Coordinate Geometry JWR Tuesday September 6, 2005 Contents 1 Introduction 3 2 Some Fallacies 4 2.1 Every Angle is a Right Angle!? . . . . . . . . . . . . . . . . . 5 2.2 Every Triangle is Isosceles!? . . . . . . . . . . . . . . . . . . . 6 2.3 Every Triangle is Isosceles!? -II . . . . . . . . . . . . . . . . . 7 3 Affine Geometry 8 3.1 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 3.2 Affine Transformations . . . . . . . . . . . . . . . . . . . . . . 12 3.3 Directed Distance . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.4 Points and Vectors . . . . . . . . . . . . . . . . . . . . . . . . 20 3.5 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 3.6 Parallelograms . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.7 Menelaus and Ceva . . . . . . . . . . . . . . . . . . . . . . . . 24 3.8 The Medians and the Centroid . . . . . . . . . . . . . . . . . . 26 4 Euclidean Geometry 30 4.1 Orthogonal Matrices . . . . . . . . . . . . . . . . . . . . . . . 30 4.2 Euclidean Transformations . . . . . . . . . . . . . . . . . . . . 31 4.3 Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 4.4 Similarity Transformations . . . . . . . . . . . . . . . . . . . . 33 4.5 Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4.6 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 4.7 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 1 4.8 Addition of Angles . . . . . . . . . . . . . . . . . . . . . . . . 39 5 More Euclidean Geometry 43 5.1 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 5.2 The Circumcircle and the Circumcenter . . . . . . . . . . . . . 44 5.3 The Altitudes and the Orthocenter . . . . . . . . . . . . . . . 44 5.4 Angle Bisectors . . . . . . . . . . . . . . . . . . . . . . . . . . 46 5.5 The Incircle and the Incenter . . . . . . . . . . . . . . . . . . 46 5.6 The Euler Line . . . . . . . . . . . . . . . . . . . . . . . . . . 47 5.7 The Nine Point Circle . . . . . . . . . . . . . . . . . . . . . . 47 5.8 A Coordinate Proof . . . . . . . . . . . . . . . . . . . . . . . . 49 5.9 Simson’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 50 5.10 The Butterfly . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 5.11 Morley’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 54 5.12 Bramagupta and Heron . . . . . . . . . . . . . . . . . . . . . . 54 5.13 Napoleon’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 54 5.14 The Fermat Point . . . . . . . . . . . . . . . . . . . . . . . . . 54 6 Projective Geometry 55 6.1 Homogeneous coordinates . . . . . . . . . . . . . . . . . . . . 55 6.2 Projective Transformations . . . . . . . . . . . . . . . . . . . . 57 6.3 Desargues and Pappus . . . . . . . . . . . . . . . . . . . . . . 60 6.4 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 6.5 The Projective Line . . . . . . . . . . . . . . . . . . . . . . . . 64 6.6 Cross Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 6.7 A Geometric Computer . . . . . . . . . . . . . . . . . . . . . . 67 7 Inversive Geometry 69 7.1 The complex projective line . . . . . . . . . . . . . . . . . . . 69 7.2 Feuerbach’s theorem . . . . . . . . . . . . . . . . . . . . . . . 69 8 Klein’s view of geometry 70 8.1 The elliptic plane . . . . . . . . . . . . . . . . . . . . . . . . . 70 8.2 The hyperbolic plane . . . . . . . . . . . . . . . . . . . . . . . 70 8.3 Special relativity . . . . . . . . . . . . . . . . . . . . . . . . . 70 A Matrix Notation 71 B Determinants 73 2 C Sets and Transformations 75 1 Introduction These are notes to Math 461, a course in plane geometry I sometimes teach at the University of Wisconsin. Students who take this course have com- pleted the calculus sequence and have thus seen a certain amount of analytic geometry. Many have taken (or take concurrently) the first course in linear algebra. To make the course accessible to those not familiar with linear al- gebra, there are three appendices explaining matrix notation, determinants, and the language of sets and transformations. My object is to explain that classical plane geometry is really a subset of algebra, i.e. every theorem in plane geometry can be formulated as a theorem which says that the solutions of one system of polynomial equations satisfy another system of polynomial equations. The upside of this is that the criteria for the correctness of proofs become clearer and less reliant on pictures. The downside is evident: algebra, especially complicated but elementary algebra, is not nearly so beautiful and compelling as geometry. Even the weakest students can appreciate geometric arguments and prove beautiful theorems on their own. For this reason the course also includes synthetic arguments as well. I have not reproduced these here but