i(3 −

√

√ √ i( 6 − 2) . 2 Since i is sent to 0, we don’t have to ﬁnd the image of the other three points. They must be the points an equidistance away from 0 on the two axes. That is, they are √ √ √ √ √ √ 6− 2 2− 6 i( 2 − 6) , , . 2 2 2 =

Let us now use these to create a tiling. Script Ex08014a. This script finds the vertices of the square centered at O that tiles the plane with six squares per vertex. The input is O at the origin and P at one. Script Ex08014b. This script draws the Poincaré line through two points in the disc model. The input is the two points and the circle that bounds the disc. Script Ex08014c. This script tiles the disc using six squares per vertex. Ex08014a is used to find the original square. Ex08014b is used to draw lines through various points. Ex08013b is used to reflect points through some of these Poincaré lines (to get the vertices of new tiles). We tile around one vertex, then reflect the results in a diagonal, then horizontal and vertical axis. Incidently, the tilings in the text were not produced this way – they were produced using Maple. Exercise 8.15*. Find a if A = ia is the vertex of a regular n-gon in H centered at i and of the appropriate shape to tile H with m tiles at each vertex. [S]

217

8.4 TILINGS IN HYPERBOLIC SPACE

Exercise 8.16. Tile the disc with four hexagons to a vertex, but with one hexagon centered at 0. Solution. We must first find a vertex. Look closely at the answer to Exercise 8.15. The answer is symmetric in m and n. Thus, if A = ia is a vertex of a hexagon centered at i that tiles with four to a vertex, then it is also a vertex of the square centered at i that tiles with six to a vertex. Thus, we have already found the point, both in H and in D. Let us use this information to come up with a tiling. Script Ex08015. After discovering that the vertex of the square centered at the origin which tiles with six to a vertex is the same as the vertex of the hexagon centered at the origin which tiles with four to a vertex, we run Ex08014a to find one of these vertices. The rest are then easy to find. We tile a region using Ex08014b and Ex08013b, and reflect several times to get a decent tiling. Exercise 8.17*. Tile the disc with three squares and three triangles at each vertex. Solution. The clever part of this exercise is to realize that we want to tile the disc with octagons, three per vertex. Joining every second vertex gives triangles and squares, as required. The next step is to find the coordinates of the point that will generate an appropriate octagonal tile. Using the solution to Exercise 8.15 (page 360 of the text – note that the last two lines are incorrect in the first printing), the Poincaré radius r of the circumcircle of the tile satisfies cosh r = cot(π/8) cot(π/3). The easy one to evaluate is 1 cot(π/3) = √ . 3 For the other, we note that 1 2 2 cos(π/4) = √ = 2 cos (π/8) − 1 = 1 − 2 sin (π/8). 2 Thus

√ cos(π/8) =

√

√ 2 √ 2− 2 sin(π/8) = √ 2 √ 2+ 2 cot(π/8) = √ √ 2− 2

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CHAPTER 8. TILINGS AND LATTICES

= =

(2 + √

√

2)2

2 2 + 1. √

Thus,

2+1 cosh r = √ . 3 Ultimately, we want the value tanh(r/2) (see page 168 of the text) and tanh( 2) = e − 1 = cosh r + sinh r − 1 r/ . er + 1 cosh r + sinh r + 1 r

For simplicity, let us set s = sinh r and c = cosh r. Note that s = Thus, (c + s − 1)(c + 1 − s) tanh(r/2) = (c + 1 + s)(c + 1 − s) 2 (c − (s − 1)2) = (c + 1)2 − s2 2 2 c − s + 2s − 1 = c2 + 2c + 1 − s2 2s = 2 + 2c s . = 1+c

√

c2 − 1.

Thus, we ﬁrst ﬁnd √ sinh r =

cosh r − 1 =

√ ( 2 + 1)2

−1=

√ 2 2 . 3

√ √ 242 . √ 3+ 2+1 Let us now construct this point and tile the disc using Sketchpad.

Then

tanh(r/2) = √

Script Ex08017a. This exercise constructs the vertices of the octagon cente√ red at A that ti√ les the plane. We do this in steps. √ We√construct a = 2√ . The√n b = 4 2, using Ex04008. Then c = ab = 2 4 2. Then d = 1 + 2+ 3. We use Ex04007d to ﬁnd 1/d, and Ex04007c to ﬁnd c/d. This is the radius of the circle on which the eight vertices lie. Script Ex08017b. – line segments in the Poincaré disc. This script constructs the Poincaré line segment AB in the disc with boundary c1. I had intended including another script to construct a partial tiling, but there is a rather small limit on the number of steps in a script (1500). The sketch I was working from is shown in Figure 8.3.

8.4 TILINGS IN HYPERBOLIC SPACE

219

Figure 8.3 Exercise 8.18*. Let the regular pentagon ABCDE in H be centered at i and let A = ir. If the length r is chosen so that this pentagon can tile H with four pentagons to a vertex, then r is constructible. Find it, and use it to tile around the vertex A. Exercise 8.19. Let the Poincaré disc model be drawn in the complex plane centered at 0. Let A1A2...An be an n-gon with center 0, A1 real, and such that the hyperbolic plane can be tiled with these n-gons with m such tiles at a vertex. Show that A1 and A2 are constructible (in the Euclidean plane) if and only if both the n-gon and m-gon can be constructed. Solution. This exercise, as worded in the ﬁrst printing, is probably true, but I don’t see how to do it. The wording above appears in the second printing. If the m-gon and n-gon can be constructed using a straightedge and compass in the Euclidean plane, then we can construct cot(m/π) and cot(n/π), so we can construct ai ∈ H, a vertex of the tile centered at i in H. Since A1 =

a−1 , a+1

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CHAPTER 8. TILINGS AND LATTICES

we can construct A1 too. Since we can construct the regular n-gon, we can construct A2. For the other direction, we ﬁrst note that since we have A1 and A2, we can use these points to construct the regular n-gon inscribed in the circle centered at O and going through A1. To show that we can construct the m-gon too, we note that A1 =

1 −1 a 1 1

1 1 A1 = a −1 1 a = A1 + 1 . −A1 + 1 r

Thus, a is constructible. Since a = e , we can also construct 2

cot(π/m) cot(π/n) = cosh r =

a +1 er + e−r . = 2a 2

Since we know we can construct the regular n-gon, we can construct cot(π/n), so we can divide by this to isolate cot(π/m). Now construct a right angle triangle with short sides of length 1 and cot(π/m). Then the angle opposite the side of length 1 is π/m. We can use this to construct the regular m-gon. Exercise 8.20*. The notion of constructibility is equally valid in hyperbolic geometry. The hyperbolic straightedge is used to make the hyperbolic line through two constructed points, and the hyperbolic compass is used to make the circle centered at a constructed point and going through a constructed point. Suppose that we begin with the points 0 and 1/2 in the Poincaré disc D. Show that every point which is constructible in D using the hyperbolic tools is also constructible using Euclidean tools. Are all points which are in D and are constructible using Euclidean tools also constructible using hyperbolic tools?

8.5

Tilings in Art

Exercise 8.21. What is the area of each ﬁsh in Escher’s Circle Limit III (Figure 8.16 [of the text])? [S] Exercise 8.22. Create an artistic tiling that has translational symmetry in two directions. The example in Figure 8.13 [of the text] was created by Sasha Neugebauer. Exercise 8.23. Create an artistic tiling that has translational and rotational symmetry. Exercise 8.24*. Create an artistic tiling in the Poincaré disc.

Chapter 9

Foundations 9.1

Theories

9.2

The Real Line

9.3

The Plane

9.4

Line Segments and Lines

Exercise 9.1. Show that if |PR |+ RQ | |= P| Q ,| then R lies on the line segment P Q. Hint: Since this is not true in spherical geometry, any solution will require Axiom 1 or 2. Solution. Suppose R is not on the line segment P Q. We will show that the segmented path PR united with RQ satisﬁes the deﬁnition of a line segment, thereby contradicting Axiom 1, whence we conclude R is on P Q. We must check two properties. Suppose R1 and R2 are points on this segmented path. Suppose R1 and R2 both lie on PR. Then, with a suitable reordering, |P Ri | + |Ri Rj | + |Rj R| + |RQ| = |P R| + |RQ| = |P Q|. Similar arguments can be made wherever R1 and R2 are on PR or RQ. For the second property, if 0 ≤ r ≤ |P R|, then since PR is a segment, there exists an R′ on PR such that |PR′| = r. If |PR| < r < |P Q|, then there exists an R′ on RQ such that |RR′| = |PR|− r. Then |PR ′ |≤ |P R| + |RR′| = r, but if |PR′| < r, then we note that |R′Q| = |P Q|− r, so |P Q|≤ |P R′| + |R′Q| < r + (|P Q|− r) = |P Q|. 221

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CHAPTER 9. FOUNDATIONS

Thus, we must have |PR′| = r, as required. Thus, the segmented path PRQ is a segment. Since segments are unique (by Axiom 1), R must be on P Q.

9.5

Separation Axioms

Exercise 9.2. We postulate that there exists a point P not on l. Show that there exists a point Q on the other side of l. Solution. Let R be a point on l and consider the line that contains the line segment PR (as guaranteed by Axiom 2). Let r =| PR |. Then there exists exactly two points P and P ′ such that r =| PR| = |P ′R |(as guaranteed by the definition of a line). We wish to show that R is on the line segment PP ′, so P ′ is on the other side of l. This is an important enough result that we will state it as a lemma: Lemma. Let P be a point on a line l and suppose R and R′ are two distinct points on l such that |P R| = |P R′ |. Then P lies on the line segment RR′ . Proof. Suppose P is not on RR′. Let S be the midpoint of RR′ and let s = PS | .| Since R and R′ are on l, the line segment RR′ is a subset of l (by definition) so S is on l. If s = r, then there are there points, R, R′, and S, that are a distance R away from P . If s > r, then there exists a point R′′ on PS such that| PR′′ = | r. Note that R′′ is on l too, so again we have three points on l a distance r away from l, so R′′ = R or R′. Suppose R′′ = R. Then R is on the segment PS, and |PR′| + |R′S| = |PR| + |RS| = |PS|, so R′ is on PS too. That is, there are two points on PS a distance r away from P , a contradiction of the definition of a line segment. Thus, we must have s < r. If r −s < s, then there exists a point T on PS such that ST | |= r −s, so there are three points on l (R, R′, and T ) a distance r − s from S. Note that T cannot be R or R′, since | PT | = 2s − r < r =| PR| = ′ (recall, s < r). If r ′ are all a distance s PR s = s, then P , R, and R | | − from S, a contradiction. If r− s > s, then there exists a point T on SR a distance s from S. Since P is also a distance s from S, T must be on SR′ too. Then we have the same situation as with S, only | PT| = r− 2s. If r — 2s s,≤then we continue as before. Otherwise, we find a T2 such that | PT2 |= r −3s, .... There exists an n such that r − ns < s, giving us the requisite contradiction. Thus, our assumption that P is not on RR′ is false. Exercise 9.3 (Pasch’s Theorem). Suppose a line l intersects ∆ABC and that none of A, B, or C lie on l. Show that l intersects exactly two of the three sides of ∆ABC. [S]

9.5 SEPARATION AXIOMS

223

Exercise 9.4. Deﬁne the interior of an angle ∠BAC.

[A]

Exercise 9.5. Deﬁne the interior of a triangle. Answer. We say that P is inside ∆ABC if P is inside ∠ABC, ∠BCA, and ∠CAB. Equivalently, P is inside ∆ABC if P and A are on the same side of BC, P and B are on the same side of AC, and P and C are on the same side of AB. Exercise 9.6. Suppose P is inside ∠BAC, and that Q is on the ray AP . Prove that Q is inside ∠BAC. Solution. Suppose Q is not inside ∠BAC. Then, without loss of generality, we may say that Q and B are on opposite sides of AC. Since P and B are on the same side of AC, that means P and Q are on opposite sides of AC. Thus the line segment PQ intersects the line AC. Two lines intersect in at most one point and since PQ intersects AC at A, we get that A is on the segment P Q. But that means Q is not on the ray AP , a contradiction. Thus, Q is inside the angle ∠BAC. Exercise 9.7. Suppose D is on the line segment BC. Prove that D is inside ∠BAC. Solution. This is not quite true. Neither B nor C are inside ∠BAC, but both are on the line segment BC. Otherwise ... since D is on BC, the line DB intersects AC at C. Furthermore, D is between B and C, so C is not on BD. Thus, D and B are on the same side of AC. Similarly, D and C are on the same side of AB, so D is inside ∠BAC. Exercise 9.8. Suppose the segments BB′ and CC ′ intersect at A. Prove that the interiors of the four angles ∠BAC, ∠CAB′, ∠B′AC′, and ∠C′AB are disjoint. Solution. Suppose there exists a point P inside both ∠BAC and ∠CAB′. Since P is inside ∠BAC, P is on the same side of AC as B. Since P is inside ∠CAB′, P and B′ are on the same side of AC. Thus, B and B′ are on the same side of AC, so the segment BB′ does not intersect AC. This is a contradiction, since the segment BB′ intersects AC at A. Thus, such a P cannot exist. Thus, the interiors of ∠BAC and ∠CAB′ are disjoint. Similarly, the interiors of ∠BAC and ∠BAC′ are disjoint. Now, suppose there exists a point P inside ∠BAC and ∠B′AC′. Since P is inside ∠BAC, P and C are on the same side of AB. Since P is inside ∠B′AC′, P and C′ are on the same side of AB′ = AB. Since B and B′ are on opposite sides of AB, we get that P is on the opposite side of AB as P , a contradiction. Thus, the interiors of the other three angles are disjoint from the interior of ∠BAC. Since ∠BAC is arbitrary, the four interiors are disjoint.

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CHAPTER 9. FOUNDATIONS

Exercise 9.9. Suppose the segments BB′ and CC ′ intersect at A. Suppose P is inside ∠BAC. Let P ′ be a point on the line AP such that A is between P and P ′. Show that P ′ is inside ∠B′AC′. Solution. Since P is inside ∠BAC, P and C are on the same side of AB. Note that C and C′ are on opposite sides of AB (since the segment CC ′ intersects AB at A), and P and P ′ are on opposite sides of AB. Thus, C′ and P ′ are on the same side of AB = AB′. Similarly, B′ and P ′ are on the same side of AC = AC′, so P ′ is inside ∠B′A′C′. Exercise 9.10**. Suppose A lies on l and that there exists a point P on l such that P is inside ∆ABC. Prove that l intersects the side BC. [S] Exercise 9.11**. Come up with a deﬁnition of when two nondegenerate triangles have the same orientation. Prove that, by your deﬁnition, orientation is an equivalence relation with two equivalence classes on the set of nondegenerate triangles. [H][A] Exercise 9.12*. Suppose f is an isometry, and f preserves the orientation of a nondegenerate triangle ∆ABC. Prove that f is a direct isometry. That is, prove that f preserves the orientation of all nondegenerate triangles. Solution. Suppose f preserves the orientation of ∆ABC. Let P be any point on the same side of AB as C. Then ∆ABC and ∆ABP have the o same orientation. That is, ∆ABC ∼ ∆ABP . Let f (A) = A′ , f (B) = B ′ , f (C) = C′, and f (P ) = P ′. If P ′ and C′ are on opposite sides of A′B′, then the segment P ′C′ intersects the line A′B′ at some point Q′. But then 1 Q = f − (Q′) is on the line segment PC and the line AB, so P and C are on opposite sides, a contradiction. Thus P ′ and C′ are on the same side o of A′ B ′ , so ∆A′ B ′ C ′ ∼ ∆A′ B ′ P ′ . That is, f preserves the orientation of ∆ABP . We have therefore shown the following: Lemma. If f preserves the orientation of a triangle ∆ABC, and P is any point on the same side of AB as C, the f preserves the orientation of ∆ABP . Now, suppose ∆ABC ∼o ∆P QR. Then there exist two triangles ∆1 and ∆1 such that ∆ABC and ∆1 share two vertices; ∆1 and ∆2 share two o o o vertices; ∆2 and ∆P QR share two vertices; and ∆ABC ∼ ∆1 ∼ ∆2 ∼ o ∆P QR (see the hint to the last exercise). By the lemma, ∆A′B′C′ ∼ o o ∆′1, ∆′ 1 ∼ ∆′ 2, and ∆′ 2 ∼ ∆P ′Q′R′, where ∆′ 1 = f (∆ ,1 etc. Since same orientation is an equivalence, we get that ∆A′B′C′ o∼ ∆P ′Q′R′. Since o o ∆ABC ∼ ∆A′ B ′ C ′ , we get ∆P QR ∼ ∆P ′ Q′ R′ . That is, f preserves the o orientation of ∆P QR. Finally, we note that ∆A′C′B′ ∼ ∆ACB, so if o ∆P QR is arbitrary, then ∆P QR ∼ ∆ABC or ∆ACB, and in either case, o we get ∆P QR ∼ ∆P ′ Q′ R′ , so f preserves the orientation of all triangles. That is, f is a proper isometry.

