Barycentric Coordinates Or Barycentrics

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Barycentric coordinates or Barycentrics

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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

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

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′). Thus the equation of this line is

∣푏′ − 푐′ 푐′ − 푎′ 푎′ − 푏′∣ ∣ ∣ ∣ 푎 푏 푐 ∣ = 0, ∣ ∣ ∣ 푥 푦 푧 ∣

and the coefficients of this line, written in theform 푝푥 + 푞푦 + 푟푧 = 0, are given by:

(푝, 푞, 푟) = (푎 + 푏 + 푐)(푎′, 푏′, 푐′) − (푎푎′ + 푏푏′ + 푐푐′)(1, 1, 1). (11)

Thus, the line 휂 can be written as a linear combination of the given line 휀 and the line at infinity 휀∞ , in the form

′ ′ ′ 휂 ∶ (푎 + 푏 + 푐)휀 − (푎푎 + 푏푏 + 푐푐 )휀∞ = 0 ⇔ 휂 ∶ (푎 + 푏 + 푐)(푎′푥 + 푏′푦 + 푐′푧) − (푎푎′ + 푏푏′ + 푐푐′)(푥 + 푦 + 푧) = 0. (12)

4 Ceva’s and Menelaus’ theorems in barycentrics

Three points {퐴′, 퐵′, 퐶′} on the sides of the triangle of reference 퐴퐵퐶 are the “traces” of a point 푃(푥, 푦, 푧) if and only if they can be written in the form:

퐴′(0, 푦, 푧), 퐵′(푥, 0, 푧), 퐶′(푥, 푦, 0). (13)

Their ratios on the triangle’s sides satisfy then the equation:

B'(x,0,z)

C'(x,y,0) P(x,y,z)

B A'(0,y,z) C

Figure 3: Ceva’s theorem

퐴′퐵 퐵′퐶 퐶′퐴 ⋅ ⋅ = −1. (14) 퐴′퐶 퐵′퐴 퐶′퐵 5 Trilinear polar in barycentrics 6

As in the preceding section, this follows from the expressions of the ratios 퐴′퐵 푧 퐵′퐶 푥 퐶′퐴 푦 = − , = − , = − . 퐴′퐶 푦 퐵′퐴 푧 퐶′퐵 푥 Multiplying the three ratios gives the value −1. The inverse is also obvious. If the product of the three ratios is −1, then {푥, 푦, 푧} can be found such that {퐴′, 퐵′, 퐶′} have the shown co‑ ordinates. This defines uniquely 푃(푥, 푦, 푧). For another more conventional view of Ceva’s theorem look at the file Ceva’s theorem. It is obvious, that the points {퐴′, 퐵′, 퐶′} on the sides of the triangle of reference 퐴퐵퐶 are on a line 푝푥 + 푞푦 + 푟푧 = 0, if and only if they can be written in the form:

(0, −푟, 푞), (푟, 0, −푝), (−푞, 푝, 0).

Their ratios satisfy then the equation: 퐴′퐵 퐵′퐶 퐶′퐴 ⋅ ⋅ = 1. (15) 퐴′퐶 퐵′퐴 퐶′퐵 This follows from the expression of the ratio 퐷퐵/퐷퐶 = −푧/푦, for 퐷 = 푦퐵 + 푧퐶, seen in

y=0 px+qy+rz=0

z=0 B'(r,0,-p) C'(-q,p,0)

x=0 A'(0,-r,q) B C

Figure 4: Menelaus’ theorem section 2. By this 퐴′ = −푟퐵 + 푞퐶 ⇒ 퐴′퐵/퐴′퐶 = 푞/푟 etc. (See Figure 4). For another more conventional view of Menelaus’ theorem look at the file Menelaus’ theorem.

5 Trilinear polar in barycentrics

퐶퐴 퐷퐴 In this section the notion of “cross ratio” (퐴퐵, 퐶퐷) = 퐶퐵 ∶ 퐷퐵 of four collinear points is needed and the associated notion of harmonic conjugate points {퐶, 퐷} w.r.t. {퐴, 퐵}, de‑ fined by the condition (퐴퐵, 퐶퐷) = −1. When four points satisfy the last condition we say 퐷 is “harmonic conjugate” to 퐶 w.r.t {퐴, 퐵} and write 퐷 = 퐶(퐴퐵). More on this can be found in the file Cross Ratio. The “trilinear polar” or “tripolar” 휀푃 of the point 푃 w.r.t. the triangle 퐴퐵퐶 is the line containing the “harmonic conjugates” of the traces {퐴′, 퐵′, 퐶′} of 푃 on the sides of 퐴퐵퐶 ∶

퐴″ = 퐴′(퐵퐶), 퐵″ = 퐵′(퐴퐶), 퐶″ = 퐶′(퐵퐴).

That these three points {퐴″, 퐵″, 퐶″} are on a line follows by writing 푃 = 푢퐴 + 푣퐵 + 푤퐶 . ′ Then the traces are given by {퐴 = 푣퐵 + 푤퐶, 푃퐵 = 푤퐶 + 푢퐴, 푃퐶 = 푢퐴 + 푣퐵}. By section 2 the harmonic conjugates of the traces are then

퐴″ = 푣퐵 − 푤퐶 , 퐵″ = 푤퐶 − 푢퐴 , 퐶″ = 푢퐴 − 푣퐵 5 Trilinear polar in barycentrics 7

C'' B'' A B'(u,0,w) C'(u,v,0)

trilinear polar of P P(u,v,w)

A''

B A'(0,v,w) C

Figure 5: Trilinear polar of 푃 and satisfy 퐴″ + 퐵″ + 퐶″ = 0 showing that 퐶″ is on the line of 퐴″퐵″ (see figure 5). Alternatively, the barycentrics of {퐴″, 퐵″, 퐶″} are respectively

퐴″(0, 푣, −푤), 퐵″(−푢, 0, 푤), 퐶″(푢, −푣, 0), which up to a multiplicative factor can be written in the form

퐴″(0, 1/푤, −1/푣), 퐵″(−1/푤, 0, 1/푢), 퐶″(1/푣, −1/푢, 0), implying that they are points of the line described by the next theorem:

Theorem 2. For every point 푃(푢, 푣, 푤) of the plane, different from the vertices of the triangle of reference 퐴퐵퐶, the corresponding “trilinear polar” 푡푟(푃) is described by the equation: 푥 푦 푧 + + = 0. (16) 푢 푣 푤 Conversely, every line 휀 ∶ 푝푢 + 푞푣 + 푟푤 = 0 of the plane, not passing through a vertex of 퐴퐵퐶 is the trilinear polar 푡푟(푃) of the point

1 1 1 푃 ( , , ), (17) 푝 푞 푟 which is called “tripole” of the line 휀.

Remark 4. The map 푃 ↦ 푡푟(푃) is one‑to‑one for all points of the plane except those lying on the side‑lines of the triangle of reference 퐴퐵퐶. For these points 푃, if they are different from a vertex and lie on a side‑line 휀, the corresponding line 푡푟(푃) coincides with 휀. For the vertices of 퐴퐵퐶 the trilinear polar cannot be uniquely defined.

D' B P C ε tr(P) D P'

Figure 6: Trilinear polars for 푃 ∈ 휀 ∋ 퐶 pass all through 퐷′ = 퐷(퐴퐵) 6 Relation to cartesian coordinates, inner product 8

Remark 5. Using the definition of the trilinear polar, it is easy to see, that for the points 푃 of a fixed line 휀 passing through a vertex, 퐶 say, of the triangle of reference 퐴퐵퐶, the corresponding trilinear polar 푡푟(푃) passes through the “harmonic conjugate” point 퐷′ = 퐷(퐴퐵) , where 퐷 = 휀 ∩ 퐴퐵 (see figure 6). Further, for the point 푃′ = 푡푟(푃) ∩ 휀 the cross ratio (퐶퐷, 푃푃′) = −2 is constant and, as the point 푃 approaches 퐶 , the cor‑ responding trilinear polar 푡푟(푃) tends to coincide with the “harmonic conjugate” line 퐶퐷′ of 퐶퐷 = 휀 w.r.t. the line‑pair (퐶퐴, 퐶퐵).

Remark 6. From its proper definition follows that the trilinear polar ofthe “centroid” with equation 휀∞ ∶ 푢 + 푣 + 푤 = 0, is the “line at infinity” of the plane.

6 Relation to cartesian coordinates, inner product

Denoting by (푥, 푦) the cartesian coordinates and by (푢, 푣, 푤) the “absolute barycentrics” of the same point 푃, the relation between the two is expressed in matrix form by translating equation 1: 푥 푢 퐴1 퐵1 퐶1 푢 ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜푦⎟ = 푀 ⋅ ⎜푣⎟ = ⎜퐴2 퐵2 퐶2⎟ ⋅ ⎜푣⎟ . (18) ⎝1⎠ ⎝푤⎠ ⎝ 1 1 1 ⎠ ⎝푤⎠ This matrix and its inverse, studied further in sections 18 and 19, can be used to express other geometric objects through relations of the barycentrics. The equation of a line, which in cartesian coordinates is 푥 ⎜⎛ ⎟⎞ 푎푥 + 푏푦 + 푐 = (푎, 푏, 푐) ⋅ ⎜푦⎟ = 0 ⎝1⎠ translates into barycentrics to the equation

푎′푢 + 푏′푣 + 푐′푤 = 0 with (푎′, 푏′, 푐′) = (푎, 푏, 푐) ⋅ 푀.

