Geometry JWR. Monday September 29, 2003

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1 Geometry JWR Monday September 29, Foundations In this section we see how to view geometry as algebra. The ideas here should be familiar to the reader who has learned some analytic geometry (including vectors) and linear algebra. 1. A transformation T : R 2 R 2 is called affine iff it has the form (x, y ) = T (x, y) x = ax + by + p y = cx + dy + q. where ad bc 0. In matrix notation this takes the form T (X) = AX + B, where [ x T (X) = y ] [ x, X = y ] [ a b, A = c d ] [ p, B = q ], and det(a) 0. The affine transformation T is called linear iff B = 0; a translation iff A = I (the identity matrix); orientation preserving iff det(a) > 0; orientation reversing iff det(a) < 0; Euclidean iff A 1 = A T (the transpose of A); 1

2 a rotation iff A 1 = A T, B = 0, and det(a) > 0; a similarity transformation iff A 1 = µ 2 A T for some µ R; a dilation iff A = µi (for some µ > 0) and B = 0. Theorem 2. The set of all affine transformations is a group, i.e. (1) the identity transformation I(x, y) = (x, y) is affine, (2) the composition T 1 T 2 of two affine transformations T 1 and T 2 is affine, and (3) the inverse T 1 of an affine transformation T is affine. Similarly each of the following sets is a group: the set of linear transformations, the set of orientation preserving affine transformations, the set of Euclidean affine transformations, the set of conformal affine transformations, the set of translations, the set of rotations, and the set of dilations. Example 3. For each real number α the matrix [ ] cos α sin α R α = sin α cos α represents a rotation. Using the trigonometric addition formulas it is easy to see that R 0 = I, R α R β = R α+β, Rα 1 = R α. 4. A line is a subset l of R 2 of form l = {(x, y) R 2 : ax + by + c = 0} where either a 0 or b 0 (or both). Given two distinct points P and Q there is a unique line containing both; it is given by (See Theorem 8 below.) The set l = {tp + (1 t)q : t R}. QP := {tp + (1 t)q : 0 t 1} 2

3 is called the line segment joining Q and P and a set of form ρ = {tp + (1 t)q : t 0} is called a ray emanating from Q. Three points are called collinear iff they lie on a common line; three lines are called concurrent iff the pass through a common point. Two lines are called parallel iff they do not intersect. When three distinct points P, Q, R are collinear, we say that R lies between P and Q iff R QP and R P, Q. Theorem 5. An affine transformations maps lines to lines, i.e. if T is an affine transformation and l is a line, then T (l) := {T (P ) : P l} is a line. Theorem 6. Three points P i = (x i, y i ) are collinear if and only if x 1 y 1 1 det x 2 y 2 1 = 0. x 3 y 3 1 Three distinct lines l i = {(x, y) : a i x + b i y + c i = 0} are either concurrent or parallel if and only if a 1 b 1 c 1 det a 2 b 2 c 2 = 0. a 3 b 3 c 3 Proof. Hint: Let M be a 3 3 matrix. When does the homogeneous linear system MZ = 0 have a nonzero solution Z? Corollary 7. Let P = (x 1, y 1 ) and Q = (x 2, y 2 ) be distinct points, then there is a unique line containing P and Q. I Proof. Hint: A point P = (x, y) lies on this line if and only if the points P 1, P 2, P are collinear. Find a, b, c by expansion by minors in Theorem 6. Show that two lines l = {(x, y) R 2 : ax + by + c = 0}, l = {(x, y) R 2 : a x + b y + c = 0} are the same if and only the coefficients are proportional, i.e. (a, b, c ) is a nonzero multiple of (a, b, c). 3

