M 460 Supplement: nalytic geometry Donu rapura In the 1600 s Descartes introduced cartesian coordinates which changed the way we now do geometry. This also paved for subsequent developments such as calculus. Here we revisit some parts of Euclidean geometry from this perspective. (Note that we are not trying to be historically accurate here, since we will make use of concepts such as vector sums and dot products which came much later.) Finally, we will end with Klein s conception of what Euclidean geometry actually is. Notation: To avoid confusion, we label theorems by letters here. Numbered theorems refer to theorems in version 5 of McClure s notes. Given a pair of points and B, B may refer the line through B, the line segment B, or the length of this line segment depending on the context which will be made clear. 1 Cartesian Coordinates We choose a point on the plane O called the origin, and draw two perpendicular lines call the x-axis and the y-axis. t the risk of sounding pedantic, we should also choose directions along these axes which tells us whether we are to the left or right of O along the x-axis, or above or below O along the y-axis. Given a point on the plane, we can construct a line through parallel to the y-axis. This line will meet the x-axis at a point, say P. We can measure the distance OP. The x-coordinate a 1 of is +OP if P is to the right of O, and OP if P is the left of O. We also construct a line Q parallel to the x-axis with Q on y-axis. The y-coordinate a 2 of is defined similarly as +OQ if Q is above O and OP otherwise. The quadrilateral P OQ is automatically a rectangle by Theorem. parallelogram with one right angle is a rectangle. Proof. You should check this yourself! (It s too easy to be a homework problem.) 1
y axis Q a 2 O a 1 P x axis The point is determined by the pair (a 1, a 2 ). Conversely, any pair of real numbers corresponds to a point in the plane by reversing this process. To summarize: Theorem B. There is a one to one correspondence between points on the plane and the set of pairs of real numbers denoted by R 2. We will simply regard a point as equal to the pair (a 1, a 2 ) of its coordinates. Note that O = (0, 0). The set of points where a 1 0 and a 2 0 is called the first quadrant. The remaining quadrants are where some or all these numbers are negative. Theorem C. Given two points = (a 1, a 2 ) and B = (b 1, b 2 ). 1. If a 1 = b 1, the distance from to B is b 2 a 2. 2. If a 2 = b 2, the distance from to B is b 1 a 1. Proof. We treat case 2. The line l through perpendicular to the y-axis contains B. By swicthing and B if necessary, we can assume that b 1 a 1, which means that B is to the right of on l. There are 3 cases: (a) and B are to the right of O, that is 0 a 1 b 1. (b) O is in between and B, that is a 1 0 b 1. (c) and B are to the left of O, that is a 1 b 1 0. In case (a), we have b 1 = BO = O + B = a 1 + B. Therefore B = b 1 a 1 = b 1 a 1 In case (b), B = O + BO = a 1 + b 1 = b 1 a 1 2
Finally, in case (c), b 1 = BO = O + B = a 1 + B. Therefore B = a 1 b 1 = b 1 a 1 a 1 b a 1 1 O B a 1 b 1 O B a b 1 1 b 1 B O Theorem D. The distance between = (a 1, a 2 ) and B = (b 1, b 2 ) is (b 1 a 1 ) 2 + (b 2 a 2 ) 2. Proof. Draw the right triangle BC, where C = (b 1, a 2 ). B By the previous theorem, C = b 1 a 1 and BC = b 2 a 2. Then by Pythagoreas theorem (theorem 9), B 2 = C 2 + BC 2 = (b 1 a 1 ) 2 + (b 2 a 2 ) 2 C so that B = (b 1 a 1 ) 2 + (b 2 a 2 ) 2 In the future, we will sometimes write this as d(, B). 3
