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CEVIANS of RANK (K,L,M) in TRIANGLE INTERNATIONAL JOURNAL OF GEOMETRY Vol. 1 (2012), No. 2, 22 - 33 CEVIANS OF RANK (k,l,m) IN TRIANGLE NICU¸SOR MINCULETE and CAT¼ ALIN¼ I. BARBU Abstract.In this article we want to do a characterization of some im- portant points from the triangle chosen from C. Kimberling’s Encyclopedia of triangle centers. A series of these points being points of concurrence of cevians of rank (k; l; m), of the triangle. Also, we present several equalities from these points. 1. Introduction For the beginning we will make a short presentation of barycentric coordi- nates, these being necessary for us in subsequent approaches and specifying the fact that they were introduced in 1827 by Möbius as [3] : Barycentric coordinates are triplets of numbers (t1; t2; t3) corresponding to the placed masses in the vertices of ABC triangle. These masses deter- mine a point P , which is named geometric center of weight or barycentric center with coordinates (t1; t2; t3) : Note that the triangle areas BP C; CP A and AP B are proportional with barycentric coordinates t1; t2 and t3. For more details we refer to the monographs of C. Bradley [3], C. Coanda¼ [4], C. Co¸sni¸ta¼ [5], C. Kimberling [7], S. Loney [8] and to the papers of O. Bottema [2], J. Scott [11], H. Tanner [12], and P. Yiu [13]. Denote by a; b; c the lenghts of the sides in the standard order, by s the semiperimeter of triangle ABC; by [ABC] the area of the triangle ABC: ————————————– Keywords and phrases: Triangle, cevians of rank (k; l; m); barycentric coordinates (2010)Mathematics Subject Classi…cation: 51M04, 51M25, 51M30 Received: 21.02.2012. In revised form: 4.04.2012. Accepted: 18.05.2012. Ceviansofrank(k,l,m)intriangle 23 2. About the cevians of rank (k; l; m) For start we will do the link between barycentric coordinates and the reports in which the triangle sides are divided from the lines passing through the triangle’spoint and through the barycentric center. Lemma 2.1. Let ABC be a triangle and a point P which has the barycentric coordinates (t1; t2; t3) and AP BC = A0 ; BP CA = B0 and CP AB = C : Then the equalities\ f g \ f g \ f 0g BA t CB t AC t (1) 0 = 3 ; 0 = 1 ; 0 = 2 A0C t2 B0A t3 C0B t1 hold. Proof. Consider a point P in interior of the triangle ABC (see Figure 1). If P is not in the triangle’sinterior, the demonstration is done similarly. We have [ABP ] AB AP sin BAA [ABA ] BA = 0 = 0 = 0 ; [ACP ] AC AP sin CAA [ACA ] A C 0 0 0 but as the areas of the triangles P BC; P CA; P AB are proportional with barycentric coordinates t1; t2 and t3, it follows that [ABP ] t = 3 ; [ACP ] t2 from where, we deduce the relation BA t 0 = 3 : A0C t2 The other relations are demonstrated similarly. Theorem 2.1. The internal angle bisectors of a triangle are concurrent; the point of concurrence I is the incenter. Theorem 2.2. ( [1] ; [6] ; [9]): The lines joining the vertices of a triangle to the points of contact of the opposite sides with the excircles relative to these sides are concurrent. Proof. Let D; E; F be the contact points of the excircles with sides opposite to vertices A; B; C (see Figure 2). The next equalities are true: BD = AE = s c; DC = AF = s b; CE = FB = s a [6] : 24 Nicu¸sorMinculeteandCat¼ alin¼ Barbu Therefore BD s c CE s a AF s b = ; = ; = DC s b EA s c FB s a hence BD CE AF = 1; DC EA FB and from the converse of Ceva’s theorem, we get that lines AD; BE and CF are concurrent. The point N common to the lines AD; BE; CF is often referred to as the Nagel point of the triangle ABC: Theorem 2.3. ( [9]): Let I be the incenter of triangle ABC. Through point B draw a parallel to IC which intersects the parallel drawn through C at IC in point A0 . Similarly are obtained the points B0 and C0. Then the lines AA0, BB0, CC0 are concurrent. Proof. Denote by D; E; F the intersections of the lines AA0 with BC; BB0 with AC, and CC0 with AB; respectively (see Figure 3). Because r r BA0 = CI = C ;CA0 = BI = B ; sin 2 sin 2 it follows that C BD [BAA0] BA0 AB sin 2 + B = = B = DC [CAA0] CA CA sin + C 0 2 Ceviansofrank(k,l,m)intriangle 25 sin C + B cos C sin A + sin B a + b (2) 2 2 = = sin B + C