Geometry. Class Examples (July 29) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

Transcription

1 Geometry lass Examples (July 29) Paul Yiu Department of Mathematics Florida tlantic University c a Summer 2014

2 1 The Pythagorean Theorem Theorem (Pythagoras). The lengths a <cof the sides of a right triangle satisfy the relation a = c 2. Proof. a a a a a 3 1 a Theorem (onverse of Pythagoras theorem). If the lengths of the sides of satisfy a = c 2, then the triangle has a right angle at. Y a c a Z X Proof. onsider a right triangle XY Z with Z =90, YZ = a, and XZ =. y the Pythagorean theorem, XY 2 = YZ 2 +XZ 2 = a = c 2 = 2. It follows that XY =, and XY Z y the SSS test, and = Z =90.

3 2 Theorem (The law of sines). Let R denote the circumradius of a triangle. 2R = a sin α = sin β = c sin γ. α F O E O α D D Since the area of a triangle is given y Δ= 1 2 c sin α, the circumradius can e written as R = ac 4Δ.

4 3 The law of cosines Given a triangle, we denote y a,, c the lengths of the sides,, respectively. Theorem (The law of cosines). c 2 = a a cos γ. c c X a a X Proof. Let X e the altitude on. c 2 = X 2 + X 2 =(a cos γ) 2 +(sin γ) 2 = a 2 2a cos γ + 2 (cos 2 γ +sin 2 γ) = a a cos γ.

5 4 pollonius Theorem Theorem. Given triangle, let D e the midpoint of. The length of the median D is given y =2(D 2 + D 2 ). D Proof. pplying the law of cosines to triangles D and D, and noting that cos D = cos D,wehave 2 = D 2 + D 2 2D D cos D; 2 = D 2 + D 2 2D D cos D, 2 = D 2 + D 2 +2D D cos D. The result follows y adding the first and the third lines. If m a denotes the length of the median on the side, m 2 a = 1 4 (22 +2c 2 a 2 ).

6 ngle isector theorem Theorem (ngle isector theorem). The isectors of an angle of a triangle divide its opposite side in the ratio of the remaining sides. If X and X respectively the internal and external isectors of angle, then X : X = c : and X : X = c :. Z c Z X X Proof. onstruct lines through parallel to the isectors X and X to intersect the line at Z and Z. (1) Note that Z = X = X = Z. This means Z =. learly, X : X = : Z = : = c :. (2) Similarly, Z =, and X : X = : Z = : = c :.

7 6 The lengths of the isectors Theorem. (a) The lengths of the internal and external isectors of angle are respectively t a = 2c + c cos α 2 and t a = 2c c sin α 2. c t a t a X X Remarks. (1) 2c +c is the harmonic mean of and c. It can e constructed as follows. If the perpendicular to X at X intersects and at Y and Z, then Y = Z = 2c +c. c t a Z X Y (2) pplying Stewart s Theorem with λ = c and μ = ±, we also otain the following expressions for the lengths of the angle isectors: ( ( ) ) 2 a t 2 a = c 1, + c ( ( ) 2 t 2 a a = c 1). c

8 7 The incircle The internal angle isectors of a triangle are concurrent at the incenter of the triangle. This is the center of the incircle, the circle tangent to the three sides of the triangle. Let the isectors of angles and intersect at I. onsider the pedals of I on the three sides. Since I is on the isector of angle, IX = IZ. Since I is also on the isector of angle, IX = IY. It follows IX = IY = IZ, and the circle, center I, constructed through X, also passes through Y and Z, and is tangent to the three sides of the triangle. s a s a Y Y Z I Z I s c s X s X s c This is called the incircle of triangle, and I the incenter. Let s e the semiperimeter of triangle. The incircle of triangle touches its sides,, at X, Y, Z such that Y =Z = s a, Z =X = s, X =Y = s c. The inradius of triangle is the radius of its incircle. It is given y r = 2Δ a + + c = Δ s.

9 8 The excircles The internal isector of each angle and the external isectors of the remaining two angles are concurrent at an excenter of the triangle. n excircle can e constructed with this as center, tangent to the lines containing the three sides of the triangle. I I c X Y r a Z r a r a The exradii of a triangle with sides a,, c are given y r a = Δ s a, r = Δ s, r c = Δ s c. I a The areas of the triangles I a, I a, and I a are 1 2 ar a, 1 2 r a, and 1 2 cr a respectively. Since Δ= ΔI a +ΔI a +ΔI a, we have Δ= 1 2 r a( a + + c) =r a (s a), from which r a = Δ s a.

