Chapter 14. Cevian nest theorem Trilinear pole and polar Trilinear polar of a point

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1 hapter 14 evian nest theorem 14.1 Trilinear pole and polar Trilinear polar of a point Given a point ith traces and on the sidelines of triangle let = Y = Z =. These points Y Z lie on a line called the trilinear polar (or simply tripolar) of. Y Z

2 410 evian nest theorem If = (u : v : ) then = (u : 0 : ) and = (u : v : 0). The line has equation x u + y v + z = 0. It intersects the sideline at the point = (0 : v : ). Similarly = (0 : v : ) and the points Y Z are Y = ( u : 0 : ) Z = (u : v : 0). The line containing the three points Y Z is This is the tripolar of. x u + y v + z = 0.

3 14.1 Trilinear pole and polar Tripole of a line Given a line L intersecting at Y Z respectively let = Y Z = Z = Y. The lines and are concurrent. The point of concurrency is the tripole of L. Z Y learly is the tripole of L if and only if L is the tripolar of.

4 412 evian nest theorem 14.2 nticevian triangles The vertices of the anticevian triangle of a point = (u : v : ) cev 1 () : a = ( u : v : ) b = (u : v : ) c = (u : v : ) are the harmonic conjugates of ith respect to the cevian segments and i.e. : = a : a ; similarly for b and c. This is called the anticevian triangle of since is the cevian triangle in a b c. It is also convenient to regard a b c as a harmonic quadruple in the sense that any three of the points constitute the harmonic associates of the remaining point onstruction of anticevian triangle If the trilinear polar L of intersects the sidelines at Y Z respectively then the anticevian triangle cev 1 () is simply the triangle bounded by the lines Y and Z. Y c a Z b

5 14.2 nticevian triangles 413 nother construction of anticevian triangle Here is an alternative construction of cev 1 (). Let H H H be the orthic triangle and the reflection of ina then the intersection of the lines H ando is the harmonic conjugate a of in : a a =. H H a a roof. Let be the reflection of in. ith transversal H a e have pplying Menelaus theorem to triangle This gives a H a H = 1. a a = = = shoing that a and divide harmonically.

6 414 evian nest theorem Examples of anticevian triangles (1) The anticevian triangle of the centroid is the superior triangle bounded by the lines through the vertices parallel to the opposite sides. (2) The anticevian triangle of the incenter is the excentral triangle hose vertices are the excenters. (3) The vertices of the tangential triangle being = ( a 2 : b 2 : c 2 ) = (a 2 : b 2 : c 2 ) = (a 2 : b 2 : c 2 ) these clearly form are the anticevian triangle of a point ith coordinates (a 2 : b 2 : c 2 ) hich e call the symmedian point K. (4) The anticevian triangle of the circumcenter. Here is an interesting property of cev 1 (O). Let the perpendiculars to and at intersect at b and c respectively. We call b c an orthial triangle of. The circumcenter of b c is the vertex O a of cev 1 (O); similarly for the other to orthial triangles. (See??). O c O b b c O a

7 14.3 evian quotients evian quotients The cevian nest theorem Theorem For arbitrary points and Q the cevian triangle cev() and the anticevian triangle cev 1 (Q) are alays perspective. If = (u : v : ) and Q = (u : v : ) then (i) the perspector is the point (cev()cev 1 (Q)) = (u ( u v + (ii) the perspectrix is the line L (cev()cev 1 (Q)) ith equation ) 1 ( u u v + x = 0. cyclic ) ( ) ( )) : v v v + + u : u + u v Y Z Y Z M Q roof. (i) Let cev() = YZ and cev 1 (Q) = Y Z. Since = (0 : v : ) and = ( u : v : ) the line has equation ( ) 1 u v x 1 v v y + 1 z = 0. The equations of YY and ZZ can be easily ritten don by cyclic permutations of (uv)(u v ) and(xyz). It is easy to check that the line contains the point ) ( ) ( )) (u ( u v + : v v v + + u : u + u v

