CEVIANS of RANK (K,L,M) in TRIANGLE

Home , Lemoine point, Mittenpunkt, Symmedian

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 coordinates, 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 determine a point P , which is named geometric center of weight or barycentric center with coordinates (t1; t2; t3) : Note that the triangle areas BPC;CPA 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 . De…nition 5. If on side (BC) of a ABC unisosceles triangle a point D is taken, so that BD a + b m = DC a + c m R , then AD is called S - cevian of rank m, and if D BC [BC] , so m 2 BD a+b 2 n that = , m R, then AD is called S - excevian of rank m or S DC a+c 2 - exterior cevian of rank m: If ABC triangle ABC is isosceles (AB = AC), then, through convention, S - cevian of rank m from A is the median from A. To approach unitary these types of cevians we will de…ne a more general type of cevian, namely the cevian of rank (k; l; m) as it follows: De…nition 6. If on side (BC) of a ABC unisosceles triangle a point D is taken, so that: BD c k s c l a + b m (4) = DC b s b a + c k; l; m R, then AD is called cevian of rank (k; l; m); and if D BC [BC], 2 l m 2 n BD c k s c a+b so that DC = b sb a+c ; k; l; m R, then AD is called excevian 2 of rank (k; l; m ) or exterior cevian of rank (k; l; m): If ABC triangle is isosceles (AB = AC), then, through convention, the cevian of rank (k; l; m) is the median from A. Theorem 2.8. In a triangle the cevians of rank (k; l; m) are concurrent. Proof. It’s easy to see that if we write the relations analogous to the DB EC FA relationship (4) and then we multiply them, we get DC EA FB = 1, so, from Ceva’sconverse theorem, we deduce that the ceviens of rank (k; l; m), AD; BE; CF are concurrent. Remark 2.

a) The median is cevian of rank zero (k = 0), and I0 = G; b) The bisector is cevian of rank one (k = 1), and I1 = I; c) The symmedian (the median’s isogonal) is cevian of rank two (k = 2) and I2 = K (Lemoine’s point); d) The antibisector (the angle bisector’sisotomic) is cevian of rank ( 1); e) Antibisector’sisogonal is cevian of rank three; f) Symmedian’sisotomic is cevian of rank ( 2); g) Isogonal of a cevian of rank k is a cevian of rank (2 k); 28 Nicu¸sorMinculeteandCat¼ alin¼ Barbu

h) Isotomic of a cevian of rank k is a cevian of rank ( k); i) The exsymmedian is cevian of rank 2; j) We note with Ia (k) the intersection point of cevian of rank k which leaves from A and the excevians of rank k which leaves from B and C. Similarly are de…ned Ib (k) and Ic (k); k) Note with I(k; l; m) the point of intersection of cevians of rank (k; l; m). l) Barycentric coordinates of I(k; l; m) are ak(s a)l(b + c)m : bk(s b)l(c + a)m : ck(s c)l(a + b)m: Theorem 2.9. (Gergonne [1] ; [6] ; [9]): In a triangle the cevians that unite the vertices of a ABC triangle with contact points of the incircle with opposite sides are concurrent. The demonstration results immediate, considering the fact that incircle and excircles determine on every side of the triangle, pairs of isotomic points. The point of concurrence of the cevians considered is called Gergonne’spoint or X(7) by [7]. Remark 3. According to Theorem 16 we can say that: 1) The bisector is cevian of rank (1; 0; 0) ; 2) The median is cevian of rank (0; 0; 0) ; 3) The symmedian is cevian of rank (2; 0; 0) ; 4) The antibisector is cevian of rank ( 1; 0; 0) ; 5) Antibisector’sisogonal is cevian of rank (3; 0; 0) ; 6) The symmedian’sisotomic is cevian of rank ( 2; 0; 0) ; 7) A cevian’sisogonal of rank (k; 0; 0; ) is a cevian of rank (2 k; 0; 0) ; 8) A cevian’sisotomic of rank (k; 0; 0) is a cevian of rank ( k;0; 0) ; 9) The exsymmedian is cevian of rank (2; 0; 0) : Considering Lemma 1 and the barycentric coordinates of the point pro- vided by [7], we will analize which of these points are the intersections of cevians of rank (k; l; m). Starting from the idea of a problem [10], we obtain the following : Theorem 2.10. Let ABC be a triangle. Denote by D;E and F respectively, the point of intersection of the cevians of rank (k; l; m) from A; B; C with the opposite sides. Let P be a point on the sides of the triangle and X;Y and Z respectively, the perpendicular feet of P on the side BC;CA and AB:There are the following relations: a) If P [EF ] ; then we have 2 x y z (5) = + ; ak 1(s a)l(b + c)m bk 1(s b)l(a + c)m ck 1(s c)l(a + b)m where PX = x; PY = y; PZ = z; b) Ifj P j [FDj]; thenj wej havej 2 y x z (6) = + bk 1(s b)l(a + c)m ak 1(s a)l(b + c)m ck 1(s c)l(a + b)m c) If P [DE]; then we have 2 z x y (7) = + ck 1(s c)l(a + b)m ak 1(s a)l(b + c)m bk 1(s b)l(a + c)m Ceviansofrank(k,l,m)intriangle 29

