Barycentric coordinates or Barycentrics A file of the Geometrikon gallery by Paris Pamfilos The artist finds a greater pleasure in painting than in having completed the picture. Seneca, Letter to Lucilius Contents (Last update: 09‑06‑2021) 1 Preliminaries and definition 2 2 Traces, ratios, harmonic conjugation 3 3 Lines in barycentric coordinates 4 4 Ceva’s and Menelaus’ theorems in barycentrics 5 5 Trilinear polar in barycentrics 6 6 Relation to cartesian coordinates, inner product 8 7 The circumcircle of ABC in barycentrics 9 8 Displacement vectors, inner product, distance 9 9 Orthogonality of lines, orthocentroidal circle 11 10 Distance of a point from a line, distance of two parallels 12 11 Meaning of line coefficients 15 12 Power of a point, general circle, Euler circle 16 13 Centroid, Incenter, Circumcenter, Symmedian point 17 14 Euler line, Orthocenter, center of Euler’s circle 18 15 Triangle Area in Barycentrics 19 16 Circumcevian triangle of a point 20 17 Circle through three points, Brocard circle 21 18 The associated affine transformation 22 19 Relations between barycentrics and cartesian coordinates 25 20 Affine transformations represented in barycentrics 26 21 Remarks on working with barycentrics 28 1 Preliminaries and definition 2 1 Preliminaries and definition “Barycentric coordinates” or “barycentrics” do not use distances of points, but only ratios of lengths and areas. Thus, they belong to the geometry of the “affine plane” ([2, p.191]), which deals with parallels and ratios of lengths of collinear segments without to use a notion of distance between two points, as it does euclidean geometry. In the subsequent discussion points {퐴, 퐵, …} of the plane are identified with two di‑ mensional vectors {(푎1, 푎2), (푏1, 푏2) …}. Next properties are easily verified ([7, p.30]): 1. Three points of the plane {퐴, 퐵, 퐶} are collinear, if and only, there are three numbers {푥, 푦, 푧} not all zero, such that 푥퐴 + 푦퐵 + 푐푍 = 0 and 푥 + 푦 + 푧 = 0. 2. Let the points of the plane {퐴, 퐵, 퐶} be non‑collinear. Then, for every other point 퐷 of the plane there are unique numbers {푥, 푦, 푧} with 푥 + 푦 + 푧 = 1 and 퐷 = 푥퐴 + 푦퐵 + 푧퐶. 3. For non‑collinear points {퐴, 퐵, 퐶} the point 퐷 , defined by the equation 푂퐷 = 푥 ⋅ 푂퐴 + 푦 ⋅ 푂퐵 + 푧 ⋅ 푂퐶, with 푥 + 푦 + 푧 = 1, (1) is independent of the choice of origin 푂 and depends only on {푥, 푦, 푧}. 4. The numbers {푥, 푦, 푧} with 푥 + 푦 + 푧 = 1 expressing 퐷 in the preceding equation are equal to the quotients of signed areas: A E D B C O Figure 1: Barycentric coordinates (푥, 푦, 푧) of 퐸 퐴푟푒푎(퐷퐵퐶) 퐴푟푒푎(퐷퐶퐴) 퐴푟푒푎(퐷퐴퐵) 푥 = , 푦 = , 푧 = . (2) 퐴푟푒푎(퐴퐵퐶) 퐴푟푒푎(퐴퐵퐶) 퐴푟푒푎(퐴퐵퐶) Nr‑1. If 푥 + 푦 + 푧 = 0 and 푥퐴 + 푦퐵 + 푧퐶 = 0, then (−푦 − 푧)퐴 + 푦퐵 + 푧퐶 = 0 ⇒ 푦(퐵 − 퐴) + 푧(퐶 − 퐴) = 0 i.e. {퐴, 퐵, 퐶} are collinear. For the converse, simply