LOCI OF CENTERS IN PENCILS OF TRIANGLES IN THE ISOTROPIC PLANE

. In this paper we consider a triangle pencil in an isotropic plane consisting of those triangles that have two fixed vertices, while the third vertex is moving along a line. We study the curves of centroids, Gergonne points, symmedian points, Brocard points and Feuerbach points for such a pencil of triangles.


Introduction
In [7] the authors considered a triangle pencil in an isotropic plane consisting of the triangles that have the same circumcircle. They studied the loci of their centroids, Gergonne points and symmedian points, while in [6] the loci of the first and second Brocards points where observed.
In this paper we will do a similar study for the triangles that have two fixed vertices and a vertex moving along a line. Furthermore, we will extend the study with the locus of Feuerbach points.
Let us start by recalling some basic definitions and facts about the isotropic plane. It is a real projective plane where the metric is induced by a real line f and a real point F incident with it. All lines through the absolute point F are called isotropic lines, and all points incident with the absolute line f are called isotropic points. Two lines are parallel if they are incident with the same isotropic point, and two points are parallel if they lie on the same isotropic line. In the affine model of the isotropic plane where the coordinates of points are defined by x = x 1 x 0 , y = x 2 x 0 , the absolute line has the equation x 0 = 0 and the absolute point has the coordinates (0, 0, 1). For two non-parallel points A = (x A , y A ) and B = (x B , y B ) the distance is defined by d(A, B) = x B − x A , and for two non-parallel lines p and q, given by the equations y = k p x + l p and y = k q x + l q , the angle is defined by ∠(p, q) = k q − k p (see [8], [9]). The midpoint of points A and B is given by 1 2 (y A + y B ) , while the bisector of lines p and q is given by the equation y = 1 2 (k p + k q )x + 1 2 (l p + l q ). A circle is defined as a conic touching the absolute line at the absolute point and it has an equation of the form y = ax 2 + bx + c, a, b, c ∈ R.

Pencil of triangles
A pencil consisting of the triangles that have two fixed vertices A, B and a vertex C moving along a line p is observed. Without loss of generality, by a suitable choice of coordinate system, we may assume that A and B have coordinates (−1, 0) and (1,0). We have to distinguish between two cases: (i) p is a non-isotropic line, (ii) p is an isotropic line. In every section we shall first focus on the case (i) and assume that p is given by the equation y = kx+l (with k, l ∈ R), and than we will give the results for the case (ii) when p is the isotropic line with the equation Let a triangle ABC with vertices be given. Its sides AB, BC, CA have the equations respectively. Therefore, the midpoints of its sides are respectively. This pencil contains four special (singular) triangles: when C coincides with the point parallel to A or B, when C is the intersection point of p and AB, and if C is the isotropic point of the line p.
In the case (ii) when the isotropic line p is given by the equation x = l, l ∈ R, the vertices of the triangle ABC have the coordinates and the sides have the equations

Loci of centroids and symmedian points
It follows from (2.1) and (2.3) that the three medians AA m , BB m and CC m of ABC in the case (i) are given by respectively, and they intersect at the point Instead of observing one triangle, we will observe the whole pencil of triangles. When C runs along the line p, the centroid X 2 runs along the curve k X2 which is already parametrized by (3.2), Figure 1. Its equation y = kx+ l 3 is obtained by eliminating c from equations Therefore, the following statement holds: The symmedians are the reflections of medians in the bisectors. Therefore, we can easily calculate their equations They intersect in the symmedian point of ABC By eliminating c from the equations of the locus k S of symmedian points is obtained, Figure 1. Thus, we have proved: In the isotropic plane the center of the conic is defined as the pole of the absolute line with respect to the conic, while the axis of the conic is defined as the polar of the absolute point (see [2]). Therefore, the conic k S has the center in the point (0, l 3 ) and the line with equation y = k 2 x + l 3 is its axis. Indeed, The center (0, l 3 ) of the locus k S obviously lies on the locus k X2 with the equation y = kx + l 3 . The intersection points of the conic with its axis are called foci of the conic (see [9, p. 72]). Thus, the foci of k S are points with coordinates Let us notice that the ellipse k S passes through the midpoint (0, 0) of the line segment AB. The tangent line t at that point, given by the equation y = k 2 x, is parallel to the axis of the ellipse. If the given line p passes through the midpoint (0, 0) of the line segment AB, i.e., l = 0, the locus of symmedian points k S becomes a singular conic with the equation (kx − 2y) 2 = 0, the line t with multiplicity two.
We shall now observe the special case of pencil of triangles having two fixed vertices A = (−1, 0), B = (1, 0), and a vertex C = (l, c) running along the isotropic line p, and prove: Proof. Let a triangle ABC with vertices (2.5) be given. Using similar method as in the case (ii) we get the centroid and the symmedian point to be respectively. Therefore, the curves k X2 and k S have equations x = l 3 and , respectively.

