cevians of rank (k,l,m) in triangle - ijgeometry.com 2.1. the internal angle bisectors of a triangle...

INTERNATIONAL JOURNAL OF GEOMETRYVol. 1 (2012), No. 2, 22 - 33

CEVIANS OF RANK (k,l,m) IN TRIANGLE

NICUSOR MINCULETE and C¼AT¼ALIN 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 Encyclopediaof triangle centers. A series of these points being points of concurrence ofcevians of rank (k; l;m), of the triangle. Also, we present several equalitiesfrom 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 specifyingthe 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 barycentriccenter with coordinates (t1; t2; t3) : Note that the triangle areas BPC;CPAand APB are proportional with barycentric coordinates t1; t2 and t3. Formore details we refer to the monographs of C. Bradley [3], C. Coand¼a [4], C.Cosnit¼a [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, 51M30Received: 21.02.2012. In revised form: 4.04.2012. Accepted: 18.05.2012.

Cevians of rank (k,l,m) in triangle 23

2. About the cevians of rank (k; l;m)

For start we will do the link between barycentric coordinates and thereports in which the triangle sides are divided from the lines passing throughthe triangle�s point and through the barycentric center.

Lemma 2.1. Let ABC be a triangle and a point P which has the barycentriccoordinates (t1; t2; t3) and AP \ BC = fA0g ; BP \ CA = fB0g and CP \AB = fC 0g : Then the equalities

(1)BA0

A0C=t3t2;CB0

B0A=t1t3;AC 0

C 0B=t2t1

hold.

Proof. Consider a point P in interior of the triangle ABC (see Figure 1).If P is not in the triangle�s interior, the demonstration is done similarly.

We have� [ABP ]

� [ACP ]=AB �AP � sinBAA0AC �AP � sinCAA0 =

� [ABA0]

� [ACA0]=BA0

A0C;

but as the areas of the triangles PBC;PCA;PAB are proportional withbarycentric coordinates t1; t2 and t3, it follows that

� [ABP ]

� [ACP ]=t3t2;

from where, we deduce the relation

BA0

A0C=t3t2:

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 tothe points of contact of the opposite sides with the excircles relative to thesesides are concurrent.

Proof. Let D;E; F be the contact points of the excircles with sides oppositeto 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 Nicusor Minculete and C¼at¼alin Barbu

ThereforeBD

DC=s� cs� b ;

CE

EA=s� as� c ;

AF

FB=s� bs� a

henceBD

DC� CEEA

� AFFB

= 1;

and from the converse of Ceva�s theorem, we get that lines AD; BE andCF are concurrent. The point N common to the lines AD;BE;CF is oftenreferred to as the Nagel point of the triangle ABC: �Theorem 2.3. ( [9]): Let I be the incenter of triangle ABC. Through pointB draw a parallel to IC which intersects the parallel drawn through C at ICin point A0 . Similarly are obtained the points B0 and C 0. Then the linesAA0, BB0, CC 0 are concurrent.

Proof. Denote by D;E; F the intersections of the lines AA0 with BC; BB0

with AC, and CC 0 with AB; respectively (see Figure 3).

BecauseBA0 = CI =

r

sin C2; CA0 = BI =

r

sin B2;

it follows that

BD

DC=� [BAA0]

� [CAA0]=BA0 �AB � sin

�C2 +B

�CA0 � CA � sin

�B2 + C

� =


(2)sin�C2 +B

�� cos C2

sin�B2 + C

�� cos B2

=sinA+ sinB

sinA+ sinC=a+ b

a+ c

Similarly:

(3)CE

EA=b+ c

b+ a;AF

FB=c+ a

c+ b

From relations (2) and (3) we get

BD

DC� CEEA

� AFFB

= 1;

which means that, by using Ceva�s converse theorem, the lines AD;BE;CFare concurrent. �Denote by S the intersection of the lines AD;BE; and CF . Because

BDDC =

a+ba+c ,

CEEA =

b+cb+a and

AFFB =

c+ac+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 thecenter of the inscribed circle in the median triangle corresponding to triangleABC ([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 2 BC)are called isogonals in relation with angle ^BAC; if ^BAE � ^CAF and^BAF � ^CAE , otherway said, if cevians AE and AF are symmetricalin relation with the angle bisector of A. Also, AE will be called the isogonalof AF and reverse.