9.6 CIRCLES

225

Exercise 9.13. We are in the process of deﬁning Euclidean geometry in the plane. How should the separation axioms read in three dimensions? How should they read in one dimension? Solution. We say two points P and Q not on a plane p are on the same side of p if the line segment PQ does not intersect p. If there is a point of intersection, we say P and Q are on opposite sides of p. We postulate that if P and Q are on the same side of P , and Q and R are on the same side of P , then P and R are on the same side of P . We further postulate that for any plane p, there exists a point P not on p, and that if P and Q are on opposite sides of p, and Q and R are on opposite sides of p, then P and R are on the same side of P .

9.6

Circles

Exercise 9.14. Using the above deﬁnition, what is a ‘circle’ in three dimensions? What is it in one dimension? Solution. A ‘circle’ in three dimensions is a sphere. In one dimension, it is just two points. Exercise 9.15*. Let CP (r) be a circle and suppose R and S are points such that |PR| < r and |PS| > r. Show that the line segment RS intersects the circle CP (r). The solution is in the text. Exercise 9.16*. We say three values a, b, and c are triangular if there exists a triangle with sides of length a, b, and c. That is, a, b, and c are triangular if a ≤ b + c, b ≤ a + c, and c ≤ a + b. Suppose C P (r) and C Q(s) are two circles and that r, s, and| P Q|are triangular. Show that these two circles intersect. [H] Exercise 9.17. Fill in the details of the second case in the proof of Theorem 9.6.1. Exercise 9.18. Suppose C and C′ are on the same side of AB, and that A and B are on opposite sides of CC ′ . Prove that either C or C′ is inside the triangle formed by the other three points. Solution. Let CC′ intersect AB at P . Without loss of generality, we may assume C′ and P are on the same side of C. Then C′ and P are on the same side of AC. Since CC ′ intersects the segment AB, the point P is between A and B, so P and B are on the same side of AC. Thus, C′ and B are on the same side of AC. Similarly, C′ and A are on the same side of BC, and of course, we are given that C and C′ are on the same side of AB. Thus, C′ is inside ∆ABC.

226

9.7

CHAPTER 9. FOUNDATIONS

Isometries and Congruence

Exercise 9.19. We have not yet formally defined angles. Come up with a definition. Does your definition work in the hyperbolic plane? Does it work on the sphere? Come up with a refinement which works in these cases. Answer. We haven’t defined a ray either .... Given points A and B, we call the ray AB the set of points P on the line AB such that P and B are on the same side of A. This works in Euclidean and hyperbolic geometry, but not in spherical geometry. In spherical geometry, we have a concept of antipodal points, those points at either end of a diameter. Let A and B be points on the sphere that are not antipodal. For the moment, let us define the line segment AB on the sphere to be the shortest path between A and B. Let us define the ray AB to be the set of points P such that P is on the line segment AB, or B is on the line segment AP . In particular, the antipodal point to A is not on the ray AB. We are now ready to define the angle ∠ABC in all three geometries. The angle ∠ABC is the union of the rays BA and BC. Exercise 9.20. Prove that any two right angles are congruent. Thus, we do not need to state Euclid’s fourth postulate as an axiom. Exercise 9.21. We have not formally defined angular measure. Come up with a definition. Answer. We define the measure of an angle m∠BAC to be the function on angles with the following properties: 1. The measure of a right angle is π/2. 2. If ∠BAC = ∠B′A′C′, then their measures are equal (m∠BAC = m∠B′A′C′). 3. If P is a point inside ∠BAC, then m∠BAC = m∠BAP + m∠P AC. This uniquely defines the measure of an angle. For an arbitrary angle ∠BAC, one can measure it by deciding whether B is inside a right angle. If not, think of it as the sum of a right angle and an acute angle. Otherwise, we can think of B as being inside a right angle ∠P1AC. Let P2 be a point inside ∠P1AC such that ∠P1AP2 ≡∠P2AC. Then these angles have the same measure, so each measure π/4. Decide which angle B is in, and bisect that angle. Keep doing this. We get a sequence of numbers pn such that pn

∠ BAC <

<m 2 nπ

pn + 1 . 2nπ

This sequence converges at the measure of ∠BAC. Exercise 9.22. How should Axiom 6 read in one dimension?

9.8 THE PARALLEL POSTULATE

227

Answer. Given any point P on the line l, there exists an isometry f of l such that f (P ) = P and f (Q) Q for any Q /= P . Exercise 9.23. How should Axiom 6 read in three dimensions? Is the full group of isometries in three dimensions guaranteed to exist using this revised version of Axiom 6? Answer. Given a plane p in three dimensions, there exists an isometry f such that f (P ) = P for all P∈ p and f (Q) /= Q for all Q∈ / p. These isometries generate all the isometries of Euclidean three space. Exercise 9.24. Prove Lemma 9.7.4. [A line and circle intersect in either zero, one, or two points. There is exactly one point of intersection if and only if the radius to the point of intersection is perpendicular to the line.]

9.8

The Parallel Postulate

Exercise 9.25. Let l1, l2, and l3 be three distinct lines. Prove that if l1 is parallel to l2, and l2 is parallel to l3, then l1 is parallel to l3. [H] Solution. Suppose l1 and l3 are not parallel. Then they intersect. Let this point of intersection be P . Then l1 is a line through P that is parallel to l2, and l3 is a line through P that is parallel to l2. By the parallel postulate, there can be only one such line, so l1 = l3. That is, the three lines are not distinct, a contradiction. Thus, l1 and l3 must be parallel.

9.9

Similar Triangles

Exercise 9.26. Prove Lemma 9.9.2. [Let ABCD be a parallelogram. Then |AB| = |CD|.] [H] Solution. Let us draw the diagonal AC. The transversal AC cuts the parallel lines AB and CD, so ∠BAC = ∠ACD. Since AC cuts the parallel lines AD and BC, we get ∠ACB = ∠CAD. By ASA, ∆ABC ∼ = ∆CDA. In particular, |AB| = |CD| and |BC| = |DA|. Exercise 9.27. Prove Lemma 9.9.4. [Let ABCD be a trapazoid with parallel sides AD and BC. Let B′ be the midpoint of AB and let the line through B′ parallel to BC intersect DC at C′. Then C′ is the midpoint of CD.] Solution. Without loss of generality, we may assume | | BC ≤ |AD |, as in Figure 9.1. Let D′ be the point on AD such that BD′ is parallel to CD. Draw BD′ and D′C. Let the line l through B′ parallel to BC intersect BD′ at B′′ and D′C at C′′. Since l is parallel to BC, it is parallel to AD′, so by Lemma 9.9.3, it bisects BD′. By applying Lemma 9.9.3 to ∆BD′C, C′′ is the midpoint of D′C. Applying Lemma 9.9.3 again to ∆D′CD, we ﬁnd that C′ is the midpoint of CD, as desired.

228

CHAPTER 9. FOUNDATIONS C

B B′

B′′

C′′

D′

C′

Figure 9.1 Exercise 9.28. Prove Lemma 9.9.5. [Let B′ be a point on AB in ∆ABC AB'| p such that | = .n Let the line through B′ and parallel to BC intersect |AB| 2 ' AC | AB'| AC at C′. Then | = | .] |AC|

|AB|

Chapter 10

Spherical Geometry 10.1

The Area of Triangles

Exercise 10.1. What is the area of a quadrilateral ABCD on the sphere? What is the area of a polygon on the sphere? Solution. Draw the diagonal AC. Then the quadrilateral ABCD can be thought of as the union of two triangles, the area is the sum of the areas of the two triangles, and the sum of the angles of ABCD is the sum of the angles of the two triangles. Thus, the area of ABCD is 2π − A − B − C − D. In general, the area of an n-gon is (n − 2)π less the sum of the angles. Exercise 10.2. Show that the area A of a circle of radius r on the sphere of radius ρ is r 2 2 . A = 4πρ sin [H] 2ρ Solution. Let us describe the sphere by revolving the circle centered at the 2 origin and with radius ρ about the x-axis. Let P = (ρ, 0) be the point where the circle intersects the x-axis. Let us describe this circle in terms of θ, the angle made with the x-axis. The portion of the circle traced out by ∆θ has length ρ∆θ. This portion is revolved about the x-axis in a circle with radius ρ sin θ, so the portion of the surface of revolution represented 2 by ∆θ has an area of 2πρ sin θ∆θ. In terms of θ, the circle centered at P with radius r corresponds to θ ranging from 0 to r/ρ. Thus, the area of this circle is ∫ r/ρ 2 2πρ sin θdθ A= 0

r/ρ

−2πρ cos θ 0

= −2πρ cos(r/ρ) + 2πρ 2

= 2πρ (1 − cos(r/rho)). 2

229

230

CHAPTER 10. SPHERICAL GEOMETRY

Recall that cos(2θ) = 1 − 2 sin θ, 2

so 2

A = 4πρ sin

r 2ρ

Exercise 10.3. Compare the formula for the area of a circle in spherical geometry with the one in hyperbolic geometry, which is A = 4πk2 sinh 2

. 2k (We derived this for k = 1 in Exercise 7.88.) Show that the formula for the area on the sphere gives the formula for the area inH if we substitute ρ = ik. (Hint: Use Euler’s formula to show that i sin θ = sinh iθ.) Solution. Recall that eiθ − e−iθ 2i ex − e−x sinh x = , 2 sin θ =

so i sin θ =

eiθ − e−iθ = sinh iθ. 2

Let us plug this into the formula for the area of a circle on a sphere of ‘radius’ ρ = ik: r 2 2 A = 4π(ik) sin 2ik −ir 2 2 = −4πk sin 2k 2 =4 r 2 πk sinh . 2k Exercise 10.4. Substitute the linear approximation for sin θ for θ near zero into the formula for the area of a circle on the sphere (see Exercise 10.2). This shows that this formula is consistent with the Euclidean formula for small circles on the sphere. This should be no surprise, since locally, spherical and Euclidean geometry are alike. Solution. If we substitute θ ≈sin θ for small r into the area of a circle, we get 2 r 2 2 r 2 2 A = 4πρ sin = πr . ≈ 4πρ 2ρ 2ρ

10.2 THE GEOMETRY OF RIGHT TRIANGLES

10.2

231

The Geometry of Right Triangles

Exercise 10.5. Let ∆ABC be a right angle triangle on the unit sphere with right angle at C. Prove that cos A = cos a sin B. What is the corresponding result in Euclidean geometry?

[A]

Solution. Note that cos A sin b cos a sin c = = cos a, sin B sin c sin b so cos A = cos a sin B. The corresponding result in Euclidean geometry is what happens when a, b, and c are very small. In this case, cos a ≈ 1, so cos A ≈ sin B. That is, the corresponding result in Euclidean geometry is cos A = sin B = sin(π/2 − A). Exercise 10.6. Let ∆ABC be a right angle triangle on the unit sphere with right angle at C. Prove that cot A cot B = cos c. Solution. We note that cot A =

cos A sin b cos a sin c sin b cos a = = . sin A sin c sin a sin a

Thus, cot A cot B =

sin b cos a sin a cos b sin a

= cos a cos b = cos c.

sin b

Exercise 10.7. Show that the formulas for sin A and cos A give the usual deﬁnitions for the sine and cosine if the lengths a, b, and c are very small. Solution. For small x, we have sin x ≈ x and cos x ≈ 1, so for small values of a, b, and c, we have sin a a ≈ , sin c c sin b cos a b cos A = ≈ . sin c c sin A =

232

CHAPTER 10. SPHERICAL GEOMETRY

Exercise 10.8. What is the Pythagorean theorem on a sphere of radius ρ? What are the trigonometric formulas? Solution. We have derived the Pythagorean theorem for a sphere of radius 1. If the radius is ρ, then we must normalize by dividing lengths by ρ. Thus, the conclusion of the Pythagorean theorem is cos(c/ρ) = cos(a/ρ) cos(b/ρ). Similarly, the trigonometric formulas become sin(a/ρ) sin(c/ρ) sin(b/ρ) cos(a/ρ) cos A = . sin(c/ρ) sin A =

10.3

The Geometry of Spherical Triangles

Exercise 10.9. Show that the spherical Law of Sines is consistent with the Euclidean Law of Sines for very small sides a, b, and c. Solution. Recall that sin x = x + ..., where the ellipses (the ‘...’) represents higher order terms. For small x, the sum of these higher order terms is small compared to x. Thus, the spherical Law of Sines is a + ... b + ... c + ... = = , sin A sin B sin C which gives the Euclidean Law of Sines if we ignore terms of degree two or higher. Exercise 10.10. Show that the spherical Law of Cosines is consistent with the Euclidean Law of Cosines for very small sides a, b, and c. x2

Solution. Recall that sin x = x + ... and cos x = 1 − 2+ ..., where the ellipses (the ‘...’) represent terms of degree three or higher. Thus, the spherical Law of Cosines says b2 1 c2 a2 )(1 − — 2 + ... = (1 − 2 2 ) + (a + ...)(b + ...) cos C c2 a2 b2 1 − 2 + ... = 1 − 2 − + ... + (ab + ...) cos C 2 2 2 2 c = a + b − 2ab cos C + .... Thus, if we ignore all terms of degree three or higher, we get the Euclidean Law of Cosines. Exercise 10.11. What are the Law of Sines and the Laws of Cosines on a sphere of radius ρ?

