On Some Properties of the First Brocard Triangle in the Isotropic Plane

: In this paper we introduce the ﬁrst Brocard triangle of an allowable triangle in the isotropic plane and derive the coordinates of its vertices in the case of a standard triangle. We prove that the ﬁrst Brocard triangle is homologous to the given triangle and that these two triangles are parallelogic. We consider the relationships between the ﬁrst Brocard triangle and the Steiner axis, the Steiner point, and the Kiepert parabola of the triangle. We also investigate some other interesting properties of this triangle and consider relationships between the Euclidean and the isotropic case.


Introduction and Motivation
The isotropic (or Galilean) plane is a projective-metric plane, where the absolute consists of one line-the absolute line ω A , and one point on that line-the absolute point Ω A .
When using homogeneous coordinates in the projective plane, P = (x 0 : x 1 : x 2 ), x 2 0 + x 2 1 + x 2 2 = 0, then we choose the absolute point Ω A = (0 : 1 : 0) and the absolute line ω A having the equation x 2 = 0. Points incident to the absolute line ω A are called isotropic points and lines incident to the absolute point Ω A are called isotropic lines. We will mention a few well known metric quantities in the isotropic plane for which we assume that x = x 0 x 2 and y = x 1 x 2 . Two lines are called parallel if they have the same isotropic point. Points which lie on the same isotropic line are said to be parallel.
For two non-parallel points P 1 = (x 1 , y 1 ) and P 2 = (x 2 , y 2 ) the isotropic distance is defined as d(P 1 , P 2 ) := x 2 − x 1 . Notice that the isotropic distance is directed. For two parallel points P 1 = (x 1 , y 1 ) and P 2 = (x 1 , y 2 ), the isotropic span is defined as s(P 1 , P 2 ) := y 2 − y 1 . The midpoint of the points P 1 = (x 1 , y 1 ) and P 2 = (x 2 , y 2 ) is defined as M = x 1 +x 2 2 , y 1 +y 2 2 . The angle formed by non-isotropic lines l 1 and l 2 given by y = m 1 x + b 1 and y = m 2 x + b 2 is defined by ϕ = ∠(l 1 , l 2 ) := m 2 − m 1 , and it is directed. The bisector of the lines l 1 and l 2 is given by the equation y = m 1 +m 2 2 x + b 1 +b 2 2 . A normal line to a line l at a point P is the isotropic line n passing through P.
All projective transformations that preserve the absolute figure are of the form x = a + px, a, b, c, p, q ∈ R, y = b + cx + qy, pq = 0, and form the 5-parametric group G 5 known as the group of similarities of the isotropic plane (see [1]).
Distances, spans and angles are kept invariant under the subgroup G 3 of G 5 which consists of transformations of the form G 3 is called the motion group of the isotropic plane. Metric quantities and all the facts related to the geometry of the isotropic plane can be found in [1,2].
A triangle is called allowable if none of its sides are isotropic [1]. As it is explained in [3], according to [1], to any allowable triangle in the isotropic plane there is exactly one circumscribed circle. The equation of this circle is of the form y = ux 2 + vx + w, u = 0. Choosing a suitable coordinate system and applying the group of similarities, we may assume that the equation of this circle is y = x 2 , and that the vertices of the allowable triangle ABC are A = (a, a 2 ), B = (b, b 2 ), C = (c, c 2 ), where a, b, and c are mutually different numbers. For convenience, we will frequently use abbreviations abc = p and ab + bc + ca = q. Choosing, without loss of generality, that a + b + c = 0, the diameter of the circle circumscribed to the triangle ABC, passing through its centroid G = a+b+c 3 , a 2 +b 2 +c 2 3 = 0, − 2 3 q , lies on the y-axis, while the x-axis is tangent to this circle at the endpoint of that diameter.
For each allowable triangle ABC, one can, in the described way, achieve that its circumscribed circle has the equation y = x 2 , and its vertices are of the form A = (a, a 2 ), B = (b, b 2 ), and C = (c, c 2 ), with a + b + c = 0. We shall say that such a triangle is in the standard position, or shorter, that the triangle ABC is the standard triangle. To prove geometric facts for allowable triangles, it is sufficient to give a proof for a standard triangle. Its sides BC, CA, and AB have equations y = −ax − bc, y = −bx − ca, and y = −cx − ab. Using the mentioned notations it can be proved that q = bc − a 2 and (c − a)(a − b) = 2q − 3bc.
The tangential triangle of a given triangle ABC is the triangle A t B t C t determined by the three tangents to the circumscribed circle of the triangle ABC at its vertices. It can be proved that the lines AA t , BB t , and CC t are symmetric, with respect to bisectors of the angles A, B, and C, to the medians AG, BG, and CG of the triangle ABC. The lines AA t , BB t , and CC t meet at the point K which is called the symmedian center of the triangle ABC.
Let ABC be a standard triangle and let A, B, and C, respectively A , B , and C , be lines through the points A, B, and C such that ∠(AB, A) = ∠(BC, B) = ∠(CA, C) =: ϕ, ∠(A , AC) = ∠(B , BA) = ∠(C , CB) =: ψ. In [4] it is proved that the lines A, B, and C pass through a common point, say Ω 1 , if and only if ϕ = ω, and the lines A , B , and C pass through a common point, say Ω 2 , if and only if ψ = ω, where ω is given by . The points Ω 1 and Ω 2 are called Crelle-Brocard points, and ω is called the Brocard angle.
The isotropic analogue of Brocard's theorem was first obtained in [2]. The standard triangle ABC has, by [4], the symmedian center K and Crelle-Brocard points Ω 1 and Ω 2 given by where p 1 = 1 3 (bc 2 + ca 2 + ab 2 ), p 2 = 1 3 (b 2 c + c 2 a + a 2 b). One can prove that p 1 + p 2 + p = 0, p 2 1 + p 1 p 2 + p 2 2 = − q 3 9 , p 2 + pp 1 + p 2 1 = − q 3 9 , and p 2 + pp 2 + p 2 2 = − q 3 9 . These three points lie, according to [5], on the Brocard circle of the triangle ABC (see Figure 1), given by Triangles A B C are the so-called Kiepert triangles of the triangle ABC, and the line T is the axis of homology of the triangle ABC and the corresponding Kiepert triangle A B C . Axes of homology of an allowable triangle ABC and its Kiepert triangles envelope a parabola which is called the Kiepert parabola [6].
The inscribed and the circumscribed Steiner's ellipses of an allowable triangle have the same nonisotropic axis, which passes through the centroid G of that triangle and which in the case of a standard triangle has equation y = − 3p 2q x − 2 3 q. This axis is called the Steiner's axis of the considered triangle. In [7], the Steiner point of the allowable triangle ABC is defined as the fourth (the first three being A, B, and C) common point S of the circumscribed circle and the circumscribed Steiner ellipse of that triangle. If ABC is a

