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3 Analytic

3.1 The Cartesian Co-ordinate System Pure in the style of and Hilbert is what we call synthetic: axiomatic, with- out co-ordinates or explicit formulæ for , , , etc. Nowadays, the practice of ele- mentary geometry is almost entirely analytic: reliant on , co-ordinates, vectors, etc. The major breakthrough came courtesy of Rene´ Descartes (1596–1650) and (1601/16071–1655), whose introduction of an axis, a fixed reference ruler against which objects could be measured using co-ordinates, allowed them to apply the Islamic invention of algebra to geometry, resulting in more efficient computations. The new geometry was revolutionary, so much so that Descartes felt the need to justify his argu- ments using , lest no-one believe his work! This attitude persisted for some time: when Issac Newton published his groundbreaking Principia in 1687, his presentation was largely syn- thetic, even though he had used co-ordinates in his derivations. Synthetic geometry is not without its benefits—many results are much cleaner, and presents its own logical difficulties— but, as time has passed, its study has become something of a fringe activity: co-ordinates are simply too useful to ignore! Given that Cartesian geometry is the primary form we learn in grade-school, we merely sketch the familiar ideas of co-ordinates and vectors.

• Assume everything necessary about on the real . continuity y 3 • axes meet at the origin. P 2 • The Cartesian co-ordinates of a P are measured by project- ing onto the axes: in the picture, P has co-ordinates (1, 2), often 1 written simply as P = (1, 2).

• are defined using . For instance, the Pythagorean 2 1 1 2 3 function says that the of 1 centered at the − − 1 x origin may be described by the x2 + y2 = 1. − 2 − In the 1600’s this was not considered a new , but rather a collection of computational tools built on top of Euclid. We may therefore assume anything from Euclid and mix strategies as appropriate. To see this at work, consider a simple result. B Lemma 3.1. Let O = (0, 0), A = (x, y) and B = (v, w) where O, A, B are non-collinear. If C = (x + v, y + w), then the quadri- C lateral OACB is a .

p Proof. Calculate : BC = x2 + y2 = OA , etc. | | | | Now use side-side-side to see that OAC = CBO. The usual 4 ∼ 4 discussion of lines/alternate from Euclid O opposite sides to be parallel. A

1There is some argument over Fermat’s birth given that he possibly had a deceased older brother, also named Pierre.

1 Vector Geometry Vectors come to us courtesy of several mathematicians, most prominently (1805–1865), Oliver Heaviside (1850–1925) and J. Willard Gibbs (1839–1903). Hamilton had stumbled upon the algebra of quaternions when attempting to extend to three the contemporary use of complex numbers to describe planar geometry.2 Heaviside and Gibbs independently developed vector . The revolution was rapid; by 1900 vector calculations were dominant in .

Definition 3.2. A directed −→AB is a segment together with an orientation.a The position vector of a point A is the directed line segment OA−→ where O is the origin. A vector is an equivalence class of directed line segments where two segments are equivalent if and only if they are congruent and oriented in the same direction.

aWe write −→AB for the directed line segment so as to distinguish it from the ray −→AB of Hilbert’s Euclidean geometry.

A vector has length and direction, but no fixed location. All directed line segments with the same length and direction represent the same vector. The standard representation of a vector involves placing its tail at the origin: we can then describe the vector by giving the co-ordinates of its head.

Example 3.3. In the picture, the standard representation of a vector y 3 v is shown in blue. The green arrows are other representations of the same vector. Various common notations include 2 2 v = ~v = v = OA−→ = = 2, 1 = 2i + j 1 A 1 h i O The usual convenient abuse of terminology is at work: strictly OA−→ v 1123 ∈ − 1 since v is an equivalence class, but no-one writes this. − x Addition and Scalar Multiplication are defined algebraically in the familiar manner:

v  w  v + w  λv  v + w = 1 + 1 = 1 1 , λv := 1 v v2 w2 v2 + w2 λv2 w Vector addition can be visualized by placing representative segments nose- to-tail. In view of Lemma 3.1, the commutativity of addition v + w = v + w w w + v is often known as the . Indeed the Lemma may be rephrased in this language: the parallelogram OACB is spanned by O v x  v  OA−→ = and OB−→ = y w The scalar multiple λv may be viewed as stretching or shrinking v, and reversing its direction when λ < 0.