Thus, the equation of the line at infinity, described in cartesians by 푧 = 0 , corresponds by the preceding rule, to the equation in barycentrics:

푢 ⎜⎛ ⎟⎞ (0, 0, 1)푀 ⎜푣⎟ = 푢 + 푣 + 푤 = 0. (19) ⎝푤⎠ The “inner product” in cartesian coordinates is also expressible through a “bilinear form”:

푥′ 푢′ ′ ′ ⎜⎛ ′⎟⎞ 푡 ⎜⎛ ′ ⎟⎞ 푥푥 + 푦푦 = (푥, 푦, 1) ⎜푦 ⎟ − 1 = (푢, 푣, 푤)푀 푀 ⎜푣 ⎟ − 1. (20) ⎝ 1 ⎠ ⎝푤′⎠

The matrix 푀푡푀 is easily seen to be

퐴2 퐴 ⋅ 퐵 퐴 ⋅ 퐶 1 1 1 푡 ⎜⎛ 2 ⎟⎞ ⎜⎛ ⎟⎞ 푀 푀 = ⎜퐴 ⋅ 퐵 퐵 퐵 ⋅ 퐶⎟ + ⎜1 1 1⎟ . (21) ⎝퐴 ⋅ 퐶 퐵 ⋅ 퐶 퐶2 ⎠ ⎝1 1 1⎠ 7 The circumcircle of ABC in barycentrics 9

Taking into account that 푢 + 푣 + 푤 = 푢′ + 푣′ + 푤′ = 1, we find that

푢′ 퐴2 퐴 ⋅ 퐵 퐴 ⋅ 퐶 ′ ′ ⎜⎛ ′ ⎟⎞ ⎜⎛ 2 ⎟⎞ 푥푥 + 푦푦 = (푢, 푣, 푤)푁 ⎜푣 ⎟ , with 푁 = ⎜퐴 ⋅ 퐵 퐵 퐵 ⋅ 퐶⎟ . (22) ⎝푤′⎠ ⎝퐴 ⋅ 퐶 퐵 ⋅ 퐶 퐶2 ⎠

7 The circumcircle of ABC in barycentrics

By the discussion in section 1 we can take the origin of cartesian coordinates anywhere we like. Selecting it at the “circumcenter” 푂 of the circumcircle of the triangle of reference 퐴퐵퐶, we find that

퐴2 = 퐵2 = 퐶2 = 푅2, 퐴 ⋅ 퐵 = 푅2 cos(2훾), 퐴 ⋅ 퐶 = 푅2 cos(2훽), 퐵 ⋅ 퐶 = 푅2 cos(2훼), where {훼, 훽, 훾} the angles of the triangle opposite respectively to {퐵퐶, 퐶퐴, 퐴퐵} and 푅 is the circumradius of the triangle of reference 퐴퐵퐶. Setting cos(2훾) = 1 − 2 sin2(훾), … we find 1 1 1 0 sin2(훾) sin2(훽) ⎛ ⎞ ⎛ ⎞ 2 ⎜ ⎟ 2 ⎜ 2 2 ⎟ 푁 = 푅 ⎜1 1 1⎟ − 2푅 ⎜sin (훾) 0 sin (훼)⎟ ⇔ 2 2 ⎝1 1 1⎠ ⎝sin (훽) sin (훼) 0 ⎠ 1 1 1 0 푐2 푏2 1 ⎛ ⎞ 1 ⎛ ⎞ 푁 = 푅2 ⋅ 푁 − 푁 = 푅2 ⎜1 1 1⎟ − ⎜푐2 0 푎2⎟ , (23) 1 2 2 ⎜ ⎟ 2 ⎜ ⎟ ⎝1 1 1⎠ ⎝푏2 푎2 0 ⎠ where {푎 = |퐵퐶|, 푏 = |퐶퐴|, 푐 = |퐴퐵|} the side lengths of the triangle. Taking into account the condi‑tion {푢 + 푣 + 푤 = 1, …} for “absolute barycentric coordinates”, we find that the ex‑ pression of the inner product 푥푥′ + 푦푦′ in terms of the corresponding absolute barycentric coordinates is: 푢′ ′ ′ ⎜⎛ ′ ⎟⎞ 푥푥 + 푦푦 = (푢, 푣, 푤)푁 ⎜푣 ⎟ (24) ⎝푤′⎠ 1 = 푅2 − (푐2(푢푣′ + 푢′푣) + 푏2(푤푢′ + 푤′푢) + 푎2(푣푤′ + 푣′푤)). 2 Having the center of coordinates at the circumcenter of 퐴퐵퐶, the equation of the “cir‑ cumcircle” 휅 of 퐴퐵퐶 in barycentric coordinates can be found from the corresponding equation in cartesian coordinates:

푅2 = 푥2 + 푦2 = 푅2 − (푐2푢푣 + 푏2푤푢 + 푎2푣푤) ⇔ 푎2푣푤 + 푏2푤푢 + 푐2푢푣 = 0. (25)

8 Displacement vectors, inner product, distance

Displacement vectors are differences of absolute barycentric coordinates of two points

(푝, 푞, 푟) = (푢, 푣, 푤) − (푢′, 푣′, 푤′).

Since 푢 + 푣 + 푤 = 푢′ + 푣′ + 푤′ = 1, they satisfy

푝 + 푞 + 푟 = 0 8 Displacement vectors, inner product, distance 10 and represent points at infinity. Also through their corresponding cartesian coordinate vectors, defined by equation18 ( ), they represent the usual displacement from one point to the other. The inner product of the corresponding cartesian coordinate vectors can be calculated using the same bilinear form (23). Thus, introducing the vectorial notation 푥 = (푥, 푦) and 푃 = (푝, 푞, 푟) for cartesian, respectively barycentric cordinates, and also de‑ noting by 푥푃 = 푀푃, we have:

푥푃푄 = 푥푄 − 푥푃 = 푀(푄 − 푃) = 푀 ⋅ 푃푄 and from this, the expression of the usual inner product: 2 2 ′ 푡 0 푐 푏 푝 ′ ′ 1 ⎜⎛ 2 2⎟⎞ ⎜⎛ ′⎟⎞ 푥 ⋅ 푥 ′ ′ = 푃푄 ⋅ 푁 ⋅ 푃 푄 = − (푝, 푞, 푟) ⎜푐 0 푎 ⎟ ⎜푞 ⎟ , 푃푄 푃 푄 2 ⎜ ⎟ ⎜ ⎟ ⎝푏2 푎2 0 ⎠ ⎝푟′⎠ where we set 푃푄 = 푄 − 푃 = (푝, 푞, 푟) and 푃′푄′ = 푄′ − 푃′ = (푝′, 푞′, 푟′). Expanding this, we get: 1 2 ′ ′ 2 ′ ′ 2 ′ ′ 푥 ⋅ 푥 ′ ′ = − [푎 (푞푟 + 푞 푟) + 푏 (푟푝 + 푟 푝) + 푐 (푝푞 + 푝 푞)]. (26) 푃푄 푃 푄 2 This expresses the fundamental relation allowing the computation of euclidean distances and angles in terms of barycentric coordinates. Thus, for example, the distance of two points represented in absolute barycentric coordinates {푃, 푄} can be expressed through their “displacement vector” 푃푄 = (푝, 푞, 푟) and the formula: |푃푄|2 = −푎2푞푟 − 푏2푟푝 − 푐2푝푞. (27) Introducing the “Conway symbols” 2 2 2 2 2 2 2 2 2 푆퐴 = (푏 + 푐 − 푎 )/2, 푆퐵 = (푐 + 푎 − 푏 /2), 푆퐶 = (푎 + 푏 − 푐 )/2, where {푎, 푏, 푐} are the side‑lengths of the triangle of reference, and calculating the expres‑ sion on the right of equation 26 we find, using conditions {푝 + 푞 + 푟 = 푝′ + 푞′ + 푟′ = 0}, that it is equivalent to ′ ′ ′ 푥푃푄 ⋅ 푥푃′푄′ = 푆퐴푝푝 + 푆퐵푞푞 + 푆퐶푟푟 . (28) Consequently, formula 27 is also equivalent to: 2 2 2 2 |푃푄| = 푆퐴푝 + 푆퐵푞 + 푆퐶푟 , (29) where {푝 = 푢′ − 푢, 푞 = 푣′ − 푣, 푟 = 푤′ − 푣} and {푃(푢, 푣, 푤), 푄(푢, 푣, 푤)} are in absolute bary‑ centrics. Using the last formula we can easily find the equation of the “medial line” of two points {푃, 푄}, i.e. the points 푋(푋푢, 푋푣, 푋푤) whose distances |푋푃| = |푋푄|. For this consider the quadratic form defined by the right side of the last formula

푓 (푋, 푌) = 푆퐴푋푢푌푢 + 푆퐵푋푣푌푣 + 푆퐶푋푤푌푤 . The equality {|푋푃|2 = |푋푄|2} by formula (29) is equivalent to 푓 (푃 − 푋, 푃 − 푋) = 푓 (푄 − 푋, 푄 − 푋) ⇔ 푓 (푃, 푃) − 2푓 (푃, 푋) + 푓 (푋, 푋) = 푓 (푄, 푄) − 2푓 (푄, 푋) + 푓 (푋, 푋) ⇔ 푓 (2(푃 − 푄), 푋) + 푓 (푄, 푄) − 푓 (푃, 푃) = 0 ⇔

푆퐴(푃푢 − 푄푢)푋푢 + 푆퐵(푃푣 − 푄푣)푋푣 + 푆퐶(푃푤 − 푄푤)푋푤 + 푘/2 = 0 (30) with 2 2 2 2 2 2 푘 = 푓 (푄, 푄) − 푓 (푃, 푃) = 푆퐴(푄푢 − 푃푢) + 푆퐵(푄푣 − 푃푣) + 푆퐶(푄푤 − 푃푤). 9 Orthogonality of lines, orthocentroidal circle 11

Theorem 3. Equation (30) represents the medial line of two points {푃, 푄} in absolute barycentrics. Remark 7. It is easy verifiable that the “Conway symbols” satisfy the conditions:

2 2 2 푆퐵 + 푆퐶 = 푎 , 푆퐶 + 푆퐴 = 푏 , 푆퐴 + 푆퐵 = 푐 , (31) 2 푆퐴푆퐵 + 푆퐵푆퐶 + 푆퐶푆퐴 = 푆 , (32) 2 2 2 2 푆퐴푆퐵 + 푐 푆퐶 = 푆퐵푆퐶 + 푎 푆퐴 = 푆퐶푆퐴 + 푏 푆퐵 = 푆 . (33) 2 2 2 2 푎 푆퐴 + 푏 푆퐵 + 푐 푆퐶 = 2푆 . (34) where 푆 denotes twice the area of the triangle of reference 퐴퐵퐶 ([8, p.33]). More general, for an angle of measure 휙 we can define the compatible to the preceding symbol 푆휙 = 푆 ⋅ cot(휙), (35) used in several formulas involving calculations with barycentrics ([8]). More on this can be found in the file Conway triangle symbols. Theorem 4. Two displacement vectors {(푝, 푞, 푟), (푝′, 푞′, 푟′)} are orthogonal, if and only if they satisfy the equation: ′ ′ ′ 푆퐴푝푝 + 푆퐵푞푞 + 푆퐶푟푟 = 0. (36)

9 Orthogonality of lines, orthocentroidal circle

The parametric form of a line represented in barycentrics by the equation

푝푢 + 푞푣 + 푟푤 = 0, can be given in the form

(푢, 푣, 푤) = (푞, −푝, 0) + 푡(푟 − 푞, 푝 − 푟, 푞 − 푝).

