# Gob’s Circles of a Triangle in the Isotropic Plane

Authors: Ružica Kolar-Šuper
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we consider Gob’s circles of an allowable triangle in the isotropic plane and derive their equations in the case of a standard triangle. We prove that the potential axes of the circumscribed circle and Gob’s circles are the angle bisectors of the given triangle. We investigate some other interesting properties of Gob’s circles and some other circles related to the given triangle.


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## Body

## 1. Introduction

LetABC be a given triangle in Euclidean plane. Denote by Ab the point on the half-line BA and by Ac the point on the half-line CA such that BAb=CAc=BC. In a similar way, construct the points Bc on the half-line CB and Ba on the half-line AB such that CBc=ABa=CA, and Ca on the half-line AC and Cb on the half-line BC such that ACa=BCb=AB. The three circles AAbAc, BBcBa, and CCaCb are the so-called Gob's circles of the triangle ABC [1, 2]. The radii of these circles are equal to OI, where O is the center of the circumscribed circle and I is the center of the inscribed circle of the triangle ABC. The tangents at points A, B, and C to these three circles meet at one point on the circumscribed circle of the triangle ABC, and lines AbAc, BcBa, and CaCb are perpendicular to the line OI. These and some other properties of Gob’s circles in Euclidean plane have their analogues in the isotropic plane, and that is what we will study here.The isotropic plane is the projective-metric plane, where the absolute consists of a line, the absolute lineω, and a point on that line, the absolute point Ω. Lines through the point Ω are isotropic lines, and points on the line ω are isotropic points.Thedistance between two points Pi=xi,yii=1,2 in the isotropic plane is defined by dP1,P2=x2−x1, and if x1=x2, we say that P1 and P2 are parallel. For two parallel points P1 and P2, we define their span by sP1,P2=y2−y1. The angle between two lines y=kix+lii=1,2 is k2−k1, and if k1=k2, we say that they are parallel. Any isotropic line is perpendicular to any nonisotropic line. Facts about the isotropic plane can be found in [3, 4].We say that a triangle isallowable if none of its sides is isotropic. If we choose the coordinate system in such a way, the circumscribed circle of an allowable triangle ABC has the equation y=x2, and therefore, its vertices are the points A=a,a2,B=b,b2, and C=c,c2, while a+b+c=0, and we say that the triangle ABC is in standard position or shorter triangle ABC is a standard triangle. Its sides BC, CA, and AB have equations y=−ax−bc,y=−bx−ca, and y=−cx−ab. In order to prove geometric facts for any allowable triangle, it suffices to prove them for a standard triangle [5].Denotingq=bc+ca+ab, some equalities have been proved in [5], e.g., a2−bc=b2−ca=c2−ab=−q and a2+ab+b2=b2+bc+c2=c2+ca+a2=−q.