Locus of Gergonne points
In order to determine the Gergonne point of the triangle ABC in case (i), we should first calculate the equation of its incircle k i : It touches the sides BC, CA, AB at the points (4.2) respectively. The lines AA g , BB g , CC g having the equations respectively, intersect at one point, the Gergonne point G of the triangle ABC (see [3]) with coordinates When C moves along the line p, the Gergonne point G moves along the curve k G whose equation This leads to:  The cubic k G intersects the absolute line at a real point and a pair of conjugate imaginary points. To prove this fact, we switch to homogeneous coordinates by setting x = x1 x0 , y = x2 x0 . Now, (4.5) takes the form After inserting x 0 = 0 we get solutions x 2 = kx 1 and x 2 = 3k± √ 3li 6 x 1 , which correspond to the isotropic points (0, 1, k) and (0, 1, 3k± √ 3li 6 ). The line p touches k G at the isotropic point (0, 1, k).
After doing some elementary calculations and using the tools of differential geometry, we come to the conclusion that the cubic k G has a double point in the point Figure 2. The locus k G of Gergonne for the pencil of triangles with two fixed vertices A, B, and the third vertex C moving along the line p.
and that D is a node, an isolated double point or a cusp depending on whether l 2 − k 2 is greater than, less than, or equal to zero. If p passes through A (l = k), the cusp coincides with B, and if p passes through B (l = −k), the cusp coincides with A.
If the given line p passes through the midpoint (0, 0) of the line segment AB, i.e., l = 0, two isotropic conjugate imaginary points of k G coincide with the real isotropic point (0, 1, k 2 ), the isolated double point of k G .
In the special case of pencil of triangles having two fixed vertices A, B, and a vertex C running along the isotropic line p given by the equation x = l, l ∈ R, the following result is obtained:

Locus of Feuerbach points
In [1] and [10, pp. 109-115] the authors in different ways proved that in a triangle in the isotropic plane the midpoints of the sides and the feet of the altitudes lie on a circle, so-called Euler circle. They also proved that the incircle k i and Euler circle k e touch each other in a point which is called the Feuerbach point of the triangle ABC.
Let us now determine the equation of the Euler circle k e for the triangle with vertices (2.1) and midpoints (2.3). It can be easily checked that k e has the equation It follows from (4.1) and (5.1) that the Feuerbach point is It can be noticed that Φ is parallel to the centroid X 2 given by (3.2). By eliminating c from the equations  Proof. To determine the intersection points of the absolute line and k Φ , we can write (5.3) in homogenous coordinates as Since the absolute line is given by x 0 = 0 we get two solutions, a double solution x 1 = 0 and solution x 2 = 4 3 kx 1 . Therefore, the absolute point F = (0, 0, 1) is the intersection point with the intersection multiplicity 2, and M = (0, 1, 4 3 k) is the intersection point with the intersection multiplicity 1. The absolute point F can have intersection multiplicity 2 in two ways, either it is a double point of k Φ or it is a regular point in which k Φ touches the absolute line. We will prove that the first case is true and we will determine the tangents of k Φ at F . Every line through F has the equation of the form x = m, or in terms of homogeneous coordinates x 1 = mx 0 . It intersects the cubic (5.4) in the points whose coordinates (x 0 , x 1 , x 2 ) satisfy equation (5.5) x 2 0 4m 2 (3km + l)x 0 + (1 − 9m 2 )x 2 = 0. Therefore, x 0 = 0 is a double root and F is the intersection point counted twice. The solution x 0 = 0 will be a triple root and F , the intersection point, counts three times if, and only if, m = ± 1 3 . We can conclude that F is a node at which k Φ has tangents x = ± 1 3 . The absolute point takes the role of the Feuerbach point of two special triangles in the pencil when C is the point parallel to A or B, i.e., c = ∓1, respectively.
Let us now determine the tangent line of k Φ at its isotropic point M = (0, 1, 4 3 k). Every line through M has equation of the form y = 4 3 kx + n, i.e., x 2 = 4 3 kx 1 + nx 0 . It intersects the cubic (5.4) in the points whose coordinates satisfy equation When n = 4l 9 , x 0 = 0 is a double root. Thus, the line with equation y = 4 3 kx + 4l 9 is the tangent line at the isotropic point M . The curve k Φ intersects the fixed line AB in the points satisfying equations (5.3) and y = 0. From 4x 2 (3kx + l) = 0 we get intersection points C m = (0, 0) counted twice, and the point (− l 3k , 0). When C approaches the point in which p meets AB, the Feuerbach point of the triangle ABC approaches the obtained point (− l 3k , 0). In the cases when p passes through A or B, the cubic k Φ splits onto an isotropic line and a special hyperbola. To prove this fact, we will assume that p passes through A which is precisely when l = k. The equation ( In the case (ii) when the vertices of the triangle ABC have coordinates given by (2.5), using similar method as above, we get the Feuerbach point to be .
The locus of Feuerbach points is, therefore, given by the equation and the following theorem holds: Theorem 5.2. Let the points A and B and the isotropic line p be given. The curve of Feuerbach points of all triangles ABC such that C lies on p is an isotropic lines.
We can see that the isotropic lines k X2 and k ϕ coincide.