10 9 Heron s formula for the area of a triangle onsider a triangle with area Δ. Denote y r the inradius, and r a the radius of the excircle on the side of triangle. It is convenient to introduce the semiperimeter s = 1 2 (a + + c). I a I r a r Y Y (1) From the similarity of triangles IY and I Y, r = s a. r a s (2) From the similarity of triangles IY and I Y, r r a =(s )(s c). (3) From these, (s a)(s )(s c) r =, s s(s )(s c) r a =. s a Theorem (Heron s formula). Δ= s(s a)(s )(s c). Proof. Δ=rs.

11 10 Menelaus theorem Theorem (Menelaus). Given a triangle with points X, Y, Z on the side lines,, respectively, the points X, Y, Z are collinear if and only if X X Y Y Z Z = 1. Y Z W X Proof. (= ) Let W e the point on such that W//XY. Then, X X = WY Y, and Z Z = Y YW. It follows that X X Y Y Z Z = WY Y Y Y Y YW = Y Y Y Y WY YW = 1. ( =) Suppose the line joining X and Z intersects at Y. From aove, X X Y Y Z = 1 =X Z X Y Y Z Z. It follows that Y Y = Y Y. The points Y and Y divide the segment in the same ratio. These must e the same point, and X, Y, Z are collinear.

12 11 eva s theorem Theorem (eva). Given a triangle with points X, Y, Z on the side lines,, respectively, the lines X, Y, Z are concurrent if and only if X X Y Y Z Z =+1. Z Y P X Proof. (= ) Suppose the lines X, Y, Z intersect at a point P. onsider the line PY cutting the sides of triangle X. y Menelaus theorem, Y Y P PX X = 1, or Y Y P XP X =+1. lso, consider the line PZ cutting the sides of triangle X. y Menelaus theorem again, Z Z X XP P = 1, or Z Z X XP P =+1. Multiplying the two equations together, we have ( =) Exercise. Y Y Z Z X X =+1.

13 1 arycentric coordinates on a line w X v Let and e two fixed points. point X on the line is defined y one of the following: (1) the division ratio X X = w v, (2) distriuting a unit mass at and so as to alance at X: the asolute arycentric coordinates X = v + w v + w, (3) the homogeneous arycentric coordinates relative to : X = v : w.

14 2 arycentric coordinates with respect to a triangle Given a reference triangle, we put at the vertices,, masses u, v, w respectively, and determine the alance point. (u) (u) v + w P (u + v + w) w v u (v) (w) X(v + w) X The masses at and can e replaced y a single mass v + w at the point X = v + w. v + w Together with the mass at, this can e replaced y a mass u + v + w at the point P which divides X in the ratio P : PX = v + w : u. This is the point with asolute arycentric coordinates provided u + v + w 0. u + v + w, u + v + w We also say that the alance point P has homogeneous arycentric coordinates (u : v : w) with reference to.

15 13 evian triangle Theorem (eva). Let X, Y, Z e points on the lines,, respectively. The lines X, Y, Z are collinear if and only if the given points have coordinates of the form X = (0 : y : z), Y = (x : 0 : z), Z = (x : y : 0), for some x, y, z. If this condition is satisfied, the common point of the lines X, Y, Z is P =(x : y : z). Z Y P P P P X P (1) The points X, Y, Z are called the traces of P. We also say that XY Z is the cevian triangle of P (with reference to triangle ). Sometimes, we shall adopt the more functional notation for the cevian triangle and its vertices: cev(p ): P =(0:y : z), P =(x :0:z), P =(x : y :0). (2) The point P divides the segment X in the ratio PX : X = x : x + y + z.

16 8 Theorem. The homogeneous arycentric coordinates of P are in the proportion of the (signed) areas of triangles P, P, P. P P P P P X Proof. Since PX : X = x : x + y + z, ΔPX : ΔX = x : x + y + z, and ΔPX : ΔX = x : x + y + z. Therefore, Similarly, ΔP = From these, ΔP = ΔPX +ΔPX x = x + y + z ΔX + x x + y + z ΔX x = (ΔX +ΔX) x + y + z x = x + y + z Δ. y x + y + z Δ, ΔP = z x + y + z Δ. ΔP : ΔP : ΔP = x : y : z.

