Advanced Euclidean Geometry

Transcription

1 dvanced Euclidean Geometry Paul iu Department of Mathematics Florida tlantic University Summer 2016 July 11 Menelaus and eva Theorems

2 Menelaus theorem Theorem 0.1 (Menelaus). Given a triangle with points,, on the side lines,, respectively, the points,, are collinear if and only if = 1. W 1

3 Menelaus theorem Proof. (= ) W Let W be the point on such that W//. Then, = W, and = W. It follows that = W W = W W = 1. 2

4 Menelaus theorem ( =) Suppose the line joining and intersects at. From above, = 1 =. It follows that =. The points and divide the segment in the same ratio. These must be the same point, and,, are collinear. 3

5 Example: The external bisectors The external angle bisectors of a triangle intersect their opposite sides at three collinear points. c b a Proof. If the external bisectors are,, with,, on,, respectively, then = c b, = a c, = b a. It follows that = 1 and the points,, are collinear. 4

6 Example Given triangle and points on, on, and on the extension of, such that,, are collinear. If = x, = z, and = y, and two of these lengths are given, calculate the remaining one. x a b y z c 5

7 x a b y z c (a, b, c) x y z (3, 4, 5) (3, 5, 6) 1 4 (3, 5, 7) 1 6 (4, 5, 6) 3 4 (4, 5, 7) 2 4 (5, 6, 7) 1 6 (6, 7, 8) 4 6 6

8 (2, 3, 4) triangle is a triangle with a =2, b =3, c =4. transversal intersects the sidelines at,, such that = = = t. alculate t

9 (7, 12, 18) triangle is a triangle with a =7, b =12, c =18. transversal intersects the sidelines at,, such that = = = t. alculate t

10 (9, 10, 12) triangle is a triangle with a =9, b =10, c =12. transversal intersects the sidelines at,, such that = = = t. alculate t

11 Example Given three circles with centers,, and distinct radii, show that the exsimilicenters of the three pairs of circles are collinear. 10

12 Line with equal intercepts on sidelines of a given triangle Given triangle, construct a line intersecting at externally, at and at internally so that = =. x x x 11

13 Solution Given triangle, construct a line intersecting at externally, at and at internally so that = =. x r r I If = x, by Menelaus theorem, we require a + x x x b x x c x = 1. From this, Note that x = bc a + b + c. bc sin x = 2s sin = Δ s 1 sin = r sin. This means that I is parallel to, and suggests the following simple construction of the line. 12

14 onstruction x r r I (1) onstruct the incenter I of triangle. (2) onstruct a line through I parallel to, to intersect at. (3) onstruct a circle with center, radius, to intersect externally at and internally at. Then,, are collinear with = = = r sin. 13

15 Line with equal intercepts on sidelines of a given triangle Given triangle, construct a line intersecting at externally, at and at internally so that = =. y y y 14

16 Solution Given triangle, construct a line intersecting at externally, at and at internally so that = =. c I c c y y y c If = = = y, by Menelaus theorem, we require a + y y y b y c y y = 1. From this, ca y = a + b c = r c sin, where r c is the radius of the excircle on the side. 15

17 eva s theorem Theorem 0.2 (eva). Given a triangle with points,, on the side lines,, respectively, the lines,, are concurrent if and only if =+1. P 16

18 Proof P Proof. (= ) Suppose the lines,, intersect at a point P. onsider the line P cutting the sides of triangle. y Menelaus theorem, P P = 1, or P P =+1. lso, consider the line P cutting the sides of triangle. y Menelaus theorem again, P P = 1, or P P =+1. Multiplying the two equations together, we have =+1. ( =) Exercise. 17

19 Example Given triangle and points on, on, on such that the cevians,, are concurrent. If = x, = z, and = y, and two of these lengths are given, calculate the remaining one. a x b y z c (a, b, c) x y z (3, 4, 6) (3, 5, 6) 1 4 (3, 5, 7) 1 3 (4, 5, 6) 3 4 (4, 5, 7) 1 4 (5, 6, 7) 3 4 (6, 7, 9)

20 (3, 4, 6) triangle is a triangle with a =3, b =4, c =6.,, are points on,, respectively such that = = = t. If the cevians,, are concurrent, calculate t