Proof. Since AD is the cevian of rank (k; l; m); implies the relation BD c k s c l a + b m = ; DC b s b a + c so ack(s c)l(a + b)m BD = bk(s b)l(a + c)m + ck(s c)l(a + b)m In the analogous way, we deduce the relations bck(s c)l(a + b)m AE = ak(s a)l(b + c)m + ck(s c)l(a + b)m and cbk(s b)l(a + c)m AF = ak(s a)l(b + c)m + bk(s b)l(a + c)m

Therefore, we have the relation (see Figure 5) [AF E] = [AF P ] + [AP E]; which is equivalent to (bc)k+1 [(s b)(s c)]l[(a + b)(a + c)]m sin A = [ak(s a)l(b + c)m + ck(s c)l(a + b)m][ak(s a)l(b + c)m + bm(s b)l(a + c)m] 2 (8) cbk(s b)l(a + c)m z bck(s c)l(a + b)my + 2[ak(s a)l(b + c)m + bk(s b)l(a + c)m] 2[ak(s a)l(b + c)m + ck(s c)l(a + b)m] bc sin A But [ABC] = 2 and 2[ABC] = ax + by + cz; and using relation (8), we obtain: k 1 l m (bc) [(s b)(s c)] [(a + b)(a + c)] (ax + by + cz) = k 1 k l m k l m l m zb [a (s a) (b + c) + c (s c) (a + b) ](s b) (a + c) + k 1 k l m k l m l m yc [a (s a) (b + c) + b (s b) (a + c) ](s c) (a + b) ; so k 1 l m k k 1 l m xa(bc) [(s b)(s c)] [(a+b)(a+c)] +yb c [(s b)(s c)] [(a+b)(a+c)] + k k 1 l m zc b [(s b)(s c)] [(a + b)(a + c)] = k k 1 l m k 1 k l m za b [(s a)(s b)] [(b+c)(a+c)] +zb c [(s c)(s b)] [(a+b)(a+c)] + k k 1 l m k k 1 l m ya c [(s a)(s c)] [(b+c)(a+b)] +yb c [(s b)(s c)] [(a+b)(a+c)] 30 Nicu¸sorMinculeteandCat¼ alin¼ Barbu

Dividing by (abc)k 1 [(s a)(s b)(s c)]l [(a + b)(b + c)(c + a)]m; we deduce the relation (5). Similary, we …nd relations (6) and (7). Remark 4. The notation cevians of rank (k; l; m) can be extend to notation the cevians of rank (ku; ku+1; :::; kw) if relation (4) can be written

BD w is c ki (9) = DC is b i=u Y where u w; u; w Z; ki R; for all i u; ::; w : For u = 0; w = 2; k0 = 2 2 2 f g k; k1 = l; k2 = m; we deduce relation (4). Therefore relation (5) becomes

ax by cz (10) w = w + w (is a)ki (is b)ki (is c)ki i=l i=l i=l Y Y Y

3. Particular cases

C. Kimberling, in [7], presents a set of points, which are written as X(q): If we take P X(q); where the point X(q) is a point of type I(k; l; m), then we obtain a series of equalities for several particular cases in relation (5).