The First Brocard Triangle of a Triangle in the Isotropic Plane
In this section we will define the first Brocard triangle of a triangle in the isotropic plane.
Theorem 1. Given a standard triangle ABC, the lines through its symmedian center K and parallel to its sides BC, CA, and AB meet the Brocard circle, besides the point K, at points which lie on bisectors of the sides BC, CA, and AB, respectively (see Figure 1).
Proof. The point A 1 obviously lies on the bisector of BC, and with x = − a 2 , from (2) and we get the ordinate of the point A 1 . Therefore, A 1 is the intersection of that line and the Brocard circle (2). This line is parallel to BC and passes through K, see (1).
The points A 1 , B 1 , and C 1 from Theorem 1 determine the first Brocard triangle of the triangle ABC (see Figure 1).

Theorem 2.
The sides of the first Brocard triangle A 1 B 1 C 1 of the standard triangle ABC are given by Proof. The point B 1 satisfies the first equation in (5) because and so does the point C 1 .

Theorem 3.
A triangle and its first Brocard triangle have the same centroid (see Figure 1).
Proof. According to [3], the triangle ABC has the centroid G = 0, − 2 3 q . The triangle A 1 B 1 C 1 with vertices (3), has the same centroid because a + b + c = 0 and

The First Brocard Triangle and Some Other Significant Elements
In this section we consider the relationships between the first Brocard triangle and some other objects related to a triangle in the isotropic plane. Theorem 4. Let G be the centroid and A 1 B 1 C 1 the first Brocard triangle of an allowable triangle ABC. Then the pairs of lines GA, GA 1 ; GB, GB 1 ; and GC, GC 1 have the same bisector. This bisector is the Steiner axis of the triangle ABC ( Figure 1).
Proof. The lines GA and GA 1 have slopes The sum of slopes is equal to − 3p q , therefore the bisector of these lines has the slope − 3p 2q . The Steiner axis of the triangle ABC is given by , which passes through the centroid G = 0, − 2 3 q , and coincides with this bisector.
Theorem 5. The first Brocard triangle A 1 B 1 C 1 of an allowable triangle ABC is homologous with this triangle, and the center of homology is K -the point reciprocal to the symmedian center of the triangle ABC (see Figure 2). For the Euclidean case see [9]. Proof. The line passes through the point A = (a, a 2 ) and also through the point A 1 , hence it is the line AA 1 .
In addition, this line passes through the point which is, by [5], reciprocal to the symmedian center K and anticomplementary to the midpoint of Crelle-Brocard points Ω 1 and Ω 2 . The same also holds for analogous lines BB 1 and CC 1 .