233

10.3 THE GEOMETRY OF SPHERICAL TRIANGLES

Solution. A sphere of radius ρ is a sphere where all distances have been scaled by a factor of ρ from the unit sphere. Thus, the Law of Sines becomes sin(a/ρ) sin(b/ρ) sin(c/ρ) = = , sin A sin B sin C and the Law of Cosines becomes cos(c/ρ) = cos(a/ρ) cos(b/ρ) + sin(a/ρ) sin(b/ρ) cos C. Exercise 10.12* (The Extended Law of Sines on the Sphere). Let R be the radius of the circumcircle of an arbitrary triangle ∆ABC on the unit sphere. Prove that tan R =

tan(a/2) cos

B+C−A 2

tan(b/2) cos

A+C−B 2

tan(c/2) cos

A+B−C 2

Verify that, for small lengths a, b, and c, this gives the extended Law of Sines in Euclidean geometry. A γ β

C′

B′ O R β

γ α B

α a — 2

A′

Figure 10.1 Solution. The diagrams in the text take a while to make, so we will go with an ‘abstract’ drawing, the one that was used to show the extended Law of Sines in Euclidean geometry (Figure 10.1). The proof of the result in Euclidean geometry makes use of the Star Trek lemma, which in turn makes use of the result that the sum of angles is 180◦. We do not have such a result in spherical geometry. The substitute

234

CHAPTER 10. SPHERICAL GEOMETRY

result that we will make use of is the observation that the triangle ∆BOC is an isosceles triangle. Let the base angles of ∆BOC be α, let those in ∆COA be β, and let those in ∆AOB be γ. We note that A = β + γ, B = α + γ, and C = α + β, so B+C−A 2 A+C−B β= 2 A+B−C γ= . 2

α=

In light of sums like these, let us introduce an angular analogue of the semiperimeter – let A+B+C S= , 2 so that α = S —A, β = S −B, and γ = S − C. Let us now look at ∆A′OB, where A′ is the midpoint of BC. Since ∆BOC is isosceles, the angle ∠OA′ B is a right angle (use SSS to show ∆OA′ B ∼ = ∆OA′ C). Let ′ h = |OA | and apply some spherical trig: cos α =

sin(a/2) cos h . sin R

We can eliminate cos h by using the Pythagorean theorem, which gives cos R = cos h cos(a/2). Combining, we get cosα = cos(S − A ) =

sin(a/2) cos R . cos(a/2) sin R

Separating R and expressions involving a and A, we get tan R =

tan(a/2) cos(S − A)

Since there is nothing special about the angle α, we have similar results for the angles β and γ. That is, tan R =

10.4

tan(a/2) cos(S − A)

tan(b/2) tan(c/2) = . cos(S − B) cos(S − C)

Menelaus’ Theorem

Exercise 10.13. Derive the Pythagorean theorem by applying Menelaus’ theorem to the triangle ∆ABC and the transversal DEF in Figure 10.2.

235

10.4 MENELAUS’ THEOREM

C a

π/2 −a

π/2 −c

π/2 −b

π/2 −A F

A E

Figure 10.2 Solution. By Menelaus’ theorem applied to ∆ABC and the transversal DEF in Figure 10.2, we get sin(π/2) sin(b − π/2) sin sin(a − π/2)sin(π/2) = −1 sin π/2 sin(c − π/2) 1 (− cos a) (− cos b) = −1 1 1 (− cos c) cos a cos b = cos c. Note that the sign of the angles represent whether the point F is between A and B, and is accurate whether c > π/2 or not. Similarly for the points D and E. Exercise 10.14. Apply Menelaus’ theorem to the triangle ∆AEF and the transversal CBD in Figure 10.2. Use the Pythagorean theorem to reduce the obtained expression into one of the formulas of Theorem 10.2.2. Solution. Applying Menelaus’ theorem, as instructed, we get sin |AC| sin |ED| sin |FB| = −1 sin |CE| sin |DF | sin |BA| sin b sin(π/2) sin(π/2 − c) = −1 sin(π/2 − b) sin(A − π/2) sin c

236

CHAPTER 10. SPHERICAL GEOMETRY sin b 1 (− cos c) = −1 cos b cos A sin c sin b cos c = cos b sin c cos A sin b cos a cos b = cos b sin c cos A sin b cos a cos A = . sin c

Exercise 10.15. Apply Menelaus’ theorem to the triangle ∆DEC and the transversal ABF in Figure 10.2. Use the results of Exercises 10.13 and 10.14 to reduce the obtained expression into the other formula of Theorem 10.2.2. Solution. Applying Menelaus’ theorem as instructed, we get sin |DF | sin |EA| sin |CB| = −1 sin |FE| sin |AC| sin |BD| sin(π/2 − A) sin(π/2) sin a = −1 sin A sin(−b) sin(π/2 − a) cos A 1 sin a = −1 sin A (− sin b) cos a cos A sin a = sin A sin b cos a sin b cos a sin a = sin A sin b cos a sin c sin a sin A = . sin c Exercise 10.16. Prove Lemma 10.4.2 for the case shown in Figure 10.3(b). A

F A′

F′

A B′

B F O

(a)

(b)

Figure 10.3

F′

237

10.4 MENELAUS’ THEOREM

Solution. The proof is the same. Let the perpendicular at A intersect OF at A′ and let the perpendicular at B intersect OF at B′. Then ∆AA′F ′ ∼ ∆BB′F ′, so |AF ′| |AA′| = . |F ′B| |BB′| Let us not worry about the sign yet. Note that sin ∠AOF =

|AA′| ′ 1 = |AA |.

Similarly, |BB′| = sin ∠FOB. Thus, as a ratio of lengths or values, |AF ′| |F ′B|

sin ∠AOF sin ∠FOB

Let us now check the signs. In the case we are looking at, F ′ is not between A and B, so the sign of the left hand ratio is negative. For the right hand side, ∠AOF and ∠FOB have opposite orientations, so that side is negative too. Note that F is one of the points of intersection of OF ′ and the circle, and it does not matter which one it is (in terms of the orientations of the angles). Exercise 10.17. Prove Menelaus’ theorem on the sphere in the cases where one or more of OD, OE, and OF are parallel to the plane through A, B, and C. Solution. Suppose ﬁrst that OF is parallel to the plane through A, B, and C, but that neither OD nor OE is. Let OD and OE intersect this plane at D′ and E′, respectively. Then the spherical line DE goes through F if and only if the Euclidean line D′E′ does not intersect the Euclidean line AB. That is, D′E′ is parallel to AB, so by Theorems 1.7.2 and 1.7.3, |CA| |CB| = . |CD′| |CE′| If we write |CB| = |CD′ | ± |D′ B| and |CA| = |CE ′ | ± |E ′ A|, then we get |D′ B| |CD′|

|E ′ A| |CE′|

Note that this equality holds even if we think of these as signed ratios. By Lemma 10.4.2, this gives us sin ∠DOB| sin ∠EOA = . sin ∠COD sin ∠COE Using spherical lengths and rearranging, we get sin |DB| sin |CE| = 1. sin |CD| sin |EA|

238

CHAPTER 10. SPHERICAL GEOMETRY

Now note that ∠AOF and ∠FOB are supplementary, so sin ∠AOF = sin ∠FOB, and that the two angles have opposite orientations, so the signed ratio sin ∠AOF = −1. sin ∠FOB Thus, we have DE goes through F if and only if sin |BD| sin |CE| sin |AF | = —1. sin |DC| sin |EA| sin |FB| This deals with the case when OF is parallel to the plane through A, B, and C, but that neither OD nor OE is. Now, suppose that both OE and OF are parallel to the plane through A, B, and C. Then the spherical line EF is the intersection of the sphere with the plane through O and parallel to the plane through A, B, and C. Thus, D is on the spherical line EF if and only if OD is parallel to the plane through A, B, and C. This happens if and only if the angles ∠BOD and ∠COD are supplementary. Since these angles have opposite orientations, they are supplementary if and only if the signed ratio satisﬁes sin ∠BOD = —1. sin ∠DOC ∠ As before, the signed ratiossin ∠FOB and ∠ are both −1, so D is on sin ∠EOC EF if and only if sin

AOF

sin

AOE

sin |AF | sin |BD| sin |CE| = —1. sin |FB| sin |DC| sin |EA| Exercise 10.18 (Ceva’s Theorem in Spherical Geometry). State and prove Ceva’s theorem on the sphere. A

F E

Figure 10.4 Solution. We note that the proof of Ceva’s theorem given in the text is merely a double application of Menelaus’ theorem. Since we have a spherical

10.4 MENELAUS’ THEOREM

239

version of Menelaus’ theorem, we should be able to apply it in the same way. So, suppose D, E, and F are points on the extended sides BC, AC, and AB, respectively, of a spherical triangle ∆ABC. Suppose that the Cevians AD, BE, and CF intersect at a common point P . Then we can apply Menelaus’ theorem to ∆ABE and the line F PC (see Figure 10.4, which is just Figure 1.47 of the text, and should be thought of as an abstract representation of the situation). We get sin |AF | sin |BP| sin |EC| = —1. sin |FB| sin |PE| sin |CA| We also apply it to ∆BCE and the line DPA to get sin |BD| sin |CA| sin |EP| = —1. sin |DC| sin |AE| sin |PB| Multiplying, we get sin |AF | sin |BP| sin |EC| sin |BD| sin |CA| sin |EP| = 1. sin |FB| sin |PE| sin |CA| sin |DC| sin |AE| sin |PB| We cannot cancel terms without ignoring the signed ratios. We do so, to get sin |AF | sin |EC| sin |BD| = ±1. sin |FB| sin |AE| sin |DC| We note that either one or three of these ratios are positive, so we get +1. We now know how the spherical version of Ceva’s theorem should be stated: Theorem (Ceva’s theorem in spherical geometry). Suppose D, E, and F are points on the sides BC, AC, and AB, respectively, of a spherical triangle ∆ABC. Then the lines AD, BE, and CF are coincident at a point P (and its antipodal point P ′) if and only if sin |AF | sin |BD| sin |CE| = 1. sin |FB| sin |DC| sin |EA| We have shown one direction. For the other direction, we proceed as in the proof in Euclidean geometry. Suppose sin |AF | sin |BD| sin |CE| = 1. sin |FB| sin |DC| sin |EA| and that these Cevians are not coincident. Let P be one of the points of intersection of AD and BE, and let F ′ be one of the points where CP intersects AB. Then, by the direction that we have already shown, we have sin |AF ′| sin |BD| sin |CE| = 1, sin |F ′B| sin |DC| sin |EA|

240 whence

CHAPTER 10. SPHERICAL GEOMETRY sin |AF ′| sin |F ′B|

sin |AF |

. sin |FB|

This means F = F ′ or its antipodal point. To see this, think of F ′ as sin AF'| being ﬁxed, so that sin | | 'FB|= k is a constant. Then determine how many solutions there are to the equation sin |AF | sin |FB|

= k.

It is clear that this has exactly two solutions, since the left hand side in′ creases for | FB | (0, ∈ π) and decreases on (π, 2π). Since F = F and F equal to the point antipodal to F ′ are two distinct solutions, they can be the only ones. Finally if DE goes through the point antipodal to F ′, then it also goes through F ′. Thus, DE goes through F . Exercise 10.19. Suppose ACBD is a quadrilateral with right angles at A, B, and C. Let a = |BC| and b = |AC|. Prove cos D = − sin a sin b. Solution. Whoops, the sign was wrong in the ﬁrst printing. One should realize that D must be greater than π/2 since the area of the quadrilateral is 3π/2 + D —2π = D π/2 − and is positive, so cos D < 0, while the sines are both positive, since a, b ∈ [0, π). One might also realize that this is a spherical version of Exercise 7.104 and that the proof is the same. We think of the quadrilateral ACBD as the union of two triangles ∆ABC and ∆ABD. See Figure 7.20 for a picture (that that picture is in H is not important). Let A′ and B′ be the angles at A and B of ∆ABC. Let A′′ and B′′ be their complements (since the angles at A and B in the quadrilateral are right angles). Note that A′′ and B′′ are the angles at A and B of ∆ABD. Let us apply the Law of Cosines for angles to ∆ABD. Then cos D = − cos A′′ cos B′′ + sin A′′ sin B′′ cos c. Since A′ and A′′ are complementary, cos A′′ = sin A′, and so on, so we get cos D = − sin A′ sin B′ + cos A′ cos B′ cos c. The angle at C is a right angle, so we can use the trigonometry rules to ﬁnd sin A′, etc., in terms of a, b, and c: sin a sin b sin b cos a sin a cos b cos D = − + cos c sin c sin c sin c sin c sinh a sinh b = ( −1 + cos a cos b cos c) . 2 sinh c

241

10.5 HERON’S FORMULA

Since the angle at C is a right angle, we have, by the Pythagorean theorem, that cos a cos b = cos c, so cos D =

sin a sin b 2

( −1 + cos c)

sin c sin a sin b = ( −sin2 c) 2 sin c = − sin a sin b.

10.5

Heron’s Formula

Exercise 10.20. Let us begin with a question of plausibility. Verify that for small lengths a, b, and c, Heron’s formula on the sphere is approximated by Heron’s formula in Euclidean geometry. Solution. On the left hand side, we should use a degree two approximation for cos x since the one’s cancel, but in the denominator of the right hand side, the one’s are not cancelled, so they dominate all other terms and a degree one approximation is enough. We therefore get ∆2 1− 1− 2

≈

4s(s − a)(s − b)(s − c) (2)(2)(2)

∆ ≈ s(s − a)(s − b)(s − c) √ ∆ ≈ s(s − a)(s − b)(s − c), 2

which is Heron’s formula in Euclidean geometry. Thus, the formula is at least plausible. Exercise 10.21. Let ∆ABC be a right angle triangle with right angle at C and area ∆. Prove that sin a sin b 1 + cos c cos a + cos b cos ∆ = . 1 + cos c sin ∆ =

Solution. We note that sin ∆ = sin(A + B + π/2 − π) = − sin(π/2 − A − B) = − cos(A + B) = − cos A cos B + sin A sin B − sin b cos a sin a cos b + sin a sin b = 2 sin c sin a sin b = ( —cos acos b+ 1) 1 − cos2 c

242

CHAPTER 10. SPHERICAL GEOMETRY sin a sin b (1 —cos c) (1 − cos c)(1 + cos c) sin a sin b = . 1 + cos c =

Also, cos ∆ = cos(A + B − π/2) = cos(π/2 − A − B) = sin(A + B) = sin A cos B + sin B cos A 2

sin a cos b + sin b cos a 2 sin c 2 (1 − cos a) cos b + (1 − cos2 b) cos a = 1 − cos2 c (cos a + cos b)(1 − cos a cos b) = 1 − cos2 c (cos a + cos b)(1 − cos c) = (1 − cos c)(1 + cos c) cos a + cos b = . 1 + cos c =

Exercise 10.22. Let ∆ABC be a spherical triangle. Let the altitude at A be AD and let h = |AD|. Prove that 1 − cos ∆ =

sin h(1 − cos a) . (1 + cos b)(1 + cos c) 2

What formula in Euclidean geometry is approximated by this formula for small values of a, b, and c? [H][A] Solution. Note the correction (made in the second printing). Let x = ±|BD |where we take ‘+’ if D and C are on the same side of B, and —’ otherwise. Then| DC |= (a ± x), − where the sign depends on whether x < a or not (which corresponds to whether D and B are on the same side of C or not). Let ∆1 = ±| ∆ABD | and ∆2 = ±|∆ADC where the sign ‘ ± ’ | is determined by whether the triangle has the same orientation as ∆ABC or not. Then ∆ = ∆1 + ∆2 and cos ∆ = cos(∆1 + ∆2) = cos ∆1 cos ∆2 − sin ∆1 sin ∆2 =

(cos x + cos h)(cos(a − x) + cos h) (sin x sin h sin(a − x) sin h) − . (1 + cos b)(1 + cos c) (1 + cos b)(1 + cos c)

Note that the signs are correct. Though x or a − x might be negative (so we may have incorrectly used Exercise 10.22), their signs are the same as

243

10.5 HERON’S FORMULA the signs of ∆1 and ∆2, respectively. Continuing, we have =

cos x cos(a − x) + cosh h cos(a − x) + cos x cos h + cos2 h (1 + cos b)(1 + cos c) 2 sin x sin(a − x) sin h — (1 + cos b)(1 + cos c)

cos x cos(a − x) cos2 h + cos x cos(a − x) sin h (1 + cos b)(1 + cos c) 2 2 cos b + cos c + 1 − sin h − sin x sin(a − x) sin h + (1 + cos b)(1 + cos c) 2 2 cos b cos c + cos a sin h + cos b + cos c + 1 − sin h = (1 + cos b)(1 + cos c) 2 sin h(cos a − 1) =1+ . (1 + cos b)(1 + cos c) 2

Thus,

sin h(1 − cos a) . (1 + cos b)(1 + cos c) 2

1 − cos ∆ =

For small a, b, and c, we have small ∆, so we get 1

(h + ...) (1 − 1 + a /2 + ...) (1 ∆2 + ...) = — − (1 + 1 + ...)(1 + 1 + ...) 2 2 2 ∆ (h + ...)2(a + ...) + ... = . 2 (4 + ...) 2

Ignoring the higher order terms, we get 2

∆ = ∆=

1 2

ha 2

ha.

