International Journal of Computer Discovered Mathematics (IJCDM) ISSN 2367-7775 c IJCDM November 2015, Volume 0, No.0, pp.32-48. Received 12 July 2015. Published on-line 15 September 2015 web: http://www.journal-1.eu/ c The Author(s) This article is published with open access1. Barycentric Coordinates Francisco Javier Garc´ia Capitan´ I.E.S. Alvarez Cubero, Priego de C´ordoba,Spain e-mail: [email protected] web: http://garciacapitan.99on.com/ Abstract. We give a short introduction to barycentric coordinates. Keywords. barycentric coordinates, triangle geometry. Mathematics Subject Classification (2010). 51-04. 1. Barycentric coordinates with respect to a triangle. 1.1. Homogeneous barycentric coordinates. Let ABC be a triangle and u; v; w 2 such that u + v + w 6= 0. For any point O, let P be the point R −! −! −−! −! on the plane such that (u + v + w)OP = uOA + vOB + wOC. We can prove that −−! −−! −−! −−! P does not depend on O. For, if (u + v + w)O0P 0 = uO0A + vO0B + wO0C, then −−! −−! −−! −! −−! −! −−! −! (u + v + w)(O0P 0 − OP ) =u(O0A − OA) + v(O0A − OA) + w(O0A − OA) = −−! =(u + v + w)O0O; −−! −−! −! −−! therefore O0P 0 = O0O + OP = O0P and P 0 = P . Hence we can define P as the center of mass of the system formed by the points A; B; C with weights u; v; w. The barycentric coordinates of a point P with respect to the triangle ABC is a list (x : y : z) of three numbers such that x : y : z = (P BC):(PCA):(P AB): Now we prove that P is the center of mass of the system A; B; C with weights x; y; z. 1This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited. 32 Francisco Javier Garc´ıa Capitan´ 33 A −−! −! −−! −! m −−! AD =AB + BD = AB + m+n BC; −! s −−! s −! sm −−! s AP = s+t AD = s+t AB + (s+t)(m+n) BC; −! −! −−! −! −! −−! P OP − OA = s OB − OA + sm OC − OB ; s+t (s+t)(m+n) t −! −! −−! −! OP = t OA + sn OB + sm OC: m n s+t (s+t)(m+n) (s+t)(m+n) B D C Then P is the center of mass of the system formed by A; B; C with weights t(m + n), sn and sm. But, sn DC (ADC) (P DC) (ADC) − (P DC) 4(PCA) = = = = = : sm BD (ABD) (P BD) (ABD) − (P BD) (P AB) This proves that y : z = (PCA):(P AB). In the same way, we can prove that x : y : z = (P BC):(PCA):(P AB). Examples. A A A O G I B C B C B C (1) The centroid G has homogeneous barycentric coordinates (1 : 1 : 1), be- cause the areas GBC, GCA and GAB are equal to each other. (2) The incenter I has coordinates a : b : c, because the areas of the trian- 1 1 1 gles IBC, ICA e IAB are respectively 2 ar, 2 br and 2 cr, where r is the inradius. (3) The circuncenter O. If R is the circumradius, the coordinates of O are (OBC):(OCA):(OAB) = 1 2 1 2 1 2 = 2 R sin 2A : 2 R sin 2B : 2 R sin 2C = = sin A cos A : sin B cos B : sin C cos C = b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 =a · = b · = c · = 2bc 2ac 2ab =a2(b2 + c2 − a2): b2(c2 + a2 − b2): c2(a2 + b2 − c2): (4) The points on line BC have coordinates of the form (0 : y : z). In the same way, points on lines CA and AB have coordinates (x : 0 : z) and (x : y : 0), respectively. Exercises. (1) Show that the sum of coordinates of the circumcenter equals to 4S2, where S is twice the area of ABC. 34 Barycentric coordinates We consider the Heron formula for the area of the triangle and do some algebraic manipulation: a2(b2 + c2 − a2) + b2(c2 + a2 − b2) + c2(a2 + b2 − c2) = =2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4 = =(a + b + c)(−a + b + c)(a − b + c)(a + b − c) = a + b + c −a + b + c a − b + c a + b − c =4 · 4 · · · · = 2 2 2 2 =4S2: (2) Find the coordinates of the excenters. We consider the following figure that shows the triangle ABC and the excenter Ib. A rb I b BC The barycentric coordinates of Ib are (IbBC):(IbCA):(IbAB) = arb : −brb : crb = a : −b : c: We observe that the orientation of triangle IbCA is opposite the orienta- tion of the other two, giving the negative second coordinate. In the same way, we have Ia = (−a : b : c) and Ic = a : b : −c. 1.2. Absolute barycentric coordinates. Let P be a point with (homogeneous