4 Theorem 8. Let P = (x 1, y 1 ) and Q = (x 2, y 2 ) be distinct points. Then the set l = {tp + (1 t)q : t R} is the unique line containing P and Q. 9. Define the inner product U, V between two vectors U = (u 1, u 2 ) and V = (v 1, v 2 ) by U, V := u 1 v 1 + u 2 v 2. Define the norm U of the vector U by U 2 = U, U. The distance between the two points P = (x 1, y 1 ) and Q = (x 2, y 2 ) is the norm of the difference: P Q := ((x 1 x 1 ) 2 + (y 1 y 2 ) 2. Theorem 10. A Euclidean affine transformation T preserves distance, i.e. for P, Q R 2. T (P ) T (Q) = P Q Proof. Hint: Show that AU, V = U, A T V. Then consider AU, AU. Corollary 11. A similarity transformation T preserves ratios of distances, i.e. T (P 1 ) T (P 2 ) T (P 3 ) T (P 4 ) = P 1 P 2 P 3 P 4 for P 1, P 2, P 3, P 4 R 2. Theorem 12. Any affine transformation preserves ratios of distances of collinear points, i.e. T (P 1 ) T (P 2 ) T (P 3 ) T (P 4 ) = P 1 P 2 P 3 P 4 whenever P 1, P 2, P 3, P 4 R 2 are collinear. Proof. Hint: P i = t i P 1 (1 t i )P 2. Theorem 13 (Parallel postulate). Given a point P and a line l not containing P there is a unique line parallel to l and containing P. 4

5 2 Angles Definition 14. For three points O = (x 0, y 0 ), P = (x 1, y 1 ), Q = (x 2, y 2 ) of R 2 with P, Q O define cos P OQ = (x 1 x 0 )(x 2 x 0 ) + (y 1 y 0 )(y 2 y 0 ) P O Q O sin P OQ = (x 1 x 0 )(y 2 y 0 ) (y 1 y 0 )(x 2 x 0 ) P O Q O In this section the equation P O Q = P OQ is an abbreviation for the two equations cos P O Q = cos P OQ and sin P O Q = sin P OQ. Theorem 15. ( cos P OQ ) 2 + ( sin P OQ ) 2 = 1. Proof. Hint: Prove (u 1 v 1 + u 2 v 2 ) 2 + (u 1 v 2 u 2 v 1 ) 2 = (u u 2 2)(v v 2 2) and take u 1 = x 1 x 0, v 1 = y 1 y 0, u 2 = x 2 x 0, v 2 = y 2 y 0. Remark 16. With these substitutions, the formulas in Definition 14 take the form u v = cos θ u v, u v = sin θ u v k where θ is the angle between the vectors u = u 1 i + u 2 j and v = v 1 i + v 2 j. (The operations u v and u v are the dot product and cross product from second semester calculus.) Definition 17. Henceforth we view an angle as a pair (c, s) of numbers such that c 2 + s 2 = 1. Thus three points P, O, Q with P, Q O determine the angle (c, s) where c = cos P OQ, s = sin P OQ. Given an angle θ = (c, s) we write cos θ = c, sin θ = s. In this notation 0 = (1, 0), 90 = (0, 1), and 180 = ( 1, 0). 5

6 Theorem 18. A similarity transformation T preserves angles. More precisely, if P, Q O and P = T (P ), O = T (O), Q = T (Q), then cos P O Q = cos P OQ and sin P O Q = ± sin P OQ where we take the plus sign if T is orientation preserving and the minus sign if T is orientation reversing. Theorem 19. The angle P OQ depends only on the rays emanating from O through P and Q. More precisely, if P, Q O and if P = tp + (1 t)o and Q = sq + (1 s)o, where t, s > 0, then P OQ = P OQ. Theorem 20 (Vertical angles are equal). If line P Q and P Q intersect in a point O such that O is between P and Q and O is between P and Q, then P OQ = P OQ. Theorem 21 (Alternate angles are equal). If lines OP and O P are parallel, the points O, O, Q are collinear, and cos OP Q and cos O P Q have the same sign, then OP Q = O P Q. Theorem 22 (Trigonometric addition formulas). Let O, P, Q, R be points with P, Q, R O. Let Then α = P OQ, β = QOR, γ = P OR. cos γ = cos α cos β sin α sin β, cos γ = sin α cos β + cos α sin β. Definition 23. The formula γ = α + β means that α, β, γ are related as in Theorem 22. Also if θ = (c, s) we define θ = (c, s). With these notations angles may be added and subtracted and the associative and commutative laws hold. Also the law θ + 0 = θ holds. 3 Triangles 24. The notation used in geometry books is generally less precise then the notation of Sections 1 and 2. For example, depending on the context, the 6