X(q) I(k; l; m) Point description P X(q) in relation (5) X(1) I(1; 0; 0) incenter x = y + z [10] X(2) I(0; 0; 0) centroid ax = by + cz 1 1 1 X(6) I(2; 0; 0) Lemoine point a x = b y+ c z X(7) I(0; 1; 0) Gergonne point a(s a)x = b(s b)y + c(s c)z a b c X(8) I(0; 1; 0) Nagel point s a x = s b y+ s c z 1 1 1 X(9) I(1; 1; 0) mittenpunkt s a x = s b y+ s c z a b c X(10) I(0; 0; 1) Spieker point b+c x = c+a y+ a+b z X(1);X(5) harmonic a(s a) b(s b) c(s c) X(12) I(0; 1; 2) f g 2 x = 2 y+ 2 z conjugate of X(11) (b+c) (c+a) (a+b) (b+c) (c+a) (a+b) X(21) I(1; 1; 1) Schi• er point s a x = s b y+ s c z 1 1 1 X(31) I(3; 0; 0) 2nd power point a2 x = b2 y+ c2 z 1 1 1 X(32) I(4; 0; 0) 2rd power point a3 x = b3 y+ c3 z crosspoint of X(37) I(1; 0; 1) 1 x = 1 y+ 1 z incenter and centroid b+c c+a a+b X(6) Ceva conjugate 1 1 1 X(41) I(3; 1; 0) a2(s a) x = b2(s b) y+ c2(s c) z of X(31) crosspoint of X(42) I(2; 0; 1) 1 x = 1 y+ 1 z incenter and Lemoine point a(b+c) b(c+a) c(a+b) 1 1 1 X(55) I(2; 1; 0) insimilicenter a(s a) x = b(s b) y+ c(s c) z Ceviansofrank(k,l,m)intriangle 31

s a s b s c X(56) I(2; 1; 0) exsimilicenter x = y+ z a b c X(57) I(1; 1; 0) isogonal conjugate of X(9) (s a)x = (s b)y + (s c)z X(58) I(2; 0; 1) isogonal conjugate of X(10) b+c x = c+a y+ a+b z a b c orthocenter of s a s b s c X(65) I(1; 1; 1) x = y+ z the intouch triangle b+c c+a a+b X(76) I( 2; 0; 0) 3rd Brocard point a3x = b3y + c3z Cevapoint of incenter X(81) I(1; 0; 1) (b + c)x = (c + a)y + (a + b)z and Lemoine point X(85) I( 1; 1; 0) isotomic conjugate of X(9) a2(s a)x = b2(s b)y + c2(s c)z Cevapoint of incenter X(86) I(0; 0; 1) a(b + c)x = b(c + a)y + c(a + b)z and centroid 1 1 a a(s a)x = X(174) I ; ; 0 Y¤ center of congruence 2 2 b b(s b)y + c c(s c)z p a a x = 1 1 2nd mid-arc point of p s a p X(188) I ; ; 0 2 2 anticomplementary triangle bq c b s b y + c s c z 1 X(8) Ceva conjugate q (s a)2 xq= X(200) I(1; 2; 0) 1 1 of X(9) (s b)2 y+ (s c)2 z 1 X(10) Ceva conjugate (s a)(b+c) x = X(210) I(1; 1; 1) 1 1 of X(37) (s b)(c+a) y+ (s c)(a+b) z 1 X(6) Ceva conjugate a2(b+c) x = X(213) I(3; 0; 1) 1 1 of X(42) b2(c+a) y+ (s c)(a+b) z 1 X(9) Ceva conjugate a(s a)2 x = X(220) I(2; 2; 0) 1 1 of X(55) b(s b)2 y+ c(s c)2 z X(7) Ceva conjugate a(s a)x b(s b)y c(s c)z X(226) I(0; 1; 1) = + of X(65) b+c c+a a+b 1 1 a b c X(259) I 2 ; 2 ; 0 isogonal conjugate of X(174) s a x = s b y+ s c z q a(qs a)x =q X(266) I 1 ; 1 ; 0 isogonal conjugate of X(188) 2 2 b(s b)y+ c(s c)z p a(s a)2x = X(269) I(0; 2; 0) isogonal conjugate of X(200) p p b(s b)2y + c(s c)2z a2(b + c)x = X(274) I( 1; 0; 1) isogonal conjugate of X(213) 2 b2(c + a)y + c (a + b)z a(s a)2x = X(279) I( 1; 2; 0) isogonal conjugate of X(220) b(s b)2y + c(s c)2z a3(b + c)x = X(310) I( 2; 0; 1) isotomic conjugate of X(42) 3 b3(c + a)y + c (a + b)z a2 b2 c2 X(312) I( 1; 1; 0) isotomic conjugate of X(57) s a x = s b y+ s c z a3 b3 c3 X(313) I( 2; 0; 1) isotomic conjugate of X(58) b+c x = c+a y + a+b z a2(b+c) s a x = X(314) I( 1; 1; 1) isotomic conjugate of X(65) 2 2 b (c+a) c (a+b) s b y + s c z a2 b2 c2 X(321) I( 1; 0; 1) isotomic conjugate of X(81) b+c x = c+a y+ a+b z a(b+c) b(c+a) c(a+b) X(333) I(0; 1; 1) Cevapoint of X(8) and X(9) s a x = s b y+ s c z a2 b2 c2 X(341) I( 1; 2; 0) isotomic conjugate of X(269) (s a)2 x = (s b)2 y+ (s c)2 z 32 Nicu¸sorMinculeteandCat¼ alin¼ Barbu

