Let ABC be a fundamental triangle with the area ∆. For a circle K the powers of vertices A,B,C with regard to K divided by 2∆ are said to be the barycentric coordinates of K with respect to triangle ABC. This paper gives some theory and applications of these coordinates

V. Volenec

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Mathematical Communications 9(2004), 79-89 79Circles in barycentric coordinatesVladimir Volenec∗Abstract. Let ABC be a fundamental triangle with the area .For a circle K the powers of vertices A, B, C with regard to K dividedby 2 are said to be the barycentric coordinates of K with respect totriangle ABC. This paper gives some theory and applications of thesecoordinates.Key words: barycentric coordinates, circleAMS subject classifications: 51N20Received February 6, 2004 Accepted April 20, 2004In [4] the most important facts about metrical relations in barycentric coordi-nates are proven. Some of these facts are necessary in the present paper and weenumerate them.Let ABC be the fundamental triangle with sidelengths a = |BC|, b = |CA|,c = |AB|, measures A, B, C of the opposite angles and with area . Every(finite) point has the absolute barycentric coordinates x, y, z for which the equalityx+y+z = 1 holds and we write P = (x, y, z). For any k ∈ R\{0} numbers x′ = kx,y′ = ky, z′ = kz are the relative barycentric coordinates of point P and we writeP = (x′ : y′ : z′).Fact 1 ([1, Cor.3]). Two points Pi = (xi, yi, zi) (i = 1, 2) have the distance|P1P2| given by|P1P2|2 = 2[α(x1 − x2)2 + β(y1 − y2)2 + γ(z1 − z2)2],where α = cotA, β = cotB, γ = cotC.Fact 2 ([1, Cor.4]). For any point P = (x, y, z) we have the equalities|AP |2 = 2[α(1− x)2 + βy2 + γz2],|BP |2 = 2[αx2 + β(1 − y)2 + γz2],|CP |2 = 2[αx2 + βy2 + γ(1− z)2].∗Department of Mathematics, University of Zagreb, Bijenicˇka 30, HR-10 000 Zagreb, Croatia,e-mail: volenec@math.hr80 V.VolenecFact 3 ([1, Th.2]). For the point P = (x, y, z) and any point S we have|SP |2 = x · |SA|2 + y · |SB|2 + z · |SC|2 − a2yz − b2zx− c2xy.Fact 4 ([1, Cor.6]). For any point P = (x, y, z) and the circumcircle (O,R)of triangle ABC we have the equality |OP |2 = R2 − 2Π, where 2Π = a2yz +b2zx+ c2xy.Fact 5 ([1, Th.4]). Let points Pi = (xi : yi : zi) (i = 1, 2) have the sumssi = xi + yi + zi of coordinates. If the point P = (x : y : z) satisfies the equalityP = µP1 + νP2, then these three points have the ratioP1PP2P= (P1P2P ) = − νµ· s2s1.Any straight line P has the barycentric coordinates X, Y, Z determined up toproportionality and we write P = (X : Y : Z). The point (x : y : z) lies on the line(X : Y : Z) iff xX + yY + zZ = 0. The line at infinity is the line N = (1 : 1 : 1)and every point at infinity is of the form (x : y : z), where x + y + z = 0. Theline (X : Y : Z) has the point at infinity ((Y − Z) : (Z −X) : (X − Y )). The linethrough two points Pi = (xi, yi, zi) (i = 1, 2) has the point at infinity ((x1 − x2) :(y1 − y2) : (z1 − z2)).Fact 6 ([1, Cor.16]). Two lines with the points at infinity (xi : yi : zi) (i = 1, 2)are orthogonal iff αx1x2 + βy1y2 + γz1z2 = 0.Fact 7 ([1], formulas (21) and (8)). If α = cotA , β = cotB , γ = cotC, thenβγ + γα+ αβ = 1anda2 = 2(β + γ), b2 = 2(γ + α), c2 = 2(α+ β).Fact 8 ([1, Th.7]). The fundamental triangle ABC has the orthocenter H=(βγ, γα, αβ).Fact 9 ([1, Th.8]). The oriented angle ϑ of oriented lines Pi = (Xi : Yi : Zi)(i = 1, 2) is given bycotϑ =1k(αx1x2 + βy1y2 + γz1z2),where xi = Yi − Zi, yi = Zi −Xi, zi = Xi − Yi (i = 1, 2) andk =∣∣∣∣∣∣1 1 1X1 Y1 Z1X2 Y2 Z2∣∣∣∣∣∣ .Using Fact 1 for the distance of two points S = (xo, yo, zo) and P = (x, y, z) weobtain the equation of the circle K with the center S and the radius  in the formα(x − xo)2 + β(y − yo)2 + γ(z − zo)2 − 22 = 0. (1)Circles in barycentric coordinates 81If P = (x, y, z) is any point, then pP,K = |SP |2 − 2 is the power of point Pwith respect to the circle K. Numbersλ =12pA,K, µ =12pB,K, ν =12pC,K (2)are said to be the barycentric coordinates of the circle K with respect tothe fundamental triangle ABC. According to Facts 3 and 7 and the equalitiesx+ y + z = 1 and (2) we obtainpP,K = |SP |2 − 2 (3)= x(|SA|2 − 2) + y(|SB|2 − 2) + z(|SC|2 − 2)− a2yz − b2zx− c2xy= x · 2λ+ y · 2µ+ z · 2ν − 2(β + γ)yz − 2(γ + α)zx− 2(α+ β)xy= 2[(λx+ µy + νz)(x+ y + z)− (γ + α)yz − (α+ β)zx− (α+ β)xy].The point P lies on the circle K iff pP,K = 0. Therefore the following theoremholds.Theorem 1. The circle with the barycentric coordinates λ, µ, ν has the equation(λx + µy + νz)(x+ y + z)− (β + γ)yz − (γ + α)zx− (α+ β)xy = 0 (4)and the point P = (x, y, z) has the power pP,K with respect to this circle given by12pP,K = λx + µy + νz − (β + γ)yz − (γ + α)zx− (α+ β)xy, (5)i.e. pP,K = 2(λx+ µy + νz)− a2yz − b2zx− c2xy.Because of Theorem 1 it follows that any circle K is uniquely determined by itsbarycentric coordinates λ, µ, ν and we write K = (λ, µ, ν).The equation of the circle K in the form (1) gives immediately the coordinatesof the center S = (xo, yo, zo) and the radius  and the equation in the form (4)gives the barycentric coordinates of the circle K = (λ, µ, ν). Therefore, it is usefulto know how to pass from the first equation to the second and vice versa. Thesepassages are given by Theorems 2 and 3.Theorem 2. If the circle (λ, µ, ν) has the center S = (xo, yo, zo) and theradius , then we haveλ = α(1 − xo)2 + βy2o + γz2o −22 ,µ = αx2o + β(1− yo)2 + γz2o −22 , (6)ν = αx2o + βy2o + γ(1− zo)2 −22 .Proof. According to Fact 2 equalities (2) imply e.g.λ =12(|SA|2 − 2) = α(1 − xo)2 + βy2o + γz2o −22 .