Loci of Brocard points
It was shown in [6] that for every triangle in the isotropic plane there exist the first and second Brocard point, and they are unique. The first Brocard point B 1 is defined as the point such that its connections with the vertices A, B, C form equal angles with the sides AB, BC, and CA, respectively. The angle h is called the first Brocard angle. Analogously, the second Brocard point B 2 is defined as the point such that its connection lines with the vertices A, B, C form equal angles with the sides AC, BA, and CB, respectively. The angle is called the second Brocard angle and equals −h. The concept of Brocard points was introduced in the isotropic plane as an analogue of the concept of Brocard points in Euclidean plane given in [4].
If the vertices of the triangle ABC are given by (2.1), the first Brocard point is and the second Brocard point is Indeed, some elementary calculations show that The expressions present the parametrization of the locus of the first Brocard points, an obviously rational quartic curve k B1 . By eliminating c we get an implicit equation of k B1 : Analogously, it can be shown that the equation of the locus of the second Brocard points is  Proof. We will prove the theorem for the curve of the first Brocard points. The curve k B1 given by (6.3) is obviously a curve of degree 4. Therefore, it intersects every line in 4 points. This particularly holds for the absolute line f . To study the behavior of k B1 at the absolute line, we need to switch to homogenous coordinates, similarly as we did in the proof of Theorem 5.1.
After inserting x 0 = 0 (intersection with the absolute line) into the equation of k B1 , we get (3k 2 + l 2 )x 4 1 = 0. Thus, x 1 = 0 is fourfold solution and F = (0, 0, 1) is the only intersection point of k B1 and f . This proves that k B1 is an entirely circular quartic. Detailed studies of circular quartics in the isotropic plane were given in [5].
It can be easily checked that every line through A, i.e., the line with the equation of the form y = mx + m, intersects k B1 in A counted two times. Indeed, after inserting y = mx + m into (6.3), we get This leads to the conclusion that x = −1 is a double solution. It is a triple solution precisely when m = l−k 2 . Thus, the tangent line at the cusp A is given by y = l−k 2 x + l−k 2 . The line through B parallel to p has the equation y = kx − k. From (6.3) we get Thus, B(1, 0) is the intersection point with intersection multiplicity 2. We conclude that the tangent line of k B1 at B is given by y = kx − k.
It follows that the line p touches k B1 in the point (1, k + l) parallel to B. This completes the proof.
Let us also notice that in the four special cases when C becomes parallel to A, when C becomes parallel to B, when C becomes the intersection point of p and AB, and when C becomes an isotropic point, the first Brocard point of ABC is the point A, the point parallel to B (the contact point of k B1 and p), the intersection point of k B1 and AB, and the point B, respectively.
We will now consider the cases when the loci of Brocard points degenerate. When p passes through B, i.e., l = −k, the equation (6.3) turns into 4(kx 2 − 2y − k) 2 = 0.
Thus, the curve k B1 of the first Brocard points degenerates onto a circle with multiplicity two. The line p is the tangent line of the circle at B. Similarly, if p passes through A, the curve k B2 of the second Brocard points degenerates onto a circle with multiplicity two.
At the and observe the special case of pencil of triangles having two fixed vertices A = (−1, 0), B = (1, 0), and a vertex C = (l, c) running along the isotropic line p given by the equation x = l, l ∈ R.
Theorem 6.2. Let the points A and B and the isotropic line p be given. The curves of Brocard points of all triangles ABC such that C lies on p are isotropic lines.