cos B sin A + sin C a + c 2 2 Similarly: CE b + c AF c + a (3) = ; = EA b + a FB c + b From relations (2) and (3) we get BD CE AF = 1; DC EA FB which means that, by using Ceva’sconverse theorem, the lines AD; BE; CF are concurrent. Denote by S the intersection of the lines AD; BE; and CF . Because BD a+b CE b+c AF c+a DC = a+c , EA = b+a and FB = c+b , according to Lemma 1, it results that point S has the barycentric coordinates b + c : c + a : a + b; corresponding to Spieker’s point (Spieker’s point of a triangle ABC is the center of the inscribed circle in the median triangle corresponding to triangle ABC ([1],[7],[9])). Still more we present some results regarding concurrence lines. De…nition 1. In a triangle ABC , the ceviens AE and AF (E; F BC) are called isogonals in relation with angle BAC; if BAE CAF2 and ^ ^ ^ ^BAF ^CAE , otherway said, if cevians AE and AF are symmetrical in relation with the angle bisector of A. Also, AE will be called the isogonal of AF and reverse. Theorem 2.4. (Steiner [1] ; [6] ; [9]): In a triangle ABC, the isogonals AE and AF , where E; F BC, determine the following relation 2 BE BF AB2 = : CE CF AC2 Proof. Denote by x and y measures of angles BAE, and BAF; respectively. We have ^BAE = ^CAF = x and ^BAF = ^CAE = y (see Figure 4). We use the formula of the area of triangle ABC, expressed so: AB AC sin A [ABC] = 2 26 Nicu¸sorMinculeteandCat¼ alin¼ Barbu Therefore, BE [ABE] AB AE sin x AB sin x = = = ; CE [AEC] AC AE sin y AC sin y and BF [ABF ] AB AF sin y AB sin y = = = : CF [AF C] AC AF sin x AC sin x The conclusion results by multiplying member with member of the previous relations. From Theorem 6, we deduce easily the following: Theorem 2.5. (Mathieu [6] ; [9]). If three lines from the vertices of a triangle are concurrent, their isogonals are also concurrent. If the point M is the intersection of cevians AD; BE; CF and M 0 is an intersection point for cevians AD0; BE0;CF 0, then we say that M 0 is the isogonal point to M in relation with triangle ABC: De…nition 2. In a triangle ABC, cevians AE and AF (E; F (BC)) are called isotomics if points E and F are symmetrical to the midpoint2 of (BC). The line AE will be called the isotomic of AF and reverse. Theorem 2.6. (Neuberg [1] ; [6] ; [9]): In a triangle the isotomics of three concurrent cevians are concurrent. If M is the intersection point of the cevians AD, BE, CF and M 0 is the intersection point of the isotomics AD0, BE0, CF 0, then we say that M 0 is the isotomic of M in relation to the triangle ABC. De…nition 3. If on the side (BC) of a unisosceles triangle ABC a point D is taken, so that BD c k = DC b k R, then AD is called cevian of rank k; and if D BC [BC], so 2 BD c k 2 n that DC = b , k R, then AD is called excevian of rank k or exterior cevian of rank k: If the2 ABC triangle is isosceles (AB = AC), then, through convention, the cevian of rank k is a median from A . Theorem 2.7. ( [9] ). In a triangle, the cevians of rank k corresponding to the points A; B; C are concurrent. Remark 1. a) Note with Ik the intersection point of cevians of rank k: b) If M is the intersection point of cevians AD; BE; CF and M 0 is the isogonal point of M in relation to triangle ABC, then point I is situated on MM 0, where I is the intersection point of the angle bisectors. c) Also, the cevian of rank k from A and the excevians of rank k from B and C are concurrent. Similarly with cevians of rank k, which were de…ned starting from the c angle bisector using relation b , we will de…ne two types of cevians, namely, the N-cevian of rank l starting from the cevian determined by Nagel’spoint and S - cevian of rank m starting from the cevian determined by Spieker’s point. Ceviansofrank(k,l,m)intriangle 27 De…nition 4. If on side (BC), of a unisosceles ABC triangle a point D is taken, so that BD s c l = DC s b l R, then AD is called N - cevian of rank l, and if D BC [BC], so 2 l 2 n BD s c that DC = sb , l R, then AD is called N - cevian of rank l or N 2 - exterior cevian of rank l: If ABC triangle is isosceles (AB = AC), then, through convention, N - cevian of rank l from A is the median from A .
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