✷82 V.VolenecCorollary 1. Two circles (λ, µ, ν) and (λ′, µ′, ν′) with the radii  and ′ areconcentric iffλ− λ′ = µ− µ′ = ν − ν′ = 12 (′2 − 2).Any circle concentric with the circumcircle (0, 0, 0) of the triangle ABC has theform (κ, κ, κ) for some κ ∈ R.Corollary 2. For a circle (λ, µ, ν) with the center (xo, yo, zo) we have theequalitiesλ− α+ 2αxo = µ− β + 2βyo = ν − γ + 2γzo.This Corollary can be used if three coordinates λ, µ, ν and one of three coordi-nates xo, yo, zo are known, e.g. if we know that the center lies on one of the linesBC, CA, AB, or if all coordinates xo, yo, zo and one of coordinates λ, µ, ν areknown, e.g. if the circle passes through point A and therefore we have λ = 0. If wehave xo = 1, yo = zo = 0, then λ+α = µ−β = ν−γ, i.e. µ = λ+α+β , ν = λ+γ+αand any circle with center A has the form (λ, λ+α+β, λ+ γ+α) for some λ ∈ R.Analogously, any circle with center B has the form (µ + α + β, µ, µ + β + γ) forsome µ ∈ R and any circle with center C has the form (ν + γ + α, ν + β + γ, ν) forsome ν ∈ R.Theorem 3. The center (xo, yo, zo) and the radius  of the circle (λ, µ, ν) aregiven by equalities2xo = −(β + γ)λ+ γµ+ βν + 1− βγ,2yo = γλ− (γ + α)µ+ αν + 1− γα, (7)2zo = βλ+ αµ− (α + β)ν + 1− αβ,22 = α(µ− ν)2 + β(ν − λ)2 + γ(λ− µ)2 − 2(λ+ µ+ ν) + 2(βγλ+ γαµ+ αβν)+α+ β + γ − αβγ. (8)Proof. Because of Theorem 2 we must prove that substitutions of xo, yo, zo and2 from (6) and (8) into the right-hand side of (5) give λ, µ, ν, respectively. But,we have at firstαx2o + βy2o + γz2o −22 = βγλ+ γαµ+ αβν − αβγ (9)because the coefficients of λ2, µν, λ and 1 after these substitutions are respectively14[α(β + γ)2 + βγ2 + γβ2 − β − γ] = 14(β + γ)[α(β + γ) + βγ − 1]= 0,14[2αβγ − 2αβ(γ + α)− 2γα(α+ β) + 2α]=α2(1 − βγ − γα− αβ) = 0,Circles in barycentric coordinates 8314[− 2α(β + γ)(1− βγ) + 2βγ(1− γα) + 2βγ(1− αβ) + 2− 2βγ]=12(1 + βγ − γα− αβ) = βγ,14[α(1 − βγ)2 + β(1− γα)2 + γ(1− αβ)2 − (α+ β + γ) + αβγ]=14(αβ2γ2 + α2βγ2 + α2β2γ − 5αβγ) = 14αβγ(βγ + γα+ αβ − 5)= −αβγ,where we have used the equality βγ+γα+αβ = 1 from Fact 7, and analogously thecoefficients of µ2, ν2, νλ, λµ, µ, ν are 0, 0, 0, 0, γα, αβ, respectively. After that,by (9) and (6) we get e.g.αx2o + βy2o + γz2o −22 − 2αxo + α = βγλ+ γαµ+ αβν − αβγ + α(β + γ)λ−γαµ− αβν − α+ αβγ + α= (βγ + γα+ αβ)λ = λ.✷Specially, the circumcircle (0, 0, 0) of triangle ABC has the centerO =(12(1 − βγ), 12(1− γα), 12(1− αβ))and the radius R given by the equality2R2 = α+ β + γ − αβγ. (10)¿From (8) it follows immediately: the circle (λ, µ, ν) is real iff the sum on theright side of (8) is positive.The following theorem is very useful.Theorem 4. If Pi = (xi, yi, zi) (i = 1, 2), then the circle KP1P2 = (λ, µ, ν)with the diameter P1P2 is given byλ = αx1x2 + βy1y2 + γz1z2 + α(1 − x1 − x2),µ = αx1x2 + βy1y2 + γz1z2 + β(1 − y1 − y2),ν = αx1x2 + βy1y2 + γz1z2 + γ(1− z1 − z2).Proof. The segment P1P2 has the midpointPo =(12(x1 + x2),12(y1 + y2),12(z1 + z2))84 V.Volenecand Facts 1 and 2 imply e.g.λ =12(|APo|2 − 14|P1P2|2)= α(1− x1 + x22)2+ β(y1 + y22)2+ γ(z1 + z22)2−α(x1 − x22)2− β(y1 − y22)2− γ(z1 − z22)2= αx1x2 + βy1y2 + γz1z2 + α(1 − x1 − x2).