2Quaternions are objects of the form a + bi + cj + dk where a, b, c, d R, i2 = j2 = k2 = 1 and i, j, k multiply as if ∈ − using the cross-product (ij = k = ji, etc.). Hamilton couldn’t make the three-dimensional part (a = 0) into a suitable − algebra, but a fourth fixed things. Hamilton eventually realized that if he dropped his requirement of having a well-defined multiplication, he could apply his vector approach to the study of geometry in any dimension.

2 We finish with a famous result showing how easy it can be to work in analytic geometry.

Theorem 3.4. The medians of a meet at a point 1/3 of the way along each median.

Proof. Given OAB, let a = OA−→ and b = OB−→. 4 1 A If M is the of AB, then OM−−→ = 2 (a + b). 2 Let G lie 3 of the distance along OM−−→: that is 2 1 1 OG−→ = (a + b) = (a + b) 3 · 2 3 The point 1 of the way along the median through has position M 3 A a vector 1 1  1  1 OB−→ + OA−→ OB−→ = (a + b) G 2 3 − 2 3

Similarly, the point 1 of the way along the median through B has 3 1 ( + ) position vector 3 a b O b B 1 1  1  1 OA−→ + OB−→ OA−→ = (a + b) 2 3 − 2 3

All three points have the same position vector, and are therefore the same point G!

Compare this to the argument/exercise using Ceva’s at the end of our discussion of Eu- clidean Geometry! The proof shows another important aspect of analytic geometry. The standard approach is to place the origin and orient a figure in a manner which makes calculations simple. This is essentially Eu- clid’s (sketchy) superposition principle, or Hilbert’s , but can be more rigorously grounded in a full discussion of isometries.

Exercises. 3.1.1. Given a ABCD, let W, X, Y, Z be the of the sides AB, BC, CD and DA respectively. Use vectors to prove that WXYZ is a parallelogram.

3.1.2. (a) Let A = (a1, a2) and B = (b1, b2) be given. Show that any point on the line segment joining A and B has co-ordinates ((1 t)a1 + tb1, (1 t)a2 + tb2) where 0 t 1. − − 1 1 ≤ ≤ (b) Show that the midpoint of A and B has co-ordinates 2 (a1 + b1), 2 (a2 + b2) . (You’ll probably find it easiest to think in terms of vectors for both parts) 3.1.3. (a) Perform a pure co-ordinate proof (no vectors) of Theorem 3.4. (b) Descartes and Fermat did not, in fact, have a fixed perpendicular second axis! Their approach was equivalent to choosing a a second axis whose to the first was chosen to make a problem as easy as possible. Given OAB, where B = (b, 0), choose the second 4 axis to point along OA−→ so that A has co-ordinates (0, a). Now give an even simpler proof of the centroid theorem.

3 3.2 Angles in Planar Analytic Geometry We define angle measure a little differently: this time we use and extend to any angle.

Definition 3.5. Suppose A, B, C are distinct and draw any circle centered at A. The -measure ]BAC is the ratio of the arc-length to the radius measured counter-clockwise from −→AB to −→AC.

Being ratios, radians are naturally unitless. θ − In analytic geometry it is common to label angles using their C θ radian measure. In part this is since. . . B Angles are congruent if and only if their radian measures are equal or negative each other modulo 2π. A In the picture, θ = BAC and CAB = 2π θ. This last is 2π θ ] ] − often taken instead to be θ. − −

Definition 3.6. Let P = (x, y) lie on a circle of radius r such that the P segment OP−→ makes angle θ radians measured counter-clockwise from the positive x-axis. The cosine and sine of θ are defined by r y θ x = r cos θ, y = r sin θ O x The AAA theorem for similar says we can view these as well- defined functions of radian-measure, not merely angle!