Theorem 6.
In case of the standard triangle ABC, the axis of homology of triangles ABC and A 1 B 1 C 1 from Theorem 5, has the equation Proof. It is enough to prove e.g., that the point q 6a , − q 6 − bc lies on lines BC and B 1 C 1 , and on the line defined by (7).
In the discussion following Theorem 2 in [6] it is shown that the Kiepert triangle A B C of a triangle ABC, with t = −ω, coincides with the first Brocard triangle A 1 B 1 C 1 of the triangle ABC and the axis of homology of triangles ABC and A 1 B 1 C 1 touches the Kiepert parabola of the triangle ABC (see Figure 2). Theorem 7. The triangle ABC and its first Brocard triangle A 1 B 1 C 1 are three-homologous, and the centers of homologies are the point K and Crelle-Brocard points Ω 1 and Ω 2 of that triangle. Triangles ABC and K Ω 1 Ω 2 have the same centroid G [9] (see Figure 2).
Proof. According to [4], the lines AΩ 1 and AΩ 2 have equations The point C 1 lies on the first line, and the point B 1 lies on the second one because Therefore, points Ω 1 and Ω 2 lie on lines AC 1 and AB 1 , respectively, the point Ω 1 lies on lines BA 1 and CB 1 , and the point Ω 2 lies on lines BC 1 and CA 1 . Points Ω 1 , Ω 2 and K have the centroid G = 0, − 2 3 q because 1 3 The line AS has the slope a − 3p q , where S is the Steiner point of the triangle ABC [7]. AS B 1 C 1 because the line B 1 C 1 has the same slope as AS. Similarly, we get BS C 1 A 1 and CS A 1 B 1 . Because of Theorem 1, we have A 1 K BC, B 1 K CA, and C 1 K AB. Therefore, triangles ABC and A 1 B 1 C 1 have the property that lines through the vertices of the first triangle parallel to the corresponding sides of the second triangle pass through a common point, and lines through the vertices of the second triangle parallel to the corresponding sides of the first triangle pass through another common point. These two triangles are called parallelogic, and the two mentioned points are the centers of parallelogy of these triangles. So we have the following theorem.

Theorem 8.
A triangle and its first Brocard triangle are parallelogic with the centers of parallelogy at the Steiner point and at the symmedian center of this triangle (see Figure 3). Theorem 9. Let A 1 B 1 C 1 be the first Brocard triangle of an allowable triangle ABC. Then the lines parallel to B 1 C 1 , C 1 A 1 , and A 1 B 1 through C, A, and B, through B, C, and A, pass through a common point S 1 , respectively S 2 . In addition, the lines parallel to BC, CA, and AB through C 1 , A 1 , and B 1 , respectively B 1 , C 1 , and A 1 , pass through a common point K 1 , respectively K 2 (see Figure 3). In the case of the standard triangle ABC these points are given by Proof. According to (5), the slope of the line B 1 C 1 is a − 3p q , and the line through C parallel to it is given by y − c 2 = a − 3p q (x − c), i.e., qy = (aq − 3p)x + (c 2 − ca)q + 3cp. The point S 1 lies on this line because Substitutions B ↔ C and b ↔ c imply the substitution p 1 → p 2 , and using this substitution, the previous proof shows that the line through B, parallel to B 1 C 1 , passes through S 2 . Cyclic permutations a → b → c → a imply C 1 A 1 AS 1 CS 2 , and A 1 B 1 BS 1 AS 2 . The line y = −ax + 1 6q (9cp + 3c 2 q − 3caq − 2q 2 ) is parallel to BC, and it passes through C 1 . Let us show that this line also passes through K 1 , i.e., that − 5 6 q = −a · 3p 1 2q + 1 6q (9cp + 3c 2 q − 3caq − 2q 2 ), i.e., 3cp − 3ap 1 + c 2 q − caq + q 2 = 0, which, because q + c 2 = ab and p = abc, after dividing by a, becomes equivalent to However, we have Other statements of the theorem are proved in a similar way.
Using the obvious term three-parallelogy, statements of Theorems 8 and 9 can be briefly summarized as the following corollary. Corollary 1. The allowable triangle ABC and its first Brocard triangle A 1 B 1 C 1 are three-parallelogic, and in the case of a standard triangle ABC, centers of parallelogy are the Steiner point S of the triangle ABC and points S 1 , S 2 , and the symmedian center K of the triangle ABC and points K 1 , K 2 (see Figure 3).
In the Euclidean case, the statement about three-parallelogy, without proof, can be found in [10]. Theorem 10. Points S, S 1 , and S 2 from Theorem 9 lie on the circumscribed Steiner ellipse of the triangle ABC (see Figure 3). [7]. For the point S 1 = (x, y) we have