The last vector on the right is a formal “displacement vector”, satisfying 푢 + 푣 + 푤 = 0, i.e. representing in barycentrics the “direction” of the line. Notice that it is also the “point at in‑ finity” of that line, i.e. its intersection with the “line at infinity”, represented in barycentrics by the equation 푢 + 푣 + 푤 = 0. Thus, the lines {푝푢 + 푞푣 + 푟푤 = 0, 푝′푢 + 푞′푣 + 푟′푤 = 0} are orthogonal, if and only if their “directions”

′ ′ ′ ′ ′ ′ ′ ′ ′ (푝푑, 푞푑, 푟푑) = (푟 − 푞, 푝 − 푟, 푞 − 푝), (푝푑, 푞푑, 푟푑) = (푟 − 푞 , 푝 − 푟 , 푞 − 푝 ) are orthogonal displacements, i.e. they satisfy equation (36). Latter equation, together with the valid for points at infinity 푢 + 푣 + 푤 = 0, imply the explicit form of the “direction” of the orthogonal line:

′ ′ ′ (푝푑, 푞푑, 푟푑) = ( (푆퐵 ⋅ 푞푑 − 푆퐶 ⋅ 푟푑) , (푆퐶 ⋅ 푟푑 − 푆퐴 ⋅ 푝푑) , (푆퐴 ⋅ 푝푑 − 푆퐵 ⋅ 푞푑)). (37) Alternatively to this, i.e. using the line‑coefficients themselves and not their differences, we arive after a short calculation at the theorem ([4, II,p.57]): Theorem 5. The lines 휀 ∶ 푝푥 + 푞푦 + 푟푧 = 0 and 휀′ 푝′푥 + 푞′푦 + 푟′푧 = 0 are orthogonal if and only if their coefficients satisfy the equation

2 ′ 2 ′ 2 ′ ′ ′ ′ ′ ′ ′ 푎 푝푝 + 푏 푞푞 + 푐 푟푟 − 푆퐴(푞푟 + 푞 푟) − 푆퐵(푟푝 + 푟 푝) − 푆퐶(푝푞 + 푝 푞) = 0. (38) 10 Distance of a point from a line, distance of two parallels 12

Remark 8. The bilinear form involved in equation (38) is degenerate, since the corre‑ sponding matrix 퐻 satisfies:

2 −푎 푆퐶 푆퐵 ⎜⎛ 2 ⎟⎞ (1, 1, 1)퐻 = (1, 1, 1) ⎜ 푆퐶 −푏 푆퐴 ⎟ = 0. (39) 2 ⎝ 푆퐵 푆퐴 −푐 ⎠ Remark 9. Last equation conforms to the fact that if lines {휀 = 0, 휀′ = 0} are orthogonal ′ the same is true for {휀 + 푘 ⋅ 휀∞, 휀 }, where 휀∞ is the line at infinity. By section 3‑nr‑5 the lines {휀 + 푘 ⋅ 휀∞ = 0} represent, for variable 푘, all the parallels to line 휀 = 0 .

U U 1 2

Figure 7: Circle on diameter 푈1푈2 in absolute barycentrics

As an application of the orthogonality of displacement vectors we obtain the equation of the circle having for diameter the segment defined by the points

푈1 = (푢1, 푣1, 푤1) and 푈2 = (푢2, 푣2, 푤2). The points 푈(푢, 푣, 푤) of this circle are characterized by the orthogonality of the displace‑ ment vectors {푈푈1, 푈푈2} implying the theorem:

Theorem 6. The points 푈(푢, 푣, 푤) of the circle on diameter 푈1푈2 satisfy the equation in abso‑ lute barycentrics:

푆퐴(푢 − 푢1)(푢 − 푢2) + 푆퐵(푣 − 푣1)(푣 − 푣2) + 푆퐶(푤 − 푤1)(푤 − 푤2) = 0. (40) An application of this leads to a compact expression for the “orthocentroidal” circle of a triangle 퐴퐵퐶, defined to be the circle with diametral points the centroid 퐺 and the orthocenter 퐻 of the triangle. Taking into account formula (65) for 퐻 and doing some calculation we find its expression in barycentrics:

2 2 2 3(푆퐴푢 + 푆퐵푣 + 푆퐶푤 ) − (푆퐴푢 + 푆퐵푣 + 푆퐶푤) = 0 , (41) where (푢, 푣, 푤) are assumed to be absolute coordinates.

10 Distance of a point from a line, distance of two parallels

Representing the coordinates by 푈 = (푢, 푣, 푤), the “direction” of the line

휀 ∶ 푓 (푈) = 푝푢 + 푞푣 + 푟푤 = 0 (42) is its intersection (푟 − 푞, 푝 − 푟, 푞 − 푝) (43) with the line at infinity as explained in section 9. The orthogonal direction to this can be calculated using theorem 4 and leads to a multiple of the displacement vector 푉 ∶

(푣1 = 푆퐵(푝 − 푟) − 푆퐶(푞 − 푝), 푣2 = 푆퐶(푞 − 푝) − 푆퐴(푟 − 푞), 푣3 = 푆퐴(푟 − 푞) − 푆퐵(푝 − 푟)). 10 Distance of a point from a line, distance of two parallels 13

Which in matrix notation is represented by the equation:

푣1 0 푆퐵 −푆퐶 푟 − 푞 ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ 푉 = ⎜푣2⎟ = ⎜−푆퐴 0 푆퐶 ⎟ ⋅ ⎜푝 − 푟⎟ . (44) ⎝푣3⎠ ⎝ 푆퐴 −푆퐵 0 ⎠ ⎝푞 − 푝⎠

′ The line 휀 , passing through the point 푈0 = (푢0, 푣0, 푤0), and orthogonal to 휀 has the

U 0

U ε t

Figure 8: Calculate the distance |푈0푈1| in barycentrics parametric rerpesentation, the point 푈0 in absolute barycentrics:

′ 휀 ∶ 푈푡 = 푈0 + 푡푉.

Its intersection point 푈푡 with 휀 satisfies the equation of line 휀 ∶

푓 (푈0) 푓 (푈0 + 푡푉) = 0 ⇔ 푓 (푈0) + 푡푓 (푉) = 0 ⇔ 푡 = − . 푓 (푉)

Taking into account equation (44), we find that

0 푆퐵 −푆퐶 푟 − 푞 ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ 푓 (푉) = (푝, 푞, 푟) ⎜−푆퐴 0 푆퐶 ⎟ ⋅ ⎜푝 − 푟⎟ ⇔ ⎝ 푆퐴 −푆퐵 0 ⎠ ⎝푞 − 푝⎠ 2 2 2 푓 (푉) = 푆퐴(푟 − 푞) + 푆퐵(푝 − 푟) + 푆퐶(푞 − 푝) . (45)

Thus, the displacement vector 푈푡 − 푈0 and its length is, according to equation (27):

푓 (푈 ) 푓 (푈 )2 푈 − 푈 = − 0 ⋅ 푉 ⇒ |푈 − 푈 |2 = 0 (−푎2푣 푣 − 푏2푣 푣 − 푐2푣 푣 ). (46) 푡 0 푓 (푉) 푡 0 푓 (푉)2 2 3 3 1 1 2

The parenthesis on the right is

2 2 0 푐 푏 푣1 2 2 2 2 1 ⎜⎛ 2 2⎟⎞ ⎜⎛ ⎟⎞ 푉 = −푎 푣2푣3 − 푏 푣3푣1 − 푐 푣1푣2 = − (푣1, 푣2, 푣3) ⎜푐 0 푎 ⎟ ⎜푣2⎟ , 2 2 2 ⎝푏 푎 0 ⎠ ⎝푣3⎠ which, taking into account equation (44), leads to

푟 − 푞 1 ⎛ ⎞ 푉2 = − (푟 − 푞, 푝 − 푟, 푞 − 푝) ⋅ 퐾 ⋅ ⎜푝 − 푟⎟ , (47) 2 ⎜ ⎟ ⎝푞 − 푝⎠ where 퐾 is the matrix 2 2 0 −푆퐴 푆퐴 0 푐 푏 0 푆퐵 −푆퐶 ⎜⎛ ⎟⎞ ⎜⎛ 2 2⎟⎞ ⎜⎛ ⎟⎞ 퐾 = ⎜ 푆퐵 0 −푆퐵⎟ ⎜푐 0 푎 ⎟ ⎜−푆퐴 0 푆퐶 ⎟ . 2 2 ⎝−푆퐶 푆퐶 0 ⎠ ⎝푏 푎 0 ⎠ ⎝ 푆퐴 −푆퐵 0 ⎠ 10 Distance of a point from a line, distance of two parallels 14

Carrying out the computation we find that

2 2 −푎 푆퐴 푆퐴푆퐵푆퐶 푆퐴푆퐵푆퐶 ⎜⎛ 2 2 ⎟⎞ 퐾 = 2 ⎜푆퐴푆퐵푆퐶 −푏 푆퐵 푆퐴푆퐵푆퐶⎟ . (48) 2 2 ⎝푆퐴푆퐵푆퐶 푆퐴푆퐵푆퐶 −푐 푆퐶 ⎠ This, using the relations in section 9 of the Conway symbols becomes

2 2 2 −푎 푆퐴 + 푆퐴푆퐵푆퐶 = 푆퐴(푆퐵푆퐶 − 푎 푆퐴) = 푆퐴(푆퐵푆퐶 − (푆퐵 + 푆퐶)푆퐴)

= 푆퐴(2푆퐵푆퐶 − (푆퐵푆퐶 + 푆퐶푆퐴 + 푆퐴푆퐵)) 2 = 2푆퐴푆퐵푆퐶 − 푆퐴푆 ⇒ 2 2 2 −푎 푆퐴 = 푆퐴푆퐵푆퐶 − 푆퐴푆 . Thus, matrix 퐾 can be written in the form

1 1 1 푆 0 0 퐾 ⎛ ⎞ ⎛ 퐴 ⎞ = 푆 푆 푆 ⎜1 1 1⎟ − 푆2 ⎜ 0 푆 0 ⎟ . (49) 2 퐴 퐵 퐶 ⎜ ⎟ ⎜ 퐵 ⎟ ⎝1 1 1⎠ ⎝ 0 0 푆퐶⎠

Introducing this into equation (47) we find that

2 2 2 2 2 푉 = 푆 (푆퐴(푟 − 푞) + 푆퐵(푝 − 푟) + 푆퐶(푞 − 푝) ). (50)

This introduced to equation (46) leads finally to the expression for the distance of thepoint

푈0 from the line 휀 ∶ 푓 (푈) = 푝푢 + 푞푣 + 푟푤 = 0 ∶

Theorem 7. The distance 푑푖푠푡(푈0, 휀) of the point 푈0(푢0, 푣0, 푤0) given in absolute barycentrics from the line 휀 ∶ 푓 (푈) = 푝푢 + 푞푣 + 푟푤 = 0 is given by the formula:

푆2(푝푢 + 푞푣 + 푟푤 )2 푑푖푠푡(푈 , 휀)2 = |푈 푈 |2 = 0 0 0 . 0 푡 0 2 2 2 (51) 푆퐴(푟 − 푞) + 푆퐵(푝 − 푟) + 푆퐶(푞 − 푝) Measuring the distance 푑푖푠푡(퐴, 휀) of the vertex 퐴(1, 0, 0) from the line 휀 and doing this also for the other vertices, we find the following property:

Theorem 8. The coefficients of the line 휀 ∶ 푝푢 + 푞푣 + 푟푤 = 0 are proportional to the distances of the vertices of the triangle of reference 퐴퐵퐶 from the line:

푑푖푠푡(퐴, 휀)2 푑푖푠푡(퐵, 휀)2 푑푖푠푡(퐶, 휀)2 푆2 = = = . 2 2 2 2 2 2 (52) 푝 푞 푟 푆퐴(푟 − 푞) + 푆퐵(푝 − 푟) + 푆퐶(푞 − 푝)

Β' Α'G' ε C'

G C B

Figure 9: Distance of centroid from the line 휀 11 Meaning of line coefficients 15

Remark 10. Another aspect of the ratio on the right side of the last formula is found by considering a line not passing through the the centroid 퐺(1/3, 1/3, 1/3) of △퐴퐵퐶, i.e. satisfying 푝 + 푞 + 푟 ≠ 0. Equation (51) leads in this case to: 푆2(휎/3)2 푑푖푠푡(퐺, 휀)2 = , 휎 = 푝 + 푞 + 푟 ⇔ 2 2 2 with 푆퐴(푟 − 푞) + 푆퐵(푝 − 푟) + 푆퐶(푞 − 푝) 2 푆2 3푑푖푠푡(퐺, 휀) = ( ) 2 2 2 (53) 푆퐴(푟 − 푞) + 푆퐵(푝 − 푟) + 푆퐶(푞 − 푝) 휎 From theorem 7 and the preceding remark follows immediately the form of the equa‑ tion of the bisector lines of the angle of two lines and the form of the distance of two parallel lines: Theorem 9. The bisector lines of the angle of two intersecting lines, which do not pass through the centroid 퐺 ∶ 휀 ∶ 푓 (푢, 푣, 푤) = 푝푢 + 푞푣 + 푟푤 = 0 , 휀 ∶ 푓 ′(푢, 푣, 푤) = 푝′푢 + 푞′푣 + 푟′푤 = 0 , with normalized coefficients satisfying {푝 + 푞 + 푟 = 푝′ + 푞′ + 푟′ = 1} are given by the equa‑ tions 푑푖푠푡(퐺, 휀)푓 (푢, 푣, 푤) ± 푑푖푠푡(퐺, 휀′)푓 ′(푢, 푣, 푤) = 0 . (54) Theorem 10. The distance of two parallel lines {휀, 휀′}: 푓 (푢, 푣, 푤) = 푝푢 + 푞푣 + 푟푤, 푓 ′(푢, 푣, 푤) = 푓 (푢, 푣, 푤) + 푘 with 휎 = 푝 + 푞 + 푟 ≠ 0, is given by 3|푘| ⋅ 푑푖푠푡(퐺, 휀) 푑푖푠푡(휀, 휀′) = . (55) |휎|

11 Meaning of line coefficients

The meaning of line coefficients, expressed through equation (52), can be explained ge‑ ometrically and more directly, than it was done in section 10, whose purpose was the

C'' B'(-r,0,p) C'(-q,p,0) B'' A'' pu+qv+rw=0 A'(0,-r,q)

B C

퐴퐴″ 퐵퐵″ 퐶퐶″ Figure 10: Coefficients proportional to distances 푝 = 푞 = 푟 general formula of distance of a point from a line, as described by equation (51). For this it suffices to use the results of section 4. In fact, the intersection points {퐴′, 퐵′, 퐶′} of the line 휀 ∶ 푝푢 + 푞푣 + 푟푤 = 0 with the sides triangle of reference (See Figure 10), have corresponding coordinates 퐴′(0, −푟, 푞), 퐵′(−푟, 0, 푝), 퐶′(−푞, 푝, 0). 12 Power of a point, general circle, Euler circle 16

And the ratio of the distances {퐴퐴″, 퐵퐵″, 퐶퐶″} is 퐵퐵″ 퐴′퐵 푞 퐵퐵″ 퐶퐶″ = = ⇒ = . 퐶퐶″ 퐴′퐶 푟 푞 푟

Analogously is seen that 퐵퐵″/푞 = 퐴퐴″/푝. Notice that in order to have valid equations

퐴퐴″ 퐵퐵″ 퐶퐶″ = = , 푝 푞 푟 the distances must be signed and their signs must be properly chosen.

12 Power of a point, general circle, Euler circle

The expression in barycentrics of the “power” of a point 푃(푢, 푣, 푤) relative to a circle, can be found from the corresponding expression in cartesian coordinates. In fact, in cartesian coordinates the power of the point 푃 w.r.t. the circle 휅(푂, 푟) is

푃푂2 − 푟2, which, using equation (27) for points {푂(푢0, 푣0, 푤0), 푃(푢, 푣, 푤)} in absolute barycentrics becomes:

2 2 2 2 2 2 −(푃푂 − 푟 ) = 푎 (푣0 − 푣)(푤0 − 푤) + 푏 (푢0 − 푢)(푤0 − 푤) + 푐 (푢0 − 푢)(푣0 − 푣) + 푟 .

The equation of the circle 휅(푂, 푟) results by equating this to zero:

푎2푣푤 + 푏2푤푢 + 푐2푢푣 2 2 2 2 +푎 푣0푤0 + 푏 푢0푤0 + 푐 푢0푣0 + 푟 (56) 2 2 2 −푎 (푣0푤 + 푣푤0) − 푏 (푤0푢 + 푤푢0) − 푐 (푢0푣 + 푢푣0) = 0.

Taking into account that 푢 + 푣 + 푤 = 1, the two last rows are seen to sum into the linear term: (푟2 − 푂퐴2)푢 + (푟2 − 푂퐵2)푣 + (푟2 − 푂퐶2)푤. (57) Thus, the equation of the circle 휅(푂, 푟) becomes

푎2푣푤 + 푏2푤푢 + 푐2푢푣 + (푟2 − 푂퐴2)푢 + (푟2 − 푂퐵2)푣 + (푟2 − 푂퐶2)푤 = 0, which represents the equation of the circle 휅 as a sum of the corresponding expression of the circumcircle of the triangle of reference and a certain line. Thus, the difference of the expressions of the two circles is the expression of the “radical line” of the two circles and the following theorem is valid.

Theorem 11. The equation

(푟2 − 푂퐴2)푢 + (푟2 − 푂퐵2)푣 + (푟2 − 푂퐶2)푤 = 0 (58) represents the radical axis of the circle 휅(푂, 푟) and the circumcircle 휅0 of the triangle of reference 퐴퐵퐶 . The expression of the equation of the general circle in absolute barycentric coordinates is the sum the corresponding expressions of 휅0 and this line:

휅(푂, 푟) ∶ 푎2푣푤 + 푏2푤푢 + 푐2푢푣 + (푟2 − 푂퐴2)푢 + (푟2 − 푂퐵2)푣 + (푟2 − 푂퐶2)푤 = 0. (59) 13 Centroid, Incenter, Circumcenter, Symmedian point 17

λ B''

Orthic axis B' H N O Euler line

B C

Figure 11: The Euler circle of the triangle 퐴퐵퐶

Remark 11. By multiplying with 1 = 푥 + 푦 + 푧, we can “homogenize” the last equation, thus leading to the homogeneous one:

푎2푣푤 + 푏2푤푢 + 푐2푢푣 + ((푟2 − 푂퐴2)푢 + (푟2 − 푂퐵2)푣 + (푟2 − 푂퐶2)푤)(푢 + 푣 + 푤) = 0 . (60)

As an example application of this equation we consider the “Euler circle” 휆 of the triangle of reference 퐴퐵퐶. The powers of the vertices {퐴, 퐵, 퐶} w.r.t. 휆(푁, 푟) are easily seen (see figure 11) to be

2 2 ′ ″ 푂퐴 − 푟 = 퐴퐵 ⋅ 퐴퐵 = 푏푐 cos(퐴)̂ = 푆퐴/2 and the analogous expressions obtained by the cyclic permutations of {퐴, 퐵, 퐶}. Here {퐵′, 퐵″} are respectively the traces on 퐴퐶 of the median and altitude from 퐵. Thus, equa‑ tions (59) resp. (60) become in this case

2 2 2 2(푎 푣푤 + 푏 푤푢 + 푐 푢푣) − (푆퐴푢 + 푆퐵푣 + 푆퐶푤) = 0 ⇔ (61) 2 2 2 2 2 2 푎 푣푤 + 푏 푤푢 + 푐 푢푣 − (푆퐴푢 + 푆퐵푣 + 푆퐶푤 ) = 0 . (62)

Figure 11 illustrates the case showing also the radical axis of the Euler circle and the cir‑ cumcircle. This is the “orthic axis” of the triangle, identified by theorem 11 with the line

푆퐴푢 + 푆퐵푣 + 푆퐶푤 = 0, (63) and coinciding with the trilinear polar of the orthocenter 퐻 of 퐴퐵퐶.

13 Centroid, Incenter, Circumcenter, Symmedian point

These remarkable points are the simplest examples of “triangle centers” of the triangle of reference 퐴퐵퐶, whose barycentric coordinates can be easily calculated. The first, the “centroid” 푋(2) in Kimberling’s notation ([3]), is the unit point of the projective base {퐴, 퐵, 퐶, 퐺} defining the system of “barycentrics” and has coordinates (1, 1, 1). The incen‑ ter 푋(1), using the definition of barycentrics through areas, is easily seen tobe (푎, 푏, 푐). The circumcenter 푋(3) of the triangle 퐴퐵퐶 is also calculated using the area definition of barycentrics. In fact, the area of the triangle 푋(3)퐵퐶 is

1 푅 (푋(3)퐵퐶) = 퐵퐶2 cot(훼) = ... = 푎2푆 ⇒ 푋(3) = (푎2푆 , 푏2푆 , 푐2푆 ), 4 4푎푏푐 퐴 퐴 퐵 퐶 14 Euler line, Orthocenter, center of Euler’s circle 18 with corresponding absolute barycentrics: 1 푋(3) = (푎2푆 , 푏2푆 , 푐2푆 ). (64) 2푆2 퐴 퐵 퐶 The symmedian point 푋(6) (see file Symmedian point of the triangle) is characterized by its property to have distances from the sides analogous to these sides. Thus, the ratio of the areas (푋(6)퐴퐵) 푐2 = ⇒ 푋(6) = (푎2, 푏2, 푐2). (푋(6)퐴퐶) 푏2 The equalities 푋(푘) = (푢, 푣, 푤) for barycentrics must be understood in a wider sense and often is used the symbol 푋(푘) = (푢 ∶ 푣 ∶ 푤) to stress the fact that the barycentrics are de‑ fined up to multiplicative constants, so that only their relative ratios are uniquely defined. Alternatively, we use also the symbol 푋 ≅ (푢, 푣, 푤).