## 2. Main Results

LetBa, Ca; Cb, Ab; and Ac, Bc be pairs of points on the lines BC, CA, and AB, respectively, such that(1)dAc,B=dC,Ab=dB,C,dBa,C=dA,Bc=dC,A,dCb,A=dB,Ca=dA,B.The circlesGa, Gb, and Gc through three points A, Ab, Ac; B, Bc, Ba; and C, Ca, Cb, respectively, will be called, by analogy with the Euclidean case, Gob’s circles of the allowable triangle ABC (Figure 1).Figure 1
Gob’s circlesGa, Gb, and Gc of an allowable triangle ABC. Visualization of the statement of Theorem 4.Theorem 1.
Gob’s circlesGa, Gb, and Gc of a standard triangle ABC have equations(2)Ga…3y=x2+ax+a2,Gb…3y=x2+bx+b2,Gc…3y=x2+cx+c2.Proof.
LetAb and Ac be points on the lines CA and AB with abscissae 2c−b and 2b−c, respectively. Then,(3)dAc,B=b−2b−c=c−b=dB,C,dC,Ab=2c−b−c=c−b=dB,C.
From equationy=−bx−ca of the line CA, the ordinate of the point Ab turns out to be(4)y=−b2c−b−ca=−q−2bc,and similarly, the point Ac has the same ordinate −q−2bc; i.e., we have(5)Ab=2c−b,−q−2bc,Ac=2b−c,−q−2bc.
The pointA=a,a2 lies on the circle Ga and, by (2), the points Ab and Ac from (5) obviously lie on this circle because, for the former one, we get(6)2c−b2−b+c2c−b+b+c2=3b2−bc+c2=3−q−2bc.In the isotropic plane, the circle with the equation y=ux2+vx+w has radius equal to 1/2u. Therefore, the Gob's circles, equation (2), have equal radii.Theorem 2.
The potential axis of the circumscribed circleK of an allowable triangle ABC, and Gob’s circle Ga is the angle bisector of the angle A. The other intersection of these two circles is the intersection of the same bisector with the perpendicular bisector of the side BC. Analogous statements hold for K and circles Gb and Gc (Figure 2).Figure 2
Visualization of statements of theorems2 and 3.Proof.
From the equationy=x2 of the circumscribed circle K and the equation (2) of the circle Ga, eliminating x2, we get 2y=ax+a2, which is, by [6], the equation of the angle bisector of the angle A. The point −a/2,a2/4 lies on this bisector, and it also lies on the bisector x=−a/2 of the side BC. It also lies on the circles K and Ga.Theorem 3.
The tangents of Gob’s circles of an allowable triangle at its vertices pass through the dual Feuerbach point of that triangle (Figure2).Proof.
The liney=ax passes through the point Φ′=0,0 which is, by [7], the dual Feuerbach point of the triangle ABC. Inserting y=ax into the first equation (2), we get the equation x2−2ax+a2=0 with the double solution x=a, and this line touches the circle Ga at the point A.Theorem 4.
The intersections of two Gob’s circles of an allowable triangle are the intersections of its orthic line with the corresponding altitudes (Figure1).Proof.
From the second equation (2), for x=a,(7)3y=a2+ab+b2=−q,i.e., y=−q/3, and the same result follows from the third equation (2). So, the circles Gb and Gc intersect at the point a,−q/3, which lies on lines x=a and y=−q/3. The former one is the altitude of the triangle ABC through the vertex A, and the latter one is, by [5], the orthic line ℋ of that triangle.
There are some more interesting circles of a triangle which are related to Gob’s circles.
From the equationy=−ax−bc of the line BC, for x=a, we obtain y=−a2−bc=q−2bc, so the foot of the altitude from the vertex A is the first one of three points:(8)Ah=a,q−2bc,Bh=b,q−2ca,Ch=c,q−2ab,and the other two are feet of altitudes from vertices B and C.Theorem 5.
IfAh, Bh, and Ch are feet of altitudes from vertices A, B, and C of a standard triangle ABC, then the circles ℋa, ℋb, and ℋc through three points A, Bh, Ch; B, Ch, Ah; and C, Ah, Bh, respectively, (Figure 3) have equations(9)ℋa…y=−x2+ax+a2,ℋb…y=−x2+bx+b2,ℋc…y=−x2+cx+c2.Figure 3
Visualization of statements of Theorems5–7Proof.
The pointA=a,a2 obviously lies on the circle ℋa from (9), and so do also points Bh and Ch from (8) because, e.g., for Bh, we get(10)−b2+ab+a2=−b2−ca=q−2ca.Theorem 6.
Using the notation from Theorems1 and 5, the circumscribed circle of the triangle ABC belongs to three pencils of circles determined by the circle pairs Ga, ℋa; Gb, ℋb; and Gc, ℋc, whose potential axes are the angle bisectors of angles A, B, and C (Figure 3).Proof.
By Theorem2, it is enough to prove, e.g., that the angle bisector of angle A having the equation 2y=ax+a2 is the potential axis of the circle ℋa from (9) and the circumscribed circle y=x2, and this is obtained by adding the equations of these two circles.
The circlesGa and ℋa pass through the point A and through points Ab, Ac and Bh, Ch, respectively; hence, it is interesting to consider circles through the point A and through points Bh, Ac and Ch, Ab. Let us denote these circles by ℐbc and ℐcb. There are four more analogous circles ℐca, ℐac, ℐab, and ℐba (Figure 3).Theorem 7.
In a standard triangleABC, the circles ℐbc, ℐcb, ℐca, ℐac, ℐab, and ℐba have the following equations:(11)ℐbc…y=x2−b−cx+ab−c,ℐcb…y=x2+b−cx−ab−c,ℐca…y=x2−c−ax+bc−a,ℐac…y=x2+c−ax−bc−a,ℐab…y=x2−a−bx+ca−b,ℐba…y=x2+a−bx−ca−b.Proof.
The circleℐbc with the first equation (11) obviously passes through the point A=a,a2, and it also passes through the points Bh from (8) and Ac from (5) (Figure 3) because we get(12)b2−b−cb+ab−c=bc−ca+ab=q−2ca,2b−c2−b−c2b−c+ab−c=2b2−bc−b2−c2=b2−bc+c2=−q−2bc.
The equation of the circleℐcb follows from the equation of ℐbc by substituting b↔c, and the remaining two equations (11) are obtained by cyclically permuting a⟶b⟶c⟶a.Theorem 8.
Using the notation from Theorem7, the midpoint of the line segment joining the intersections of any isotropic line with two circles ℐbc and ℐcb lies on the circle circumscribed to the triangle ABC; i.e., these two circles are inverse one another with respect to the circumscribed circle. The same holds true for the other two circle pairs: ℐca,ℐac and ℐab,ℐba.Proof.
The equationy=x2 of the circumscribed circle is the arithmetic mean of the first two equations (11). According to [8], the second statement of the theorem follows.Theorem 9.
Using the notation from Theorem7, the centroid of three points at which an isotropic line intersecting three circles ℐbc, ℐca, and ℐab lies on the circle circumscribed to the triangle ABC. The same is also true for circles ℐcb, ℐba, and ℐac.Proof.
The equationy=x2 is the arithmetic mean of the first, third, and fifth equations in (11).


## 3. Conclusion

In this work, Gob’s circles of a triangle in the isotropic plane are introduced, and their connections with some other well-known concepts related to the triangle in the isotropic plane are studied. Some of the most interesting statements are the following: the potential axes of the circumscribed circle and Gob’s circles are the angle bisectors of the given triangle, the tangents of Gob’s circles of a triangle at its vertices pass through the dual Feuerbach point of the triangle, and the intersections of two Gob’s circles of a triangle are the intersections of its orthic line with the corresponding altitudes. The approach used in this paper is introduced and developed in [5].



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