7 notation AB might denote the line through the points A and B, the line segment joining A and B, or the distance from A to B. When no confusion can result the angle ABC might be denoted by B. Also usually we don t distinguish the angles ABC and CBA so that two angles B and B are deemed equal if cos B = cos B. (It then follows that sin B = ± sin B.) 25. A triangle is a triple of three non collinear points. The points are called the vertices of the triangle. The notation ABC is used to denote the triangle whose vertices are A, B, and C. The line segments connecting the vertices are called the sides of the triangle. The lines extending these line segments are also called the sides of the triangle or sometimes the extended sides to emphasize the difference. Two triangles ABC and A B C are said to be congruent iff there is a Euclidean affine transformation T such that A = T (A), B = T (B), C = T (C), Two triangles ABC and A B C are said to be similar iff there is a similarity transformation T such that A = T (A), B = T (B), C = T (C). Theorem 26. The angles of a triangle sum to 180. Lemma 27. (i) For any two points P and Q there is an orientation preserving Euclidean transformation such that T (P ) = (0, r), T (Q) = (0, 0) where r = P Q. (ii) For any triangle ABC there is a Euclidean transformation T and numbers a, b, c such that T (A) = (a, 0), T (B) = (b, 0), T (C) = (0, c). (iii) For any triangle ABC there is an affine transformation T such that T (A) = (1, 0), T (B) = (0, 1), T (C) = (0, 0). Theorem 28 (SSS,SAS,ASA). For triangles ABC and A B C following conditions are equivalent: the (i) the triangles ABC and A B C are congruent; (ii) AB = A B, BC = B C, and CA = C A ; (iii) AB = A B, BC = B C and B = B ; 7

8 (iv) A = A, B = B, and AB = A B. Theorem 29 (AAA,S:S:S,S:A:S). For triangles ABC and A B C the following conditions are equivalent: (i) the triangles ABC and A B C are similar; (ii) A = A, B = B, and C = C ; (iii) the sides are proportional, i.e. A B AB = B C BC = C A CA ; (iv) A = A and A B AB = C A CA. 4 Ceva and Menelaus 30. Let P, Q, R be points on (extended) sides BC, CA, AB of triangle ABC. Thus there are numbers p, q, r with P = pb + (1 p)c, Q = qc + (1 q)a, R = ra + (1 r)b. The six distances BP, P C, CQ, QA, AR, RB in Theorems 31 and 32 below are signed, i.e. XY = Y X. The sign convention is such that the signs of BP, P C, CQ, QA, AR, RB are the same as the signs of 1 p, p, 1 q, q, 1 r, r. Theorem 31 (Menelaus). The points P, Q, and R are collinear if and only if BP P C CQ QA AR RB = 1 Proof. (See also [1] page 146 and [2] page 66.) By Theorem 5 the condition that the lines AP, BQ, and CR are concurrent is preserved by affine transformations. By Theorem 12 the ratios (and hence product of the ratios) is preserved by affine transformations. Hence by Theorem 27 we may assume that A = (1, 0), B = (0, 1) and C = (0, 0) so P = (0, p), Q = (1 q, 0) and R = (r, 1 r). By Theorem 6 the points P, Q, R are collinear if and only if det 0 p 1 1 q 0 1 r 1 r 1 8 = 0.

9 The condition on the ratios is 1 p p 0 1 q 2(1 r) q 0 = 1. 2(r 0) Both conditions simplify to 1 p q r + pq + qr + rp = 0. Theorem 32 (Ceva). The lines AP, BQ, and CR are concurrent if and only if BP P C CQ QA AR RB = 1. Proof. (See also [1] page 126 and [2] page 4.) As in Theorem 31 we may assume that A = (1, 0), B = (0, 1), C = (0, 0), P = (0, p), Q = (1 q, 0), R = (r, 1 r). The lines AP, BQ, CR have equations px + y = p, x + (1 q)y = 1 q, (r 1)x + ry = 0. By Theorem 6 these lines are concurrent if and only if p 1 p det 1 1 q q 1 = 0. r 1 r 0 The condition on the ratios is 1 p p 0 1 q 2(1 r) q 0 = 1. 2(r 0) Both conditions simplify to 1 p q r + pq + qr + rp = 2pqr. 33. Ceva s Theorem has the following four corollaries. The medians of a triangle are the lines connecting the vertices to the midpoints of the opposite sides. The triangle formed by joining this midpoints is called the medial triangle. The altitudes of a triangle are the lines through the vertices perpendicular the opposite sides. The triangle formed by joining the feet of the altitudes is called the orthic triangle. The perpendicular bisector of a segment is the line perpendicular to the segment and passing through its midpoint. The bisector of an angle P OQ is the line OR such that P OR = ROQ. Corollary 34. The medians of a triangle are concurrent. 9