π 3π The definitions work for any angles: if, say, 2 < θ < 2 , simply take x < 0. As co-ordinates, it is no problem for x or y to be negative. Basic relationships should also be obvious from the picture: e.g.

cos( π θ) = sin θ = sin(π θ) = y  2 − − r Moreover, by drawing an isosceles right-triangle and an , the most well-known values of the sine and cosine are easily recovered:

sin φ π π 4 6 √2 2 √ cos φ 1 3 1 sin θ

φ π π θ 4 3 cos θ 1 1 cos( π θ) = sin θ sin π = cos π = 1 sin π = cos π = √3 2 − 4 4 √2 3 6 2 sin( π θ) = cos θ sin π = cos π = 1 2 − 6 3 2

4 Solving Triangles In analytic geometry, a triangle is described by six values: three side and three angle measures. Euclid’s triangle congruence (SAS, ASA, SSS, SAA) say that knowing three of these in a suitable combination should be enough to recover the remainder. The way this is done is using the cosine and sine rules.

Theorem 3.7 (Cosine Rule). In a trianglea ABC, we have c2 = a2 + b2 2ab cos γ 4 − aWe follow the convention that a is the length of the side opposite A, which has (interior) angle measure α.

Proof. Let the base be b and drop a perpendicular from B to AC. B Call the height h and split b = x + z, where x = a cos γ. Applying ’ twice, we have:

2 2 2 2 2 2 a c a = h + x , c = h + z h

Eliminating the h terms yields γ c2 = a2 x2 + z2 = a2 + (z + x)(z x) − − b D AC = a2 + b(x + z 2x) = a2 + b(b 2x) − − x > 0 z > 0 = a2 + b2 2bx = a2 + b2 2ab cos γ − − The argument works for all possible arrangements: just be careful with the signs of x and z!

B B

c a h a c h γ γ z < 0 D x < 0 b AC b AC D z > 0 x > 0

Example 3.8. The SSS congruence corresponds to solving a triangle using the cosine rule. For instance, the given triangle has angles satisfying α 62 + 72 32 19 6 cos α = − = 25° 2 6 7 21 ≈ · · 32 + 72 62 11 cos β = − = 58° 7 2 3 7 21 ≈ · · γ 32 + 62 72 1 cos γ = − = − 96° 2 3 6 9 ≈ β · · 3 We use measure for clarity. In modern times computing inverse cosines is easy, but historically this required large data tables.

5 Theorem 3.9 (Sine Rule). If d is the of the circumcircle of ABC, then 4 sin α sin β sin γ 1 = = = a b c d

Proof. Draw the circumcircle and construct BCD with diameter BD. A 4 This is right-angled at C by Thales’ Theorem. There are two cases: α

) B 1. If A lies on the major arc BC, then A and D share the same arc, d whence ]BDC = α and so a α a D sin α = sin ]BDC = d C ) 2. If A lies on the minor arc BC, then the quadrilateral ABDC lies on a circle whence opposite angles A, B are supplementary. Thus B a d sin α = sin(π α) = sin ]BDC = − d a α π α D − The two other angle-side combinations are similar, or can be seen sim- A ply by permutation. C The sine and cosine rules, together with the fact that angles sum to a straight edge, allow one to solve any triangle. For instance, given an angle-side-angle combination (α, c, β): 1. Compute the remaining angle γ = π α β; − − γ a 2. Use the sine rule twice to compute b β sin α sin β a = c, b = c α sin γ sin γ c

Multiple-angle formulæ A quick appeal to the and vector notation recovers the usual multiple-angle formulæ.

Definition 3.10. The dot product of vectors u = ( u1 ) and v = ( v1 ) is u v = u v + u v . u2 v2 · 1 1 2 2

Theorem 3.11. If θ is the angle (either one!) between u and v, then u v = u v cos θ. · | | | | The proof is an easy exercise following the cosine rule.