Thus, we get the ‘one half base times height’ formula for the area of a triangle. Exercise 10.23. Using the proof of Heron’s formula in Euclidean geometry as a guide (see page 39 [of the text]), use the Law of Cosines and Exercise 10.22 to ﬁnd a formula for 1 − cos ∆ which depends only on the sides a, b, and c. Solution. Using the proof of Theorem 1.10.4 as a guide, we ﬁrst express h in terms of b and sin A. Note that sin A =

sin h sin b

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CHAPTER 10. SPHERICAL GEOMETRY

so sin h = sin b sin A and sin b sin A(1 − cos c) (1 + cos a)(1 + cos b) 2 2 sin b sin A(1 − cos c)(1 + cos c) = (1 + cos a)(1 + cos b)(1 + cos c) 2 2 2 sin b sin A sin c . = (1 + cos a)(1 + cos b)(1 + cos c) 2

1 − cos ∆ =

But, using the Law of Cosines, sin A = 1 − cos A cos a − cos b cos c =1− sin b sin c 2

so 1 − cos ∆ =

sin b sin c − (cos a − cos b cos c)2 . (1 + cos a)(1 + cos b)(1 + cos c) 2

(10.1)

Since this formula depends only on a, b, and c, we are done. Exercise 10.24. Prove the following trig identity, which is found in the inside front or back cover of most calculus texts: cos α − cos β = −2 sin

α+β 2

sin

α−β . 2

Solution. This follows from the angle sum formula. cos(r + s) = cos r cos s − sin r sin s cos(r − s) = cos r cos s + sin r sin s. Taking the diﬀerence, we get cos(r + s) − cos(r − s) = −2 sin r sin s. Thus, we set r + s = α and r − s = β, which gives α−β α+β and s= , r= 2 2 so α−β α+β sin cos α − cos β = −2 sin 2 2

Exercise 10.25. Finish the proof of Heron’s formula on the sphere by reducing the answer found in Exercise 10.23. Use the proof of Heron’s formula in Euclidean geometry as a guide, and when that is no longer helpful, use the trig identity in Exercise 10.24.

10.6 TILINGS OF THE SPHERE

245

Solution. Continuing where we left oﬀ in (10.1), we have sin b sin c − (cos a − cos b cos c)2 (1 + cos a)(1 + cos b)(1 + cos c) (sin b sin c − cos a + cos b cos c)(sin b sin c + cos a − cos b cos c) = (1 + cos a)(1 + cos b)(1 + cos c) (cos(b − c) − cos a)(cos a − cos(b + c)) = (1 + cos a)(1 + cos b)(1 + cos c) 4 sin b−c+a sin b−c−a sin a−b−c sin a+b+c 2 2 2 2 = (1 + cos a)(1 + cos b)(1 + cos c) 4 sin(s − c) sin(b − s) sin(a − s) sin s = (1 + cos a)(1 + cos b)(1 + cos c) 4 sin s sin(s − a) sin(s − b) sin(s − c) = . (1 + cos a)(1 + cos b)(1 + cos c)

1 − cos ∆ =

10.6

Tilings of the Sphere

Exercise 10.26. What is the length of the edges of the triangles and pentagons in the semiregular tiling of the unit sphere which has two triangles and two pentagons at each vertex? [A][S] Exercise 10.27. Find the percentage of the area of the sphere that is covered by triangles if the sphere has a semiregular tiling with two triangles and two pentagons at each vertex. [S] Exercise 10.28. What percentage of the sphere is covered by squares in the semiregular tiling with representation (3, 4, 3, 4)? Solution. Let s be the lengths of the sides of each tile and let r be the circumradius of the squares. Note that the equator through a diagonal of a square is the union of four diagonals, so 8r = 2π. Thus, by the Pythagorean theorem (since the angles between the diagonals are right angles), we have 1 2 2 cos s = cos r = cos (π/4) = . 2 Thus s = π/3. Let θ be a base angle of one of the isosceles triangles with base s and equal sides r. Since the other angle is π/2, we can use trig, so √ sin r 2 sin θ = = √ . sin s 3 The angles in the square are 2θ, so the area of each square is 8θ − 2π. There are six squares and the sphere has area 4π, so the ratio of the sphere covered by squares is 6(8θ − 2π) θ = 12 − 3 ≈ 64.9%. 4π π

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CHAPTER 10. SPHERICAL GEOMETRY

Exercise 10.29*. Find the percentage of the area of the soccer ball which is black. That is, ﬁnd the percentage of the area of the soccer ball which is covered with pentagons. Solution. The ﬁrst step is to get out a soccer ball and notice that, if we let the ‘North Pole’ be the center of a pentagon, then the equator bisects two sides of a hexagon that are one side apart, and that the equator is the union of ten of these segments. In the absence of a soccer ball, we refer the reader to Figure 5.2. An abstract rendering of these segments is shown in Figure 10.5 – abstract in the sense that the diagram is Euclidean but is supposed to represent a spherical hexagon. . C γ

c O

Figure 10.5 With the labeling of Figure 10.5,| AB| = π/5. Let D be the midpoint of AB, so AD | =| π/10. Let O be the center of the hexagon, so ∠ODA = π/2. Further, ∠AOD = π/3, so sin(π/10) sin c sin(π/10) sin c = . sin(π/3)

sin(π/3) =

We are interested in the angle γ = ∠ACO, which we now have enough information to calculate using the Law of Cosines for angles. Note that ∠CAO = π/2 and ∠COA = π/6, so cos γ = − cos(π/2) cos(π/6) + sin(π/2) sin(π/6) cos c =

1 2

sin (π/10) − 2 sin (π/3)

γ ≈ 1.085. The angles in the hexagons are 2γ, so the angles in the pentagon are 2(π − 2γ) = 2π − 4γ. Thus, the percentage of the soccer ball that is black is 12(5(2π − 4γ) − 3π) 3(7π − 20γ) γ = = 21 − 60 ≈ 28.2%. 4π π π

247

10.7 THE AXIOMS

10.7

The Axioms

Exercise 10.30*. Show that every direct isometry of the sphere has at least two fixed points. Conclude that there are no translations on the sphere. Exercise 10.31. Find an isometry of the sphere which has no fixed points. Write this isometry as a product of reflections and rotations. Solution. One such isometry is the map f (x, y, z) = (−x, −y, −z), 2

where (x, y, z) is a point on the sphere x + y + z = 1. That is, f sends a point to its antipodal point. This map is neither a reﬂection or rotation, but we can think of f as the composition f = g ◦ h where g(x, y, z) = (x, y, −z) h(x, y, z) = (−x, −y, z). The map g is reﬂection through the equator, and the map h is rotation through 180◦ about the North Pole (0, 0, 1).

10.8

Elliptic Geometry

Exercise 10.32. Do the separation axioms follow from the other axioms of spherical geometry? That is, are they necessary or superfluous? Exercise 10.33. What is the difference between spherical and elliptic geometry which makes Axiom 3 (with the given definition of ‘same side’) succeed in one and fail in the other? The isometries on P are those induced by the isometries on S. Exercise 10.34. Find a nontrivial isometry of S which induces the identity on P. Solution. The map is f (x, y, z) = (−x, −y, −z) 2

on the sphere x + y + z = 1. It is the map that sends a point P on S to its antipodal point. It is not the identity on S, but is the identity on P since P is constructed by identifying antipodal points on the sphere. That is, P and its antipodal point P ′ on S are the same point on P. Exercise 10.35. Suppose Φ is an improper isometry of S which induces the isometry φ of P. Show that there exists a proper isometry Φ ′ of S which induces the same isometry φ of P.

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CHAPTER 10. SPHERICAL GEOMETRY

Solution. Suppose Φ is an improper isometry. Note that f (x, y, z) = (−x, −y, −z) is also an improper isometry of S, since it can be thought of as the product of a reﬂection and a rotation (see Exercise 10.31). But f is the identity on P (see Exercise 10.34), so if Φ is an improper isometry on S that induces φ on P , then f ◦ Φ is a proper isometry of S that induces the same map φ on P.

Chapter 11

Projective Geometry 11.1

Moving a Line to Inﬁnity

Exercise 11.1. Prove Desargues’ theorem by moving a line to inﬁnity. P

A′ R

C′ B′ A

S C B T

Figure 11.1

Solution. Let us move the line RS to infinity and consider the resulting figure in Euclidean geometry (that is, with a metric). Since R is at infinity, the lines BC and B′C′ are parallel. Thus, by considering ∆PBC and 249

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CHAPTER 11. PROJECTIVE GEOMETRY

∆PB ′ C ′ , we have

|PC| |PB| = . ′ |PB | |PC ′ | Since S is at infinity, the lines AC and A′C′ are parallel, so looking at ∆P AC and ∆P A′C′, we see that |PC| |P A| = . ′ |P A | |PC ′ | Combining these two results, we get |PB| |P A| = , ′ |PB′| |P A | so AB is parallel to A′B′. That is, the point T is at infinity, so the line through R and S also goes through T . Exercise 11.2. Given a convex quadrilateral ABCD in a plane and a point O not in the plane, show that there are points A′ on OA, B′ on OB, C′ on OC, and D′ on OD such that A′B′C′D′ is a parallelogram. Solution. This question really asks whether the reader understands how we are moving a line to infinity. Let AB and CD intersect at P , and let AD and BC intersect at Q. If we move PQ to infinity, then the lines AB and CD become parallel, and the lines AD and BC become parallel. What does this really mean? We view the line AB as representing the plane that contains O, A, and B. The plane that contains O, C, and D represents the line CD. These two planes intersect in a line that contains O and P . Similarly, the plane through O, A, and D, and the plane through O, B, and C, intersect in the line OQ. The two lines OP and OQ define a plane p. Let us choose any plane p′ /= p that is parallel to p. This has the effect of moving PQ to infinity. Note that neither OP nor OQ intersect p′, since these lines are parallel to p′. Let OA intersect p′ at A′, let OB intersect p′ at B′, etc. Then the lines A′B′ and C′D′ are parallel, since OP does not intersect p′. Similarly, A′D′ and B′C′ are parallel. Hence A′B′C′D′ is a parallelogram, as desired. We did not use the fact that ABCD is convex. I’m not sure why I included it. Perhaps I was originally thinking of asking for points on the rays OA, OB, OC, and OD. Then we do need convexity.

11.2

Pascal’s Theorem

11.3

Projective Coordinates

Exercise 11.3. In this exercise, we are asked to render a perspective drawing of a cube. The points (x, y, z) where x ∈ {2, 3}, y ∈ {3, 4}, and z ∈ {1, 2} form a cube. Project these points on the plane x = 1.

251

11.3 PROJECTIVE COORDINATES

(0,0)

(1,0)

Figure 11.2 Solution. Note that the projection of the point (x, y, z) onto the plane x = 1 is the point y z . 1, , x x Thus, the square projected onto this plane has coordinates {(1, 3/2, 1/2), (1, 2, 1/2),(1, 3/2, 1), (1, 2, 1), (1, 1, 1/3), (1, 4/3, 1/3), (1, 1, 2/3), (1, 4/3, 2/3)}. Dropping the x-coordinate, we get the coordinates in the plane {(3/2, 1/2), (2, 1/2), (3/2, 1), (2, 1), (1, 1/3), (4/3, 1/3), (1, 2/3), (4/3, 2/3)}. The appropriate edges between these points are plotted in Figure 11.2. Exercise 11.4. Prove that three points P , Q, and R ∈ P are collinear if and only if P0 P1 P2 det Q0 Q1 Q2 = 0. R0 R1 R2 2

Solution. Three points in P are collinear if they lie in the same plane through the origin. If we think of their coordinates as representing vectors, then these three points are collinear if and only if the vectors are dependent. That is, they are collinear if and only if there exists a nontrivial solution to the equation aP + bQ + cR = (0, 0, 0) P0 P1 P2

Q0 Q1 R2

R0 a 0 R1 b = 0 . Q2 c 0

Such a matrix equation has a nontrivial solution if and only if the matrix is not invertible. That is, if and only if its determinant is zero.

