Mathematical Communications 8(2003), 55-68 55 Metrical relations in barycentric coordinates Vladimir Volenec∗ Abstract.Let∆ be the area of the fundamental triangle ABC of barycentric coordinates and let α =cotA, β =cotB, γ =cotC.The vectors vi =[xi,yi,zi](i =1, 2) have the scalar product 2∆(αx1x2 + βy1y2 + γz1z2). This fact implies all important formulas about metrical relations of points and lines. The main and probably new results are Theorems 1 and 8. Key words: barycentric coordinates AMS subject classifications: 51N20 Received December 20, 2002 Accepted April 2, 2003 Let ABC be a given triangle with the sidelengths a = |BC|,b= |CA|,c= |AB|, A, B, C P P the measures of the opposite angles and the area ∆. For any point−→ let be PQ Q−P the radiusvector of this point with respect to any origin. Then−→ we have−→ =−→ . There are uniquely determined numbers y,z ∈ R so that AP = y · AB + z · AC, i.e. P − A = y(B − A)+z(C − A). If we put x =1− y − z, i.e. x + y + z =1, (1) then we have P = xA + yB + zC. (2) Numbers x, y, z, such that (1) and (2) are valid, are uniquely determined by point P and triangle ABC, i.e. these numbers do not depend on the choice of the origin. We say that x, y, z,aretheabsolute barycentric coordinates of point P with respect to triangle ABC and write P =(x, y, z). Obviously A =(1, 0, 0), B =(0, 1, 0), C =(0, 0, 1). Actually, point P is the barycenter of the mass point system of points A, B, C with masses x, y, z, respectively. The centroid of triangle ABC is point G =(1/3, 1/3, 1/3). Any three numbers x ,y ,z proportional to the coordinates x, y, z are said to be relative barycentric coordinates of point P with respect to triangle ABC and we write P =(x : y : z ). Here we have x + y + z =0.Point P is uniquely ∗Department of Mathematics, University of Zagreb, Bijeniˇcka 30, HR-10 000 Zagreb, Croatia, e-mail: [email protected] 56 V. Volenec determined by its relative barycentric coordinates x , y , z because its absolute barycentric coordinates are x y z x = ,y= ,z= . x + y + z x + y + z x + y + z If point P divides two different points P1 =(x1,y1,z1)andP2 =(x2,y2,z2)in −−→ −−→ the ratio (P1P2P )=λ, i.e. if P1P = λ · P2P ,thenfromP − P 1 = λ(P − P 2)with P i = xiA + yiB + ziC (i =1, 2) we obtain (1 − λ)P =(x1 − λx2)A +(y1 − λy2)B +(z1 − λz2)C. Because of xi + yi + zi =1(i =1, 2) we have 1 (x1 − λx2 + y1 − λy2 + z1 − λz2)=1 1 − λ and therefore x1 − λx2 y1 − λy2 z1 − λz2 P = , , , (3) 1 − λ 1 − λ 1 − λ P =((x1 − λx2):(y1 − λy2):(z1 − λz2)) . (4) Specially, with λ = −1, point P is the midpoint of the points P1 and P2 with x1 + x2 y1 + y2 z1 + z2 P = , , , 2 2 2 1 1 i.e. P = 2 P1 + 2 P2.SidesBC, CA, AB have the midpoints 1 1 1 1 1 1 0, , =(0:1:1), , 0, =(1:0:1), , , 0 =(1:1:0). 2 2 2 2 2 2 If λ = 1, then equality (3) has no sense, but equality (4) obtains the form P =((x1 − x2):(y1 − y2):(z1 − z2)) (5) and represents the point at infinity of the straight line P1P2. For this point P we have equality (P1P2P ) = 1 and the relative coordinates in (5) have the zero sum. Therefore, this point does not have the absolute coordinates. Because of P1 = P2 point P at infinity cannot be of the form (0 : 0 : 0). Specially, straight lines BC, CA, AB have the points at infinity (0 : 1 : −1), (−1:0:1),(1:−1 : 0), respectively. v y z v −→For any−→ vector numbers and are uniquely determined such that = y · AB + z · AC, i.e. v = y(B − A)+z(C − A). If we put x = −(y + z), then we have v = xA + yB + zC,x+ y + z =0. (6) Numbers x, y, z are uniquely determined and are said to be the barycentric coordi- nates of vector v with respect to triangle ABC.Wewritev =[x, y, z]. For two points Pi =(xi,yi,zi)(i =1, 2) we have P i = xiA + yiB + ziC and therefore −−→ P1P2 = P 2 −P 1 =(x2 −x1)A+(y2 −y1)B +(z2 −z1)C =[x2 −x1,y2 −y1,z2 −z1]. Metrical relations in barycentric coordinates 57 −→ −→ −→ Specially, BC =[0, −1, 1], CA =[1, 0, −1], AB =[−1, 1, 0]. We conclude that the (relative) barycentric coordinates of the point at infinity of a straight line are pro- portional to the barycentric coordinates of any vector parallel to this line. Therefore, parallel lines have the same point at infinity. The formulas for metrical relations can be written in a more compact form if we use numbers α =cotA, β =cotB, γ =cotC (7) We have e.g. b2 + c2 − a2 =2bc cos A =2bc sin A cot A =4∆α and therefore b2 + c2 − a2 =4∆α, c2 + a2 − b2 =4∆β, a2 + b2 − c2 =4∆γ. Adding, we obtain a2 =2∆(β + γ),b2 =2∆(γ + α),c2 =2∆(α + β). (8) Now, let us prove the most important theorem about metrical relations in barycentric coordinates. Theorem 1. The scalar product of two vectors vi =[xi,yi,zi](i =1, 2) is given by v1 · v2 =2∆(αx1x2 + βy1y2 + γz1z2), where ∆ is the area of the fundamental triangle ABC and numbers α, β, γ are given by (7). −→ 2 2 Proof. Squaringthe equality AB = B − A we obtain c2 = A + B − 2A · B, i.e. 2A · B = A2 + B2 − c2 and analogously 2A · C = A2 + C2 − b2 and 2B · C = 2 2 2 B + C − a . Owingto the equalities xi + yi + zi =0(i =1, 2) and (8) we obtain successively 2v1 · v2 =2(x1A + y1B + z1C)(x2A + y2B + z2C) 2 2 2 2 2 2 =2x1x2A +2y1y2B +2z1z2C +(x1y2 + y1x2)(A + B − c ) 2 2 2 2 2 2 +(x1z2 + z1x2)(A + C − b )+(y1z2 + z1y2)(B + C − a ) 2 2 2 =(x1 + y1 + z1)(x2A + y2B + z2C ) 2 2 2 +(x2 + y2 + z2)(x1A + y1B + z1C ) 2 2 2 −a (y1z2 + z1y2) − b (z1x2 + x1z2) − c (x1y2 + y1x2) = −2∆[(β + γ)(y1z2 + z1y2)+(γ + α)(z1x2 + x1z2) +(α + β)(x1y2 + y1x2)] = −2∆{α [x1(y2 + z2)+(y1 + z1)x2]+β [y1(z2 + x2)+(z1 + x1)y2] +γ[z1(x2 + y2)+(x1 + y1)z2]} = −2∆[α(−2x1x2)+β(−2y1y2)+γ(−2z1z2)] =4∆(αx1x2 + βy1y2 + γz1z2). ✷ 58 V. Volenec Corollary 1. The length of the vector v =[x, y, z] is given by |v|2 =2∆(αx2 + βy2 + γz2). Corollary 2. The angle between two vectors vi =[xi,yi,zi](i =1, 2) is given by 1 cos ∠(v1, v2)= (αx1x2 + βy1y2 + γz1z2), Ω1Ω2 where 2 2 2 2 Ωi = αxi + βyi + γzi (i =1, 2). Corollary 3. Two points Pi =(xi,yi,zi)(i =1, 2) have the distance |P1P2| given by 2 2 2 2 |P1P2| =2∆[α(x1 − x2) + β(y1 − y2) + γ(z1 − z2) ]. Specially, with P1 = P =(x, y, z)andP2 = A =(1, 0, 0) or P2 = B =(0, 1, 0) or P2 = C =(0, 0, 1) we obtain further: Corollary 4. For any point P =(x, y, z) we have equalities |AP | =2∆[α(1 − x)2 + βy2 + γz2], |BP| =2∆[αx2 + β(1 − y)2 + γz2], |CP| =2∆[αx2 + βy2 + γ(1 − z)2]. Theorem 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. Proof. Let S be the origin. Squaring the equality P = xA + yB + zC and usingthe equalities from the proof of Theorem 1 we get |SP|2 = P 2 = x2A2 + y2B2 + z2C2 + yz(B2 + C2 − a2) 2 2 2 2 +zx(C + A − b2)+xy(A + B − c2) =(x + y + z)(xA2 + yB2 + zC2) − a2yz − b2zy − c2xy and because of x +y + z = 1 the statement of Theorem 2 follows. ✷ 1 1 1 With P = G = 3 , 3 , 3 we obtain: Corollary 5 [Leibniz]. For centroid G of triangle ABC and for any point P we have 1 3 ·|SG|2 = |SA|2 + |SB|2 + |SC|2 − (a2 + b2 + c2). 3 If S is the circumcenter O of triangle ABC,then|OA| = |OB| = |OC| = R and by(8)itfollows: Metrical relations in barycentric coordinates 59 Corollary 6. For any point P =(x, y, z) and the circumscribed circle (O, R) of triangle ABC the equality |OP|2 = R2 − a2yz − b2zx − c2xy holds, i.e. |OP|2 = R2 − 2∆Π,where 1 Π=(β + γ)yz +(γ + α)zx +(α + β)xy = (a2yz + b2zx + c2xy). (9) 2∆ With S = P Theorem 2 implies: Corollary 7. For the point P =(x, y, z) the equality x ·|AP |2 + y ·|BP|2 + z2|CP|2 =2∆Π holds, where the number Π is given by (9). The equalities from Corollary 4 can be written in another form because of (9), (8) and the equality x + y + z =1.Weobtaine.g. 1 |AP |2 = α(1 − x)2 + βy2 + γz2 2∆ = α − 2αx + αx(1 − y − z)+βy(1 − z − x)+γz(1 − x − y) = α − αx + βy + γz − (β + γ)yz − (γ + α)zx − (α + β)xy α y z βy γz − 1 γ α zx α β xy − = ( + )+ + Π=x[( + ) +( + ) ] Π 1 − β γ yz − 1 − x − β γ yz = x[Π ( + ) ] Π=x[Π(1 ) ( + ) ] 1 a2 = [Π(y + z) − yz].