2 Vector Sum We construct a geometric operation on points called the vector sum. (Incidentally, notions involving vectors evolved in the 19th century long after Descartes.) Given distinct points and B, we define + B as the vertex C of the parallelogram OCB. It is clear that this does not depend on the order, so that + B = B +. C=+B B O Theorem E. If = (a 1, a 2 ) and B = (b 1, b 2 ) are distinct, then +B = (a 1 +b 1, a 2 +b 2 ). In other words, the coordinates of + B are the sum of the coordinates of and B. Proof. We treat the case where and B both lie in the first quadrant. Other cases will be considered in the homework. Let C = + B = (c 1, c 2 ), we have to show that c 1 = a 1 + b 1. By interchanging and B if necessary, we can assume that a 1 b 1. Consider the picture below. 4
H C B G O D E F x axis We assume that D, BE and CF are parallel to the y-axis. Therefore we have a 1 = OD, b 1 = OE and c 1 = OF. We also let G be the point on CF such that BG is parallel to the x-axis. Therefore BEF G is a parallelogram because the sides are parallel. It also follows that BGC = 90. Since D and the x-axis are perpendicular, we have that ( ) BGC = OD Since CBO is a parallelogram, theorem 11 implies that O = BC By BF5, we see that OD = OHF and also that OHF = BCG. Therefore ( ) OD = BCG 5
Since the sum of the angles of OD and BCG are both 180, we deduce from this together with (*) and (**) that OD = CBG. Therefore we may conclude from S (BF3) that OD and BCG are congruent. Thus BG = OD = a 1. Since BEF G is a parallelogram (in fact a rectangle), we deduce that EF = BG = a 1. Consequently The proof that c 2 = a 2 + b 2 is similar. c 1 = OF = OE + EF = b 1 + a 1 We can extend the sum operation to allow = B, by defining + = (2a 1, 2a 2 ) We also write this as 2. More generally, for any real number let t = (ta 1, ta 2 ) Theorem F. The distance from O to t is t O. If t 0, then t lies on the ray O. If t < 0, t lies on the line O on the opposite side from (i.e. not on the ray O). Proof. This will be left as homework. We define subtraction by B = + ( 1)B = (a 1 b 1, a 2 b 2 ) The geometric interpretation is explained by the picture below. B B B 6
3 Midpoints Theorem G. If = (a 1, a 2 ) and B = (b 1, b 2 ) are two distinct points, then M = 1 2 + 1 ( 2 B = a1 + b 1, a ) 2 + b 2 2 2 is the midpoint of B. Proof. We treat the case a 1 b 1 and a 2 b 2. Let M = (m 1, m 2 ) denote the midpoint of B and let C = (b 1, a 2 ), D = (m 1, a 2 ) and E = (b 1, m 2 ). B M E D Since MD and BC are both perpendicular to the x-axis, the x-axis forms a transversal such that the corresponding angles are the same. Therefore M D and BC are parallel. Therefore by theorem 17 of McClure s notes, D is the midpoint of C. Therefore m 1 a 1 = D = 1 2 (b 1 a 1 ) C and so m 1 = 1 2 (b 1 + a 1 ) Similarly, E is a midpoint of BC. Therefore m 2 a 2 = CE = 1 2 (b 2 a 2 ) and so m 2 = 1 2 (b 2 + a 2 ) Theorem H. If = (a 1, a 2 ), B = (b 1, b 2 ) and C = (c 1, c 2 ) are three distinct points, then C is on the line segment B if and only if C = (1 t) + tb, for some 0 t 1. Proof. We have to show that c 1 = (1 t)a 1 + tb 1 and c 2 = (1 t)a 2 + tb 2 for some 0 t 1. ssume for simplicity that a 1 b 1 and a 2 b 2. Let D = (b 1, a 2 ) and let E = (c 1, a 2 ) 7
B C E D Let t = E D 0 Since E is a part of D, E D so that t 1. By theorem D, E = c 1 a 1, CE = c 2 a 2 D = b 1 a 1, BD = b 2 a 2 The triangles BD and CE share the angle. We have CE = BD = 90. Therefore CE = 180 CE = 180 BD = BD It follows that BD and CE are similar, since they have equal angles. Therefore by BF4. By substitution, we obtain CE BD = E D = t c 1 a 1 b 1 a 1 = t By algebra, c 2 a 2 b 2 a 2 = t c 1 = a 1 + t(b 1 a 1 ) = (1 t)a 1 + tb 1 c 2 = a 2 + t(b 2 a 2 ) = (1 t)a 2 + tb 2 8