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CHAPTER 11. PROJECTIVE GEOMETRY

Exercise 11.5. Prove that the equation of the line going through P and Q is x y z det P0 P1 P2 = 0. Q0 Q1 Q2 Solution. By the previous exercise, three points P , Q, and (x, y, z) in P are collinear if and only if x det P0 Q0

y P1 Q1

z P2 = 0. Q2

Thus, this is the equation of the line going through P and Q. Exercise 11.6. Prove that the intersection of two lines A0x + A1y + A2z = 0 B0 x + B1 y + B2 z = 0 is the point P=

det

A1 B1

A2 A2 , det B2 B2

A0 A , det 0 B0 B0

A1 B1

Solution. Let us think of these lines as planes in R . Their intersection is a F = (A0 , A1 , A2 ) is perpendicular to the line l through the origin. Note that A F . Similarly, l is perpendicular ﬁrst plane, so the line l is perpendicular to A F to B = (B0 , B1 , B2 ). Thus, l is parallel to a vector that is perpendicular F and B F . The cross product A F×B F is one such vector. Their are to both A a number of characterizations of the cross product, one of which is F ×B F = A

det

A1 B1

A A2 , det B2 B2 2

A0 A0 B0 , det B0

A1 B1

Finally, the two lines intersect at a point P in P that has homogeneous coordinates equal to the coordinates of any non-zero point on the line l, so F ×B F. P can be represented by A Exercise 11.7. What is the general form of a homogeneous polynomial 2 equation of degree two in P ? Solution. The general form of a homogeneous polynomial of degree two in 2 P is 2 2 2 F (x, y, z) = ax + bxy + cxz + dy + eyz + fz . Exercise 11.8. The equation f (x, y) = y − x + x + 1 = 0 2

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11.3 PROJECTIVE COORDINATES

is a polynomial equation in the affine plane. Find a homogeneous polynomial F (x, y, z) with the property that F (x, y, 1) = f (x, y). Does this new equation have any points at infinity? Decide how to find the tangent line of F (x, y, z) at some point P on the curve F (x, y, z) = 0. What is(are) the tangent line(s) to this curve at the point(s) at infinity? Solution. Note the clarification f (x, y) = 0, included in the second printing. The homogeneous form of f is F (x, y, z) = y z − x + xz + z . 2

The aﬃne plane is represented by z = 1, and the line at inﬁnity is given by z = 0. When z = 0, we get F (x, y, 0) = −x = 0, 3

so x = 0. Thus, this equation has the solution (0, 1, 0) in P corresponding to the point at infinity. Let us ponder how one might find the tangent line to F (x, y, z) = 0 2 3 in P . In R , we have a surface. We note that the tangent plane to this surface always goes through the origin, since F has homogeneous coordi2 nates. Thus, the tangent plane describes a line in P . So this might be a good way of defining the tangent line. Let us find this line for a point 2 (x0, y0, z0) ∈ P . To find the tangent plane, let us first find the gradient of F , which we know is perpendicular to the tangent plane: ∂F ∂F ∂F , , ∂x ∂y ∂z

∇F = =

−3x + z , 2yz, y + 2xz + 3z

Thus, the equation of the tangent plane at (x0, y0, z0) is 2

−3x0 + z0 , 2y0z0, y0 + 2x0z0 + 3z0 · (x − x 0 ,y − y 0 ,z − z0) = 0 2 2 2 2 (−3x + z )(x − x0) + 2y0z0(y − y0) + (y + 2x0z0 + 3z )(z − z0) = 0. 0

(11.1)

This doesn’t look like it’s homogeneous, as expected, so let us check that this plane really goes through (0, 0, 0): 3x 0 − z 0x0 − 2y 0z0 − y z00 − 2x0z − 0 3z = 0 3x −0 3x0z −03y z0 0− 3z 3

= −3(y0z0 − x 0 + x0z 0− z )0 = −3F (x0, y0, z0) = 0. 2

Let us now check that this definition is consistent with the definition of the tangent line for f (x, y) = 0. To find the tangent line at (x0, y0) to f (x, y) = 0, we first use implicit differentiation: 2yy′ − 3x + 1 = 0 2

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CHAPTER 11. PROJECTIVE GEOMETRY 2

y′ =

3x − 1 . 2y

Thus, the equation of the tangent line at (x0, y0) is given by 2 3x − 1 y − y0 = 0 2y (x − x0). 0

To compare this with the answer above, let us set z0 = 1 and z = 1. Then (11.1) gives us (−3x0 + 1)(x − x0) + 2y0(y − y0) = 0 2

y − y0 =

3x 0 − 1 (x − x0). 2y0 2

Thus, the tangent line to F (x, y, z) = 0 in P at the point (x0, y0, 1) is the same as the tangent line to the affine equation f (x, y) = 0 at (x0, y0). But now, we have a way of finding the tangent line at infinity. Recall, the only point at infinity on F (x, y, z) = 0 is the point (0, 1, 0). The tangent line there is 0(x − 0) + 0(y − 1) + (z − 0) = z = 0. That is, the tangent line at infinity is the line at infinity. Exercise 11.9. Prove that the definition for the cross ratio given in this section is consistent with the definition given in Chapter 7. Solution. Let us first check this for affine values a, b, c, and d. In projective coordinates, these are A = (a, 1), B = (b, 1), C = (c, 1), and D = (d, 1), so the projective cross ratio is (A, B; C, D) = ((a − c)(b − d), (a − d)(b − c)), which in affine coordinates gives (a − c)(b − d) a − c , b − c = = (a, b; c, d), (a − d)(b − c) a−d b−d provided a d and b /= c. If either a = d or b = c, then (a − c)(b − d) = 0 since at least three of a, b, c, and d are distinct. Then (A, B; C, D) = (1, 0), which is what we call ∞. In the affine case, if a = d or b = c, then (a, b; c, d) = ∞, so they agree. Now suppose a = ∞. Such a symbol has coordinates (1, 0). Then (A, B; C, D) = (1(b − d), 1(b − c)), which in affine coordinates is b−d . b−c The affine cross ratio is ∞− c ,b − c b−d (∞, b; c, d) = = = (A, B; C, D). ∞− d b − d b−c The rest of the cases are similar.

255

11.4 DUALITY

11.4

Duality

Exercise 11.10. Find the dual theorem to Pappus’ theorem. Solution. The dual of Pappus’ theorem is the following (follow along in Figure 11.3):

p2 p3 P

A2 B1

A1 B2

A3 Q q3

q2 q1

Figure 11.3 Theorem (Dual of Pappus’ theorem). Suppose three lines p1, p2, and p3 go through a point P, and three lines q1, q2, and q3 go through a point Q. Let p2 and q3 intersect at A1; let p3 and q1 intersect at A2, and let p1 and q2 intersect at A3. Let p3 and q2 intersect at B1; let p1 and q3 intersect at B2, and let p2 and q1 intersect at B3. Then the lines A1B1, A2B2, and A3B3 are coincident. The point of coincidence is labeled R in Figure 11.3. Exercise 11.11. Come up with new definitions of rotations and translations that are duals of each other, are well defined in Euclidean, hyperbolic, and elliptic geometry, and are consistent with our definitions of rotations and translations in Euclidean geometry. [A] Exercise 11.12* (Sylvester’s Problem). Suppose a set of n points in the plane has the property that a line through any two points goes through a third. Prove that the points are collinear.

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CHAPTER 11. PROJECTIVE GEOMETRY

Solution. Though it is possible to prove this result directly, there is a proof that uses duality. Some technical details first. We can dualize a problem in projective geometry. This problem is presumably meant to be in Euclidean geometry, though it is equally valid in projective geometry. Its dual reads: “Suppose a set of n lines in the projective plane has the property that the point of intersection of any two is on another line. Then the lines all go through a single point.” We will solve this dual problem by thinking of it as a problem in Euclidean geometry. To do so, we must first remove a line, but we should be careful about which line we remove. Since there are only a finite number of points of intersection, it is possible to find a line (not in the set) that goes through none of these points. Let us remove this line. That is, we are letting it be the line at infinity. Now our dual problem reads: “Suppose a set of n lines in the (Euclidean) plane has the property that the point of intersection of any two is on another line and no two lines are parallel. Then the lines all go through a single point.” This seemingly additional property that no two lines are parallel arises because we chose the line at infinity to be a line that contains no points of intersection. One reason we want to look at this dual problem in Euclidean geometry is because we can then use the metric. Let us first assume that there exists a line and a point of intersection not on that line. Since there are a finite number of lines and a finite number of points of intersection, there must be a line l and a point A such that the distance between A and l is minimal (but nonzero). Let Q be the point on l such that AQ is perpendicular to l. There are at least three lines through A. Let these lines intersect l at B, C, and D (recall, none of these lines are parallel to l, so they all intersect it). There are two possibilities that we will investigate. Either two of these points are on one side of Q, or Q is one of these points and the other two points are on either side of Q. Let us first suppose that B and C are on the same side of Q, and that B is between Q and C, as in Figure 11.4. Then B is closer to the line AC than A is to l, contradicting A

Figure 11.4 the way we chose A and l. Now, suppose B = Q and that B is between C and D. Since there are at least three lines (in the set) that pass through B, there must be a line through B that is neither AB nor l. This line enters

11.5 DUAL CONICS AND BRIANCHON’S THEOREM

257

∆ACD, so must intersect either AC or AD. Without loss of generality, we may assume it intersects AC. But then B is closer to AC than A is to l, again contradicting how we chose A and l. Thus, such an A and l cannot exist. Hence, the points of intersection must all lie on a single line. Let us now assume that there are at least two points of intersection. Since there are at least three lines through each point and because every pair of lines intersect, there is another point not on the line through these two points. Thus, there cannot be two points of intersection. That is, all the lines go through a single point.

11.5

Dual Conics and Brianchon’s Theorem

Exercise 11.13. This question requires some knowledge of linear algebra. We may write a general polynomial equation of degree two in the form 2

F (x, y, z) = ax + by + cz + 2exy + 2fyz + 2gxz = 0, which we may in turn write in the form x

a e g x e b f y = 0. g f c z

Let us deﬁne a e g A= e b f . g f c Let us also write X = (x, y, z), so T

F (x, y, z) = X AX. Since A is symmetric, we know that it is diagonalizable. Prove that the 2 curve F (X) = 0 describes a nondegenerate conic in P (R) if and only if two of the eigenvalues of A are positive and one is negative, or if one is positive and two are negative. Prove that the tangent line at P = (P0, P1, P2) is the line T P AX = 0. Finally, prove that the dual conic to F (X) = 0 is the conic T

X A− X = 0. Solution. First, let me apologize for using f in two different ways in the first printing (as a function and as a coefficient of the polynomial). The exercise was also slightly modified in the second printing. Suppose the eigenvalues of A are λ1, λ2, and λ3. Then there exists a change of basis matrix Q such that λ1 0 0 T Q AQ = 0 λ2 0 . 0 0 λ3

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CHAPTER 11. PROJECTIVE GEOMETRY 1

Let U = Q− X = (u, v, w). With these new coordinates, 2

F (x, y, z) = λ1u + λ2v + λ3w . For nontrivial U , this can be zero if and only if at least two of the eigenvalues have opposite signs. The equation F = 0 may have solutions when one of the eigenvalues is zero but in this case, the conic degenerates into two lines and A is not invertible. To find the tangent line at P , we must first find the gradient of F : ∂F ∂F ∂F , , ∂x ∂y ∂z = (2ax + 2ey + 2gz, 2ex + 2by + 2fz, 2gx + 2fy + 2cz)

∇F =

= 2AX. Thus, the gradient at P is 2AP and the tangent line at P is given by (2AP ) (X − P ) = 0 T

P A (X − P ) = 0 T

P AX − P AP = 0 T P AX = 0. T

In the simpliﬁcation above, we have used the facts that A is symmetric so T T A = A, and that P is a solution to F so P AP = 0. The dual of a conic is, by deﬁnition, the locus of the duals of the tangent T lines to the conic. The dual of the tangent line P AX = 0 is the point T T (P A)T = A P = AP . We note that T

P AP = 0, so T

P AA− AP = 0 T

(AP ) A− (AP ) = 0. Thus, the dual conic is X A− X = 0.

11.6

Areal Coordinates

Exercise 11.14. What are the areal coordinates of I, the incenter of ∆ABC? The solution is in the text.

259

11.6 AREAL COORDINATES A

C′

B′ G

D′

D D′′

Figure 11.5 Exercise 11.15. Show that the centroid G has areal coordinates G = (1/3, 1/3, 1/3).

Solution. In ∆ABC, let G be the centroid, let D be the base of the altitude at A, and let C′ be the midpoint of AB, as in Figure 11.5. In ∆C′BC, let D′ be the base of the altitude at C′, and in ∆GBC, let D′′ be the base of the altitude at G. Then, since ∆ABD ∼ ∆C′BD′ with |AB| = 2|C′B|, we get |AD| = 2|C′D′|. Since ∆C′CD′ ∼ ∆GCD′′ with |C′C| = (3/2)|GC|, we get |C′D′| = (3/2)|GD′′|. Thus, 1 |GD′′| = 3 |AD|, so

1 1 1 |∆GBC| = 2 |GD′′||BC| = 6 |AD||BC| = 3 |∆ABC|.

Similarly, |∆AGC| = coordinates of G are

|∆ABC| and |∆ABG| =

|∆ABC|, so the areal 3

11 1 , , 3 3 3

Exercise 11.16. Show directly that the orthocenter H has areal coordinates H = (cot B cot C, cot A cot C, cot A cot B). Conclude that, if A + B + C = 180◦, then tan A + tan B + tan C = tan A tan B tan C.

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CHAPTER 11. PROJECTIVE GEOMETRY

Solution. Let the orthocenter be H, and let the bases of the altitudes at A, B, and C be D, E, and F , respectively. We ﬁrst note that ∆BHD ∼ BCE, since they share the angle at B and both have right angles. Thus |HD| |BD|

|BC| |BE|

= cot C,

so the altitude of ∆BCH is |HD| = |BD| cot C. We note that |BD| = |AB| cos B, so |∆HBC| |∆ABC| =

2 |BC||HD| 1

2 |AD||BC|

|HD| = |AD| .

Note that the ratio on the left hand side is a signed ratio, so we should think of the ratio on the right hand side as a signed ratio too. This ratio is positive if H is between A and D, and negative otherwise. We will deal with the sign later. For now, we note that |HD| |BD| cot C |AB| cos B cot C = = = cot B cot C. |AD| |AD| |AB| sin B Note that the left hand side is positive if and only if both B and C are acute, in which case the right hand side is positive too. If either B or C is obtuse, then the other is acute, so the right hand side is negative, as desired. Thus, |∆HBC| = cot B cot C, |∆ABC| where the equality is valid for the signed ratio of areas. Thus, the ﬁrst areal coordinate of H is cot B cot C. Similarly, the other two coordinates are cot A cot C and cot A cot B, so H = (cot B cot C, cot A cot C, cot A cot B). This has the following interesting consequence. If A + B + C = 180◦, then there exists a triangle with A, B, and C as angles (take directed angles if necessary). Thus, by looking at the areal coordinates for H for this triangle, we get the relation cot B cot C + cot A cot C + cot A cot B = 1 cot A cot B cot C cot A cot B cot C cot A cot B cot C + + =1 cot A cot B cot C tan A + tan B + tan C = tan A tan B tan C. Exercise 11.17. Prove that the areal coordinates of the circumcenter O is O = u(sin 2A, sin 2B, sin 2C), where 1/u = sin 2A + sin 2B + sin 2C.

261

11.6 AREAL COORDINATES A

C′

B′ O R a — 2

A′

Figure 11.6 Solution. Let the circumradius of ∆ABC be R and follow along in Figure 11.6. By the Star Trek lemma, ∠BOA′ = ∠A, so 1 1 2 1 |∆OBC| = 2 |BC||OA′| = 2 (2 R sin A)R cos A = 2 R sin 2A. We have a similar result for| ∆OCA| and ∆OAB . Thus, in projective | | coordinates, O = (sin 2A, sin 2B, sin 2C). Normalizing so that the sum of the coordinates is one, we get the areal coordinates for O: O=

1 (sin 2A, sin 2B, sin 2C), u

where u = sin 2A + sin 2B + sin 2C. Exercise 11.18. What are the areal coordinates of the excenter Ia?