14 Euler line, Orthocenter, center of Euler’s circle

2 2 2 Having the barycentrics {(1, 1, 1), (푎 푆퐴, 푏 푆퐵, 푐 푆퐶)} of the centroid and the circumcen‑ ter, the Euler line, which passes through them, is easily seen (section 3) to have coefficients 2 2 2 2 2 2 Euler line : (푏 푆퐵 − 푐 푆퐶)푢 + (푐 푆퐶 − 푎 푆퐴)푣 + (푎 푆퐴 − 푏 푆퐵)푤 = 0. Its point at infinity 푋(30) i.e. its intersection with the line at infinity 푢 + 푣 + 푤 = 0 is then calculated to be 2 2 2 4 2 2 2 2 2 2 푋(30) = (2푎 푆퐴 − 푏 푆퐵 − 푐 푆퐶, … ) = ((2푎 − (푏 + 푐 )푎 − (푏 − 푐 ) , … ), where the dots indicate the remaining coordinates resulting by cyclic permutations of the letters {푎, 푏, 푐} and {퐴, 퐵, 퐶}. On the Euler line are located some other triangle centers, like the orthocenter 퐻 or 푋(4) and the “center of the Euler circle” 퐸 or 푋(5). The barycentrics

N G O Euler H

B C

Figure 12: Points {퐻 = 푋4, 푁 = 푋5} on the Euler line of the orthocenter are easily found using the well known ratio 퐻푂/퐻퐺 = 3/2. Taking into account the ratio rules of section 18, by which the condition 푃 = (1 − 푡)푋 + 푡푌 for points in euclidean coordinates transfers to the same relation for these points expressed in abso‑ lute barycentrics, we have: 퐺 푂 (푎2푆 ,…) 퐻 = 3 − 2 = (1, 1, 1) − 2 퐴 휎퐺 휎푂 휎푂 푎2푆 1 = (1 − 2 퐴 , … ) = (푆 푆 , … ) ∼ (푆 푆 , … ) ⇒ 2푆2 푆2 퐵 퐶 퐵 퐶 퐻 = ( 푆퐵푆퐶 ∶ 푆퐶푆퐴 ∶ 푆퐴푆퐵 ). (65) 15 Triangle Area in Barycentrics 19

where 휎푋 is the sum of barycentrics of 푋. Analogously are computed the barycentrics of the center 푁 = 푋5 of the Euler circle, which is the middle of the segment 퐻푂 ∶

2 2 2 2 2 푁 ∼ 휎푂 ⋅ 퐻 + 휎퐻 ⋅ 푂 = 2푆 ⋅ 퐻 + 푆 ⋅ 푂 ∼ ( 푆 + 푆퐵푆퐶 , 푆 + 푆퐶푆퐴 , 푆 + 푆퐴푆퐵 ). (66)

2 Remark 12. The expression of the center 푁 = 푋(5) ≅ (푆 + 푆퐵푆퐶,…) as a linear com‑ bination of the centers {퐺, 퐻} defining the Euler line generalizes for all notable triangle centers lying on that line, which can be also written as linear combinations

2 푋 = ( 푚 ⋅ 푆 + 푛 ⋅ 푆퐵푆퐶, … ).

The coefficients {푚, 푛} are called “Shinagawa coefficients” of 푋 ([3]).

15 Triangle Area in Barycentrics

Consider the triangle of reference 퐴퐵퐶 and a second △퐷퐸퐹, whose vertices have ab‑ solute bary‑centric coordinates w.r.t. 퐴퐵퐶 ∶ 퐷(푑1, 푑2, 푑3) , 퐸(푒1, 푒2, 푒3) , 퐹(푓1, 푓2, 푓3). The basic relations be‑tween barycentric and cartesian coordinates have been discussed in sec‑ tion 6. Denoting by (푋1, 푋2) the cartesian coordinates of the points 푋 of the plane and using equation (18), we obtain:

퐷1 퐸1 퐹1 퐴1 퐵1 퐶1 푑1 푒1 푓1 ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜퐷2 퐸2 퐹2⎟ = ⎜퐴2 퐵2 퐶2⎟ ⎜푑2 푒2 푓2⎟ . (67) ⎝ 1 1 1 ⎠ ⎝ 1 1 1 ⎠ ⎝푑3 푒3 푓3⎠

Taking the determinants, we obtain the relation of the areas

∣푑1 푒1 푓1∣ ∣ ∣ (퐷퐸퐹) = (퐴퐵퐶) ⋅ ∣푑 푒 푓 ∣ , (68) ∣ 2 2 2∣ ∣푑3 푒3 푓3∣ expressing the area (퐷퐸퐹) in terms of the absolute barycentrics of the vertices of the triangle. As an application, we can easily compute the area of the cevian triangle 퐴′퐵′퐶′ of a point 푃(푥, 푦, 푧) (see figure 3). The traces of 푃 on the sides of the triangle of reference 퐴퐵퐶 are 퐴′(0, 푦, 푧), 퐵′(푥, 0, 푧), 퐶′(푥, 푦, 0). With a short calculation we find then

∣0 푥 푥∣ (퐴퐵퐶) ∣ ∣ 2(퐴퐵퐶)푥푦푧 (퐴′퐵′퐶′) = ∣푦 0 푦∣ = . (69) (푦 + 푧)(푧 + 푥)(푥 + 푦) ∣ ∣ (푦 + 푧)(푧 + 푥)(푥 + 푦) ∣푧 푧 0∣

Some more effort in calculation is required for the “pedal” triangle (see figure 13) of a point 푃(푢, 푣, 푤) w.r.t. the triangle of reference 퐴퐵퐶. From the discussion in section 10 we can compute the coordinates of the projections {퐴′, 퐵′, 퐶′} on the sides:

′ 2 2 퐴 = (0 , 푎 푣 + 푢푆퐶 , 푎 푤 + 푢푆퐵), ′ 2 2 퐵 = (푏 푢 + 푣푆퐶 , 0 , 푏 푤 + 푣푆퐴), ′ 2 2 퐶 = (푐 푢 + 푤푆퐵 , 푐 푣 + 푤푆퐴 , 0).

Then, assuming absolute barycentrics, satisfying 푢 + 푣 + 푤 = 1, setting 푆 = 2(퐴퐵퐶) and 16 Circumcevian triangle of a point 20

B' C' P

BC A'

Figure 13: Pedal triangle 퐴′퐵′퐶′ of 퐴퐵퐶 w.r.t. 푃 using equation (33), we find the crucial determinant 2 2 ∣ 0 푏 푢 + 푣푆퐶 푐 푢 + 푤푆퐵∣ ∣ 2 2 ∣ 2 2 2 2 ∣푎 푣 + 푢푆퐶 0 푐 푣 + 푤푆퐴∣ = 푆 (푎 푣푤 + 푏 푤푢 + 푐 푢푣). (70) ∣ 2 2 ∣ ∣푎 푤 + 푢푆퐵 푏 푤 + 푣푆퐴 0 ∣ This, turning to general, not necessarily absolute barycentrics, implies the formula for the area (퐴′퐵′퐶′) of the pedal of the point 푃(푢, 푣, 푤) w.r.t. the triangle of reference: 푆3 푎2푣푤 + 푏2푤푢 + 푐2푢푣 (퐴′퐵′퐶′) = ⋅ . (71) 푎2푏2푐2 2(푢 + 푣 + 푤)2

16 Circumcevian triangle of a point

Given a triangle 퐴퐵퐶 and a point 푃 the “circumcevian” triangle of 푃 w.r.t. 퐴퐵퐶 is the triangle 퐴′퐵′퐶′ formed by the second intercepts {퐴′, 퐵′, 퐶′} of the cevians {퐴푃, 퐵푃, 퐶푃} of 푃 with the circum‑circle 휅 of 퐴퐵퐶 (see figure 14). The main property of the circumcevian triangle is: Theorem 12. The circumcevian triangle of 푃 is similar to the corresponding pedal triangle of 푃.

B' C' C'' B''

B A'' C

Figure 14: The circumcevian triangle 퐴′퐵′퐶′ of 푃 w.r.t. 퐴퐵퐶

Figure 14 shows the way to prove that the two triangles have the same angles. Returning to compu‑tations, the equations of the cevians are: 퐴퐴′ ∶ −푧푣 + 푦푤 = 0, 퐵퐵′ ∶ −푥푤 + 푧푢 = 0, 퐶퐶′ ∶ −푦푢 + 푥푣 = 0. (72) 17 Circle through three points, Brocard circle 21

The points {퐴′, 퐵′, 퐶′} are calculated and seen to be:

−푎2푦푧 푎2푥푧 + 푐2푥2 푎2푥푦 + 푏2푥2 ′ ⎜⎛ 2 2 2⎟⎞ ′ ⎜⎛ 2 ⎟⎞ ′ ⎜⎛ 2 2 2⎟⎞ 퐴 = ⎜푏 푦푧 + 푐 푦 ⎟ , 퐵 = ⎜ −푏 푥푧 ⎟ , 퐶 = ⎜푏 푥푦 + 푎 푦 ⎟ . (73) ⎝푐2푦푧 + 푏2푧2⎠ ⎝푐2푥푧 + 푎2푧2⎠ ⎝ −푐2푥푦 ⎠ Their determinant is found:

푑푒푡(퐴′, 퐵′, 퐶′) = (푎2푦푧 + 푏2푧푥 + 푐2푥푦)3. (74)

′ ′ ′ The determination of the area (퐴 퐵 퐶 ) requires the division with the product 휎퐴′ 휎퐵′ 휎퐶′ of the sums of the coordinates of these points:

2 2 2 2 2 2 2 2 2 2 휎퐴′ = (푏 + 푐 − 푎 )푦푧 + 푏 푧 + 푐 푦 = 푆퐴(푦 + 푧) + 푆퐵푦 + 푆퐶푧 , 2 2 2 2 2 2 2 2 2 2 휎퐵′ = (푐 + 푎 − 푏 )푧푥 + 푐 푥 + 푎 푧 = 푆퐵(푧 + 푥) + 푆퐶푧 + 푆퐴푥 , 2 2 2 2 2 2 2 2 2 2 휎퐶′ = (푎 + 푏 − 푐 )푥푦 + 푎 푦 + 푏 푥 = 푆퐶(푥 + 푦) + 푆퐴푥 + 푆퐵푦 .