4 Centroid Recall that a median of triangle is a line joining a midpoint to the opposite vertex. Theorem I. If = (a 1, a 2 ), B = (b 1, b 2 ) and C = (c 1, c 2 ) are three distinct points, then G = 1 3 + 1 3 B + 1 3 C lies on each of the medians of BC. In particular, the medians are concurrent Note that this gives a new proof of theorem 29. The formula shows that the centroid G coincides with the center of mass that you have learned about in other courses. Proof. Let M, N and P be the midpoints of BC, C and B respectively. By theorem G, M = 1 (B + C) 2 By algebra, G = 1 3 + 2 3 M By theorem H, G lies on M. By a similar argument, we can see that G lies on BN and CP. Theorem J. If = (a 1, a 2 ), B = (b 1, b 2 ) and C = (c 1, c 2 ) be the vertices of a triangle, then D lies inside BC if and only if D = r + sb + tc where 0 < r < 1, 0 < s < 1 and 0 < t < 1 and r + s + t = 1. Proof. Homework. Given a quadrilateral BCD, we can define its centroid as 1 4 + 1 4 B + 1 4 C + 1 4 D We will leave it for homework to discover what this means geometrically. 5 Dot Products We start with a converse to Pythagoreas theorem. Theorem K. Given a triangle with sides labeled a, b, c. If c 2 = a 2 + b 2, then the triangle is a right triangle with c its hypotenuse. 9
Proof. Let 1 be the angle opposite c. There are three cases 1 < 90, 1 > 90 and 1 = 90. It suffices to prove that the first two cases are impossible if c 2 = a 2 + b 2. We treat only the first in detail. Draw in the altitude from the angle opposite b. Let h be the height. The side b is divided into two nonzero parts b = d + e as pictured. a h c 1 By Pythagoreas theorem d e a 2 = d 2 + h 2 c 2 = e 2 + h 2 Since 1 < 90, we must have d 0. Therefore d 2 > 0. dding h 2 to both sides and using the first equation implies a 2 > h 2 Since b > e, we have dding these two inequalities yields which contradicts c 2 = a 2 + b 2. b 2 > e 2 a 2 + b 2 > c 2 We define the dot product of two points by (a 1, a 2 ) (b 1, b 2 ) = a 1 b 1 + a 2 b 2 Theorem L. Let, B, C, D be points, and t be a number. Then 1. B = B. 2. t( B) = (t) B = (tb) 3. ( + B) (C + D) = C + D + B C + B D. 4. ( B) ( B) equals d(, B) 2 (= the length of B squared). 10
Proof. The proof of the first three formulas is straight forward algebra. For the last item, we have ( B) ( B) = (a 1 b 1, a 2 b 2 ) (a 1 b 1, a 2 b 2 ) = (a 1 b 1 ) 2 + (a 2 b 2 ) 2 This is d(, B) 2 by theorem D. Theorem M. B = 0 if and only if O and BO are perpendicular. Proof. Consider the triangle OB. By the previous theorem, O 2 =, BO 2 = B B and B 2 = ( B) ( B) = B B + B B Therefore ( ) B 2 = O 2 + BO 2 2 B If OB = 90, then Pythagoreas theorem implies B 2 = O 2 + BO 2 When combined with ( ) this forces B = 0. Conversely, if B = 0. Then ( ) implies Therefore, by theorem K, OB = 90. By an similar argument, we get B 2 = O 2 + BO 2 Theorem N. ( C) (B C) = 0 if and only if CB = 90. 6 Circumcenter We can now give a proof of theorem 24 of McClure s notes using analytic geometry. In order to reduce the complexity of the proof, we start with a preliminary and somewhat technical result called a lemma. Lemma 1. Let D, E and X be three distinct points, and let Q be the midpoint of DE. Then XQ is a perpendicular bisector of DE if and only if 2X (E D) = E E D D Proof. By theorem N, XQ is perpendicular to DE if and only if or equivalently if and only if (X Q) (E Q) = 0 ( ) X (E Q) = Q (E Q) 11