21 Example is a right triangle. Show that the lines,, and Q are concurrent. P Q 20

22 Solution is a right triangle. Show that the lines,, and Q are concurrent Q Let intersect at 0, intersect at 0, and Q intersect at = = 2 = 1. y eva s theorem, the lines 0, 0, 0 are concurrent. 21

23 The centroid If D, E, F are the midpoints of the sides,, of triangle, then clearly F F D D E E =1. The medians D, E, F are therefore concurrent. Their intersection is the centroid G of the triangle. F G E D onsider the line GE intersecting the sides of triangle D. y the Menelaus theorem, 1 = G GD D E E = G GD It follows that G : GD =2:1. The centroid of a triangle divides each median in the ratio 2:1. 22

24 The incenter Let,, be points on,, such that,, bisect angles, and respectively. Then = b a, = c b, = a c. I It follows that = b a c b a c =+1, and,, are concurrent. Their intersection is the incenter of the triangle. 23

25 Example Given a point P, let the lines P, P, P intersect,, respectively at,,. onstruct the circle through,,, to intersect the lines,, again at,,. Then the lines,, are concurrent. 24

26 Example Suppose two cevians, each through a vertex of a triangle, trisect each other. Show that these are medians of the triangle. P 25

27 Solution Suppose two cevians, each through a vertex of a triangle, trisect each other. Show that these are medians of the triangle. P Given: Triangle with cevians and intersecting at P such that P trisects and. To prove: and are medians, i.e., and are midpoints of and respectively. Proof. (1) Since P is a trisection of and Q, P = 3 p for p =1or 2, P = q 3 for q =1or 2. (2) pplying Menelaus theorem to triangle P with transversal, we have... 26

28 (3) Similarly, by applying Menelaus theorem to triangle P with transversal, we conclude that is the midpoint of, and is also a median. 27

29 Example Let,, be cevians of intersecting at a point P. (i) Show that if bisects angle, and =, then is isosceles. (ii) Show if if,, are bisectors, and P P = P, then is a right triangle. 28

30 Example is an isosceles triangle with β = γ =40. and are points on and respectively such that bisects angle and =30. Let P be the intersection of and. Show that P =. P 29

31 Example Let be a triangle with =40, =60. Let E and F be points lying on the sides and respectively, such that E =40 and F =70. Let E and F intersect at P. Show that P is perpendicular to. F P E 30

32 Example Let,,, D, E, F be six consecutive points on a circle. Show that the chords D, E, F are concurrent if and only if D EF = DE F. F E D 31

33 Example is a regular 12-gon. Show that the diagonals 1 5, 3 6, and 4 8 are concurrent

34 Kiepert perspectors If similar isosceles triangles, and (of base angle θ) are constructed on the sides of triangle, either all externally or all internally, the lines,, and are concurrent. θ α 1 α 2 θ θ P θ β 2 β 1 θ θ γ 1 γ2 33

35 Proof Proof. pplying the law of sines to triangles and,wehave Likewise, sin β 1 sin β 2 sin α 1 = sin α 1 sin(β + θ) sin(γ + θ) sin α 2 sin(β + θ) sin(γ + θ) sin α 2 = sin(β + θ) sin(γ + θ) sin(β + θ) = sin(γ + θ). = sin(γ+θ) sin(α+θ) and sin γ 1 sin γ 2 = sin(α+θ) sin(β+θ). From these sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2 =+1. and the lines,, are concurrent. The point of intersection is called the Kiepert perspector K(θ). In particular, it is called (1) a Fermat point if θ = ±60, (2) a Napoleon point if θ = ±30, (3) a Vecten point if θ = ±45. 34

36 Trigonmetric version of the eva Theorem Let be a point on the side of triangle such that the directed angles = α 1 and = α 2. y the sine formula, = / / = sin α 1/ sin β sin α 2 / sin γ = sin γ sin β sin α 1 = c sin α 2 b sin α 1. sin α 2 α 1 α 2 β 2 β 1 Likewise, if and be points on the lines, respectively, with = β 1, = β 2 and = γ 1, = γ 2,wehave = a c sin β 1, sin β 2 = b a sin γ 1. sin γ 2 These lead to the following trigonometric version of the eva theorem. γ 2 γ 1 35

37 Theorem 0.3 (eva). The lines,, are concurrent if and only if Proof. sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2 =+1. = sin α 1 sin β 1 sin γ 1. sin α 2 sin β 2 sin γ 2 36