Gob’s Circles of a Triangle in the Isotropic Plane
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.
Let be a given triangle in Euclidean plane. Denote by the point on the half-line and by the point on the half-line such that . In a similar way, construct the points on the half-line and on the half-line such that , and on the half-line and on the half-line such that . The three circles , , and are the so-called Gob's circles of the triangle [1, 2]. The radii of these circles are equal to , where is the center of the circumscribed circle and is the center of the inscribed circle of the triangle . The tangents at points , , and to these three circles meet at one point on the circumscribed circle of the triangle , and lines , , and are perpendicular to the line . 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 line , and a point on that line, the absolute point . Lines through the point are isotropic lines, and points on the line are isotropic points.
The distance between two points in the isotropic plane is defined by , and if , we say that and are parallel. For two parallel points and , we define their span by . The angle between two lines is , and if , 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 is allowable if none of its sides is isotropic. If we choose the coordinate system in such a way, the circumscribed circle of an allowable triangle has the equation , and therefore, its vertices are the points , and , while , and we say that the triangle is in standard position or shorter triangle is a standard triangle. Its sides , , and have equations , and . In order to prove geometric facts for any allowable triangle, it suffices to prove them for a standard triangle .
Denoting , some equalities have been proved in , e.g., and .
2. Main Results
Let , ; , ; and , be pairs of points on the lines , , and , respectively, such that
The circles , , and through three points , , ; , , ; and , , , respectively, will be called, by analogy with the Euclidean case, Gob’s circles of the allowable triangle (Figure 1).
Theorem 1. Gob’s circles , , and of a standard triangle have equations
Proof. Let and be points on the lines and with abscissae and , respectively. Then,From equation of the line , the ordinate of the point turns out to beand similarly, the point has the same ordinate ; i.e., we haveThe point lies on the circle and, by (2), the points and from (5) obviously lie on this circle because, for the former one, we getIn the isotropic plane, the circle with the equation has radius equal to 1/2u. Therefore, the Gob's circles, equation (2), have equal radii.
Theorem 2. The potential axis of the circumscribed circle of an allowable triangle , and Gob’s circle is the angle bisector of the angle . The other intersection of these two circles is the intersection of the same bisector with the perpendicular bisector of the side . Analogous statements hold for and circles and (Figure 2).
Proof. From the equation of the circumscribed circle and the equation (2) of the circle , eliminating , we get , which is, by , the equation of the angle bisector of the angle . The point lies on this bisector, and it also lies on the bisector of the side . It also lies on the circles and .
Theorem 3. The tangents of Gob’s circles of an allowable triangle at its vertices pass through the dual Feuerbach point of that triangle (Figure 2).
Proof. The line passes through the point which is, by , the dual Feuerbach point of the triangle . Inserting into the first equation (2), we get the equation with the double solution , and this line touches the circle at the point .
Theorem 4. The intersections of two Gob’s circles of an allowable triangle are the intersections of its orthic line with the corresponding altitudes (Figure 1).
Proof. From the second equation (2), for ,i.e., , and the same result follows from the third equation (2). So, the circles and intersect at the point , which lies on lines and . The former one is the altitude of the triangle through the vertex , and the latter one is, by , the orthic line of that triangle.
There are some more interesting circles of a triangle which are related to Gob’s circles.
From the equation of the line , for , we obtain , so the foot of the altitude from the vertex is the first one of three points:and the other two are feet of altitudes from vertices and .
Theorem 5. If , , and are feet of altitudes from vertices , , and of a standard triangle , then the circles , , and through three points , , ; , , ; and , , , respectively, (Figure 3) have equations
Theorem 6. Using the notation from Theorems 1 and 5, the circumscribed circle of the triangle belongs to three pencils of circles determined by the circle pairs , ; , ; and , , whose potential axes are the angle bisectors of angles , , and (Figure 3).
Proof. By Theorem 2, it is enough to prove, e.g., that the angle bisector of angle having the equation is the potential axis of the circle from (9) and the circumscribed circle , and this is obtained by adding the equations of these two circles.
The circles and pass through the point and through points , and , , respectively; hence, it is interesting to consider circles through the point and through points , and , . Let us denote these circles by and . There are four more analogous circles , , , and (Figure 3).
Theorem 7. In a standard triangle , the circles , , , , , and have the following equations:
Proof. The circle with the first equation (11) obviously passes through the point , and it also passes through the points from (8) and from (5) (Figure 3) because we getThe equation of the circle follows from the equation of by substituting , and the remaining two equations (11) are obtained by cyclically permuting .
Theorem 8. Using the notation from Theorem 7, the midpoint of the line segment joining the intersections of any isotropic line with two circles and lies on the circle circumscribed to the triangle ; 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: and .
Theorem 9. Using the notation from Theorem 7, the centroid of three points at which an isotropic line intersecting three circles , , and lies on the circle circumscribed to the triangle . The same is also true for circles , , and .
Proof. The equation is the arithmetic mean of the first, third, and fifth equations in (11).
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 .
No data were used to support this study.
Conflicts of Interest
The author declares that there are no conflicts of interest regarding the publication of this paper.
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