reverse the preceding implications. Nr‑2. Extend the two‑dimensional vectors {(푥, 푦)} to three dimensional {(푥, 푦, 1)} and denote the corresponding space‑points by {퐴′, 퐵′, 퐶′, 퐷′}. Apply first nr‑1 to see that {퐴′, 퐵′, 퐶′} are independent. Then express 퐷′ in the basis {퐴′, 퐵′, 퐶′} to find the uniquely defined (푥, 푦, 푧) as required. Nr‑3. Using another point 푂′ for origin, the same equation and the analogous numbers (푥′, 푦′, 푧′) to express 퐷 ∶ 푂′퐷 = 푥′ ⋅ 푂′퐴 + 푦′ ⋅ 푂′퐵 + 푧′ ⋅ 푂′퐶 2 Traces, ratios, harmonic conjugation 3 and substracting we get: 푂퐷 − 푂′퐷 = (푥 ⋅ 푂퐴 + 푦 ⋅ 푂퐵 + 푧 ⋅ 푂퐶) − (푥′ ⋅ 푂′퐴 + 푦′ ⋅ 푂′퐵 + 푧′ ⋅ 푂′퐶) = (푥 − 푥′)퐴 + (푦 − 푦′)퐵 + (푧 − 푧′)퐶 − (푥푂 − 푥′푂′ + 푦푂 − 푦′푂′ + 푧푂 − 푧′푂′) ⇒ 푂푂′ = 푂′ − 푂 = (푥 − 푥′)퐴 + (푦 − 푦′)퐵 + (푧 − 푧′)퐶 − (푂 − 푂′) ⇒ (푥 − 푥′)퐴 + (푦 − 푦′)퐵 + (푧 − 푧′)퐶 = 0 . By nr‑1 this implies {(푥 − 푥′) = 0, (푦 − 푦′) = 0, (푧 − 푧′) = 0}, as desired. Nr‑4. Using the independence of (푥, 푦, 푧) from the position of 푂 we select 푂 = 퐷 and use the relation 푥 = 1 − 푦 − 푧 implying (1 − 푦 − 푧)퐷퐴 + 푦퐷퐵 + 푧퐷퐶 = 0 ⇒ 퐷퐴 + 푦(퐷퐵 − 퐷퐴) + 푧(퐷퐶 − 퐷퐴) = 0 ⇒ 퐴퐷 = 푦(퐴퐵) + 푧′(퐴퐶). Multiplying externally by 퐴퐵 we get 퐴퐷 × 퐴퐵 = 푧(퐴퐶 × 퐴퐵) ⇔ 푧(퐴퐵 × 퐴퐶) = 퐴퐵 × 퐴퐷. Analogous equations result also for 푦 and 푥. Taking the positive orientation in direction 퐴퐵 × 퐴퐷 we have the stated result. The numbers {푥, 푦, 푧} in (2) are called “absolute barycentric homogeneous coordinates” of the point 퐷, relative to the triangle 퐴퐵퐶. The multiples (푘 ⋅ 푥, 푘 ⋅ 푦, 푘 ⋅ 푧) by a constant 푘 ≠ 0 are called “barycentric homogeneous coordinates” or “barycentrics” of the point 퐷 ([1, p.64], [8, p.25], [5]). The signs of the numbers result from the orientations of the corresponding triangles e.g. if {퐷, 퐴} are on the same side of 퐵퐶 then 푥 is positive. Otherwise it is neg‑ ative and for 퐷 on 퐵퐶 it is zero 푥 = 0. In fact 푥 = 0 characterizes line 퐵퐶. Analogous properties are valid also for 푦 and 푧. Remark 1. Figure 1 shows a point 퐸 defined by an equation as in nr‑3, but without the restriction 휎 = 푥 + 푦 + 푧 = 1 . Point 퐷 is then obtained by dividing with 휎. Thus, the points are collinear with 푂 and defined by the equations 푥 푦 푧 푂퐸 = 푥푂퐴 + 푦푂퐵 + 푧푂퐶 and 푂퐷 = 푂퐴 + 푂퐵 + 푂퐶. 휎 휎 휎 2 Traces, ratios, harmonic conjugation Changing only the variable 푥, causes 퐷 to move along a fixed line through 퐴. For 푥 = 0, we get a point on the line 퐵퐶 ∶ 퐷퐴 = 푦퐵 + 푧퐶, called “trace” of 퐷 on the side 퐵퐶 of △퐴퐵퐶 (see figure 2). Analogously are defined the traces 퐷퐵 and 퐷퐶. The oriented ratio 푟 = 퐷퐴퐵/퐷퐴퐶, calculated using 푦 + 푧 = 1, gives the value 푟 = −(푧/푦). Theorem 1. Given a point 푃 = 푦퐵 + 푧퐶 on the line 퐵퐶 the “harmonic conjugate” of 푃 w.r. (퐵, 퐶) is 푃′ = 푦퐵 − 푧퐶. This follows from the expression of the ratio 푃퐵/푃퐶 = −푧/푦 and correspondingly for the conjugate 푃′ ∶ 푃′퐵/푃′퐶 = 푧/푦, from which follows that the “cross ratio” has the value (퐵퐶, 푃푃′) = (푃퐵/푃퐶) ∶ (푃′퐵/푃′퐶) = −1, proving that {푃, 푃′} are harmonic conjugate w.r.t. {퐵, 퐶}. 