Central line (geometry)

In geometry, central lines are certain special straight lines that lie in the plane of a triangle. The special property that distinguishes a straight line as a central line is manifested via the equation of the line in trilinear coordinates. This special property is related to the concept of triangle center also. The concept of a central line was introduced by Clark Kimberling in a paper published in 1994.

Definition
Let $△ABC$ be a plane triangle and let $x : y : z$ be the trilinear coordinates of an arbitrary point in the plane of triangle $△ABC$.

A straight line in the plane of $△ABC$ whose equation in trilinear coordinates has the form $$f(a,b,c)\,x + g(a,b,c)\,y + h(a,b,c)\,z = 0$$ where the point with trilinear coordinates $$f(a,b,c) : g(a,b,c) : h(a,b,c)$$ is a triangle center, is a central line in the plane of $△ABC$ relative to $△ABC$.

Central lines as trilinear polars
The geometric relation between a central line and its associated triangle center can be expressed using the concepts of trilinear polars and isogonal conjugates.

Let $$X = u(a,b,c) : v(a,b,c) : w(a,b,c)$$ be a triangle center. The line whose equation is $$ \frac{x}{u (a,b,c)} + \frac{y}{v(a,b,c)} + \frac{z}{w(a,b,c)}  = 0$$ is the trilinear polar of the triangle center $X$. Also the point $$Y = \frac{1}{u(a,b,c)} : \frac{1}{v(a,b,c)} : \frac{1}{w(a,b,c)}$$ is the isogonal conjugate of the triangle center $X$.

Thus the central line given by the equation $$f(a,b,c)\,x + g(a,b,c)\,y + h(a,b,c)\,z = 0$$ is the trilinear polar of the isogonal conjugate of the triangle center $$f(a,b,c) : g(a,b,c) : h(a,b,c).$$

Construction of central lines
Let $X$ be any triangle center of $△ABC$.
 * Draw the lines $AX, BX, CX$ and their reflections in the internal bisectors of the angles at the vertices $A, B, C$ respectively.
 * The reflected lines are concurrent and the point of concurrence is the isogonal conjugate $Y$ of $X$.
 * Let the cevians $AY, BY, CY$ meet the opposite sidelines of $△ABC$ at $A', B', C'$ respectively. The triangle $△A'B'C'$ is the cevian triangle of $Y$.
 * The $△ABC$ and the cevian triangle $△A'B'C'$ are in perspective and let $DEF$ be the axis of perspectivity of the two triangles. The line $DEF$ is the trilinear polar of the point $Y$. $DEF$ is the central line associated with the triangle center $X$.

Some named central lines
Let $X_{n}$ be the $n$th triangle center in Clark Kimberling's Encyclopedia of Triangle Centers. The central line associated with $X_{n}$ is denoted by $L_{n}$. Some of the named central lines are given below.



Central line associated with X1, the incenter: Antiorthic axis
The central line associated with the incenter $△ABC$ (also denoted by $I$) is $$x + y + z = 0.$$ This line is the antiorthic axis of $X_{1} = 1 : 1 : 1$.


 * The isogonal conjugate of the incenter of $△ABC$ is the incenter itself. So the antiorthic axis, which is the central line associated with the incenter, is the axis of perspectivity of $△ABC$ and  its incentral triangle (the cevian triangle of the incenter of $△ABC$).
 * The antiorthic axis of $△ABC$ is the axis of perspectivity of $△ABC$ and the excentral triangle $△ABC$ of $△I_{1}I_{2}I_{3}$.
 * The triangle whose sidelines are externally tangent to the excircles of $△ABC$ is the extangents triangle of $△ABC$. $△ABC$ and its extangents triangle are in perspective and the axis of perspectivity is the antiorthic axis of $△ABC$.

Central line associated with X2, the centroid: Lemoine axis
The trilinear coordinates of the centroid $△ABC$ (also denoted by $G$) of $X_{2}$ are: $$\frac{1}{a} : \frac{1}{b} : \frac{1}{c}$$ So the central line associated with the centroid is the line whose trilinear equation is $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 0.$$ This line is the Lemoine axis, also called the Lemoine line, of $△ABC$.