Theorem 2.4. (Steiner [1] ; [6] ; [9]): In a triangle ABC, the isogonals AEand AF , where E;F 2 BC, determine the following relation

BE �BFCE � CF =

AB2

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:

� [ABC] =AB �AC � sinA

2


Therefore,

BE

CE=� [ABE]

� [AEC]=AB �AE � sinxAC �AE � sin y =

AB � sinxAC � sin y ;

andBF

CF=� [ABF ]

� [AFC]=AB �AF � sin yAC �AF � sinx =

AB � sin yAC � sinx:

The conclusion results by multiplying member with member of the previousrelations. �From Theorem 6, we deduce easily the following:

Theorem 2.5. (Mathieu [6] ; [9]). If three lines from the vertices of atriangle are concurrent, their isogonals are also concurrent.

If the point M is the intersection of cevians AD;BE;CF and M 0 is anintersection point for cevians AD0; BE0; CF 0, then we say that M 0 is theisogonal point to M in relation with triangle ABC:

De�nition 2. In a triangle ABC, cevians AE and AF (E;F 2 (BC)) arecalled isotomics if points E and F are symmetrical to the midpoint 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 threeconcurrent cevians are concurrent.

If M is the intersection point of the cevians AD, BE, CF and M 0 is theintersection point of the isotomics AD0, BE0, CF 0, then we say that M 0 isthe isotomic of M in relation to the triangle ABC.

De�nition 3. If on the side (BC) of a unisosceles triangle ABC a pointD is taken, so that

BD

DC=�cb

�kk 2 R, then AD is called cevian of rank k ; and if D 2 BCn [BC], sothat BDDC =

�cb

�k, k 2 R�, then AD is called excevian of rank k or exteriorcevian of rank k : If the ABC triangle is isosceles (AB = AC), then, throughconvention, the cevian of rank k is a median from A .

Theorem 2.7. ( [9] ). In a triangle, the cevians of rank k corresponding tothe points A;B;C are concurrent.

Remark 1. a) Note with Ik the intersection point of cevians of rank k: b) IfM is the intersection point of cevians AD;BE;CF and M 0 is the isogonalpoint 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 cevianof 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 theangle bisector using relation c

b , we will de�ne two types of cevians, namely,the N -cevian of rank l starting from the cevian determined by Nagel�s pointand S - cevian of rank m starting from the cevian determined by Spieker�spoint.


De�nition 4. If on side (BC), of a unisosceles ABC triangle a point D istaken, so that

BD

DC=

�s� cs� b

�ll 2 R, then AD is called N - cevian of rank l , and if D 2 BCn [BC], sothat BDDC =

�s�cs�b

�l, l 2 R�, then AD is called N - cevian of rank l or N

- 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 istaken, so that

BD

DC=

�a+ b

a+ c

�mm 2 R , then AD is called S - cevian of rank m, and if D 2 BCn [BC] , sothat BDDC =

�a+ba+c

�m, m 2 R�, then AD is called S - excevian of rank m or S

- 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 fromA.

To approach unitary these types of cevians we will de�ne a more generaltype 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 istaken, so that:

(4)BD

DC=�cb

�k��s� cs� b

�l��a+ b

a+ c

�mk; l;m 2 R, then AD is called cevian of rank (k ; l ;m); and if D 2 BCn [BC],so that BDDC =

�cb

�k �� s�cs�b

�l��a+ba+c

�m; k; l;m 2 R�, then AD is called excevian

of rank (k ; l ;m) or exterior cevian of rank (k ; l ;m): If ABC triangle isisosceles (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 therelationship (4) and then we multiply them, we get DB

DC �ECEA �

FAFB = 1, so,

from Ceva�s converse 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�s isotomic) is cevian of rank (�1);e) Antibisector�s isogonal is cevian of rank three;f) Symmedian�s isotomic is cevian of rank (�2);g) Isogonal of a cevian of rank k is a cevian of rank (2� k);


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 unitethe vertices of a ABC triangle with contact points of the incircle with oppo-site sides are concurrent.