[A]

Solution. Let ra be the radius of the excircle. Then the areas of the three relevant triangles are 1 |∆Ia BC| = − ara 2 1 |∆AIaC| = bra 2 1 |∆ABIa | = cra , 2

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CHAPTER 11. PROJECTIVE GEOMETRY

so the projective coordinates of Ia are Ia = (−a, b, c). Normalizing so that the sum of the coordinates is one, we get the areal coordinates −a , b , c . Ia = s−a s−a s−a Exercise 11.19. Use areal coordinates to prove Menelaus’ theorem. Solution. Let us set up things as in Routh’s theorem: Let D, E, and F be points on the (extended) sides BC, AC, and AB respectively. Let λ=

|BD| |DC|

µ=

|CE| |EA|

and

ν=

|AF | |FB|

be signed ratios of lengths. Then, as seen in the proof of Routh’s theorem, the points D, E, and F have projective coordinates D = (0, 1, λ), E = (µ, 0, 1), F = (1, ν, 0). These three points are collinear if and only if 0 det µ 1

1 λ 0 1 =0 ν 0

(see Exercise 11.9). That is, these three points are collinear if and only if λµν + 1 = 0. Exercise 11.20. In the same way that Routh’s theorem is a generalization of Ceva’s theorem, there exists a generalization of Menelaus’ theorem. Find and prove it. Theorem (A generalizatoin of Menelaus’ theorem). Let D, E, and F be points on the (extended) sides BC, AC, and AB, respectively, of ∆ABC. Let λ=

|BD| |DC|

µ=

|CE| |EA|

and

ν=

|AF | |FB|

be signed ratios of lengths. Then the area of ∆DEF is given by |∆ DEF | =

(λµν + 1)|∆ABC| . (1 + λ)(1 + µ)(1 + ν)

263

11.6 AREAL COORDINATES

Proof. As was seen in the proof of Routh’s theorem, the points D, E, and F have projective coordinates D = (0, 1, λ), E = (µ, 0, 1), F = (1, ν, 0). We can convert to Areal coordinates by dividing the coordinates of each point by the sum of its coordinates. Thus, by Corollary 11.6.2, 0 det µ 1

1 λ 0 1 ν 0

|∆DEF | = |∆ABC| (1 + λ)(1 + µ)(1 + ν) λµν + 1 . = (1 + λ)(1 + µ)(1 + ν) This is a generalization of Menelaus’ theorem since we get as a corollary that D, E, and F are collinear if and only if the area is zero, which happens if and only if λµν = −1. Exercise 11.21. Let D, E, and F divide each of the sides BC, CA, and AB in the ratio of 2 to 1. Let P , Q, and R be the intersections of BE with CF , CF with AD, and AD with BE, respectively. Prove that the area of ∆P QR is one-seventh of the area of ∆ABC. Solution. The area of ∆P QR can be calculated using Routh’s theorem (note the error in the statement of Routh’s theorem in the ﬁrst printing). We are given that λ = µ = ν = 2, so |∆P QR| =

(8 − 1)2|∆ABC| |∆ABC| = . (4 + 2 + 1)3 7

Chapter 12

The Pseudosphere in Lorentz Space 12.1

The Sphere as a Foil

Exercise 12.1. Let P be a point on S. Define F · Fx)P F. RP (Fx) = Fx − 2(P Prove that RP (Fx) is reflection through the polar of P . Prove that RP is in O3(R) (or more precisely, show that the matrix which represents RP is in O3(R)). Conclude that O3(R) is the full group of isometries of S. Solution. Let us first show that RP is linear. Recall, this means we must show RP (aFx + Fy) = aRP (Fx) + RP (Fy). We begin with the left hand side, F · (aFx + Fy))P F RP (aFx + Fy) = (aFx + Fy) − 2(P F · Fx)P F − 2(P F · Fy)P F = aFx + Fy − 2(aP = aRP (Fx) + RP (Fy). Thus, RP is linear, so has a matrix representation, say T . To show that T T is in O3(R), we must show that T T = I, or equivalently, that RP (Fx) · RP (Fy) = Fx · Fy for all Fx and Fy. We expand: F · Fx)P F ) · (Fy − 2(P F · Fx)P F) RP (Fx) · RP (Fy) = (Fx − 2(P F · Fy)P F · Fx − 2(P F · Fx)P F · Fy + 4(P F · Fx)(P F · Fy)P F ·P F = Fx · Fy − 2(P 265

266

CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE F · Fx)(P F · Fy ) + 4(P F · Fx)(P F · Fy) = Fx · Fy − 4(P = Fx · Fy.

In the above, we used the fact that P lies on the sphere, so P· P = 1. Since RP is linear and preserves the dot product, the matrix that represents it is in O3(R). Exercise 12.2. Suppose T ∈SO3(R). Prove that T has exactly two fixed points, so is a rotation on S. Solution. First, we note that the claim is not true, since the identity I is in SO3(R), but has an infinite number of fixed points. However, this is the only exception. Suppose there exists Fx on the sphere such that T (Fx) = Fx. Then Fx is an eigenvector for T with eigenvalue 1. Furthermore, T (−Fx) = −T (Fx) = −Fx, so − Fx is also fixed by T . Thus, T has a fixed point if and only if 1 is an eigenvalue for T . Furthermore, if 1 is an eigenvalue for T , then T has at least two fixed points. Also, if T has exactly one eigenvalue equal to 1, then the vector space for this eigenvalue is one dimensional, so the intersection of this vector space with the sphere is two points. That is, if T has exactly one eigenvalue equal to 1, then T has exactly two fixed points. Suppose λ ∈ R is an eigenvalue of T . Then there exists a vector Fv on the sphere such that T (Fv) = λFv. Since T is an isometry, λFv is also on the sphere, so λ = ± 1. Thus, the eigenvalues of T are ±1 or are complex. Since T has real entries, the characteristic polynomial f (λ) = det(T − λI) has real coefficients and is of degree three. Thus, the complex roots come in conjugate pairs. Note also that f (0) = det T = 1, so the product of the roots is 1. Suppose one of the eigenvalues λ1 is complex. Then its 2 conjugate λ2 is also an eigenvalue, and λ1λ2 =|| λ1|| > 0. Thus, the third eigenvalue must be positive, so is 1. Hence, there is exactly one eigenvalue equal to 1 and T has exactly two fixed points. If none of the eigenvalues are complex, then they are all±1. If one of the eigenvalues is−1, then two of them are − 1 (since det T = 1), so again there is exactly one eigenvalue equal to 1. Finally, suppose all of the eigenvalues are 1. Then there are a couple of possibilities. Either the eigenspace is three dimensional, in which case T = I, or T is not diagonalizable. It turns out that every element of O3(R) is diagonalizable (over C), so this last case is not possible. We will use this result again in the next exercise, so let us state it as a lemma.

267

12.1 THE SPHERE AS A FOIL Lemma. Suppose T ∈ O3(R). Then T is diagonalizable over C.

Solution. Suppose T is not diagonalizable. Then there exists an eigenvalue λ of multiplicity n = 2 or 3, and whose associated eigenspace is of dimension strictly smaller than n. Since the characteristic polynomial for T is of degree three and has real coeﬃcients, the complex roots come in conjugate pairs. Thus, λ must be real. Associated to λ there is at least one eigenvector Fu. Without loss of generality, we may assume ||Fu|| = 1. Then TFu = λFu. Since T preserves the dot product (since T ∈ O3(R)), we also have TFu · TFu = Fu · Fu 2 λ = 1, whence λ = ± 1. Since the eigenspace for λ has dimension smaller than n, there must be a vector Fv such that TFv = λFv + Fu (think of the Jordan form for T ). Thus, k

k 1

T Fv = λ Fv + kλ − Fu, which is easy enough to verify using induction. The induction step is T Fv = T (λ − Fv + (k − 1)λ − Fu) k

k 1

k 2

= λ Fv + λ − Fu + (k − 1)λ − Fu k k 1 = λ Fv + kλ − Fu. k

k 1

But T preserves the dot product, so T Fv · T Fv = Fv · Fv = a , k

where a is the length of Fv. Expanding the left hand side, we get (λ Fv + kλ − Fu) · (λ Fv + kλ − Fu) = a k

k 1

2k 2

λ a + 2kλ

2k −1

k 1

Fu · Fv + λ

2k −2

2 2

=a .

Note that λ = ±1, so λ = 1. Thus, we get 2k

2kλFu · Fv = −1 −1 Fu · Fv = . 2kλ Since this must be the case for all integers k, we have a contradiction. Thus, no such Fv can exist, so the eigenspace associated to λ must have full dimension. Thus, T is diagonalizable.

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CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

Thus, if T ∈ SO3(R), then T is either the identity or it has exactly two fixed points. In either case, it is a rotation. Exercise 12.3. Suppose T ∈ O3(R) and det T = −1. Prove that T is not a rotation, so must be an improper isometry. Solution. As in the previous exercise, we look at eigenvalues. Since T is an isometry, the eigenvalues are either complex or ± 1. Since T has real coefficients, the complex eigenvalues come in conjugate pairs and the product of conjugate eigenvalues is positive. Note that T has three eigenvalues (counting multiplicity). Hence, if T has complex eigenvalues, then the third eigenvalue is negative and real (since det T = — 1), so must be− 1. In this case, there are no fixed points, so T is not a rotation. If T has no complex eigenvalues, then the eigenvalues are± 1. Since det T = −1, the product of these eigenvalues is—1, so the eigenvalues of T are 1, 1, and − 1, or they are all − 1. In the latter case, T has no fixed points, so is not a rotation. In the former case, T has a two dimensional fixed space (since T is diagonalizable – see the lemma in the previous solution) which corresponds to a line of fixed points on the sphere. Thus, T is not a rotation. Since all proper isometries on the sphere are rotations, T must be an improper isometry. Exercise 12.4. Show that the arclength element on S expressed in polar coordinates is given by 2

ds =

dr 2 2 + r dθ 2 1−r

(for r /= 1). Solution. Note the clariﬁcation made in the second printing. It’s still a little vague as to what is meant. A point on the sphere can be represented in cylindrical coordinates z, r, and θ. Since the sphere has radius one, r and z are related by 2 2 r + z = 1, so we can describe points on the upper half of the sphere using only r and θ. That this is valid on only half the sphere is a failing of this coordinate system, and this arclength element is not too useful since one cannot integrate over paths that cross the equator. The exercise is meant to be a foil for Exercise 12.15, where we have no such problems. We will ﬁnd the arclength element using these coordinates. Given small changes in x, y, and z, it is clear that the change ∆s in arclength is approximated by 2

∆s = ∆x + ∆y + ∆z 2

ds = dx + dy + dz . Recall that (x, y) = (r cos θ, r sin θ), so dx = cos θdr − r sin θdθ

269

12.1 THE SPHERE AS A FOIL dy = sin θdr + r cos θdθ, and dx + dy = cos θdr − 2r cos θ sin θdrdθ + r sin θdθ 2 2 2 2 2 + sin θdr + 2r sin θ cos θdrdθ + r cos θdθ 2

= dr + r dθ . Since 2

z + r = 1, we get 2zdz + 2rdr = 0. Thus, 2

dz = =

r dr z2 2 2 r dr 1 − r2

Combining, we get 2

ds = (dx + dy ) + dz

r dr 1 − r2 2 2 2 (1 − r + r )dr = + r2dθ2 2 1 − r 2 dr = + 2 2 r dθ . 2 1−r 2

= dr + r dθ +

Exercise 12.5. Recall that the location of a point P in R can be described using spherical coordinates. The spherical coordinates of P are (ρ, φ, θ), where | OP| = ρ, φ is the angle OP makes with the z-axis, and θ is the angle that the projection of OP onto the xy-plane makes with the x-axis. Recall also that the Cartesian coordinates (x, y, z) of P are given by (x, y, z) = (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ). Prove that the arclength element in three-space is ds =

ρ2 sin φdθ2 + ρ2dφ2 + dρ2.

Conclude that the arclength element on the sphere of radius one is ds =

sin φdθ2 + dφ2.

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CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

Solution. Since we are given x, y, and z in terms of ρ, φ, and θ, all we have to do is calculate the diﬀerentials dx, dy, and dz: dx = sin φ cos θdρ + ρ cos φ cos θdφ − ρ sin φ sin θdθ dy = sin φ sin θdρ + ρ cos φ sin θdφ + ρ sin φ cos θdθ dz = cos φdρ − ρ sin φdφ. Thus, 2

ds = dx + dy + dz

= sin φ cos θdρ + ρ cos φ cos θdφ + ρ sin φ sin θdθ + 2ρ sin φ cos φ cos θdρdφ − 2ρ sin φ cos θ sin θdρdθ 2

— 2ρ cos φ sin φ cos θ sin θdφdθ 2 2 2 2 2 2 2 2 2 2 2 + sin φ sin θdρ + ρ cos φ sin θdφ + ρ sin φ cos θdθ 2

+ 2ρ sin φ cos φ sin θdρdφ + 2ρ sin φ sin θ cos θdρdθ 2

+ 2ρ cos φ sin φ sin θ cos θdφdθ + cos φdρ + ρ sin φdφ − 2ρ cos φ sin φdρdφ 2 2 2 2 2 2 = (sin φ cos θ + sin φ sin θ + cos φ)dρ 2

+ (ρ cos φ cos θ + ρ cos φ sin θ + ρ sin φ)dφ 2

+ (ρ sin φ sin θ + ρ sin φ cos θ)dθ

+ (2ρ sin φ cos φ cos θ + 2ρ sin φ cos φ sin θ − 2ρ cos φ sin φ)dρdφ 2

+ (−2ρ sin φ cos θ sin θ + 2ρ sin φ sin θ cos θ)dρdθ 2

+ (−2ρ cos φ sin φ cos θ sin θ + 2ρ cos φ sin φ sin θ cos θ)dφdθ 2 2 2 2 2 2 = dρ + ρ dφ + ρ sin φdθ ds =

dρ2 + ρ2dφ2 + ρ2 sin φdθ2.

On the sphere, ρ = 1 and dρ = 0, so the arclength element on the sphere in spherical coordinates is ds =

dφ2 + sin φdθ2.

Exercise 12.6. The map − I is an isometry of S. Write it as a product of reﬂections in S. Solution. Note that the maps Rx(x, y, z) = (−x, y, z) Ry(x, y, z) = (x, −y, z) Rz(x, y, z) = (x, y, −z), are reﬂections through the yz-, xz-, and xy-planes, respectively. Also, RxRyRz(x, y, z) = RxRy(x, y, −z) = Rx(x, −y, −z) = (−x, −y, −z)

271

12.1 THE SPHERE AS A FOIL = −I(x, y, z) Exercise 12.7. Prove the following properties of the cross product: Fu × (Fv × w F ) = (Fu · w F )Fv − (Fu · Fv)w F (Fu × Fv) · (w F × Fx) = det

Fu · w F Fv · w F

Fu · Fx . Fv · Fx

In the following sections, we will define analogues of the dot and cross products. Solution. Note the corrections – the signs in both are wrong. These are corrected in the second printing. Let Fx = Fv × w F and Fy = u F × (Fv × w F ). Then y1 = u2x3 − u3x2 = u2(v1w2 − v2w1) − u3(v3w1 − v1w3) = (u2w2 + u3w3)v1 − (u2v2 + u3v3)w1 = (u2w2 + u3w3)v1 + u1v1w1 − (u2v2 + u3v3)w1 − u1v1w1 = (Fu · w F )v1 − (Fu · Fv)w1 . The cross and dot products are symmetric in the components by permuting the indices by the permutation (123). Fortunately, there is not much to permute, and we get y2 = (Fu · w F )v2 − (Fu · Fv)w2 y3 = (Fu · w F )v3 − (Fu · Fv)w3 , so Fu × (Fv × w F ) = (Fu · Fv)w F − (Fu · w F )Fv. For part two, let us first look at the contribution of the first components of each cross product: (u2v3 −u3v2)(w2x3 −w3x2) = u2w2v3x3 +u3w3v2x2 −u3x3v2w2 −u2x2v3w3. We permute the indices to get the contribution from the other terms, so (Fu × Fv)(w F × Fx) = (u1 w1 v2 x2 + u2 w2 v3 x3 + u3 w3 v1 x1 ) + (u1w1v3x3 + u2w2v1x1 + u3w3v2x2) — (u1x1v3w3 + u2x2v1w1 + u3x3v2w2) — (u1x1v2w2 + u2x2v3w3 + u3x3v1w1). We add and subtract the terms u1w1v1x1 and its permutations, so that this simplifies to (Fu × Fv)(w F × Fx) = (Fu · w F )(Fv · Fx) − (Fu · Fx)(Fv · w F) F = det Fu · w Fv · w F