8 416 evian nest theorem hose coordinates are invariant under the above cyclic permutations. This point therefore also lies on the lines YY and ZZ. (ii) The lines YZ andy Z have equations They intersect at the point x u + y v + z = 0 y v + z = 0. U = (u(v v ) : vv : v ). Similarly the lines pairs Z Z andy Y have intersections and V = ( uu : v(u u ) : u ) W = (uvu : uvv : (vu uv )). The three points U V W lie on the line ith equation given above. orollary If T is a cevian triangle of T and T is a cevian triangle of T then T is a cevian triangle oft. roof. With reference tot 2 the triangle T 1 is anticevian. Remark. Suppose T = cev T () and T = cev T (Q). If = (u : v : ) ith respect to T andq = (u : v : ) ith respect tot then ( u (TT ) = u (v +) : v v (+u) : ) (u+v) ith respect to triangle T. The equation of the perspectrix L (TT ) is ( 1 v + + +u + u+v ) x = 0. u u v cyclic These formulae hoever are quite difficult to use since they involve complicated changes of coordinates ith respect to different triangles. We shall simply rite and call it the cevian quotient of byq. /Q := (cev()cev 1 (Q))

9 14.3 evian quotients 417 The cevian quotients of the centroidg/ If = (u : v : ) G/ = (u( u+v +) : v( v + +u) : ( +u+v)). Some common examples of G/. G/ coordinates I M i (a(s a) : b(s b) : c(s c)) O K (a 2 : b 2 : c 2 ) K O (a 2 S α : b 2 S β : c 2 S γ ) The point M i = G/I is called the Mittenpunkt of triangle. 1 It is the symmedian point of the excentral triangle. The tangential triangle of the excentral triangle is homothetic to att. (i) We compute the symmedian point of the excentral triangle. Note that I b I 2 c = a 2 bc (s b)(s c) I ci 2 a = ab 2 c (s c)(s a) I ai 2 b = abc 2 (s a)(s b). For the homogeneous barycentric coordinates of the symmedian point of the excentral triangle I a I b I c e have = = = I b Ic 2 I a +I c Ia 2 I b +I a Ib 2 I c a 2 bc (s b)(s c) ( abc) 2(s a) + ab 2 c (s c)(s a) (a bc) 2(s b) + abc 2 (s a)(s b) (ab c) 2(s c) abc 2(s a)(s b)(s c) (a( abc)+b(a bc)+c(ab c)) abc 2(s a)(s b)(s c) (a( a+b+c)b(a b+c)c(a+b c)). From this e obtain the Mittenpunkt M i. (ii) Since the excentral triangle is homothetic to the intouch triangle at T their tangential triangles are homothetic at the same triangle center. Using the cevian nest theorem e have T = 0 (cev(g e)cev 1 (I)) = ( a : b : c s a s b s c). 1 This appears as 9 in ET.

10 418 evian nest theorem The cevian quotients of the orthocenterh/ For = (u : v : ) H/ = (u( S α u+s β v +S γ ) : v( S β v +S γ +S α u) : ( S γ +S α u+s β v)). Examples (1) H/G = (S β +S γ S α : S γ +S α S β : S α +S β S γ ) is the superior ofh. (2) H/I is a point on the OI-line dividing OI in the ratio R+r : 2r. 2 H/I = (a(a 3 +a 2 (b+c) a(b 2 +c 2 ) (b+c)(b c) 2 ) : : ). ( ) a (3) H/K = 2 S α : b2 S β : c2 S γ is the homothetic center of the orthic and tangential triangle. 3 It is a point on the Euler line. (4) H/O is the orthocenter of the tangential triangle. 4 (5) H/N is the orthocenter of the orthic triangle. 5 The cevian quotientg e /K This is the perspector of the intouch triangle and the tangential triangle 6 G e /K = (a 2 (a 3 a 2 (b+c)+a(b 2 +c 2 ) (b+c)(b c) 2 ) : : ). 2 This point appears as 46 in ET. 3 This appears as 25 in ET. 4 This appears as 155 in ET. 5 This appears as 52 in ET. 6 This appears as 1486 in ET.

11 14.3 evian quotients asic properties of cevian quotients roposition (1) / =. (2) If/Q = M thenq = /M. Y Z Y Z Z M Q Y roof. (2) Let = (u : v : )Q = (u : v : ) and M = (x : y : z). We have From these x u = u u y v = v v z = ( u ) v + ) u v v + ( u ( u v ). x u + y v + z ( )( ) u = u v v + u v x u y v + z )( ) ( = u v + u v x u + y v z )( ) ( = u u v v + u u v v +

12 420 evian nest theorem and It follos that x ( x u u + y v + z ) : = u u : v v :. u : v : = x y ( x v u y v ) + z : z ( x u + y v z ) ( x u + y v + z ) ( x : y u y v ) + z ( x : z u + y v ) z.