3 Lines in barycentric coordinates 4 A D B D C D B C D A Figure 2: Traces 퐷퐴, 퐷퐵, 퐷퐶 of 퐷 on the sides of 퐴퐵퐶 Remark 2. Since any line can be described in “parametric form” 푃 = 푦퐵 + 푧퐶 and points {퐵, 퐶} can be complemented by a point 퐴 non‑collinear with {퐵, 퐶} to a triangle of refer‑ ence, the preceding property is generally valid for any two points on a line. Remark 3. The “centroid” 퐺 of the triangle 퐴퐵퐶, i.e. the intersection‑point of its “medi‑ ans,” defines triangles {퐺퐵퐶, 퐺퐶퐴, 퐺퐴퐵} with equal areas. Thus it has “barycentric homo‑ geneous coordinates” equal to (1, 1, 1). This identifies these coordinates with the projective homogeneous coordinates w.r.t. the “projective base” {퐴, 퐵, 퐶, 퐺} with 퐺 the “unit” point (see file Projective plane). 3 Lines in barycentric coordinates The following facts concerning the representation of lines with barycentric coordinates are easily verifiable: 1. A line 휀 is represented by an equation of the form 휀 ∶ 푎푥 + 푏푦 + 푐푧 = 0. (3) 2. The intersection point of line 휀 with the line 휀′ ∶ 푎′푥 + 푏′푦 + 푐′푧 = 0 is given by the “vector product” of the vectors of the coefficients: 휀 ∩ 휀′ = (푏푐′ − 푐푏′, 푐푎′ − 푎푐′, 푎푏′ − 푏푎′). (4) 3. The “line at infinity” is represented by the equation (see equation 19) 휀∞ ∶ 푥 + 푦 + 푧 = 0. (5) 4. The point at infinity of the line 휀 ∶ 푎푥 + 푏푦 + 푐푧 = 0 i.e. its intersection with the line at infinity, identified with its “direction”, is given by 휀 ∩ 휀∞ = (푏 − 푐, 푐 − 푎, 푎 − 푏). (6) 5. Two lines {휀, 휀′} are parallel if they meet at a point at infinity, hence their coefficients satisfy: 푏푐′ − 푐푏′ + 푐푎′ − 푎푐′ + 푎푏′ − 푏푎′ = 0. (7) 6. The line passing through two given points {푃(푎, 푏, 푐), 푃′(푎′, 푏′, 푐′)} has coefficients the coordi‑nates of the vector product and can be expressed by a determinant: ∣ 푎 푏 푐 ∣ ∣ ∣ line 푃푃′ ∶ (푏푐′ − 푐푏′, 푐푎′ − 푎푐′, 푎푏′ − 푏푎′), equation: ∣푎′ 푏′ 푐′∣ = 0. (8) ∣ ∣ ∣푥 푦 푧 ∣ 4 Ceva’s and Menelaus’ theorems in barycentrics 5 7. The direction of the line passing through the points {퐴 = (푎, 푏, 푐), 퐵 = (푎′, 푏′, 푐′)}, i.e. its point at infinity is given by the weighted difference of coordinates ′ ′ ′ ′ ′ ′ 휎퐴 ⋅ (푎 , 푏 , 푐 ) − 휎퐵 ⋅ (푎, 푏, 푐) with 휎퐴 = 푎 + 푏 + 푐, 휎퐵 = 푎 + 푏 + 푐 . (9) 8. The middle 푀 of two points 퐴(푎, 푏, 푐) and 퐵(푎′, 푏′, 푐′) is found as the harmonic conjugate of the point at infinity of line 퐴퐵. Applying the preceding calculations it is found to be (the equality being for the tripples of coordinates) 푀 = (푎 + 푏 + 푐)퐵 + (푎′ + 푏′ + 푐′)퐴. (10) 9. The line 휂 parallel from 퐴(푎, 푏, 푐) to the given line 휀 ∶ 푎′푥 + 푏′푦 + 푐′푧 = 0 is the line joining 퐴 with the point at infinity of the line 휀, which is (푏′ − 푐′, 푐′ − 푎′, 푎′ − 푏′).