The demonstration results immediate, considering the fact that incircleand excircles determine on every side of the triangle, pairs of isotomic points.The point of concurrence of the cevians considered is called Gergonne�s pointor 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�s isogonal is cevian of rank (3; 0; 0) ;6) The symmedian�s isotomic is cevian of rank (�2; 0; 0) ;7) A cevian�s isogonal of rank (k; 0; 0; ) is a cevian of rank (2� k; 0; 0) ;8) A cevian�s isotomic 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 ofcevians of rank (k; l;m). Starting from the idea of a problem [10], we obtainthe 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 theopposite sides. Let P be a point on the sides of the triangle and X;Y and Zrespectively, the perpendicular feet of P on the side BC;CA and AB:Thereare the following relations:

a) If P 2 [EF ] ; then we have

(5)x

ak�1(s� a)l(b+ c)m =y

bk�1(s� b)l(a+ c)m +z

ck�1(s� c)l(a+ b)m ;

where jPXj = x; jPY j = y; jPZj = z;b) If P 2 [FD]; then we have

(6)y

bk�1(s� b)l(a+ c)m =x

ak�1(s� a)l(b+ c)m +z

ck�1(s� c)l(a+ b)m

c) If P 2 [DE]; then we have

(7)z

ck�1(s� c)l(a+ b)m =x

ak�1(s� a)l(b+ c)m +y

bk�1(s� b)l(a+ c)m


Proof. Since AD is the cevian of rank (k; l;m); implies the relation

BD

DC=�cb

�k �s� cs� b

�l �a+ ba+ c

�m;

so

BD =ack(s� c)l(a+ b)m

bk(s� b)l(a+ c)m + ck(s� c)l(a+ b)mIn the analogous way, we deduce the relations

AE =bck(s� c)l(a+ b)m

ak(s� a)l(b+ c)m + ck(s� c)l(a+ b)mand

AF =cbk(s� b)l(a+ c)m

ak(s� a)l(b+ c)m + bk(s� b)l(a+ c)m

Therefore, we have the relation (see Figure 5)

�[AFE] = �[AFP ] + �[APE];

which is equivalent to

(bc)k+1 � [(s� b)(s� c)]l[(a+ b)(a+ c)]m[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] �

sinA

2=

(8)cbk(s� b)l(a+ c)m � z

2[ak(s� a)l(b+ c)m + bk(s� b)l(a+ c)m]+bck(s� c)l(a+ b)my

2[ak(s� a)l(b+ c)m + ck(s� c)l(a+ b)m]But �[ABC] = bc sinA

2 and 2�[ABC] = ax+ by+ cz; and using relation (8),we obtain:

(bc)k�1[(s� b)(s� c)]l[(a+ b)(a+ c)]m(ax+ by + cz) =zbk�1[ak(s� a)l(b+ c)m + ck(s� c)l(a+ b)m](s� b)l(a+ c)m+yck�1[ak(s� a)l(b+ c)m + bk(s� b)l(a+ c)m](s� c)l(a+ b)m;

so

xa(bc)k�1[(s�b)(s�c)]l[(a+b)(a+c)]m+ybkck�1[(s�b)(s�c)]l[(a+b)(a+c)]m+zckbk�1[(s� b)(s� c)]l[(a+ b)(a+ c)]m =

zakbk�1[(s�a)(s�b)]l[(b+c)(a+c)]m+zbk�1ck[(s�c)(s�b)]l[(a+b)(a+c)]m+yakck�1[(s�a)(s�c)]l[(b+c)(a+b)]m+ybkck�1[(s�b)(s�c)]l[(a+b)(a+c)]m


Dividing by (abc)k�1 � [(s � a)(s � b)(s � c)]l � [(a + b)(b + c)(c + a)]m; wededuce the relation (5). Similary, we �nd relations (6) and (7). �

Remark 4. The notation cevians of rank (k; l;m) can be extend to notationthe cevians of rank (ku; ku+1; :::; kw) if relation (4) can be written

(9)BD

DC=

wYi=u

�is� cis� b

�kiwhere u � w; u;w 2 Z; ki 2 R; for all i 2 fu; ::; wg: For u = 0; w = 2; k0 =k; k1 = l; k2 = m; we deduce relation (4). Therefore relation (5) becomes

(10)ax

wYi=l

(is� a)ki=

bywYi=l

(is� b)ki+

czwYi=l

(is� c)ki

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), thenwe 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