instead refer to the excellent texts of Isaacs [4] and Coxeter & Greitzer [3] as needed. It is my hope that the course as a whole conveys the fact that the foundations of geometry can be based on algebra, but that it is not always desirable to replace traditional (synthetic) forms of argument by algebraic arguments. The following quote of a quote which I got from page 31 of [3] should serve as a warning. The following anecdote was related by E.T. Bell [1] page 48. Young Princess Elisabeth had successfully attacked a problem in elementary geometry using coordinates. As Bell states it, “The problem is a fine specimen of the sort that are not adapted to the crude brute force of elementary Cartesian geometry.” Her teacher Ren´e Descartes (who invented the coordinate method) said that “he would not undertake to carry out her solution ... in a month.” 3 The reduction of geometry to algebra requires the notion of a transfor- mation group. The transformation group supplies two essential ingredients. Firstitisusedtodefinethenotionofequivalenceinthegeometryinquestion. For example, in Euclidean geometry, two triangles are congruent iff there is distance preserving transformation carrying one to the other and they are similar iff there is a similarity transformation carrying one to the other. Sec- ondly, in each kind of geometry there are normal form theorems which can be used to simplify coordinate proofs. For example, in affine geometry every tri- angle is equivalent to the triangle whose vertices are A = (0,0), B = (1,0), 0 0 C = (0,1) (see Theorem 3.13) and in Euclidean geometry every triangle is 0 congruent to the triangle whose vertices are of form A = (a,0), B = (b,0), C = (0,c) (see Corollary 4.14). This semester the official text is [3]. In past semesters I have used [4] and many of the exercises and some of the proofs in these notes have been taken from that source. 2 Some Fallacies Pictures sometimes lead to erroneous reasoning, especially if they are not carefully drawn. The three examples in this chapter illustrate this. I got them from [6]. See if you can find the mistakes. Usually the mistake is a kind of sign error resulting from the fact that some point is drawn on the wrong side of some line. 4 2.1 Every Angle is a Right Angle!? D P C R E Q A B O Figure 1: Every Angle is a Right Angle!? Let ABCD be a square and E be a point with BC = BE. We will show that ∠ABE is a right angle. Take R to be the midpoint of DE, P to be the midpoint of DC, Q to be the midpoint of AB, and O to be the point where the lines PQ and the perpendicular bisector of DE intersect. (See Figure 2.1.) The triangles AQO and BQO are congruent since OQ is the perpendicular bisector of AB; it follows that AO = BO. The triangles DRO and ERO are congruent since RO is the perpendicular bisector of DE; it follows that DO = EO. Now DA = BE as ABCD is a square and E is a point with BC = BE. Hence the triangles OAD and OBE are congruent because the corresponding sides are equal. It follows that ∠ABE = ∠OBE −∠ABO = ∠OAD−∠BAO = ∠BAD. 5 2.2 Every Triangle is Isosceles!? A R Q O C B D Figure 2: An Isosceles Triangle!? Let ABC be a triangle; we will prove that AB = AC. Let O be the point where the perpendicular bisector of BC and the angle bisector at A intersect, D be the midpoint of BC, and R and Q be the feet of the perpendiculars from O to AB and AC respectively (see Figure 2.2.) The right triangles ODB and ODC are congruent since OD = OD and DB = DC. Hence OB = OC. Also the right triangles AOR and AOQ are congruent since ∠RAO = ∠QAO (AO is the angle bisector) and ∠AOR = ∠AOQ (the angles of a triangle sum to 180 degrees) and AO is a common side. Hence OR = OQ. The right triangles BOR and COQ are congruent since we have proved OB = OC and OR = OQ. Hence RB = QC. Now AR = AQ (as AOR and AOQ are congruent) and RB = QC (as BOR and COQ are congruent) so AB = AR+RB = AQ+QC = AC as claimed. 6 2.3 Every Triangle is Isosceles!? -II A (cid:72) (cid:0)(cid:0)(cid:68) (cid:72)(cid:72) (cid:0) (cid:68) (cid:72)(cid:72) (cid:0) (cid:68) (cid:72)(cid:72) (cid:0) (cid:68) (cid:72)(cid:72) (cid:0) (cid:68) (cid:72)(cid:72) (cid:0) (cid:68) (cid:72)(cid:72) (cid:0) (cid:68)(cid:68) (cid:72)(cid:72) B X C Figure 3: AX bisects ∠BAC In a triangle ABC, let X be the point at which the angle bisector of the angle at A meets the segment BC. By Exercise 2.2 below we have XB XC = . (1) AB AC Now ∠AXB = ∠ACX +∠CAX = ∠C + 1∠A since the angles of a triangle 2 sum to 180 degrees. By the Law of Sines (Exercise 2.1 below) applied to triangle AXB we have XB sin∠BAX sin 1∠A = = 2 (2) AB sin∠AXB sin(∠C + 1∠A) 2 Similarly ∠AXC = ∠ABX +∠BAX = ∠B + 1∠A so 2 XC sin 1∠A = 2 . (3) AC sin(∠B + 1∠A) 2 From (1-3) we get sin(∠C + 1∠A) = sin(∠B+ 1∠A) so ∠C + 1∠A = ∠B+ 2 2 2 1∠A so ∠C = ∠B so AB = AC so ABC is isosceles. 