Fu · Fx . Fv · Fx

272

12.2

CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

The Pseudosphere

Exercise 12.8. Prove that the arclength element on V is given by 1 2 ds = dx 2 + dy 2 + (ydx − xdy) 2 . [S] z2 Exercise 12.9. Let Fx(t) be a piecewise smooth curve on V + such that Fx(0) = N = (0, 0, 1) and Fx(t0 ) = P = (x0 , 0, z0 ). Prove that the arclength of Fx(t) for t ∈ [0, t0 ] is at least arccosh (z0 ). Conclude that the line through N and P is the intersection of V+ with the plane y = 0. Solution. As worded, this curve is described using rectangular coordinates. The argument is easier to see in cylindrical coordinates and using Exercise 12.15, in which we show dr2 2 2 2 ds = + r dθ . 1 + r2 Since the curve is smooth, its arclength is 2 ∫ t0 dr 1 S= ≥

∫ t0

∫ r(t0)

+ r2

1 + r2 2

1 1 + r2

dθ

dr √

≥

. 1 + r2 A technical detail: In the last step, we made the substitution r = r(t), which is valid as long as r(t) is invertible. If r(t) is not invertible, we will have to ‘chop’ up the interval [0, t0] into pieces where r(t) is invertible on each piece. The inequality above is valid on each interval, and over the entire interval, r ranges between 0 and r0, so the above inequality still holds. 2 Let us now make the substitution r = sinh θ, so dr = cosh θ and 1+r = 2 cosh θ. Thus, ∫ 0

arcsinh r0

S≥

dθ = θ 0

arcsinh r0

. 0

We note that r − z = −1, so when r = sinh θ, z = cosh θ. Thus 2

S ≥ arccosh z0. Finally, we note that the curve C given by (sinh t, 0 cosh t) for t ∈ [0, arccosh z0] is on V+, begins at 0, and ends at z0. It is the curve along the intersection of V+ with the xz-plane. Furthermore, its arclength is ∫ arccosh z0 2 2 d sinh t d cosh t dt − S= dt dt 0

273

12.2 THE PSEUDOSPHERE ∫ arccosh z0 √ 2

cosh t − sinh tdt 0

= arccosh z0. Thus every piecewise smooth curve from N to P is longer than C , so C is the line segment NP . Exercise 12.10. Let Rθ rotate V an angle θ about the z-axis. What is the + matrix representation of Rθ? Prove that Rθ ∈ OJ . [A] Solution. The matrix representation for Rθ is Rθ =

cos θ −sin θ 0

sin θ 0 cos θ 0 . 0 1

This map is in OJ since − sin θ cos θ 0

cos θ T R θ JRθ = sin θ 0 cos θ = sin θ 0

0 1 0 0 0 0 1 0 1 0 0 −1

− sin θ 0 cos θ cos θ 0 − sin θ 0 1 0 2

cos2 θ + sin θ = sin θ cos θ − cos θ sin θ 0 = J.

cos θ − sin θ 0

sin θ cos θ 0

0 0 −1

cos θ sin θ − sin θ cos θ 2 sin θ + cos2 θ 0

To check that Rθ ∈ OJ, we note that +

x Rθ y = z

x cos θ + y sin θ, −x sin θx + y cos θ, z ,

so the image of a point on V+ is again on V+. Exercise 12.11. Prove that the map Tφ =

cosh φ 0 sinh φ

0 sinh φ 1 0 0 cosh φ

is in OJ . Solution. We ﬁrst check that Tφ ∈ OJ : T

Tφ JTφ =

cosh φ 0 sinh φ

0 sinh φ 1 0 0 1 0 0 1 0 0 cosh φ 0 0 −1

cosh φ 0 sinh φ

0 0 1

0 sinh φ 1 0 0 cosh φ

0 0 −1

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CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

cosh φ 0 sinh φ 0 1 0 sinh φ 0 cosh φ 2

cosh φ 0 — sinh φ 2

cosh φ — sinh φ = 0 sinh φ cosh φ − cosh φ sinh φ = J.

0 sinh φ 1 0 0 − cosh φ

0 1 0

cosh φ sinh φ − sinh φ cosh φ 0 2 2 sinh φ − cosh φ

Thus, Tφ ∈ OJ . To show that Tφ ∈ O , Jwe need check only one point. We note that sinh φ 0 cosh φ 0 sinh φ 0 1 0 0 , 0 = sinh φ 0 cosh φ 1 cosh φ +

which is on V+ (since cosh φ > 0 for all φ). Thus, Tφ sends a point on V+ + to a point on V+, so must send all of V+ to V+. Thus, Tφ ∈ OJ . +. Use Exercises 12.10 and Exercise 12.12. Let P and Q be points on V + 12.11 to prove that there exists a∈T OJ such that TQ = N and TP = P ′ where the y component of P ′ is zero. Use Exercise 12.9 to conclude that the line NP ′ is the intersection ofV + with the plane y = 0. Conclude that the line PQ is the intersection ofV + with the plane through P , Q and the origin O.

Solution. It is clear that there exists an angle α such that the map Rα rotates Q to a point whose y-coordinate is zero. Let this new point be Q′ = (Q′x , 0, Q′z ). Recall that the surface V + can be parameterized by φ and θ via the map (cos θ sinh φ, sin θ sinh φ, cosh φ), so there exists a φ such that Q′ = (sinh φ, 0, cosh φ). We apply T−φ to Q′ to get T −φ

Q′ =

cosh φ 0 − sinh φ 0 1 0 −sinhφ 0 cosh φ

sinh φ 0 cosh φ

cosh φ sinh φ − sinh φ cosh φ 0 = 2 2 — sinh φ + cosh φ

0 = 0 = N. 1

Under the image of Rα and T−φ, the point P has been moved to a point, say, P ′′. There exists another angle β such that Rβ sends P ′′ to a point P ′ with y-component equal to zero. Note that Rβ ﬁxes N . Thus, the map T = Rβ T − φ R α

275

12.2 THE PSEUDOSPHERE

sends Q to N and P to a point P ′ with zero y-component, as desired. By Exercise 12.9, the (pseudospherical) line through N and P ′ is the intersection of V+ with the xz-plane. Now, suppose S′ is a point on this line. Since 1 T is an isometry, S = T − S′ is a point on the (pseudospherical) line P Q. Since T is a linear transformation, and S′ is on the xz-plane, the point S lies on a plane through the origin. Thus, the (pseudospherical) line PQ is the intersection of the surfaceV + and the plane through P , Q, and the origin. Exercise 12.13†. Let Fa ◦ Fa = 1. Prove that the plane Fa ◦ Fx = 0 intersects V. Deﬁne the map R→a (Fx) = Fx − 2(Fa ◦ Fx)Fa. +

Prove that the matrix which represents R→a is in OJ , and that it is reﬂection through the line described by the intersection of V + with the plane Fa ◦Fx = 0. + Conclude that, if V+ is a model of hyperbolic geometry, then OJ is the full + group of isometries of V . r2 – z2 = –1 r 2 – z2 = 1

(a1,a2,–a3)

Figure 12.1 Solution. To show that Fa ◦ VFx = 0 intersects let us use a picture of the cross section of Lorentz space that contains Fa and the z-axis. In cylindrical

276

CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

coordinates, the surface V is described by the equation r − z = −1, 2

and the surface V ∗ = Fx ◦ Fx = 1 is described by r − z = 1. 2

If Fa ◦ Fa = 1, then Fa is on V ∗ . Let us consider the equation of the plane: Fa ◦ Fx = 0 a1x + a2y − a3z = 0 (a1, a2, −a3) · (x, y, z) = 0. Thus the plane is perpendicular (in Euclidean R ) to the vector Fa∗ = (a1 , a2 , —a3 ), which also satisfies Fa∗ ◦ Fa∗ = 1. Now Figure 12.1 finishes the argument. We could explicitly find an Fx ∈ V + that satisfies Fa ◦ Fx = 0, but that wouldn’t be as enlightening. + Let us now show that R→a is in OJ . To show this, we must show that it is linear and that it preserves the Lorentz product. First, linearity: 3

R→a (Fx + rFy) = (Fx + rFy) − 2(Fa ◦ (Fx + rFy))Fa = Fx + rFy − 2(Fa ◦ Fx + rFa ◦ Fy)Fa = Fx − 2(Fa ◦ Fx)Fa + r(Fy − 2(Fa ◦ Fy)Fa) = R→a (Fx) + rR→a (Fy). The map preserves the Lorentz product: R→a (Fx) ◦ R→a (Fy) = (Fx − 2(Fa ◦ Fx)Fa) ◦ (Fy − 2(Fa ◦ Fy)Fa) = Fx ◦ Fy − 2(Fa ◦ Fx)Fa ◦ Fy − 2(Fa ◦ Fy)Fx ◦ Fa + 4(Fa ◦ Fx)(Fa ◦ Fy)Fa ◦ Fa = Fx ◦ Fy. +

Thus, the matrix that represents R→a is in O J. Furthermore, if Fx is on the line described by Fa ◦ Fx = 0, then R→a (Fx) = Fx − 2(Fa ◦ Fx)Fa = Fx, so R→a fixes Fx. Also, if R→a fixes a point Fx, then Fa ◦ Fx = 0, so Fx is on this line. A map that fixes a line and fixes no other points is called a reflection. Since all lines are described by Fa ◦ Fx = 0 for some Fa, we have a description of the full set of reflections onV+. Since reflections generate the full group + of isometries, we therefore know that OJ is the full group of isometries. +

Exercise 12.14. Suppose T∈O .J Prove that T is a proper isometry if and only if det T = 1. [H]

277

12.2 THE PSEUDOSPHERE

Solution. Let A, B, and C be three points on V+ and suppose ∆ABC is oriented counterclockwise (when looking down V+ from a point (0, 0, z0) F, B F , and C F that represent these points with z0 > 1). Then the vectors A in Lorentz space obey the right-hand rule. Recall that a linear transformation T preserves the orientation of vectors if and only if det T > 0. If after applying T , the images of these vectors still obey the right-hand rule, then the image of these points is still oriented counterclockwise. Thus, an + isometry T ∈ O preserves orientation if and only if det T > 0. Since the J determinant can be only±1, we have that T is a proper isometry if and only if det T = 1. Exercise 12.15. Show that the arclength element onVexpressed in polar coordinates is given by ds2 =

dr2

+ r 2dθ2 .

1 + r2 Solution. A point on the pseudosphere can be represented in cylindrical coordinates z, r, and θ. Since the pseudosphere has the equation x + y − z = −1, 2

r and z are related by r − z = −1, 2

so we can describe points on the upper half of the pseudosphereV + using only r and θ. We will ﬁnd the arclength element using these coordinates. The arclength element in rectangular coordinates is, by deﬁnition, ds = dx + dy − dz . 2

Recall that (x, y) = (r cos θ, r sin θ), so dx = cos θdr − r sin θdθ dy = sin θdr + r cos θdθ, and dx + dy = cos θdr − 2r cos θ sin θdrdθ + r sin θdθ 2 2 2 2 2 + sin θdr + 2r sin θ cos θdrdθ + r cos θdθ 2

= dr + r dθ . Since r − z = −1, 2

we get 2zdz − 2rdr = 0.

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CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

Thus, 2

r dr z2 2 2 r dr . = 1 + r2

dz =

Combining, we get ds = (dx + dy ) − dz 2 2 r dr 2 2 2 = dr + r dθ — 1 + r2 2 2 2 (1 + r − r )dr = + r2dθ2 1 + r2 2 dr = + 2 2 r dθ . 2 1+r 2

Exercise 12.16. Recall, the surface V can be parameterized by Fx(θ, φ) = (cos θ sinh φ, sin θ sinh φ, cosh φ). What is the arclength element on V expressed in terms of θ and φ? Solution. Since ds = dx + dy − dz , 2

we should calculate dx, dy, and dz: dx = − sin θ sinh φdθ + cos θ cosh φdφ dy = cos θ sinh φdθ + sin θ cosh φdφ dz = sinh φdφ. Thus 2

ds = sin θ sinh φdθ + cos θ cosh φdφ

— 2 sin θ cos θ sinh φ cosh φdθdφ 2 2 2 2 2 2 + cos θ sinh φdθ + sin θ cosh φdφ + 2 cos θ sin θ sinh φ cosh φdθdφ 2

— sinh φdφ 2 2 2 2 2 = (sin θ sinh φ + cos θ sinh φ)dθ 2

+ (cos θ cosh φ + sin θ cosh φ − sinh φ)dφ 2

+ (−2 sin θ cos θ sinh φ cosh φ + 2 cos θ sin θ sinh φ cosh φ)dθdφ 2 2 2 = sinh φdθ − dφ ds =

sinh φdθ2 − dφ2. 2

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12.2 THE PSEUDOSPHERE

Exercise 12.17. Suppose Fu = (u1 , u2 , u3 ) and Fv = (v1 , v2 , v3 ) are two vectors with u3 , v3 > 0 and Fu ◦ Fu < 0, Fv ◦ Fv < 0. Let us define φ ≥ 0 such that Fu ◦ Fv = ||Fu||||Fv|| cosh φ. Prove that φ is well defined and depends only on the directions of Fu and Fv, and not on their magnitudes. Solution. Note the correction (corrected in the second printing). We should first make a remark about the notation ||Fu|| =

√ Fu ◦ Fu.

√ When we write a for a > 0, we mean the positive square root. In the √ above notation, we have to choose a branch for − a when a > 0. It doesn’t matter which branch we take, as long as we are consistent, but convention is that √ √ −a = i a. Fu ◦ Fv As for the exercise, let us ﬁrst show that the quantity depends ||Fu||||Fv|| only on the direction of u and F Fv, and not on√ their lengths. To see this, consider w = F λFu for λ > 0. Note that ||w F || = λ2Fu ◦ Fu = λ||Fu||. Thus w F ◦ Fv ||w F ||||Fv||

λFu ◦ Fv λ||Fu||||v F ||

Fu ◦ Fv ||Fu||||Fv||

Hence this quantity does not depend on the length of Fu, and by symmetry, it does not depend on the length of Fv either. In particular, we can pick Fu and Fv on V + . By Exercise 12.12, there exists an isometry T such that TFu = (0, 0, 1). Let TFv = (v1′ , v2′ , v3′ ). Then Fu ◦ Fv = T (Fu) ◦ T (Fv) = −v3′ < −1. Since ||Fu||||Fv|| = −1, we get Fu ◦ Fv ||Fu||||Fv||

≥ 1.

Thus, φ is well deﬁned, since the map cosh φ for φ ∈ [0, ∞) is invertible on [1, ∞). Exercise 12.18. We have discovered that the distance between two points P and Q on V+ is given by F ◦Q F ). |P Q| = arccosh (−P Prove directly that this distance function satisﬁes the triangle inequality.