Since Q is a midpoint, Q = 1 2 D + 1 2 E by theorem G. Substituting this into the left and right side of (*) gives X (E Q) = 1 X (E D) 2 and Q (E Q) = 1 4 (E + D) (E D) = 1 (E E D D) 4 Substituting these into (*) yields or 1 2 X (E D) = 1 (E E D D) 4 2X (E D) = E E D D Theorem O. The perpendicular bisectors of the sides of a triangle BC are concurrent. Proof. Let M, N, P be the midpoints of B, C and BC respectively. Let X be the intersection of the perpendicular bisectors of B and C. Then by lemma 1, and Subtracting these equations yields which simplifies to 2X (B ) = B B 2X (C ) = C C 2X (C ) 2X (B ) = C C B B + 2X (C B) = C C B B Lemma 1 will then imply that X lies on the perpendicular bisector of BC as required. 7 rea of a triangle Theorem P. If = (a 1, a 2 ), B = (b 1, b 2 ) and O = (0, 0) are distinct points, then the area of BO is 1 2 a 1b 2 a 2 b 1 12
( You ) may recognize the expression (a 1 b 2 a 2 b 1 ) as the determinant of the matrix a1 a 2. b 1 b 2 Proof. We prove this under that a 1, a 2, b 1, b 2 0. By switching, B if necessary, we can assume that b 2 a 2. There are two cases to consider, b 2 = a 2 or b 2 > a 2. We just do the last case, and leave the first for homework. Then the line B is not parallel to the x-axis, so it must meet it a point C = (c 1, 0). B O If a 1 = b 1, then B is a vertical line, so that c 1 = a 1. Otherwise the points (x, y) on B satisfy an equation y = mx + e for constants m, c, with m 0. Substituting, B and C into this yields Some algebra shows a 2 = ma 1 + e b 2 = mb 1 + e 0 = mc 1 + e m = a 2 b 2 a 1 b 1 e = a 2 ma 1 ( ) c 1 = e m = a 1b 2 a 2 b 1 b 2 a 2 The area of BO is the area of CBO minus the area of CO, which is 1 2 b 2c 1 1 2 a 2c 1 fter substituting c 1 = a 1 = b 1 in the first case, or ( ) in the second, we get rea( BC) = 1 2 (a 1b 2 a 2 b 1 ) C 13
8 Epilogue It should already be clear by now that many results in Euclidean geometry can be approached and solved using analytic geometry. What may not be clear is that all of Euclidean geometry can be reduced to analytic geometry (in theory, although usually not in practice). We explain the rough idea. We first need to translate concepts such as congruence of triangles to the analytic side. The idea, which is already implicit in Euclid, is that two triangles are congruent if you can move one so that it lays on top of the other. This idea of moving points in the plane can be made mathematically precise by the (modern) notion of a transformation or function T : R 2 R 2. This is a process which takes a point P to a new point T (P ). We say that T is one to one and onto if for any point P, there exists a unique point Q with T (Q) = P. This implies that we can define a new transformation T 1 : R 2 R 2, called the inverse, by T 1 (P ) = Q. transformation is called an Euclidean motion if it is one to one and onto, and it preserves distance, i.e. the distance from P to Q equals the distance from T (P ) to T (Q). So now we can make the definition of congruence in this setting precise. triangle BC is simply given by the ordered triple of points, B, C. Triangles BC and DEF are congruent if there exists a Euclidean motion T such that T () = D, T (B) = E and T (C) = F. There are some wrinkles with this approach. Things like SSS and SS, which were previously regarded as axioms or basic facts, now have to be proved. We will give the proof of SSS below. We first start with some basic examples of Euclidean motions. Given a point, translation by is defined by T (X) = X +. Reflection about the y-axis is defined by F (x, y) = (x, y). Rotation counterclockwise about O through an angle θ is given by R(x, y) = (x cos θ y sin θ, x sin θ + y cos θ) Theorem Q. The above transformation T, F, R are all Euclidean motions. Proof. Suppose that = (a 1, a 2 ), P = (p 1, p 2 ), and Q = (q 1, q 2 ). By theorem D, d(t (P ), T (Q)) = ((p 1 + a 1 ) (q 1 + a 1 )) 2 + ((p 2 + a 2 ) (q 2 + a 2 )) 2 = (p 1 q 1 ) 2 + (p 2 q 2 ) 2 = d(p, Q) The Euclideanness of L is even easier, and we omit it. The proof for R is potentially pretty messy, but we use some tricks to simplify things. Set (x, y) = P Q. By theorem L, d(p, Q) 2 = (P Q) (P Q) = x 2 + y 2. While d(r(p ), R(Q)) = (R(P ) R(Q)) (R(P ) R(Q)) = R(P Q) R(P Q) = (x cos θ y sin θ) 2 + (x sin θ + y cos θ) 2 = x 2 + y 2 by algebra and the familiar identity sin 2 θ + cos 2 θ = 1. 14