Corollary 3.12. If α and β are any angles, then

cos(α β) = cos α cos β sin α sin β ± ∓ sin(α β) = sin α cos β cos α sin β ± ±

6 Proof. With respect to a circle of radius 1, the vectors making angles α, β with the positive x-axis are b cos α cos β a = b = a α sin α sin β β

Simply take dot products and apply Theorem 3.11:

cos α cos β + sin α sin β = a b = cos(α β) · − since the vectors have length 1 and the angle between them is α β. − While the picture has been drawn as if α β 0, the calculation works regardless or α, β, since ≥ ≥ analytic geometry (and cosine!) is comfortable with negative angles. The remaining rules are an exercise: use the even/oddness of cosine/sine and apply the identity sin α = cos( π α), etc. 2 − Exercises. 3.2.1. (a) As we did on page 6, describe the process for solving triangles give for the remaining combinations SAS and SAA. p (b) A triangle has an angle of 2π radians between sides of lengths √2 and 2 √3. Find the 3 − length of the remaining side, and the angles. (Hint: (√3 1)2 = 4 2√3) − − 3.2.2. Prove Theorem 3.11 by applying the cosine rule to the triangle formed by u, v and v u. − Explain why the comment ‘either one!’ is appropriate. 7π 3.2.3. Use a multiple angle to find an exact value for cos 12 . 3.2.4. Complete the proof of Corollary 3.12 by supplying an argument for each of the remaining three cases.

3.2.5. Consider the given picture. Find x, the radian measure α and the exact value of cos α. (Hint: you have similar isosceles triangles. . . ) 1 xx

α x 1 x − 3.3 The Complex Italian mathematicians of the 16th century (Tartaglia, Cardano, Bombelli, etc.) began experimenting with -roots of negative numbers, mostly out of curiosity. The application of complex num- bers to geometry starts with (1707–1783), who extended the real line of Descartes to places where equations such as x2 = 1 had solutions. This produced a two-dimensional picture − with both real and imaginary axes. The set of complex numbers C allows a description of 2D ge- ometry where basic transformations (rotations and reflections) may be expressed algebraically. This approach indeed predated vectors and, in part, provided the inspiration for their development. We briefly review complex numbers, since you should have encountered them elsewhere.

7 Definition 3.13. Let i be an abstract symbol satisfying the property z = 2 + 3i i2 = 1, and let x, y be real numbers. The z = x + iy is 3i − the point with co-ordinates (x, y). In this context, the Cartesian plane is 2i known as the Argand diagram.a i z = √13 Addition, scalar multiplication (by real numbers) and complex multiplica- | | tion follow the natural commutative and distributive laws for the real numbers, while using i2 = 1 to simplfy. i 1234 − − The complex conjugate of z is z = x iy. 2i p− The modulus of z is z = √zz = x2 + y2. − | | 3i − z = 2 3i aIn the language of , C is a vector over R with 1, i . − { }

Example 3.14. We compute a simple complex multiplication: make sure this is clear to you!

(2 + 3i)(4 + 5i) = 2 4 + 2 5i + 3i 4 + 3i 5i = 8 + 10i + 12i 15 = 7 + 22i · · · · − − The complex numbers are of interest to us because the rules scream geometry! In particular: • Addition by z translates all points by z. • Scalar multiplication scales points. • Conjugation is reflection across the x-axis: (x, y) (x, y). 7→ − • The modulus is the distance from the origin; or the length of a vector. The ability of complex multiplication to describe rotations is particularly useful. For example, iz = i(x + iy) = y + ix − π Is the result of rotating z counter-clockwise 2 radians about the origin. To obtain all rotations and reflections, we need an alternative description of a complex number.

Lemma 3.15. 1. (Euler’s Formula) For any θ R, eiθ = cos θ + i sin θ. ∈ 2. (Exponential laws) eiθeiφ = ei(θ+φ) and (eiθ)n = einθ for any n Z. ∈ Evaluating at θ = π yields the famous Euler identity eiπ = 1. − Proof. 1. This requires either power series or elementary differential equations, topics best de- scribed elsewhere.

2. Apply the multiple-angle formulæ (Corollary 3.12):

ei(θ+φ) = cos(θ + φ) + i sin(θ + φ) = cos θ cos φ sin θ sin φ + i sin θ cos φ + i cos θ sin φ − = (cos θ + i sin θ)(cos φ + i sin φ) = eiθeiφ

iθ iθ Performing an induction with θ = φ and combining with e e− = 1 gives the final result.