10 Proof. The three ratios in Ceva s Theorem are all one. Corollary 35. The altitudes of a triangle are concurrent. Proof. See [1] page 61 Theorem 2.10 and [2] page 5 Exercise 2. The lengths in Ceva s Theorem are BP = AB cos B, P C = CA cos C, CQ = BC cos C, QA = AB cos A, AR = CA cos A, RB = BC cos B. Corollary 36. The angle bisectors of a triangle are concurrent. Proof. See [1] page 21 Theorem 1.12 and [2] page 10 Theorem Remark 37. Another proof is as follows. The bisector of an angle is the locus of all points equidistant from the arms of the angle. Hence the point of intersection of two angle bisectors of ABC is equidistant from all thee sides and hence must lie on the third angle bisector. Corollary 38. The perpendicular bisectors of a triangle are concurrent. Proof. These are the altitudes of the medial triangle. Remark 39. Another proof is as follows. The perpendicular bisector of a segment is the locus of all points equidistant from the endpoints of the segment. Hence the point of intersection of two perpendicular bisectors of ABC is equidistant from the vertices and hence must lie on the third perpendicular bisector. Corollary 40. Let X BC, Y CA, Z AB be the points of tangency of the inscribed circle of triangle ABC. Then the lines AX, BY, CZ are concurrent. Their common intersection is called the Gergonne point of ABC. 5 The extended law of sines 41. We frequently use the convention that for a triangle ABC the sides are denoted by the lower case letter corresponding to the name of the opposite vertex. Thus a = BC, b = CA, c = AB. Often a, b, c denote the lengths of the sides and not the sides themselves. 10

11 Theorem 42. For triangle ABC there is a unique circle, called the circumcircle containing the vertices A, B, and C. The center of the circumcircle is called the circumcenter it is usually denoted by O and its radius is called the circumradius and is usually denoted by R. Lemma 43. Theorem 44 (Extended law of sines). For triangle ABC we have A a = B b where R is the circumradius of ABC. = C c = 2R Proof. See [2] Theorem 1.1 or [1] Theorem 2.3. The proof uses the theorem that an inscribed angle in circle is equal in degrees to half the subtended arc; see [1] Theorem Euler Line and nine point Circle 45. For any triangle ABC the circumcenter O of is the center of the circumscribed circle; it is also the intersection point of the perpendicular bisectors of the sides; the incenter I is the center of the inscribed circle; it is also the intersection point of the angle bisectors; the centroid G is the intersection of the medians; the orthocenter H is the intersection of the altitudes; the medial triangle is the triangle formed by joining the midpoints of the sides; the orthic triangle is the triangle formed by joining the feet of the altitudes; the nine point circle 1 is the circumcircle of the medial triangle; 1 The nine point circle is so called because of Theorem 50 below. It is also sometimes called the Euler circle or the Feuerbach circle. 11