X(6) I(2; 0; 0) Lemoine point 1ax =

1by+

1cz

X(7) I(0;�1; 0) Gergonne point a(s� a)x = b(s� b)y + c(s� c)zX(8) I(0; 1; 0) Nagel point a

s�ax =bs�by+

cs�cz

X(9) I(1; 1; 0) mittenpunkt 1s�ax =

1s�by+

1s�cz

X(10) I(0; 0; 1) Spieker point ab+cx =

bc+ay+

ca+bz

X(12) I(0;�1; 2) fX(1); X(5)g�harmonicconjugate of X(11)

a(s�a)(b+c)2

x = b(s�b)(c+a)2

y+ c(s�c)(a+b)2

z

X(21) I(1; 1;�1) Schi er point (b+c)s�a x =

(c+a)s�b y+

(a+b)s�c z

X(31) I(3; 0; 0) 2nd power point 1a2x = 1

b2y+ 1

c2z

X(32) I(4; 0; 0) 2rd power point 1a3x = 1

b3y+ 1

c3z

X(37) I(1; 0; 1)crosspoint of

incenter and centroid1b+cx =

1c+ay+

1a+bz

X(41) I(3; 1; 0)X(6) Ceva conjugate

of X(31)1

a2(s�a)x =1

b2(s�b)y+1

c2(s�c)z


incenter and Lemoine point1

a(b+c)x =1

b(c+a)y+1

c(a+b)z

X(55) I(2; 1; 0) insimilicenter 1a(s�a)x =

1b(s�b)y+

1c(s�c)z


X(56) I(2;�1; 0) exsimilicenter s�aa x =

s�bb y+

s�cc z

X(57) I(1;�1; 0) isogonal conjugate of X(9) (s� a)x = (s� b)y + (s� c)zX(58) I(2; 0;�1) isogonal conjugate of X(10) b+c

a x =c+ab y+

a+bc z

X(65) I(1;�1; 1) orthocenter ofthe intouch triangle

s�ab+cx =

s�bc+ay+

s�ca+bz

X(76) I(�2; 0; 0) 3rd Brocard point a3x = b3y + c3z

X(81) I(1; 0;�1) Cevapoint of incenterand Lemoine point

(b+ c)x = (c+ a)y + (a+ b)z

X(85) I(�1;�1; 0) isotomic conjugate of X(9) a2(s� a)x = b2(s� b)y + c2(s� c)z

X(86) I(0; 0;�1) Cevapoint of incenterand centroid

a(b+ c)x = b(c+ a)y + c(a+ b)z

X(174) I��12 ;

�12 ; 0

�Y¤ center of congruence

apa(s� a)x =

bpb(s� b)y + c

pc(s� c)z

X(188) I��12 ;

12 ; 0� 2nd mid-arc point of

anticomplementary triangle

aq

as�ax =

bq

bs�by + c

qcs�cz


of X(9)

1(s�a)2x =1

(s�b)2 y+1

(s�c)2 z


of X(37)

1(s�a)(b+c)x =1

(s�b)(c+a)y+1

(s�c)(a+b)z


of X(42)

1a2(b+c)

x =1

b2(c+a)y+ 1

(s�c)(a+b)z


of X(55)

1a(s�a)2x =1

b(s�b)2 y+1

c(s�c)2 z

X(226) I(0;�1; 1) X(7) Ceva conjugateof X(65)

a(s�a)xb+c = b(s�b)y

c+a + c(s�c)za+b

X(259) I�12 ;12 ; 0�

isogonal conjugate of X(174)q

as�ax =

qbs�by+

qcs�cz

X(266) I�12 ;12 ; 0�

isogonal conjugate of X(188)

pa(s� a)x =p

b(s� b)y+pc(s� c)z

X(269) I(0;�2; 0) isogonal conjugate of X(200)a(s� a)2x =

b(s� b)2y + c(s� c)2z

X(274) I(�1; 0;�1) isogonal conjugate of X(213)a2(b+ c)x =

b2(c+ a)y + c2(a+ b)z

X(279) I(�1;�2; 0) isogonal conjugate of X(220)a(s� a)2x =

b(s� b)2y + c(s� c)2z

X(310) I(�2; 0;�1) isotomic conjugate of X(42)a3(b+ c)x =

b3(c+ a)y + c3(a+ b)z

X(312) I(�1; 1; 0) isotomic conjugate of X(57) a2

s�ax =b2

s�by+c2

s�cz


b+cx =b3

c+ay +c3

a+bz

X(314) I(�1; 1;�1) isotomic conjugate of X(65)a2(b+c)s�a x =

b2(c+a)s�b y + c2(a+b)