17 Circle through three points, Brocard circle

The expression in barycentrics of the circle passing through three points {푃푖(푢푖, 푣푖, 푤푖)}, can be found from the corresponding expression in cartesian coordinates, using the trans‑ formation 푋푖 = 푀푃푖 of section 6. In fact, using cartesian coordinates {푋푖 = (푥푖, 푦푖)}, the circle through these points is represented by the equation:

2 2 2 2 ∣푋1 푋2 푋3 푋 ∣ ∣ 푥 푥 푥 푥 ∣ 0 = ∣ 1 2 3 ∣ ∣ 푦 푦 푦 푦 ∣ ∣ 1 2 3 ∣ ∣ 1 1 1 1 ∣

∣푥2 푥3 푥∣ ∣푥1 푥3 푥∣ ∣푥1 푥2 푥∣ ∣푥1 푥2 푥3∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 푋2 ∣푦 푦 푦∣ − 푋2 ∣푦 푦 푦∣ + 푋2 ∣푦 푦 푦∣ − 푋2 ∣푦 푦 푦 ∣ 1 ∣ 2 3 ∣ 2 ∣ 1 3 ∣ 3 ∣ 1 2 ∣ ∣ 1 2 3∣ ∣ 1 1 1∣ ∣ 1 1 1∣ ∣ 1 1 1∣ ∣ 1 1 1 ∣ 2 2 2 2 = 푋1|푀푃2, 푀푃3, 푀푃| − 푋2|푀푃1, 푀푃3, 푀푃| + 푋3|푀푃1, 푀푃2, 푀푃| − 푋 |푀푃1, 푀푃2, 푀푃3| 2 2 2 2 = ( 푋1|푃2, 푃3, 푃| − 푋2|푃1, 푃3, 푃| + 푋3|푃1, 푃2, 푃| − 푋 |푃1, 푃2, 푃3| ) |푀| ⇔ 2 2 2 2 푋1|푃2, 푃3, 푃| + 푋2|푃3, 푃1, 푃| + 푋3|푃1, 푃2, 푃| − 푋 |푃1, 푃2, 푃3| = 0. (75) Here |푀| denotes the determinant of the matrix 푀 and |푃, 푄, 푅| denotes the determi‑ nant of the matrix of the columns of absolute barycentric coordinates vectors {푃, 푄, 푅}. By equation (24) the inner products are:

2 2 2 2 2 푋푖 = 푅 − (푎 푣푖푤푖 + 푏 푤푖푢푖 + 푐 푢푖푣푖). Replacing with this in equation (75), we see that the terms involving 푅2 factor into

|푃2, 푃3, 푃| + |푃3, 푃1, 푃| + |푃1, 푃2, 푃| − |푃1, 푃2, 푃3| = 0, because this is the determinant of the matrix having two equal rows:

∣ 1 1 1 1∣ ∣푥 푥 푥 푥∣ ∣ 1 2 3 ∣ = 0. ∣푦 푦 푦 푦∣ ∣ 1 2 3 ∣ ∣ 1 1 1 1∣ 18 The associated affine transformation 22

From this follows that the circle through the three points is described by the equation:

2 2 2 (푎 푣1푤1 + 푏 푤1푢1 + 푐 푢1푣1)|푃2, 푃3, 푃| 2 2 2 +(푎 푣2푤2 + 푏 푤2푢2 + 푐 푢2푣2)|푃3, 푃1, 푃| 2 2 2 (76) +(푎 푣3푤3 + 푏 푤3푢3 + 푐 푢3푣3)|푃1, 푃2, 푃| 2 2 2 −(푎 푣푤 + 푏 푤푢 + 푐 푢푣)|푃1, 푃2, 푃3| = 0.

Notice that the three first rows are the expressions of line equations, and the determinant‑ equations {|푃2, 푃3, 푃| = 0, …} represent respectively the side‑lines of the triangle 푃1푃2푃3. Since the expression in the last row is the one of the equation of the circumcircle of 퐴퐵퐶, the linear part of the first three rows represents the equation of the radical axis of thecircle 휅 through the three points and the circumcircle 휅0 of the triangle of reference 퐴퐵퐶. Thus, we have the theorem

Theorem 13. The equation

2 2 2 (푎 푣1푤1 + 푏 푤1푢1 + 푐 푢1푣1)|푃2, 푃3, 푃| 2 2 2 +(푎 푣2푤2 + 푏 푤2푢2 + 푐 푢2푣2)|푃3, 푃1, 푃| (77) 2 2 2 +(푎 푣3푤3 + 푏 푤3푢3 + 푐 푢3푣3)|푃1, 푃2, 푃| = 0, represents the radical axis of the circle 휅 through the three points {푃1, 푃2, 푃3} expressed in absolute barycentrics and the circumcircle 휅0 of the triangle of reference 퐴퐵퐶. Equation (76) leads to a condition on four points to belong to the same circle by set‑ ting 푃 = 푃4. The equation can be expressed also through a 4 × 4 determinant. Doing the calculation and some simplifications the formula takes the form:

∣ 푠(푃1) 푠(푃2) 푠(푃3) 푠(푃4) ∣ ∣ 푃 푃 푃 푃 ∣ ∣ 11 21 31 41 ∣ = 0 , (78) ∣ 푃 푃 푃 푃 ∣ ∣ 12 22 32 42 ∣ ∣ 푃13 푃23 푃33 푃43 ∣ where in each column appears the expression

푎2푥 푥 + 푏2푥 푥 + 푐2푥 푥 푠(푋) = 2 3 3 1 1 2 (푥1 + 푥2 + 푥3) and the not necessarily absolute barycentrics of the involved points. Equation (78) applied to the two Brocard points Ω ( 1 ∶ 1 ∶ 1 ), Ω′ ( 1 ∶ 1 ∶ 1 ) and 푏2 푐2 푎2 푐2 푎2 푏2 the symmedian point 퐾(푎2 ∶ 푏2 ∶ 푐2), leads to the “Brocard circle” passing through these three points: 푏2푐2푢2 + 푐2푎2푣2 + 푎2푏2푤2 − 푎4푣푤 − 푏4푤푢 − 푐4푢푣 = 0 . (79)

18 The associated affine transformation

Here we re‑examine the matrix 푀 of section 6, defining the transformation from absolute bary‑centrics to cartesian coordinates.

푥 푢 퐴1 퐵1 퐶1 푢 ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜푦⎟ = 푀 ⋅ ⎜푣⎟ = ⎜퐴2 퐵2 퐶2⎟ ⋅ ⎜푣⎟ . (80) ⎝1⎠ ⎝푤⎠ ⎝ 1 1 1 ⎠ ⎝푤⎠ 18 The associated affine transformation 23

The matrix 푀 is invertible and its inverse, denoting by 푆 twice the signed area of △퐴퐵퐶, is

퐵2 − 퐶2 퐶1 − 퐵1 퐵1퐶2 − 퐵2퐶1 −1 1 ⎜⎛ ⎟⎞ 푀 = ⎜퐶2 − 퐴2 퐴1 − 퐶1 퐶1퐴2 − 퐶2퐴1⎟ 퐴1(퐵2 − 퐶2) + 퐵1(퐶2 − 퐴2) + 퐶1(퐴2 − 퐵2) ⎝퐴2 − 퐵2 퐵1 − 퐴1 퐴1퐵2 − 퐴2퐵1 ⎠ 퐵 − 퐶 퐶 − 퐵 퐵 퐶 − 퐵 퐶 1 ⎛ 2 2 1 1 1 2 2 1 ⎞ = ⋅ ⎜퐶 − 퐴 퐴 − 퐶 퐶 퐴 − 퐶 퐴 ⎟ . (81) 푆 ⎜ 2 2 1 1 1 2 2 1⎟ ⎝퐴2 − 퐵2 퐵1 − 퐴1 퐴1퐵2 − 퐴2퐵1 ⎠

3 The matrix 푀 defines an invertible linear transformation 퐿푀 (an isomorphism) of ℝ onto itself and the set of all absolute barycentrics satisfying 푢 + 푣 + 푤 = 1 represents the plane 휀 of ℝ3, which is orthogonal to the vector (1, 1, 1) ∈ ℝ3 and passes through the 1 3 ′ 3 point 3 (1, 1, 1) ∈ ℝ . The image 휀 = 퐿푀(휀) is the plane of ℝ parallel to the coordi‑ nate {푧 = 0}‑plane through the point 푧 = 1. The linear transformation 퐿푀 introduces by its restriction on plane 휀 ∶ ℳ = 퐿푀|휀 an “affine” transformation ([2, p.199]) between the planes {휀, 휀′}: ℳ ∶ 휀 ⟶ 휀′, with ℳ(푢, 푣, 푤) = (푥, 푦, 1). It is interesting to see some consequences of this interpretation as, for example, the trans‑ formation of lines to lines, the preservation of ratios on lines and the preservation of quotiens of areas. These are general properties of the affine transformations but can be also deduced here directly for this special case. In fact, a line in 휀 is the intersection of 휀 with a plane through the origin 휂 of ℝ3 ∶ represented by an equation of the form 휂 ∶ 푝푢 + 푞푣 + 푟푤 = 0. The image‑plane 휂′ = ℳ(휂) is found by writing the equation using matrix notation: 푢 푢 ⎜⎛ ⎟⎞ −1 ⎜⎛ ⎟⎞ 푝푢 + 푞푣 + 푟푤 = 0 ⇔ 0 = (푝, 푞, 푣) ⎜푣⎟ = (푝, 푞, 푣)푀 푀 ⎜푣⎟ . ⎝푤⎠ ⎝푤⎠

Hence the image of the plane 휂 is the plane represented by the equation

휂′ ∶ 푝′푥 + 푞′푦 + 푟′푧 = 0, where (푝′, 푞′, 푟′) = (푝, 푞, 푟)푀−1, (82) and the line of the plane 휀′ is the intersection 휀′ ∩ 휂′. In particular, the line at infinity of 휀, represented through its intersection with the plane 휂 ∶ 푢 + 푣 + 푤 = 0, maps to the line at infinity, which is the intersection of the plane 휀′ with the plane with coefficients

휂′ ∶ (1, 1, 1)푀−1 = (0, 0, 1) i.e the plane 푧 = 0.