isotomic conjugate a b c X(346) I(0; 2; 0) (s a)2 x = (s b)2 y+ (s c)2 z of X(279) X(365) I( 3 ; 0; 0) square root point 1 x = 1 y+ 1 z 2 pa pb pc isogonal conjugate X(366) I( 1 ; 0; 0) pax =pby+pcz 2 of X(365) X(269)-cross conjugate X(479) I(0; 3; 0) a(s a)3x = b(s b)3y + c(s c)3z of X(279) a(s a)(b + c)2x = X(552) I(0; 1; 2) point Maia I b(s b)(c + a)2y + c(s c)(a + b)2z 1 1 1 X(560) I(5; 0; 0) 4th power point a4 x = b4 y+ c4 z isogonal conjugate X(561) I( 3; 0; 0) a4x = b4y + c4z of 4th power point (b+c)2 (c+a)2 (a+b)2 X(593) I(2; 0; 2) 1st Hatzipolakis-Yiu point a x = b y+ c z isogonal conjugate X(594) I(0; 0; 2) a x = b y+ c z of X(593) (b+c)2 (c+a)2 (a+b)2 crosspoint of X(756) I(1; 0; 2) 1 x = 1 y+ 1 z X(10) and X(37) (b+c)2 (c+a)2 (a+b)2 isogonal conjugate X(757) I(1; 0; 2) (b + c)2x = (c + a)2y + (a + b)2z of X(756) crossum of X(872) I(3; 0; 2) 1 x = 1 y+ 1 z X(86) and X(274) a2(b+c)2 b2(c+a)2 c2(a+b)2 isogonal conjugate a2(b + c)2x = X(873) I(1; 0; 2) of X(872) b2(c + a)2y + c2(a + b)2z isogonal conjugate (s a)(b + c)x = X(1014) I(1; 1; 1) of X(210) (s b)(c + a)y + (s c)(a + b)z a2(s a)2x = X(1088) I( 1; 2; 0) triliniare square of X(7) b2(s b)2y + c2(s c)2z 2 2 X(1089) I( 1; 0; 2) triliniare square of X(10) a x = b y+ c z (b+c)2 (c+a)2 (a+b)2 isogonal conjugate X(1253) I(3; 2; 0) 1 x = 1 y+ 1 z of X(1088) a2(s a)2 b2(s b)2 c2(s c)2 a3 b3 c3 X(3596) I( 2; 1; 0) 1st Odehnal point s a x = s b y+ s c z

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