2 Exercise 2.1. The law of sines asserts that for any triangle ABC we have sin∠A sin∠B sin∠C = = BC CA AB Prove this by computing the area of ABC in three ways. Does the argument work for an obtuse triangle? What is the sign of the sine? Exercise 2.2. Prove (1). Hint: Compute the ratio of the area of ABX to the area of ACX in two different ways. 7 3 Affine Geometry 3.1 Lines 3.1. Throughout R denotes the set of real numbers and R2 denotes the set of pairs of real numbers. Thus a point of P ∈ R2 is an ordered pair P = (x,y) of real numbers. Definition 3.2. A line in R2 is a set of form (cid:96) = {(x,y) ∈ R2 : ax+by +c = 0} where a,b,c ∈ R and either a (cid:54)= 0 or b (cid:54)= 0 (or both). Three or more points are called collinear iff there is a line (cid:96) which contains them all. Three or more lines are called concurrent iff they have a common point. Two lines are said to be parallel iff they do not intersect. 3.3. The two most fundamental axioms of plane geometry are Axiom (1) Two (distinct nonparallel) lines intersect in a (unique) point. Axiom (2) Two (distinct) points determine a line. Axiom (1) says that two equations a x+b y +c = 0, a x+b y +c = 0 1 1 1 2 2 2 forlineshaveauniquecommonsolution(theusualcase), nocommonsolution (this means that the lines are parallel), or else define the same line (which is case if and only if the equations are nonzero multiples of one another). The latter two cases are characterized by the condition a b −a b = 0 and in the 1 2 2 1 first case the intersection point is c b −c b a c −a c 1 2 2 1 1 2 2 1 x = − , y = − . a b −a b a b −a b 1 2 2 1 1 2 2 1 Axiom(2)saysthatforanytwodistinctpointsP = (x ,y )andP = (x ,y ) 1 1 1 2 2 2 there is a unique line (cid:96) = {(x,y) : ax+by +c = 0} containing both. Remark 3.5 below gives a formula for this line. 8 Theorem 3.4. (I) Three points P = (x ,y ) are collinear if and only if i i i   x y 1 1 1 det x y 1  = 0. 2 2 x y 1 3 3 (II) Three distinct lines (cid:96) = {(x,y) : a x+b y +c = 0} are concurrent or i i i i parallel if and only if   a b c 1 1 1 det a b c  = 0. 2 2 2 a b c 3 3 3 Proof. A determinant is unchanged if one row is subtracted from another. Hence     x y 1 x −x y −y 0 1 1 1 3 1 3 det x y 1  = det x −x y −y 0 . 2 2 2 3 2 3 x y 1 x y 1 3 3 3 3 Evaluating the determinant on the right gives   x y 1 1 1 det x y 1  = (x −x )(y −y )−(x −x )(y −y ). 2 2 1 3 2 3 2 3 1 3 x y 1 3 3 Dividing by (x − x )(x − x ) shows that the determinant vanishes if and 1 3 2 3 only if y −y y −y 1 3 2 3 = . x −x x −x 1 3 2 3 This last equation asserts that the slope of the line P P equals the slope of 1 3 the line P P . Since P lies on both lines, this occurs if and only if the lines 2 3 3 are the same, i.e. if and only if the points P , P , P are collinear. 1 2 3 The above proof assumes that x ,x (cid:54)= x ; a special argument is required 1 2 3 in the contrary case. We give another proof which handles both cases at the same time. The matrix equation      x y 1 a 0 1 1  x y 1  b  =  0  2 2 x y 1 c 0 3 3 says that the points P lie on the line ax+by+c = 0. Any nonzero solution i (a,b,c) of this equation must have either a (cid:54)= 0 or b (cid:54)= 0 or both. Hence the 9 three points P are collinear if and only if this matrix equation (viewed as i a system of three homogeneous linear equations in three unknowns (a,b,c)) has a nonzero solution. Part (I) thus follows from the following Key Fact. A homogeneous system of n linear equations in n unknowns has a nonzero solution if and only if the matrix of coefficients has determinant zero. Part (II) is similar, but there are several cases. The matrix equation      a b c x 0 1 1 1 0  a b c  y  =  0  (1) 2 2 2 0 a b c 1 0 3 3 3 says that the point (x ,y ) lies on each of the three lines a x+b y +c = 0. 0 0 i i i The three lines are parallel (and not vertical) if and only if they have the same slope, i.e. if and only if −a /b = −a /b = −a /b . This happens if 1 1 2 2 3 3 and only if the matrix equation      a b c 1 0 1 1 1  a b c  m  =  0  (2) 2 2 2 a b c 0 0 3 3 3 has a solution m. The lines are vertical (and hence parallel) if and only if b = b = b = 0. This happens if and only if the matrix equation 1 2 3      a b c 0 0 1 1 1  a b c  1  =  0 . (3) 2 2 2 a b c 0 0 3 3 3 holds. This (and the above Key Fact) proves “only if”. For “if” assume that the matrix equation      a b c u 0 1 1 1  a b c  v  =  0  2 2 2 a b c w 0 3 3 3 has a nonzero solution (u,v,w). If w (cid:54)= 0, then x = u/w, y = v/w 0 0 satisfies (1). If w = 0 and u (cid:54)= 0, then m = v/u satisfies (2). If w = u = 0, then v (cid:54)= 0 so (1) holds. 10