Another consequence of the affine property of the transformation ℳ and its inverse is the pre‑servation of ratios along lines. Thus, for two points {푃, 푄} of the plane and their line

푡 푆푃 푆(푡) = (1 − 푡)푃 + 푡푄, with 푟 = equal to the signed ratio: 푟 = , (83) 푡 − 1 푆푄 the corresponding barycentrics satisfy the same relation:

푆1(푡) (1 − 푡)푃1 + 푡푄1 −1 −1 ⎜⎛ ⎟⎞ −1 ⎜⎛ ⎟⎞ −1 −1 푀 푆(푡) = 푀 ⎜푆2(푡)⎟ = 푀 ⎜(1 − 푡)푃2 + 푡푄2⎟ = (1 − 푡)(푀 푃) + 푡(푀 푄). ⎝ 1 ⎠ ⎝ 1 − 푡 + 푡 ⎠ 18 The associated affine transformation 24

Notice that the function 푟 = 푓 (푡) = 푡/(푡 − 1) has inverse 푓 −1 = 푓 , so that given the ratio 푟 = 푆푃/푆푄 and taking 푡 = 푟/(푟 − 1), and seting {푃′, 푄′, …} for the corresponding barycen‑ tric vectors of the points {푃, 푄, …}, we have that

1 푆′(푡) = (1 − 푡)푃′ + 푡푄′ ⇔ 푆′(푟) = (푃′ − 푟푄′), (84) 1 − 푟 satisfies precisely the relation 푆푃/푆푄 = 푟. If the ratio 푟 is expressed in the form 푟 = 푚/푛 , then for the point 푆푟 satisfying this condition and the corresponding barycentrics vectors we obtain ([6]): 푆푟푃 푚 ′ 1 ′ ′ = 푟 = ⇒ 푆푟 = (푛푃 − 푚푄 ). (85) 푆푟푄 푛 푛 − 푚 As application, we prove the collinearity of the incenter 퐼(푎 ∶ 푏 ∶ 푐), centroid 퐺(1 ∶ 1 ∶ 1)

I G N B C

Figure 15: The collinearity of {퐼, 퐺, 푁} and Nagel point 푁(푏 + 푐 − 푎, ...) (see file Nagel point of the triangle). The relation to verify is (see figure 15): 퐺푁/퐺퐼 = 푟 = −2 ⇒ 푡 = 푟/(푟 − 1) = 2/3. Thus, turning to absolute barycentrics by dividing the previous barycentrics vectors with the sum 휎푁 = 휎퐼 = 푎 + 푏 + 푐 = 2휎, we obtain: 푁 퐼 1 1 1 (1 − 푡) + 푡 = (푁 + 2퐼) = ((푏 + 푐 − 푎) + 2푎, ...) = (1, 1, 1) = 퐺. 2휎 2휎 6휎 6휎 3 As an application of the preceding result we obtain the property (see figure 16) of “the incenter 퐼 to be the Nagel point of the anticomplementary triangle 퐴′퐵′퐶′ of the middles of sides of 퐴퐵퐶 ”. This follows from the fact that the homothety with center 퐺 and ratio 1/2, as an affine map, preserves the barycentrics and maps 퐴퐵퐶 to 퐴′퐵′퐶′ and also maps the Nagel

C' B' I M G J

N K B C A'

Figure 16: The Nagel point of the anticomplementary point 푁 of 퐴퐵퐶 to 퐼. By the way, we notice also that the triples like the points {퐶′, 퐼, 퐽} are collinear (see exercise 3). The triple consisting of the middle of 퐴퐵, the incenter and the 19 Relations between barycentrics and cartesian coordinates 25 intersection 퐽 = 퐴′퐵′ ∩ 퐶퐿, where 퐿, the touch point of the incircle with 퐴퐵. A key fact to prove this synthetically is the equality 퐴퐿 = 푀퐵 = 휏 − 푎, where 휏 the half‑perimeter and 푎 = |퐵퐶|. A third property of the affine transformation ℳ, already seen in section 15, is the multiplication of areas by a constant, in this case expressed through equation (68).

19 Relations between barycentrics and cartesian coordinates

Here we give a second look at the relations between cartesian coordinates and barycentrics initiated in section 6 using a particular cartesian coordinate system. From now on the symbols {푋, 푌, …} will denote euclidean two‑dimensional position vectors with cartesian coordinates 푋(푋푥, 푋푦) w.r.t. to a system with origin at the centroid 퐺 of the triangle of reference 퐴퐵퐶, assumed to have the positive orientation. The absolute barycentric coor‑ dinates of the same point will be denoted by {푋푢, 푋푣, 푋푤}. The two sets of coordinates are related by the matrix 푀 representing the affine transformation 퐿푀 of the preceding section: 푋푥 푋푢 ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜푋푦⎟ = 푀 ⋅ ⎜푋푣 ⎟ . ⎝ 1 ⎠ ⎝푋푤⎠ Under the preceding assumption of the particular cartesian coordinate system, the two matrices {푀, 푀−1} take the form:

퐴푥 퐵푥 퐶푥 퐽(퐵퐶)푥 퐽(퐵퐶)푦 푆/3 ⎛ ⎞ 1 ⎜⎛ ⎟⎞ 푀 = ⎜퐴 퐵 퐶 ⎟ , 푀−1 = ⎜퐽(퐶퐴) 퐽(퐶퐴) 푆/3⎟ . (86) ⎜ 푦 푦 푦⎟ 푆 ⎜ 푥 푦 ⎟ ⎝ 1 1 1 ⎠ ⎝퐽(퐴퐵)푥 퐽(퐴퐵)푦 푆/3⎠

Here 푆 is twice the area of △퐴퐵퐶 and 퐽 denotes the positive rotation by 휋/2 acting on vectors 푋 = (푋푥, 푋푦) and transforming them to 퐽(푋) = (−푋푦, 푋푥), so that the inner product of two vectors equals their determinant:

⟨퐽(푈), 푉⟩ = −푈푦푉푥 + 푈푥푉푦 = |푈, 푉| .

This implies, that 푀−1 applied to a point has the form:

푋 |퐵퐶, 푋| 1 ⎛ 푥⎞ 1 ⎛ ⎞ 1 ⎛ ⎞ 푀−1 ⋅ ⎜푋 ⎟ = ⎜|퐶퐴, 푋|⎟ + ⎜1⎟ . (87) ⎜ 푦⎟ 푆 ⎜ ⎟ 3 ⎜ ⎟ ⎝ 1 ⎠ ⎝|퐴퐵, 푋|⎠ ⎝1⎠

It is interesting to notice the behavior of the absolute coordinates in the case of the sum of two euclidean vectors and the multiplication of a vector by a number.

(푋 + 푌)푢 (푋 + 푌)푥 푋푥 푌푥 0 ⎜⎛ ⎟⎞ −1 ⎜⎛ ⎟⎞ −1 ⎜⎛⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞⎟⎞ ⎜(푋 + 푌)푣 ⎟ = 푀 ⎜(푋 + 푌)푦⎟ = 푀 ⎜⎜푋푦⎟ + ⎜푌푦⎟ − ⎜0⎟⎟ ⇒ ⎝(푋 + 푌)푤⎠ ⎝ 1 ⎠ ⎝⎝ 1 ⎠ ⎝ 1 ⎠ ⎝1⎠⎠ 푋 푌 (푋 + 푌) 1 ⎛ 푢 ⎞ ⎛ 푢 ⎞ ⎛ 푢 ⎞ 1 ⎛ ⎞ ⎜푋 ⎟ + ⎜푌 ⎟ = ⎜(푋 + 푌) ⎟ + ⎜1⎟ , (88) ⎜ 푣 ⎟ ⎜ 푣 ⎟ ⎜ 푣 ⎟ 3 ⎜ ⎟ ⎝푋푤⎠ ⎝푌푤⎠ ⎝(푋 + 푌)푤⎠ ⎝1⎠

Showing that the sum of the barycentrics does not represent the point resulting by sum‑ ming the corresponding cartesian coordinates. Analogously for the multiplication by a 20 Affine transformations represented in barycentrics 26 number 휆 we have:

(휆푋)푢 (휆푋)푥 푋푥 0 ⎜⎛ ⎟⎞ −1 ⎜⎛ ⎟⎞ −1 ⎜⎛ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞⎟⎞ ⎜(휆푋)푣 ⎟ = 푀 ⎜(휆푋)푦⎟ = 푀 ⎜휆 ⎜푋푦⎟ + ⎜ 0 ⎟⎟ ⇒ ⎝(휆푋)푤⎠ ⎝ 1 ⎠ ⎝ ⎝ 1 ⎠ ⎝1 − 휆⎠⎠ (휆푋) 푋 1 ⎛ 푢 ⎞ ⎛ 푢 ⎞ 1 − 휆 ⎛ ⎞ ⎜(휆푋) ⎟ = 휆 ⎜푋 ⎟ + ⎜1⎟ . (89) ⎜ 푣 ⎟ ⎜ 푣 ⎟ 3 ⎜ ⎟ ⎝(휆푋)푤⎠ ⎝푋푤⎠ ⎝1⎠ In particular, we see that taking the negative barycentrics corresponds to considering the symmetric point w.r.t. the origin 퐺 (see also formula (95)):

(−푋) 푋 1 ⎛ 푢 ⎞ ⎛ 푢 ⎞ 2 ⎛ ⎞ ⎜(−푋) ⎟ = − ⎜푋 ⎟ + ⎜1⎟ . (90) ⎜ 푣 ⎟ ⎜ 푣 ⎟ 3 ⎜ ⎟ ⎝(−푋)푤⎠ ⎝푋푤⎠ ⎝1⎠ Remark 13. Equation (88) written in the form

(푋 + 푌) 푋 푌 1 ⎛ 푢 ⎞ ⎛ 푢 ⎞ ⎛ 푢 ⎞ 1 ⎛ ⎞ ⎜(푋 + 푌) ⎟ = ⎜푋 ⎟ + ⎜푌 ⎟ − ⎜1⎟ , (91) ⎜ 푣 ⎟ ⎜ 푣 ⎟ ⎜ 푣 ⎟ 3 ⎜ ⎟ ⎝(푋 + 푌)푤⎠ ⎝푋푤⎠ ⎝푌푤⎠ ⎝1⎠ can be considered as the result in barycentrics of an euclidean translation by the vector 푌. We see that this is not described by the addition of corresponding coordinates as is the case of the expression of the translation in cartesian coordinates. Also the expression of the displacement vectors (section 8), which are differences of barycentrics vectors, do not correspond exactly to the difference of the cartesian coordinates:

푌 푋 (푌 − 푋) 1 ⎛ 푢 ⎞ ⎛ 푢 ⎞ ⎛ 푢 ⎞ 1 ⎛ ⎞ ⎜푌 ⎟ − ⎜푋 ⎟ = ⎜(푌 − 푋) ⎟ − ⎜1⎟ . ⎜ 푣 ⎟ ⎜ 푣 ⎟ ⎜ 푣 ⎟ 3 ⎜ ⎟ ⎝푌푤⎠ ⎝푋푤⎠ ⎝(푌 − 푋)푤⎠ ⎝1⎠

20 Affine transformations represented in barycentrics

Using the embedding of the plane (푥, 푦) ∈ ℝ2 ↦ (푥, 푦, 1) into ℝ3, the affine transfor‑ −1 mation 퐿푀 and its inverse, represented by the matrices {푀, 푀 } in (86), we can easily arrive at the matrix representation by barycentrics of the affine transformations of the plane, which comprise the isometries and the similarities as special cases. The commu‑ tative diagram below leads to a method to express this matrix in terms of the cartesian coordinates of the involved points and transformations.

푀푓 ℝ3(cartesian coordinates) ℝ3(cartesian coordinates)

푀 푀−1

푁푓 ℝ3(absolute barycentrics) ℝ3(absolute barycentrics)

The representation of a general affine transformation 푓 of the plane by a matrix 푀푓 in cartesian coordinates has the form ′ 푥 푥 푃1 푄1 푉1 푥 ⎜⎛ ⎟⎞ 푓 ⎜⎛ ′⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜푦⎟ ↦ ⎜푦 ⎟ = ⎜푃2 푄2 푉2⎟ ⋅ ⎜푦⎟ ⇔ 푓 (푋) = 푀푓 ⋅ 푋 , ⎝1⎠ ⎝ 1 ⎠ ⎝ 0 0 1 ⎠ ⎝1⎠ 20 Affine transformations represented in barycentrics 27

with the determinant 푃1푄2 − 푃2푄1 ≠ 0. The matrix representing the same transforma‑ tion in barycentrics results as a product of matrices:

−1 푁푓 = 푀 ⋅ 푀푓 ⋅ 푀 .