Given transformations S : R 2 R 2 and T : R 2 R 2, we can define a new transformation, called the composition, ST : R 2 R 2 by ST (P ) = S(T (P )). Theorem R. 1. If T is a Euclidean motion, then so is T 1. 2. If S and T are Euclidean motions, then so is ST. 3. If BC and DEF are both congruent to XY Z, then they are congruent to each other. Proof. Suppose S and T are Euclidean, then if T (P ) = P and T (Q) = Q, we have This proves 1. Furthermore d(p, Q ) = d(t (P ), T (Q)) = d(p, Q) = d(t 1 (P ), T 1 (Q )) d(st (P ), ST (Q)) = d(t (P ), T (Q)) = d(p, Q) proves 2. Suppose S takes, B, C to X, Y, Z, and T takes D, E, F to X, Y, Z. Then T 1 S is a Euclidean motion taking, B, C to D, E, F. Theorem S. If BC and DEF have equal sides, they are congruent. Proof. We let r = B = DE, s = C = DF and t = BC = EF. We construct a new triangle XY Z as follows. Let X = O = (0, 0) and Y = (r, 0). Let C 1 be the circle centered at X with radius s, and let C 2 be the circle centered at Y with radius t. In terms of equations C 1 = {(x, y) R 2 x 2 + y 2 = s 2 } C 2 = {(x, y) R 2 (x r) 2 + y 2 = t 2 } We can determine the points of intersection by simultaneously solving the 2 equations. Subtracting the second from the first and simplifying gives so that 2rx r 2 = x 2 + y 2 (x r) 2 y 2 = s 2 t 2 x = x 0 = s2 t 2 + r 2 2r Substituting into the first equation gives y = ± s 2 x 2 0 Let Z the point (x 0, y) with y > 0. The remaining solution Z = (x 0, y) = F (Z). Since F (X) = X and F (Y ) = Y, we see that XY Z is congruent to XY Z. 15
We will show that BC and XY Z are congruent. Let T (P ) = P be the translation taking to X = O. Now T (B) is on the circle of radius r centered at O. Then let R be the rotation which sends T (B) to Y. So RT () = X and RT (B) = Y. The third point RT (C) has distance s from X and t from Y. So it must be either Z or Z. In either case, we see that BC and XY Z are congruent. By a similar argument, DEF and XY Z are congruent. Therefore BC and DEF are congruent. It should be clear that the set of Euclidean motions was a key player in the last argument. Theorem R tells us that this set is more than just a set, we have multiplication and inverse operations, and these are known to satisfy appropriate laws such as the associative law. One says that the set of Euclidean motions with these operations forms group, called the Euclidean group. In the 19th century, Felix Klein proposed a completely new viewpoint. He suggested that Euclidean geometry is the study of those properties of the plane which are invariant under the Euclidean group 1. This means that the property holds for T (X) whenever it holds for X. This is actually a pretty natural extension of the idea that a theorem in Euclidean geometry proved for one triangle should hold for any triangle congruent to it. These ideas of invariance under a group turn out to be really important in other areas of math and its applications to physics for example. (In case you ever wondered why a bicycle stays upright when you ride it, it is because of the law of conservation of angular momentum. This in turn is a consequence of the fact the total energy of the system, or more accurately the Hamiltonian function, is invariant under the group of three dimensional rotations.) 1 This isn t quite correct. We want to include additional transformations called dialations to account for similarity of triangles. 16