s�c z


b+cx =b2

c+ay+c2

a+bz

X(333) I(0; 1;�1) Cevapoint of X(8) and X(9) a(b+c)s�a x =

b(c+a)s�b y+

c(a+b)s�c z


(s�a)2x =b2

(s�b)2 y+c2

(s�c)2 z


X(346) I(0; 2; 0)isotomic conjugate

of X(279)a

(s�a)2x =b

(s�b)2 y+c

(s�c)2 z

X(365) I(32 ; 0; 0) square root point 1pax = 1p

by+ 1p

cz

X(366) I(12 ; 0; 0)isogonal conjugate

of X(365)pax =

pby+

pcz

X(479) I(0;�3; 0) X(269)-cross conjugateof X(279)

a(s� a)3x = b(s� b)3y + c(s� c)3z

X(552) I(0;�1;�2) point Maia Ia(s� a)(b+ c)2x =

b(s� b)(c+ a)2y + c(s� c)(a+ b)2zX(560) I(5; 0; 0) 4th power point 1

a4x = 1

b4y+ 1

c4z

X(561) I(�3; 0; 0) isogonal conjugateof 4th power point

a4x = b4y + c4z

X(593) I(2; 0;�2) 1st Hatzipolakis-Yiu point (b+c)2

a x = (c+a)2

b y+ (a+b)2

c z

X(594) I(0; 0; 2)isogonal conjugate

of X(593)a

(b+c)2x = b

(c+a)2y+ c

(a+b)2z


X(10) and X(37)1

(b+c)2x = 1

(c+a)2y+ 1

(a+b)2z

X(757) I(1; 0;�2) isogonal conjugateof X(756)

(b+ c)2x = (c+ a)2y + (a+ b)2z

X(872) I(3; 0; 2)crossum of

X(86) and X(274)1

a2(b+c)2x = 1

b2(c+a)2y+ 1

c2(a+b)2z

X(873) I(1; 0;�2) isogonal conjugateof X(872)

a2(b+ c)2x =

b2(c+ a)2y + c2(a+ b)2z

X(1014) I(1;�1;�1) isogonal conjugateof X(210)

(s� a)(b+ c)x =(s� b)(c+ a)y + (s� c)(a+ b)z

X(1088) I(�1; 2; 0) triliniare square of X(7)a2(s� a)2x =

b2(s� b)2y + c2(s� c)2zX(1089) I(�1; 0; 2) triliniare square of X(10) a2

(b+c)2x = b2

(c+a)2y+ c

(a+b)2z

X(1253) I(3; 2; 0)isogonal conjugateof X(1088)

1a2(s�a)2x =

1b2(s�b)2 y+

1c2(s�c)2 z

X(3596) I(�2; 1; 0) 1st Odehnal point a3

s�ax =b3

s�by+c3

s�cz

References

[1] Barbu, C., Fundamental Theorems of Triangle Geometry (Romanian), EdituraUnique, Bac¼au, 2008.

[2] Bottema, O., On the Area of a Triangle in Barycentric Coordinates, Crux. Math.,8(1982), 228-231.

[3] Bradley, C., The Algebra of Geometry: Cartesian, Areal and Projective Coordinates,Bath: Highperception, 2007.

[4] Coand¼a, C., Analytical Geometry in barycentric coordinates (Romanian), EdituraReprograph, Craiova, 2005.

[5] Cosnit¼a, C., Coordonnées barycentriques, Librairie Vuibert, Paris, 1941.[6] Coxeter, H., Introduction to Geometry, 2nd ed., New York, Wiley, 1969, 216-221.


[7] Kimberling, C., Encyclopedia of triangle centers,http://faculty.evansville.edu/ck6/encyclopedia/.

[8] Loney, S., The Elements of Coordinate Geometry, London, Macmillan, 1962.[9] Minculete, N., Theorems and Problems of Geometry (Romanian), Editura Euro-

carpatica, Sfântu Gheorghe, 2007.[10] Problem Enem3, La Gaceta de la RSME, 13(2010), 723-724.[11] Scott, J., Some examples of the use of areal coordinates in triangle geometry, The

Mathematical Gazette, 11(1999), 472-477.[12] Tanner, H., Areal Coordinates, The Mathematical Gazette, 28(1901).[13] Yiu, P., The Uses of Homogeneous Barycentric Coordinates in Plane Euclidean Geom-

etry, Internat. J. Math. Ed. Sci. Tech., 31(2000), 569-578.

"DIMITRIE CANTEMIR" UNIVERSITY107 BISERICII ROMÂNE, 500068 BRASOV, ROMANIAE-mail address: [email protected]

"VASILE ALECSANDRI" NATIONAL COLLEGEVASILE ALECSANDRI 37, 600011 BAC¼AU, ROMANIAE-mail address: [email protected]

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