17 10 rea and arycentric coordinates Theorem. If for i =1, 2, 3, P i = x i +y i +z i (in asolute arycentric coordinates), then the area of the oriented triangle P 1 P 2 P 3 is x 1 y 1 z 1 ΔP 1 P 2 P 3 = x 2 y 2 z 2 x 3 y 3 z 3 Δ.

18 9 Equations of straight lines Two-point form The area formula has an easy and extremely important consequence: the equation of the line joining two points with coordinates (x 1 : y 1 : z 1 ) and (x 2 : y 2 : z 2 ) is x 1 y 1 z 1 x 2 y 2 z 2 x y z =0, or (y 1 z 2 y 2 z 1 )x +(z 1 x 2 z 2 x 1 )y +(x 1 y 2 x 2 y 1 )z =0. Examples (1) The equations of the sidelines,, are respectively x =0, y =0, z =0. (2) Given a point P =(u : v : w), the cevian line P has equation wy vz =0; similarly for the other two cevian lines P and P. These lines intersect corresponding sidelines at the traces of P : P =(0:v : w), P =(u :0:w), P =(u : v :0). (3) The equation of the line joining the centroid and the incenter is a c x y z =0, or ( c)x +(c a)y +(a )z =0.

19 10 Intersection of two lines The intersection of the two lines p 1 x + q 1 y + r 1 z =0, p 2 x + q 2 y + r 2 z =0 is the point (q 1 r 2 q 2 r 1 : r 1 p 2 r 2 p 1 : p 1 q 2 p 2 q 1 ). Proposition 0.1. Three lines p i x + q i y + r i z =0, i =1, 2, 3, are concurrent if and only if p 1 q 1 r 1 p 2 q 2 r 2 p 3 q 3 r 3 =0.

20 11 The infinite point of a line The infinite point of a line L has homogeneous coordinates given y the difference of the asolute arycentric coordinates of two distinct points on the line. s such, the coordinate sum of an infinite point is zero. We think of all infinite points constituting the line at infinity, L, which has equation x + y + z =0. Examples 1. The infinite points of the side lines,, are (0 : 1 :1), (1 : 0 : 1), ( 1 :1:0)respectively. 2. The infinite point of the altitude has homogeneous coordinates (0 : S : S ) a 2 (1:0:0)=( a 2 : S : S ). 3. More generally, the infinite point of the line px + qy + rz =0is (q r : r p : p q).

21 12 Parallel lines Parallel lines have the same infinite point. The line through P =(u : v : w) parallel to L : px + qy + rz =0has equation q r r p p q u v w x y z =0. 1. Find the equations of the lines through P =(u : v : w) parallel to the side lines. 2. Let DEF e the medial triangle of, and P a point with cevian triangle XY Z (with respect to. Find P such that the lines DX, EY, FZ are parallel to the internal isectors of angles,, respectively. 1 1 The Nagel point P =( + c a : c + a : a + c).

22 15 Solution to Example (5, 8, 9) triangle. is a triangle with a =5, =8, and c =9. Show that the Gergonne point lies on the line joining the midpoints of the sides and. Y X 5 D G e I E 8 Solution. s =11. Y = Z = s c =2, Y = s a =6, Z = s =3. Z 9 solute arycentric coordinates: D = + 2, E = + 2. Homogeneous arycentric coordinates: X = 0 : 2 : 3 Y = 1 : 0 : 3 G e = 1 : 2 : 3 Therefore G e = The Gergonne point lies on the line DE if and only if G e = t D +(1 t)e for some t: = t + 2 This is the case with t = (1 t) + 2 = 1 t 2 + t

23 16 Example (7, 8, 9) triangle. is a triangle with a =7, =8, and c =9. Show that the Nagel point lies on the line joining the points of tangency of incircle with the sides and. 9 Z Z N a Y Y 8 7

24 17 Solution to Example (7, 8, 9) triangle. is a triangle with a =7, =8, and c =9. Show that the Nagel point lies on the line joining the points of tangency of incircle with the sides and. 9 Z Z N a Y Y 8 Solution. s =12. Z = s =4, Y = s c =3; Z =4, Y =3. solute arycentric coordinates: Y = 3+5 8, Z = Homogeneous arycentric coordinates: 7 Y = 5 : 0 : 3 Z = 5 : 4 : 0 N a = 5 : 4 : 3 Therefore N a = The Nagel point lies on the line YZif and only if N a = t Y +(1 t)z for some t: = t 3 +(1 t) 4 = This is the case with t = t 5(1 t) + + 5t

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