barycentric) coordinates (x : y : z). If x+y+z 6= 0, we get the absolute coordinates of P by normalizing the coefficients so that their sum becomes 1: x · A + y · B + z · C P = : x + y + z If we have absolute coordinates of P and Q, the point that divides the segment qP +pQ PQ in the ratio PX : XQ = p : q has absolute coordinates p+q . However, since it is more convenient avoiding denominators in our calculations, we adapt the previous formula in the following way: If P = (u : v : w) and Q = (u0 : v0 : w0) are homogeneous barycentric coordinates satisfying u + v + w = u0 + v0 + w0, then the point X that divides PQ in the ratio PX : XQ = p : q has homogeneous coordinates (qu + pu0 : qv + pv0 : qw + pw0). Exercises. (1) The orthocenter lies on Euler lines and divides OG in the ratio 3 : −2. Prove that their barycentric coordinates can be written H = (tan A : tan B : tan C); Francisco Javier Garc´ıa Capitan´ 35 or equivalently, 1 1 1 H = : : : b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 We have seen that O = a2(b2 + c2 − a2): b2(c2 + a2 − b2): c2(a2 + b2 − c2) ; G = (1 : 1 : 1); with sum 4S2 and 3 respectively. We first consider these coordinates with equal sum 12S2: O = 3a2(b2 + c2 − a2) : 3b2(c2 + a2 − b2) : 3c2(a2 + b2 − c2) ; G = (4S2 : 4S2 : 4S2): Now the first coordinate of H is (−2)3a2(b2 + c2 − a2) + 3 · 4S2 = = − 6(a2b2 + a2c2 − a4) + 3(2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4) = =3a4 − 3b4 + 6b2c2 − 3c4 = =3(a2 − b2 + c2)(a2 + b2 − c2): In the same way, the second and third coordinates are 3(−a2 + b2 + c2)(a2 + b2 − c2) and 3(−a2 + b2 + c2)(a2 − b2 + c2). If we divide them by (−a2 + b2 + c2)(a2 − b2 + c2)(a2 + b2 − c2) we get 1 1 1 H = : : : b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 Now, 1 sin A 1 tan A = 2bc = S = ; b2 + c2 − a2 b2 + c2 − a2 cos A S 2bc threfore H = (tan A : tan B : tan C): (2) Use that the nine point center N divides OG in the ratio 3 : −1 to show that their barycentric coordinates can be written as N = (a cos(B − C): b cos(C − A): c cos(A − B)): Starting from O = 3a2(b2 + c2 − a2) : 3b2(c2 + a2 − b2) : 3c2(a2 + b2 − c2) ; G = (4S2 : 4S2 : 4S2); the first coordinate of N is (−1)3a2(b2 + c2 − a2) + 3 · 4S2 = = −3a2b2 − 3a2c2 + 3a4 + 3(2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4) = 3(a2b2 + a2c2 + 2b2c2 − b4 − c4): 36 Barycentric coordinates To arrive the desired result, we use the formulas a2 + c2 − b2 cos B = ;S = ac sin B 2ac (and the corresponding ones for angle C). Then, cos(B − C) = cos B cos C + sin B sin C = a2 + c2 − b2 a2 + b2 − c2 S S = · + = 2ac 2ab ac ab (a2 + c2 − b2)(a2 + b2 − c2) + 4S2 = = 4a2bc 2a2b2 + 2a2c2 + 4b2c2 − 2b4 − 2c4 = = 4a2bc 1 a2b2 + a2c2 + 2b2c2 − b4 − c4 = · : a 2abc Thus we have N = (a cos(B − C): b cos(C − A): c cos(A − B)): 2. Conway formula 2.1. Notation. If θ is any angle and S is twice the area of ABC, we define Sθ = S cot θ. In particular, b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 S = ;S = ;S = : A 2 B 2 C 2 Given two angles θ and φ, we define Sθφ = Sθ · Sφ. In this notation, we have the formulas : 2 2 2 (1) SB + SC = a , SC + SA = b , SA + SB = c . 2 (2) SAB + SBC + SCA = S . The first one is easy. The second one comes from the identity cot A cot B + cot B cot C + cot C cot A = 1: To prove this, cot A cot B + cot B cot C + cot C cot A = = cot A (cot B + cot C) + cot B cot C = cos A cos B cos C cos B cos C = + + · = sin A sin B sin C sin B sin C cos A sin C cos B + sin B cos C cos B cos C = · + · = sin A sin B sin C sin B sin C cos A sin(B + C) cos B cos C = · + · = sin A sin B sin C sin B sin C cos A cos B cos C = + = sin B sin C sin B sin C cos B cos C − cos(B + C) = = sin B sin C sin B sin C = = 1: sin B sin C Francisco Javier Garc´ıa Capitan´ 37 Examples. (1) The orthocenter has coordinates 1 1 1 : : = (SBC : SCA : SAB) : SA SB SC (2) The circumcenter has coordinates 2 2 2 a SA : b SB : c SC = (SA(SB + SC ): SB(SC + SA): SC (SA + SB)) : 2 In this way, the sum of coordinates is 2(SAB + SBC + SCA) = 2S .