 * The isogonal conjugate of the centroid $△ABC$ is the symmedian point $X_{2}$ (also denoted by $K$) having trilinear coordinates $X_{6}$. So the Lemoine axis of $a : b : c$ is the trilinear polar of the symmedian point of $△ABC$.
 * The tangential triangle of $△ABC$ is the triangle $△ABC$ formed by the tangents to the circumcircle of $△T_{A}T_{B}T_{C}$ at its vertices. $△ABC$ and its tangential triangle are in perspective and the axis of perspectivity is the Lemoine axis of $△ABC$.

Central line associated with X3, the circumcenter: Orthic axis
The trilinear coordinates of the circumcenter $△ABC$ (also denoted by $O$) of $X_{3}$ are: $$\cos A : \cos B : \cos C$$ So the central line associated with the circumcenter is the line whose trilinear equation is $$x \cos A + y \cos B + z \cos C = 0.$$ This line is the orthic axis of $△ABC$.


 * The isogonal conjugate of the circumcenter $△ABC$ is the orthocenter $X_{3}$ (also denoted by $H$) having trilinear coordinates $X_{4}$. So the orthic axis of $sec A : sec B : sec C$ is the trilinear polar of the orthocenter of $△ABC$. The orthic axis of $△ABC$ is the axis of perspectivity of $△ABC$ and its orthic triangle $△ABC$. It is also the radical axis of the triangle's circumcircle and nine-point-circle.

Central line associated with X4, the orthocenter


The trilinear coordinates of the orthocenter $△H_{A}H_{B}H_{C}$ (also denoted by $H$) of $X_{4}$ are: $$\sec A : \sec B : \sec C$$ So the central line associated with the circumcenter is the line whose trilinear equation is $$x \sec A + y \sec B + z \sec C = 0.$$


 * The isogonal conjugate of the orthocenter of a triangle is the circumcenter of the triangle. So the central line associated with the orthocenter is the trilinear polar of the circumcenter.

Central line associated with X5, the nine-point center


The trilinear coordinates of the nine-point center $△ABC$ (also denoted by $N$) of $X_{5}$ are: $$\cos(B-C) : \cos(C-A) : \cos(A-B).$$ So the central line associated with the nine-point center is the line whose trilinear equation is $$x \cos(B-C) + y \cos(C-A) + z \cos(A-B) = 0.$$


 * The isogonal conjugate of the nine-point center of $△ABC$ is the Kosnita point $△ABC$ of $X_{54}$.  So the central line associated with the nine-point center is the trilinear polar of the Kosnita point.
 * The Kosnita point is constructed as follows. Let $O$ be the circumcenter of $△ABC$. Let $O_{A}, O_{B}, O_{C}$ be the circumcenters of the triangles $△ABC$ respectively. The lines $AO_{A}, BO_{B}, CO_{C}$ are concurrent and the point of concurrence is the Kosnita point of $△BOC, △COA, △AOB$. The name is due to J Rigby.

Central line associated with X6, the symmedian point : Line at infinity


The trilinear coordinates of the symmedian point $△ABC$ (also denoted by $K$) of $X_{6}$ are: $$a : b : c$$ So the central line associated with the symmedian point is the line whose trilinear equation is $$ax + by + cz = 0.$$


 * This line is the line at infinity in the plane of $△ABC$.
 * The isogonal conjugate of the symmedian point of $△ABC$ is the centroid of $△ABC$. Hence the central line associated with the symmedian point is the trilinear polar of the centroid. This is the axis of perspectivity of the $△ABC$ and its medial triangle.

Euler line
The Euler line of $△ABC$ is the line passing through the centroid, the circumcenter, the orthocenter and the nine-point center of $△ABC$. The trilinear equation of the Euler line is $$x \sin 2A \sin(B-C) + y \sin 2B \sin(C-A) + z \sin 2C \sin(A-B) = 0.$$ This is the central line associated with the triangle center $△ABC$.

Nagel line
The Nagel line of $X_{647}$ is the line passing through the centroid, the incenter, the Spieker center and the Nagel point of $△ABC$. The trilinear equation of the Nagel line is $$xa(b-c) + yb(c-a) + zc(a-b) = 0.$$ This is the central line associated with the triangle center $△ABC$.

Brocard axis
The Brocard axis of $X_{649}$ is the line through the circumcenter and the symmedian point of $△ABC$. Its trilinear equation is $$x \sin(B-C) + y \sin(C-A) + z \sin(A-B) = 0.$$ This is the central line associated with the triangle center $△ABC$.