Proof. The statement for Steiner point S is proved in
and this point lies on the circumscribed Steiner ellipse of the triangle ABC, which, by [7], has the equation q 2 x 2 − 9pxy − 3qy 2 − 6pqx − 4q 2 y + 9p 2 = 0. Analogous proof is for the point S 2 .
Theorem 11. Points K 1 and K 2 from Theorems 9 are reciprocal to Crelle-Brocard points Ω 1 and Ω 2 of that triangle (see Figure 3). In the Euclidean case this statement, without proof, can be found in [11].
Proof. For the point K 1 = (x, y) we get and, its reciprocal point has coordinates which is the point Ω 1 . The proof for K 2 is analogous.
Let us consider the indirect similarity given by and vice versa The similarity (10) obviously maps the point A = (a, a 2 ) to the point A 1 , and points B and C to points B 1 and C 1 . This similarity maps the Steiner point S, which is, according to [7], given by S = (− 3p q , 9p 2 q 2 ), to the point with abscissa 3p 2q and ordinate equal to is under the similarity (10) mapped onto itself. The circumscribed circle and the circumscribed Steiner ellipse of the triangle ABC have, according to [3,7], equations y = x 2 and q 2 x 2 − 9pxy − 3qy 2 − 6pqx − 4q 2 y + 9p 2 = 0, which, after substitutions (11), become 6p q x + 2y + 2 3 q = 4x 2 and After rearrangements and replacements x x, y y this can be written as which are, according to [5,7], equations of the Brocard circle and the inscribed Steiner ellipse of the triangle ABC. Let us find the equation of the fixed line, i.e., of the axis of similarity (10). The transformation x → −2x, y → 6p q x + 2y + 2 3 q maps the line to the line 6p i.e., to the line which coincides with the line (12) if and only if Therefore, and the required axis is the line which is, according to [7], the Steiner axis of the triangle ABC. The last result is in accordance with Theorem 4.
Thus, we have proved: Theorem 12. An allowable triangle ABC and its first Brocard triangle A 1 B 1 C 1 are indirectly similar. This similarity has the center at the common centroid of these two triangles, its coefficient equals − 1 2 , its axis is the Steiner axis of the triangle ABC, and it maps the points S, S 1 , and S 2 from Theorem 9 to points K, K 1 , and K 2 from the same theorem.

Corollary 2.
The symmedian center of the allowable triangle ABC is the Steiner point of its Brocard triangle A 1 B 1 C 1 .
For the Euclidean case see [9].  For the Euclidean case see [12].
Theorem 13. An allowable triangle is homologous with the complementary triangle of its first Brocard triangle.

Proof. Points B 1 and C 1 have the midpoint
The line 9aqy = (9b 2 c 2 + 6bcq − 5q 2 )x + 3pq − 9bcp − 4aq 2 passes through A = (a, a 2 ) and A 1 because which is the line AA 1m . This line also passes through the point 3p 5q , 1 Theorem 14. If A 1 B 1 C 1 is the first Brocard triangle of the allowable triangle ABC, then its Crelle-Brocard points Ω 1 and Ω 2 divide in equal proportions the pairs of segments CB 1 , BC 1 ; AC 1 , CA 1 ; BA 1 , AB 1 . For the Euclidean case, without proof, see [10].
Theorem 16. If P is the center of homology of the triangle ABC and its first Brocard triangle, and if D, E, and F are points symmetrical to P with respect to the midpoints of sides BC, CA, and AB, then the triangles ABC and DEF are symmetrical with respect to the midpoint S of the Crelle-Brocard points Ω 1 and Ω 2 .
For the Euclidean case see [13].

Conclusions
In this paper we introduce the first Brocard triangle of a triangle in the isotropic plane, and study its connections with some other objects related to the given triangle. Some of the most important statements that we prove are the following: the allowable triangle ABC and its first Brocard triangle are homologous, where the centers of homologies are the point reciprocal to the symmedian center of the triangle and the Crelle-Brocard points of that triangle; the allowable triangle ABC and its first Brocard triangle are threeparallelogic and indirectly similar, and the allowable triangle is homologous with the complementary triangle of its first Brocard triangle. The analytical approach used in this paper was introduced and developed in [3]. The obtained results are compared with similar results in the Euclidean plane.