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12.3

CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

Angles and the Lorentz Cross Product

Exercise 12.19. Suppose ||Fa|| = ||Fb|| = 1. Prove that there exists a T ∈ + OJ such that TFa = (1, 0, 0) and T Fb = Fb′ = (b′1 , b′2 , 0). [H] Solution. There exists a rotation Rα such that the image of Fa has a ycomponent equal to zero. That is, there exists an α such that RαFa = (a′1 , 0, a′3 ). Since Rα preserves the Lorentz product, we have (a′1) − (a′3) = 1. 2

Thus there exists a φ such that a′1 = cosh φ and a′3 = sinh φ. Let us apply T−φ to Fa′ to get T−φ RαFa =

cosh φ 0 − sinh φ

0 1 0

− sinh φ 0 cosh φ

cosh φ 0 sinh φ

cosh φ − sinh2 φ 1 0 = 0 0 — sinh φ cosh φ + cosh φ sinh φ 2

= Let

1 Sψ = 0

0 cosh ψ 0 sinh ψ

0 sinh ψ . cosh ψ +

Since everything is symmetric in x and y, this matrix is in O J for the same reasons that Tφ is. Let us apply Sψ to Fb′′ : 1 0 Sψ Fb′′ = 0 cosh ψ 0 sinh ψ =

b′1′ 0 sinh ψ b′2′ cosh ψ b′3′

b′1′ ′ b′2 cosh ψ + b′3′ sinh ψ . b′2′ sinh ψ + b′3′ cosh ψ

We want the last component to be zero, so we solve b′2′ sinh ψ + b′3′ cosh ψ = 0 b′2′ sinh ψ = −b′3′ cosh ψ −b′3′ tanh ψ = . b′2′

281

12.3 ANGLES AND THE LORENTZ CROSS PRODUCT

We note that tanh x is invertible on R, so there exists a solution ψ as long as b′2′ = 0. If b′2′ = 0, then the image of Fb′′ under Sψ is a vector Fb′ = (b′1 , b′2 , 0) whose z-component is zero. We also note that 1 0 Sψ T−φ RαFa = 0 cosh ψ 0 sinh ψ

0 1 1 sinh ψ 0 = 0 . cosh ψ 0 0

Thus, we set T = S ψ T− φ Rα . 1 If b′′2 = 0, then let us backtrack. We note that the y-component of T −φ Fb′′ is also zero, so we must have had (a1 , a2 ) = λ(b1 , b2 ) for some λ. Since Fa and Fb both have length one, we know λ 0. In this case, let us apply Tµ to Fa and Fb. We get the vectors

(a1 cosh µ + a3 sinh µ, a2, ∗)

and

(b1 cosh µ + b3 sinh µ, b2, ∗).

Suppose now that we still have a1 cosh µ + a3 sinh µ = λ(b1 cosh µ + b3 sinh µ). Then, for any µ /= 0, a3 sinh µ = λb3 sinh µ a3 = λb3. Hence, as long as Fa λFb, then we can ﬁrst apply Tµ (with µ /= 0) and proceed as before. If Fa = λFb, then we set T = T−φ Rα so that TFa = (1, 0, 0) and 1 1 T Fb = λ− TFa = λ− (1, 0, 0). Exercise 12.20. Prove the following properties of the Lorentz cross product: Fu ⊗ Fv = JFv × JFu Fu ⊗ Fv = −Fv ⊗Fu u1 u2 u3 Fu ◦ (Fv ⊗ w F ) = det v1 v2 v3 w1 w2 w3 Fu ⊗ (Fv ⊗ w F ) = (Fu ◦ Fv)w F − (Fu ◦ w F )Fv Fu ◦ Fx Fu ◦ w F . (Fu ⊗ Fv) ◦ (w F ⊗ Fx) = det Fv ◦ Fx Fv ◦ w F T

Solution. The slick way of doing the ﬁrst is to note that J J = I so J is 3 orthogonal (in R ), and that J is orientation reversing, so J(Fu × Fv) = −JFu × JFv = JFv × JFu.

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CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

Or one can grind it out: JFu × JFv = (u1 , u2 , −u3 ) × (v1 , v2 , −v3 ) = (−u2v3 + u3v2, −u3v1 + u1v3, u1v2 − u2v1) = −J(Fu × Fv). For the next one, we note that Fu ⊗ Fv = J(Fu × Fv) = J(−Fv × Fu) = −J(Fv × Fu) = −Fv ⊗ Fu. The next three look worse than Exercise 12.7 but, in fact, most of the work is done there. We begin with Fu ◦ (Fv ⊗ w F ) = JFu · J(Fv × w F) = u · (Fv × w F) u1 = det v1 w1

u2 v2 w2

u3 v3 . w3

For the next, we use Exercise 12.7: Fu ⊗ (Fv ⊗ w F ) = J(J(Fv × w F )) × JFu = −JFu × (Fv × w F) = −((JFu · w F )Fv − (JFu · Fv)w F) = (Fu ◦ Fv)w F − (Fu ◦ w F )Fv. And again for the next: (Fu ⊗ Fv) · (w F ⊗ Fx) = J(J(Fu × Fv)) · (Fx ⊗ w F) = (Fu × Fv) · (JFx × J w F) F = det Fu · JFx Fu · J w Fv · JFx Fv · J w F Fu ◦ Fx Fu ◦ w F . = det Fv ◦ Fx Fv ◦ w F Exercise 12.21. Suppose Fa and Fb have positive lengths. Prove that ||Fa ⊗ Fb|| = i||Fa||||Fb|| sin θ, where θ is the angle deﬁned by Fa ◦ Fb = ||Fa||||Fb|| cos θ. Solution. We use the last property in the previous exercise: (Fa ⊗ Fb) ◦ (Fa ⊗ Fb) = det Fa ◦ Fb Fa ◦ Fa Fb ◦ Fb Fb ◦ Fa

283

12.3 ANGLES AND THE LORENTZ CROSS PRODUCT 2 2 = (Fa ◦ Fb) − (||Fa||||Fb||) 2 = ||Fa||2 ||Fb||2 (cos θ − 1)

= −||Fa||2 ||Fb||2 sin θ ||Fa ⊗ Fb|| = i||Fa||||Fb|| sin θ. 2

Exercise 12.22. Construct a proof of the hyperbolic Pythagorean theorem and Theorem 7.16.2 which is in the spirit of the proofs of the spherical Pythagorean theorem and Theorem 10.2.2 given in Section 10.2. [H] Solution. Let C be the ‘North Pole’ N = (0, 0, 1). Let A be a point in the xz-plane so A = (A1, 0, A3), and let B b e a point in the yz-plane so B = (0, B2, B3). As usual, let a = |BC|, b = |AC|, and c = |AB|. Then cosh a = −B ◦ C = B3 cosh b = −A ◦ C = A3 cosh c = −A ◦ B = A3B3 = cosh a cosh b. The plane AC is described by the equation (A ⊗ C) ◦ Fx. We calculate A ⊗ C = (0, −A1, 0). 2 2 Since A ∈ V+, A satisﬁes A 1 − A 3 = −1, so A1 = sinh b. The plane AB is

described by the equation with normal

A ⊗ B = (A1, 0, A3) ⊗ (0, B2, B3) = (sinh b, 0, cosh b) ⊗ (0, sinh a, cosh a) = (− cosh b sinh a, − sinh b cosh a, − sinh b sinh a), which has length ||A ⊗ B|| = = =

√ √ √

cosh b sinh a + sinh b cosh a − sinh b sinh a 2

cosh b sinh a + sinh b 2

cosh b cosh a − cosh b + sinh b

√ 2 = cosh c − 1 = sinh c.

The angle A is the angle between AB and AC, so is given by cos A =

(0, − sinh b, 0) ◦ (cosh b sinh a, − sinh b cosh a, − sinh b sinh a)

||(0, − sinh b, 0)||||(− cosh b sinh a, − sinh b cosh a, − sinh b sinh a)|| 2 sinh b cosh a = sinh b sinh c

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CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE =

sinh b cosh a . sinh c 2

Let us use the Euclidean Pythagorean theorem (sin A +cos2 A = 1) to ﬁnd sin A: √ sin A = 1 − cos2 A 2

√

sinh b cosh a 2 1− sinh c

sinh c − sinh b cosh a sinh c √ 2 2 2 2 2 cosh a cosh b − 1 − cosh b cosh a + cosh a = sinh c sinh a = . sinh c =

Exercise 12.23. We saw that the surface V + can be parameterized by φ and θ so that a point P (φ, θ) has the coordinates P (φ, θ) = (sinh φ cos θ, sinh φ sin θ, cosh φ). The appropriate analogue on the sphere S is P (φ, θ) = (sin φ cos θ, sin φ sin θ, cos φ). What is the analogue of this parameterization in Euclidean geometry? [A]

12.4

A Diﬀerent Perspective

Exercise 12.24. Suppose the intersection of V + with Fa ◦ Fx = d is bounded and nonempty. Prove that this intersection is a (hyperbolic) circle. What are the necessary and suﬃcient conditions on Fa and d for this to happen? What is the center of this circle? [H] F be a point on Fa ◦ Fx = d. Then FaP F = d, so Solution. Let P F Fa ◦ Fx = Fa ◦ P F) = 0 Fa ◦ (Fx − P F ) = 0. JFa · (Fx − P Thus this equation describes a plane p. Let us take the cross section of Lorentz space that contains Fa and the z-axis, as in Figure 12.1. Without loss of generality, we may assume that ||Fa || is either ± 1 or 0. If ||Fa|| = 1, as in Figure 12.1, then the plane p is parallel to the line shown in Figure 12.1, and it is not too hard to see that its intersection with V+ is unbounded.

285

12.4 A DIFFERENT PERSPECTIVE

It might help to recall that +Vis the top hyperbola revolved about the zaxis, while p is a plane perpendicular to the ﬁgure. If ||Fa|| = 0, then Fa is on one of the asymptotes and the plane p is parallel to the other asymptote. Again, it is not hard to see that the intersection withV + is unbounded. Such an intersection gives a horocycle, but we will not show that. Thus, a necessary condition on Fa is that ||Fa|| < 0. If ||Fa|| = − 1, then the plane p that would appear in our cross section shown in Figure 12.1 would either miss the top hyperbola altogether (so the intersection is empty), or would cut the top hyperbola in two places. Note that revolving the asymptotes about the z-axis gives a cone and that the intersection of the cone with p gives an ellipse. The intersection of V+ with p lies entirely inside this ellipse, so is bounded. Hence, if ||Fa || < 0, then the intersection of p with + V is bounded. Now, suppose the intersection is nonempty. Then we can choose P on V + . Since ||Fa|| < 0, there exists a point A on V + represented 1 1 by Fa (so A = λ− Fa where | λ− | = −|| || Fa and the sign of λ is chosen so that its z-component is positive). Then d = Fa ◦ P = λA ◦ P = −λ cosh |AP |. Such a point P exists if and only if −d 1 λ ≥ d ≥ −λ =

−1 ||Fa||

Note that the inequality does not change directions since λ < 0. Thus the intersection of Fa ◦ V Fx−1 = d and + is bounded and nonempty if and only if ||Fa|| < 0 and d ≥ . Suppose X is a point on V + that is also on the ||Fa|| plane p. Then Fa ◦ X = d λA ◦ X = λA ◦ P −A ◦ X = −A ◦ P cosh |AX| = cosh |AP | |AX| = |AP |. Thus, X is a distance |AP | away from A, so X lies on the circle centered at A with radius |AP |. Conversely, suppose X lies on this circle. Then |AX| = |AP | Fa ◦ X = d, so X is on the plane p. Thus, the intersection of p and V+ is a hyperbolic Fa circle. The center of the circle is A = ± , where the sign is chosen so ||Fa|| that the z-component of A is positive.

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CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

Compare this with the equations of circles on spheres. These are the intersection of planes with the sphere. The centers (two antipodal points) of a circle are the points where the normal of the plane intersects the sphere.

12.5

The Beltrami-Klein Model

12.6

Menelaus’ Theorem

Exercise 12.25. Prove Menelaus’ theorem in hyperbolic geometry. Solution. Let p be the plane through A, B, and C in the pseudosphere model of hyperbolic geometry. Let D′, E′, and F ′ be the intersection of p with the lines OD, OE, and OF , respectively. Then D, E, and F are collinear if and only if OD, OE, and OF are coplanar, since lines in the + pseudosphere are the intersections of V with planes through O. But OD, OE, and OF are coplanar if and only if D′, E′, and F ′ are collinear in p. By Menelaus’ theorem in Euclidean geometry, D′, E′, and F ′ are collinear if and only if |AF ′| |BD′| |CE′| = − 1. |F ′ B| |D′ C| |E ′ A| By Lemma 12.6.2, this is equivalent to sinh |AF | sinh |BD| sinh |CE| = —1. sinh |FB| sinh |DC| sinh |EA| Exercise 12.26 (Ceva’s Theorem in Hyperbolic Geometry). State and prove Ceva’s theorem in hyperbolic geometry. A

F E

Figure 12.2 Theorem (Ceva’s Theorem in Hyperbolic Geometry). Let D, E, and F be three points on, respectively, the (extended) sides BC, AC, and

287

12.6 MENELAUS’ THEOREM

AB of ∆ABC. Then the lines AD, BE, and CF are coincident if and only if sinh |AF | sinh |BD| sinh |CE| = 1. sinh |FB| sinh |DC| sinh |EA| Proof. The statement is almost identical to the statement of Ceva’s theorem in Euclidean geometry, except for all the hyperbolic trig functions. The proof, too, is almost identical, at least for one direction. Suppose AD, BE, and CF are coincident at P , as in Figure 12.2, which is the ﬁgure used in the Euclidean version. We apply Menelaus’ theorem to ∆ABE and the line FPC, which gives us sinh |AF | sinh |BP| sinh |EC| = —1. sinh |FB| sinh |PE| sinh |CA| Applying Menelaus’ theorem to ∆BCE and the line CPF , we get sinh |BD| sinh |CA| sinh |EP| sinh |DC| sinh |AE| sinh |PB|

= —1.

Multiplying, we get sinh |AF | sinh |BP| sinh |EC| sinh |BD| sinh |CA| sinh |EP| sinh |FB| sinh |PE| sinh |CA| sinh |DC| sinh |AE| sinh |PB| sinh |AF | sinh |EC| sinh |BD| = ± . sinh |FB| sinh |AE| sinh |DC|

We have equality up to sign, since we lost our understanding of signs when we started cancelling terms. To determine the sign, we ﬁrst suppose that P is inside ∆ABC. Then all the ratios are positive, so the sign is positive. If P is outside ∆ABC, then one of the ratios must be positive, and the other two negative, so again we get a positive sign. Thus, we get sinh |AF | sinh |BD| sinh |CE| =1 sinh |FB| sinh |DC| sinh |EA| where this is now a product of signed ratios. I’m not certain about the converse. The proof in Euclidean geometry would work here too, except for one problem. The ﬁrst step in that proof is to let P be the point of intersection of AD and BE. But why should they intersect? Actually, this possibility is omitted in the proof in the text too .... Exercise 12.27. Despite the similarities, not everything in spherical geometry translates nicely to the pseudosphere. Explain why Ptolemy’s proof of the spherical Pythagorean theorem (see Exercise 10.13) cannot be adapted to the pseudosphere.

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CHAPTER 12. THE PSEUDOSPHERE IN LORENTZ SPACE

Solution. Ptolemy proves the spherical Pythagorean theorem by considering a right angle triangle ∆ABC with right angle at C and ﬁnding points D and E on BC and AC such that AE | = | CD | =| π/2. There are three nice consequences of choosing D and E this way. We get | AF| = π/2, sin π/2 = 1, and probably most importantly, sin(π/2− θ) = cos θ. In hyperbolic trigonometry, we note that sinh(a − x) = sinh a cosh x − sinh x cosh a, so if we want this to equal cosh x (perhaps multiplied by a constant), then we need to choose a such that cosh a = 0. No such a exists.