12 the nine point center N is the center of the nine point circle. Theorem 46. A triangle and its medial triangle have the same centroid. Theorem 47. The circumcenter of a triangle is the orthocenter of its medial triangle. Theorem 48. The points O, G, H are collinear with G between O and H and HG = 2GO. The line containing O, G, H is called the Euler line. Proof using transformations. Consider the similarity transformation T which fixes the centroid G, contracts the distance to G be a factor of 1/2, and rotates through 180 about G. If G = (0, 0) then T (x, y) = ( x/2, y/2). The transformation T maps the original triangle to its medial triangle and sends an altitude of the original triangle to an altitude of the medial triangle, i.e. a perpendicular bisector of the original triangle. Hence T (H) = O. But T (G) = G and for any point P, the points P, G, and P = T (P ) are collinear with 2P G = P G. Corollary 49. The nine point center N is the midpoint of the segment HO. The radius of the nine point circle is half the radius of the circumcircle. Proof. Theorem 50. The nine point circle contains the following nine points: (i) the midpoints of the sides; (ii) the feet of the altitudes; (iii) the midpoints (called the Euler points) of the segments AH, BH, CH. Proof. Denote the midpoints of the sides of ABC, the feet of the altitudes, and the Euler points by by M A, M B, M C, H A, H B, H C, E A, E B, E C, with the subscripts chosen so that the medians are AM A, BM B, CM C, the altitudes are AH A, BH B, CH C, and the points E A, E B, E C lie on these altitudes respectively. We claim that M A M B E B E A is a rectangle. This is because M A M B C is similar to ABC by SAS so M A M B is parallel to AB; E A E B H is similar to ABH by SAS so E A E B is parallel to AB; 12

13 AM B E A is similar to ACH C by SAS so M B E A is parallel to CH C and hence perpendicular to the parallel lines AB and M A M B ; BM A E B is similar to BCH C by SAS so M A E B is parallel to CH C and hence perpendicular to the parallel lines AB and M A M B. The rectangle M A M B E B E A is inscribed in a circle with diameters M A E A and M B E B. Reading B and C for A and B we obtain that the rectangle M B M C E C E B is inscribed in a circle with diameters M B E B and M C E E C. These two circles share a common diameter M B E B and so must be the same circle. As this circle contains M A, M B, M C it is the circumcircle of the medial triangle, i.e. the nine point circle. The lines E A H A and AHA A are the same as are the lines HAM A and BC so (as AH A is an altitude of ABC) E A H A M A is a right angle. Hence H A (and similarly A B and H C ) lie on the none point circle. Remark 51. Here is a coordinate proof. For the triangle ABC with A = (a, 0), B = (b, 0), and C = (0, c) we have ( ) ( a + b ab + c2 a + b O =,, G = 2 2c 3, c ) (, H = 0, ab ). 3 c In the notation of the proof of Theorem 48, ( ) a + b N = 4, c2 ab. 4c etc. Theorem 52. The orthocenter H of an acute angled triangle is the incenter of its orthic triangle. Proof. We must show H A H C H = H B H A H. (The same argument shows H C H B H = H C H B H and H C H B H = H A H C H.) Now the quadrilateral HH C BH A is cyclic (i.e. is inscribed in a circle) since HH C B = HH A B = 90. Hence HH A H C = HBH C. But HBH C = H B BA is the complement α of CBA so HH A H C = α. The same argument (reverse the roles of B and C) shows that HH A H B = α as required. Remark 53. For an obtuse triangle the argument shows that H is the center of one of the exscribed circles, not the center of the inscribed circle. 13

14 Corollary 54. The triangle of least perimeter inscribed in a given triangle ABC is the orthic triangle of ABC. Proof. For three points X, Y, Z, on lines BC, CA, AB respectively let f = f(x, Y, Z) = XY + Y Z + ZX. This function is continuous. When one or more of the points X, Y, Z are far from A, B, and C the function f is large so the minimum occurs either a point where the partial derivatives of f are zero, or else at a point where the function f is not differentiable. The function f is differentiable except where one of the three lengths is zero, i.e. when X = Y = C or Y = Z = A or Z = X = B. Theorem 55 (Feuerbach). The nine point circle is tangent to each of the four tritangent circles (i.e. the inscribed circle and the three exscribed circles). Proof. Later References [1] I. M. Isaacs: Geometry for College Students, Brooks/Cole; [2] H. S. Coxeter & S. L. Greitzer: Geometry Revisted, reissued by MAA; [3] G. A. Jennings: Modern Geometry with Applications, Springer Universitext. 14

Classical Theorems in Plane Geometry 1

BERKELEY MATH CIRCLE 1999 2000 Classical Theorems in Plane Geometry 1 Zvezdelina Stankova-Frenkel UC Berkeley and Mills College Note: All objects in this handout are planar - i.e. they lie in the usual