8 Definition 3.16. Let z = x + iy be a non-zero complex number. 2i iπ Writing x = r cos θ and y = r sin θ, we obtain the polar form z = 1 + √3i = 2e 3

z = reiθ = r(cos θ + i sin θ) p Here r = z = x2 + y2 is the modulus and θ = arg(z) the argu- | | i z = 2 ment of z; the angle measured counter-clockwise from the positive | | real axis to the vector ( x ). y π arg(z) = 3

1 2

Interpreting Lemma 3.15 in the language of the polar form, we find:

Corollary 3.17. 1. For any non-zero complex numbers z, w,

zw = z w , arg(zw) arg(z) + arg(w)(mod 2π) | | | | | | ≡ 2. The complex number eiθz is obtained by rotating z counter-clockwise about the origin through an angle θ radians.

Proof. Simply let w = reiθ and z = seiφ: the first result is immediate from

reiθseiψ = rsei(θ+ψ)

For the second part, take r = 1 and observe that eiθz has the same modulus as z, while its argument has increased by θ (modulo 2π).

√ 3π Example 3.18. Rotate z = 1 + 3i counter-clockwise by 4 2i radians around the origin. z We simply multiply by

3π 3π 1 i e3πi/4 = cos + i sin = ( 1 + i) 4 4 √2 − 3π to obtain 4 1 e3πi/4z = ( 1 + i)(1 + √3i) 2 11 √2 − − − 3πi 1  4 = 1 √3 + i(1 √3) e z √2 − − − i − Alternatively, we can keep the calculation in polar form:

πi/3 3πi/4 3πi + πi 13πi/12 z = 2e = e z = 2e 4 3 = 2e ⇒

9 Complex conjugation takes care of reflections. Since z z reflects 7→ across the real axis, we can use this together with rotations and trans- 1 lations to describe all reflections. To reflect across the line making angle θ with the positive real axis, we do three things: 2 1. Rotate by θ. − θ 2. Reflect across the real axis. 3. Rotate back by θ.

Otherwise said:

Theorem 3.19. To reflect z C across the line making angle θ with ∈ 3 the positive real axis, we compute

z eiθ(e iθz) = e2iθz 7→ −

Example 3.20. Reflect z = 2 + 3i across the line through the origin and w = √3 + i. − First compute θ = arg(w) = tan 1 1 = π . The desired point is then − √3 6 ! ! 1 √3 3√3  3 eiπ/3( 2 3i) = + i ( 2 3i) = 1 √3 + i − − 2 2 − − 2 − − 2

Corollary 3.21. 1. To rotate z by θ about a point w, compute z w + eiθ(z w). 7→ − 2. To reflect z across the line with θ through a point w, compute z w + e2iθ(z w). 7→ − Mathematicians later tried to replicate this algebraic encoding of simple geometric transformations in three and higher dimensions. The attempt led (via Hamilton’s quaternions) first to vectors and then to linear algebra/ calculations and a unified theory of basic transformations in any dimension.

Exercises. 3.3.1. (a) Express each of the following fractions as complex numbers by rationalizing the denominator (multiplying through by the complex conjugate. . . )

1 1 + i 1 , , 2i 1 i 2 + 4i − (b) Prove that C is closed under multiplicative inverses: i.e., z C 0 , prove that 1 C. ∀ ∈ \{ } z ∈ 3.3.2. Use complex numbers to compute the result of the following transformations: you can answer in either standard or polar form. (a) Rotate 3 5i counter-clockwise around the origin by 3π radians. − 4 (b) Reflect 2 i across the joining 1 + i√3 and the origin. − π (c) Reflect 1 + i across the line through the origin making angle 5 radians with the positive real axis. p p (d) Reflect 2 + 3i across the line joining the origin and the point ( 2 + √2, 2 √2). − −

10 3.3.3. By letting n = 3 in Lemma 3.15, prove that

cos 3θ = 4 cos3θ 3 cos θ − Find a corresponding trigonometric identity for sin 3θ.