The matrices {푀, 푀−1}, introduced in section 6 and having the form (86), explained in section 19, lead to the following expression for the product:

|퐵퐶, 푃| |퐵퐶, 푄| |퐵퐶, 푉| + 푆/3 1 ⎛ ⎞ 푀−1 ⋅ 푀 = ⎜|퐶퐴, 푃| |퐶퐴, 푄| |퐶퐴, 푉| + 푆/3⎟ . 푓 푆 ⎜ ⎟ ⎝|퐴퐵, 푃| |퐴퐵, 푄| |퐴퐵, 푉| + 푆/3⎠

−1 This, multiplying with 푀, leads finally to the matrix 푁푓 = 푀 ⋅ 푀푓 ⋅ 푀 expressed in terms of the euclidean coordinates of the vectors {퐴, 퐵, 퐶} w.r.t. a system having its origin at the centroid 퐺 of the triangle of reference 퐴퐵퐶. The symbol 푓 (푋) in the formula below denotes the application of the transformation 푓 to 푋(푥, 푦) i.e. 푓 (푋) = 푥푃 + 푦푄 + 푉 all this expressed as a two dimensional vector (omitting the last 1 from (푥, 푦, 1) ).

Remark 14. It is easily verified that the determinants {|퐵퐶, 푉|, ...} are related to the abso‑ lute coordi‑nates {푢, 푣, 푤} of 푉 by the simple formula:

|퐵퐶, 푉| 3푢 − 1 ⎛ ⎞ 푆 ⎛ ⎞ ⎜|퐶퐴, 푉|⎟ = ⎜3푣 − 1⎟ . (93) ⎜ ⎟ 3 ⎜ ⎟ ⎝|퐴퐵, 퐶|⎠ ⎝3푤 − 1⎠

Applying this to formula (92) we find that the matrix 푁푓 can be expressed also in the form 푓 (퐴)푢 푓 (퐵)푢 푓 (퐶)푢 ⎜⎛ ⎟⎞ 푁푓 = ⎜푓 (퐴)푣 푓 (퐵)푣 푓 (퐶)푣 ⎟ , (94) ⎝푓 (퐴)푤 푓 (퐵)푤 푓 (퐶)푤⎠ where the symbols {푓 (푋)푢, 푓 (푋)푣, 푓 (푋)푤} denote the absolute barycentrics of the point 푓 (푋). Note that this representation of the affinity by 푁푓 is not valid for a general projec‑ tivity 푓 , since this would imply that all projectivities fixing the vertices of △퐴퐵퐶 have the same matrix representation, which is not true. Affinities, though, are completely de‑ termined by their action on the vertices of △퐴퐵퐶. Formula (94) conforms to this fact and shows, that affinities, expressed in absolute barycentrics, act as linear transformations and their values are completely determined from those on the “base” consisting of the vertices {퐴, 퐵, 퐶}.

In equation (91) we saw the expression of a translation in barycentrics. Another simple example of affine transformation is the point‑symmetry. This, using formula (92) or its equivalent (94) and the rules of section 19, is easily seen to be expressible in barycentrics in the same typical form as in cartesian coordinates:

′ 푢 푢 푊푢 ⎜⎛ ′ ⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜푣 ⎟ = − ⎜푣⎟ + 2 ⎜푊푣 ⎟ , (95) ′ ⎝푤 ⎠ ⎝푤⎠ ⎝푊푤⎠ 21 Remarks on working with barycentrics 28

where {푊푢, 푊푣, 푊푤} are the barycentrics of the center 푊 of the symmetry. This conforms to the remarks on the ratio of segments along a fixed line of section 18 and could be de‑ duced trivially from the arguments used there. This remark applies also to homotheties, which generalize the point‑symmetry and in cartesian coordinates are described by the formula: 푓 (푋) = 푌 = 푊 + 푘(푋 − 푊) , (96) where {푊, 푘} are respectively the center and the homothety ratio. The corresponding for‑ mula in absolute barycentrics has the same form.

Exercise 1. Show, using formula (94), that the reflection 푓퐵퐶 in the side 퐵퐶 of △퐴퐵퐶 is rep‑ resented in barycentrics by the matrix

−1 0 0 푢 ⎜⎛ 2 ⎟⎞ ⎜⎛ ⎟⎞ 푓퐵퐶(푋) = ⎜2푆퐶/푎 1 0⎟ ⋅ ⎜푣⎟ . 2 ⎝2푆퐵/푎 0 1⎠ ⎝푤⎠ Exercise 2. Using absolute barycentrics, the formula (46) for the projection of a point on a line and formula (95), show that the reflection 푈′ of a point 푈(푢, 푣, 푤) in the line, expressed by the equation 푓 (푢, 푣, 푤) = 푝푢 + 푞푣 + 푟푤 = 0, is given by the formula

푆 (푝 − 푟) − 푆 (푞 − 푝) 푓 (푈) ⎛ 퐵 퐶 ⎞ 푈′ = 푈 − 2 푉, with 푉 = ⎜푆 (푞 − 푝) − 푆 (푟 − 푞)⎟ . (97) 푓 (푉) ⎜ 퐶 퐴 ⎟ ⎝푆퐴(푟 − 푞) − 푆퐵(푝 − 푟)⎠ Remark 15. Notice that 푉 represents in barycentrics the orthogonal direction to that of the line 푓 (푢, 푣, 푤) = 푝푢 + 푞푣 + 푟푤 = 0 . Also, if you interpret this equation as one defining a “plane” in ℝ3 through its origin and 푉 as an orthogonal vector to this plane, which in this case is a multiple of (푝, 푞, 푟), then the reflection in this plane expressed in cartesian coordinates of ℝ3 is formaly the same with equation (97).

Exercise 3. Referring to figure 16 and using formula (96), show that the barycentrics of the inter‑ section point 퐽 = 퐴′퐵′ ∩ 퐶퐿 are a non‑zero multiple of (휏 − 푏 , 휏 − 푎 , 2휏 − 푎 − 푏). Use this to show the collinearity of points {퐶′, 퐼, 퐽}.

21 Remarks on working with barycentrics

3 The vectors of barycentric coordinates {푈푃, 푈푄, 푈푅 ⋯ ∈ ℝ } representing the points of the plane {푃, 푄, 푅, ⋯ ∈ ℝ2} are not the points we see. What we see are the points of ℝ2 . This is clearly understood in the case of the triangle of reference 퐴퐵퐶. The barycentrics‑ vectors representing it are {(1, 0, 0), (0, 1, 0), (0, 0, 1)} and define the vertices of an equi‑ 3 lateral triangle lying on the plane 휀 ∶ 푢 + 푣 + 푤 = 1 of ℝ . The map 퐿푀 of section 18 is the affine transformation mapping the equilateral 퐴′퐵′퐶′ onto 퐴퐵퐶. By means of it, all properties of the triangle correspond to properties of the equilateral and vice versa. In particular, properties of the equilateral which are preserved by affine transformations map to similar properties of 퐴퐵퐶. A characteristic example is the circum‑circle 휅′ of the ′ ′ ′ ′ ′ ′ equilateral 퐴 퐵 퐶 , which carries also the symmetrics {퐴1, 퐵1, 퐶1} of the vertices w.r.t. the ′ ′ center 퐺 of the equilateral (see figure 17). The map 퐿푀 transforms 퐺 to the centroid 퐺 of 퐴퐵퐶 and the circumcircle of 퐴′퐵′퐶′ to the “Steiner (outer) ellipse” 휅 of 퐴퐵퐶, char‑ acterized by the fact to pass through the vertices of 퐴퐵퐶 and their symmetrics 퐴1, 퐵1, 퐶1 ′ ′ ′ ′ w.r.t. 퐺, point 퐺 being its center. Similarly, 퐿푀 maps the incircle 휆 of 퐴 퐵 퐶 to the “Steiner in‑ellipse” 휆 of the triangle 퐴퐵퐶, characterized by its tangency at the middles BIBLIOGRAPHY 29

A A'(1,0,0)

κ' λ κ λ' LM G'(1,1,1)/3 G

C A' B A0 B'(0,1,0) 0 C'(0,0,1)

A 1 A'1

Figure 17: We see 퐴퐵퐶 and work with 퐴′퐵′퐶′

′ ′ ′ 퐴0, 퐵0, 퐶0 of the sides of 퐴퐵퐶. The homothety of {휅 , 휆 } with center 퐺 and ratio 2 ∶ 1 translates to the homothety of {휅, 휆} with center 퐺 and the same ratio. Invertible affine transformations or affinities like 퐿푀, besides the preservation of ratios along lines, map also areas 휎 to multiples 푘휎, with 푘 a constant factor. This implies, that points 푃′ with barycentrics (푢′ ∶ 푣′ ∶ 푤′) w.r.t. 퐴′퐵′퐶′ map to corresponding points ′ ′ ′ ′ 푃 = 퐿푀(푃 ) with the same barycentrics (푢 ∶ 푣 ∶ 푤) = (푢 ∶ 푣 ∶ 푤 ) w.r.t. 퐴퐵퐶. Thus, for example, the circumcircle of the equilateral 퐴′퐵′퐶′ with side 푎 = |퐵′퐶′|, whose equation, according to section 7 is

푎2푣′푤′ + 푎2푤′푢′ + 푎2푢′푣′ = 0 ⇔ 푣′푤′ + 푤′푢′ + 푢′푣′ = 0, maps to the Steiner outer ellipse, which in barycentrics w.r.t. 퐴퐵퐶 must satisfy the same equation: 푣푤 + 푤푢 + 푢푣 = 0. (98) On the other hand, the incircle of the equilateral 퐴′퐵′퐶′, which, according to section 12, is repre‑sented in barycentrics w.r.t. 퐴′퐵′퐶′ by the equation

푎2 푎2(푣푤 + 푤푢 + 푣푤) − (푢 + 푣 + 푤)2 = 0 ⇔ 푢2 + 푣2 + 푤2 − 2(푣푤 + 푤푢 + 푢푣) = 0, 4 maps to the Steiner inner ellipse, which in barycentrics w.r.t. 퐴퐵퐶 is represented by the same equation: 푢2 + 푣2 + 푤2 − 2(푣푤 + 푤푢 + 푢푣) = 0. (99)

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Related topics 1. Cross Ratio 2. Ceva’s theorem 3. Conway triangle symbols 4. Menelaus’ theorem 5. Nagel point of the triangle 6. Projective line 7. Projective plane 8. Symmedian point of the triangle

Any correction, suggestion or proposal from the reader, to improve/extend the exposition, is welcome and could be send by e‑mail to: [email protected]