✷Corollary 3. If P = (x, y, z), then we haveKAP = (0, αx + β(1− y), αx+ γ(1− z)),KBP = (βy + α(1− x), 0, βy + γ(1− z)),KCP = (γz + α(1 − x), γz + β(1− y), 0).Corollary 4. If D = (0, d, d′), E = (e′, 0, e), F = (f, f ′, 0), then we haveKAD = (0, βd′, γd), KBE = (αe, 0, γe′), KCF = (αf ′, βf, 0). Especially, KBC =(α, 0, 0), KCA = (0, β, 0), KAB = (0, 0, γ).Corollary 5. If Pi = (0, yi, zi) resp. Pi = (xi, 0, zi) resp. Pi = (xi, yi, 0) fori = 1, 2, then we haveKP1P2 = (α+ βy1y2 + γz1z2, (β + γ)z1z2, (β + γ)y1y2),KP1P2 = ((γ + α)z1z2, αx1x2 + β + γz1z2, (γ + α)x1x2),KP1P2 = ((α+ β)y1y2, (α + β)x1x2, αx1x2 + βy1y2 + γ),respectively.In the proof of Corollary 5 we used e.g. the equalitiesβy1y2+γz1z2+β(1−y1−y2) = β(1−y1)(1−y2)+γz1z2 = βz1z2+γz1z2 = (β+γ)z1z2.The points B = (0, 1, 0) and C = (0, 0, 1) have the midpoint D = (0, 12 ,12 )whereas the point A = (1, 0, 0) and the orthocenter H = (βγ, γα, αβ) from Fact 8have the midpoint D′ = (12 (1 + βγ),12γα,12αβ). Let us find the coordinates of thecircle KDD′ using Theorem 4. With x1 = 0, y1 = z1 = 12 and x2 = 12 (1 + βγ),y2 = 12γα, z2 =12αβ we obtain αx1x2 + βy1y2 + γz1z2 =12αβγ and thereforeλ =12αβγ + α[1− 12(1 + βγ)] =α2,µ =12αβγ + β(1− 12− 12γα) =β2and analogously ν = γ2 . Because of symmetry of coordinates of this circle E =(α2 ,β2 ,γ2 ) it follows that this circle has diameters DD′, EE′, FF ′, where D, E, FandD′, E′, F ′ are the midpoints of segments BC, CA, AB and AH , BH , CH . TheCircles in barycentric coordinates 85obtained circle E is the nine–point circle of triangle ABC and it passes obviouslythrough the feet AH ∩BC, BH ∩CA, CH ∩AB of the altitudes AH , BH , CH ofthis triangle. For the center (xo, yo, zo) of circle E by Theorem 3 we obtain e.g.2xo = −(β+γ)α2 +γβ2+βγ2+1−βγ = 1− 12(γα+αβ) = 1− 12(1−βγ) = 12(1+βγ).Therefore, this nine–point center O9 of triangle ABC has the formO9 =(12(1 + βγ),12(1 + γα),12(1 + αβ)).This point is the midpoint of the orthocenter H = (βγ, γα, αβ) and the circum-center O = (12 (1−βγ), 12 (1− γα), 12 (1−αβ)) of the same triangle. The homothety(H, 12 ) maps the circumcircle (O,R) through points A, B, C onto the nine–pointcircle through points D′, E′, F ′ and therefore the nine–point circle has the radiusR2.Now, we shall prove a statement which gives another interpretation of barycen-tric coordinates of a circle.Theorem 5. The circles (λ, 0, 0) resp. (0, µ, 0), resp.(0, 0, ν) are the sets ofpoints P such thatcot∠(BP, CP ) = α−λ resp. cot∠(CP, AP ) = β−µ resp. cot∠(AP, BP ) = γ−ν.Proof. If P = (x : y : z), then we have BP = (z : 0 : −x), CP = (−y : x : 0),k =∣∣∣∣∣∣1 1 1z 0 −x−y x 0∣∣∣∣∣∣ = x(x+ y + z)and with cotϑ = cot∠(BP, CP ) = α − λ Fact 9 implies x(x + y + z)(α − λ) =αx2 − β(z + x)y − γz(x+ y), i. e.λx(x + y + z)− (β + γ)yz − (γ + α)zx− (α+ β)xy = 0. (11)But, this is the equation of the circle (λ, 0, 0). ✷Corollary 6. The set of points P with the given oriented angle of lines AP andBP for two given points A and B is a circle through these points A and B.This is the famous theorem on angles inscribed in the same arc in the versionfor directed angles as developed by R.A. Johnson [2] and [3, numbers 16–19]; for amodern approach see [1].With P = (x, y, z) formula (11) takes the form λx = Π, whereΠ = (β + γ)yz + (γ + α)zx+ (α+ β)xy =12(a2yz + b2zx+ c2xy). (12)Therefore, Theorem 5 implies two more corollaries.86 V.VolenecCorollary 7. For any point P = (x, y, z) we have the equalitiescot∠(BP, CP ) = α− Πx, cot∠(CP, AP ) = β − Πy, cot∠(AP, BP ) = γ − Πz,where Π is given by (12).Corollary 8. For any point P = (x, y, z) the circles BCP , CAP , ABP arethe circles (Πx, 0, 0),(0,Πy, 0),(0, 0,Πz),respectively, where Π is given by (12).Now, let K1 and K2 be any two circles with centers S1 and S2 and radii 1 and2. If these two circles intersect each other, then let ϑ be the angle of these circles,i.e. the directed angle of their tangents at one of their common point P . But thisangle is equal to the angle of the lines S1P and S2P . Therefore, we havecosϑ =1212(21 + 22 − |S1S2|2). (13)Specially, if ϑ = 0 or ϑ = π, i.e. if |S1S2| = |1 − 2| resp. |S1S2| = 1 + 2, thenthe circles K1 and K2 touch each other inwardly resp. outwardly. If ϑ = π2 , i.e. if|S1S2|2 = 21 + 22, (14)then the circles K1 and K2 are orthogonal.Theorem 6. Two circles Ki = (λi, µi, νi) (i = 1, 2) are orthogonal iffα(µ1 − ν1)(µ2 − ν2) + β(ν1 − λ1)(ν2 − λ2) + γ(λ1 − µ1)(λ2 − µ2)−(1− βγ)(λ1 + λ2)− (1− γα)(µ1 + µ2)− (1− αβ)(ν1 + ν2)+α+ β + γ − αβγ = 0. (15)Proof. Let Si = (xi, yi, zi) be the centers of Ki for i = 1, 2. According to Fact 1and Theorem 3 we obtain2|S1S2|2 = 4[α(x1 − x2)2 + β(y1 − y2)2 + γ(z1 − z2)2]= α[− (β + γ)(λ1 − λ2) + γ(µ1 − µ2) + β(ν1 − ν2)]2+β[γ(λ1 − λ2)− (γ + α)(µ1 − µ2) + α(ν1 − ν2)]2+γ[β(λ1 − λ2) + α(µ1 − µ2)− (α+ β)(ν1 − ν2)]2,i.e.2|S1S2|2 = α(µ1−ν1−µ2+ν2)2+β(ν1−λ1−ν2+λ2)2+γ(λ1−µ1−λ2+µ2)2 (16)Circles in barycentric coordinates 87because the coefficients of e.g. (λ1 − λ2)2 and 2(µ1 − µ2)(ν1 − ν2) areα(β + γ)2 + βγ2 + γβ2 = (β + γ)[α(β + γ) + βγ] = β + γ,αβγ − αβ(γ + α)− αγ(α + β) = −α(βγ + γα+ αβ) = −α.As we have e.g.α(µ1 − ν1)2 + α(µ2 − ν2)2 − α(µ1 − ν1 − µ2 + ν2)2 = 2α(µ1 − ν1)(µ2 − ν2),so by formula (8) for 21 and 22 and by (16) it follows2 (21 + 22 − |S1S2|2) = α(µ1 − ν1)(µ2 − ν2) + 2β(ν1 − λ1)(ν2 − λ2)+2γ(λ1 − µ1)(λ2 − ν2)− 2(1− βγ)(λ1 + λ2)−2(1− γα)(µ1 + µ2)− 2(1− αβ)(ν1 + ν2)+2(α+ β + γ − αβγ)and equality (14) is equivalent to equality (15). ✷If K1 = K = (λ, µ, ν) and K2 = (0, 0, 0), then Theorem 6 implies the followingcorollary.Corollary 9. The circle K = (λ, µ, ν) is orthogonal onto the circumcircle oftriangle ABC iff(1− βγ)λ+ (1− γα)µ+ (1− αβ)γ = α+ β + γ − αβγ. (17)The powers pP,Ki of any point P = (x, y, z) with respect to the circles Ki =(λi, µi, νi) (i = 1, 2) are given by (because of Theorem 1)12pP,Ki = λix+ µiy + νiz −Π (i = 1, 2),where Π is given by (12). The set of points P such that pP,K1 = pP,K2 is given bythe equation (λ1−λ2)x+(µ1 −µ2)y+(ν1− ν2)z = 0. This set is a line, the radicalaxis of the circles K1 and K2. We have the following theorem.Theorem 7. Two circles Ki = (λi, µi, νi) (i = 1, 2) have the radical axis P12 =((λ1 − λ2) : (µ1 − µ2) : (ν1 − ν2)).Specially, a circle (λ, µ, ν) and the circumcircle (0, 0, 0) have the radical axis(λ : µ : ν).According to Theorem 3 the point at infinity of the line S1S2 has the coordinatesβ(ν − λ) + γ(µ− λ), γ(λ− µ) + α(ν − µ), α(µ− ν) + β(λ− ν),where λ = λ1 − λ2, µ = µ1 − µ2, ν = ν1 − ν2. On the other hand, from Theorem 7it follows that the radical axis of the circles K1 and K2 has the point at infinity((µ− ν) : (ν − λ) : (λ− ν)). Obviously we haveα(µ− ν)[β(ν − λ) + γ(µ− λ)]+ β(ν − λ)[γ(λ− µ) + α(ν − µ)]+γ(λ− µ)[α(µ− ν) + β(λ − ν)]= 088 V.Volenecand by Fact 6 there follows the well–known fact that the radical axis of two circlesis orthogonal to the join of their centers. If two circles have common points, thenthese points lie on their radical axis.Theorem 8. Three circles Ki = (λi, µi, νi) (i = 1, 2, 3) have in pairs the radicalaxes P12, P13, P23 which have a common pointP =∣∣∣∣∣∣1 1 1µ1 µ2 µ3ν1 ν2 ν3∣∣∣∣∣∣ :∣∣∣∣∣∣1 1 1ν1 ν2 ν3λ1 λ2 λ3∣∣∣∣∣∣ :∣∣∣∣∣∣1 1 1λ1 λ2 λ3µ1 µ2 µ3∣∣∣∣∣∣ , (18)the radical center of the circles K1, K2, K3.Proof. The point P from (18) can be written in the formP =([µ(ν1−ν3)−ν(µ1−µ3)] : [ν(λ1−λ3)−λ(ν1−ν3)] : [λ(µ1−µ3)−µ(λ1−λ3)]),where again λ = λ1 − λ2 , µ = µ1 − µ2 , ν = ν1 − ν2 and by Theorem 7 we haveP12 = (λ : µ : ν). Because ofλ[µ(ν1−ν3)−ν(µ1−µ3)]+µ[ν(λ1−λ3)−λ(ν1−ν2)]+ν[λ(µ1−µ3)−µ(λ1−λ3)]= 0point P lies on line P12. Analogously, this point lies on lines P13 and P23. ✷Two points P and P ′ are said to be inverse to each other with respect to thecircle (S, ) if these points are collinear with the center S and if −→SP · −−→SP ′ = 2.Theorem 9. A point P (x, y, z) has the inverse pointP ′ =(1s′(2R2x− αa2Π), 1s′(2R2y − βb2Π), 1s′(2R2z − γc2Π))(19)with respect to the circumcircle of the triangle ABC, where s′ = 2R2− 4Π and Πis given by (12).Proof. Owing to Fact 7 we obtain2R2x− αa2Π+ 2R2y − βb2Π+ 2R2z − γc2Π= 2R2(x+ y + z)− 2Π[α(β + γ) + β(γ + α) + γ(α+ β)]= 2R2 − 4Π = s′and the point (19) has the sum of coordinates equal to 1. The circumcenter of thetriangle ABC can be written in the formO =(αa24 ,βb24 ,γc24).We have the equality 2R2 · P − s′ · P ′ = 4Π ·O because of e.g. 2R2x− (2R2x−αa2Π) = Π · αa2. Therefore Fact 5 implies the equalities(PP ′O) =s′2R2=R2 − 2ΠR2=|OP |2R2Circles in barycentric coordinates 89because of the equality R2 − 2Π = |OP |2 implied by Fact 4. Hence−→OP =|OP |2R2−−→OP ′and scalar multiplication by −→OP gives −→OP · −−→OP ′ = R2. ✷The author is indebted to the referee for many useful remarks.References[1] R. Fritsch, Ein axiomatischer Zugang zu einigen Winkelsa¨tzen der ebenenGeometrie, Schr. Sudet.dtsch. Akad. Wiss. Ku¨nste Forsch.beitr. Nat.wiss.Kl.16(1995), 41–57.[2] R.A. Johnson, Directed angles in elementary geometry, Amer. Math. Monthly24(1917), 313–317.[3] R.A. Johnson, Advanced Euclidean Geometry, Dover Publ. Inc., New York,1960.[4] V.Volenec, Metrical relations in barycentric coordinates, Math. Commun.8(2003), 55–68.

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