3.4 Birkhoff’s Axiomatic System for Analytic Geometry (non-examinable) Analytic geometry was originally built as an addition to Euclidean geometry. In 1932, courtesy of , it was axiomatized in its own right.

Background Assume the usual properties/ of the real numbers as a complete ordered field. Birkhoff’s system is typical of modern axiomatic systems in that it is built on top of pre-existing systems (, complete ordered fields, etc.).

Undefined terms Point, line, distance and angle measure. Let the set of points be denoted , then, S Distance is a function d : R+ S × S → 0 Angle measure is a function : [0, 2π) ] S × S × S →

Axioms

Euclidean Given two distinct points, there exists a unique line containing them.

Ruler Points on a line ` are in bijective correspondence with the real numbers in such a way that if t , t correspond to A, B `, then t t = d(A, B). A B ∈ | A − B| Protactor The rays emanating from a point O are in bijective correspondence with the set [0, 2π) so that if a, b correspond to rays OA−→, OB−→, then AOB a b (mod 2π). This correspondence is ] ≡ − continuous in A, B.

SAS 3 If triangles have a pair of angles with equal measure, and the sides adjacent to said angles are in the same ratio, then the remaining angles have equal measure and the final sides are in the same ratio.

Definitions As with Hilbert, some of these are required before later axioms make sense. In partic- ular, the definition of ray is required before the protractor .

BetweennessB lies between A and C if d(A, B) + d(B, C) = d(A, C)

Segment AB consists of the points A, B and all those between

Ray −→AB consists of the segment AB and all points C such that B lies between A and C.

Basic shapes Triangles, , etc.

3As with Hilbert, Birkhoff makes SAS an axiom: Birkhoff’s version is stronger, for it also applies to similar triangles

11 Analytic geometry as a model The axioms should feel familiar. Being shorter than Hilbert’s list, and being built on familiar notions such as the real line, it is somewhat easier for us to understand what the axioms are saying and to visualize them. There is something to prove however; indeed the major point of Birkhoff’s system!

Theorem 3.22. Cartesian Analytic Geometry is a model of Birkhoff’s axioms.

Recall what this requires: we must provide a definition of each of the undefined terms and prove that these satisfy each of Birkhoff’s axioms. Here are suitable definitions for Cartesian analytic geometry:

Point An (x, y) of real numbers. q Distanced (A, B) = (A B )2 + (A B )2 x − x y − y Line all points satisfying a ax + by + c = 0.

0 1  Angle Define vectors as ordered pairs and consider the matrix J = 1− 0 whose action on vector is π to rotate counter-clockwise by 2 . Now define angle via ( v w [0, π] v Jw 0 cos θ = · where θ ⇐⇒ · ≥ v w ∈ (π, 2π) v Jw < 0 | | | | ⇐⇒ · Cosine may be defined using power series, so no pre-existing geometric meaning is required.

Proof. (Euclidean axiom) If (x1, y1) and (x2, y2) satisfy ax + by + c = 0 then

a(x x ) + b(y y ) = 0, 1 − 2 1 − 2 whence a = y y , b = x x up to . It follows that the line has equation 1 − 2 2 − 1 (y y )x + (x x )y + x y x y = 0 1 − 2 2 − 1 1 2 − 2 1 unique up to multiplication of all three of a, b, c by a non-zero constant. The remaining axioms are exercises.

Exercises. 3.4.1. Prove that the ruler axiom is satsisfied: (a) First show that if P = Q lie on `, then any point A on the line has the form 6 t A = P + A (Q P) where t R d(P, Q) − A ∈

(b) Use this formula to compute d(A, B)2 = (t t )2. A − B 1  1 v i 3.4.2. Let I = (1, 0) and i = 0 . Given v = OA−→, define a = cos− v· as above. Clearly a is a continuous function of v and thus A. Now use the multiple-angle| formula| (Corollary 3.12) to prove that the protractor axiom is satisfied. 3.4.3. Use the cosine rule (Theorem 3.7) to prove that the SAS similarity axiom is satisfied.

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