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Geometric property of certain lines with respect to a given triangle
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.[1][2]
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 {\displaystyle 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 ) {\displaystyle 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.[2][3][4]
Central lines as trilinear polars[edit]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 ) {\displaystyle X=u(a,b,c):v(a,b,c):w(a,b,c)} be a triangle center. The line whose equation is x u ( a , b , c ) + y v ( a , b , c ) + z w ( a , b , c ) = 0 {\displaystyle {\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.[2][5] Also the point Y = 1 u ( a , b , c ) : 1 v ( a , b , c ) : 1 w ( a , b , c ) {\displaystyle 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 {\displaystyle 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 ) . {\displaystyle f(a,b,c):g(a,b,c):h(a,b,c).}
Construction of central lines[edit]Let X be any triangle center of △ABC.
Let Xn be the nth triangle center in Clark Kimberling's Encyclopedia of Triangle Centers. The central line associated with Xn is denoted by Ln. Some of the named central lines are given below.
Antiorthic axis as the axis of perspectivity of △ABC and its excentral triangle. Central line associated with X1, the incenter: Antiorthic axis[edit]The central line associated with the incenter X1 = 1 : 1 : 1 (also denoted by I) is x + y + z = 0. {\displaystyle x+y+z=0.} This line is the antiorthic axis of △ABC.[6]
The trilinear coordinates of the centroid X2 (also denoted by G) of △ABC are: 1 a : 1 b : 1 c {\displaystyle {\frac {1}{a}}:{\frac {1}{b}}:{\frac {1}{c}}} So the central line associated with the centroid is the line whose trilinear equation is x a + y b + z c = 0. {\displaystyle {\frac {x}{a}}+{\frac {y}{b}}+{\frac {z}{c}}=0.} This line is the Lemoine axis, also called the Lemoine line, of △ABC.
The trilinear coordinates of the circumcenter X3 (also denoted by O) of △ABC are: cos A : cos B : cos C {\displaystyle \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. {\displaystyle x\cos A+y\cos B+z\cos C=0.} This line is the orthic axis of △ABC.[8]
The trilinear coordinates of the orthocenter X4 (also denoted by H) of △ABC are: sec A : sec B : sec C {\displaystyle \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. {\displaystyle x\sec A+y\sec B+z\sec C=0.}
The trilinear coordinates of the nine-point center X5 (also denoted by N) of △ABC are:[9] cos ( B − C ) : cos ( C − A ) : cos ( A − B ) . {\displaystyle \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. {\displaystyle x\cos(B-C)+y\cos(C-A)+z\cos(A-B)=0.}
The trilinear coordinates of the symmedian point X6 (also denoted by K) of △ABC are: a : b : c {\displaystyle a:b:c} So the central line associated with the symmedian point is the line whose trilinear equation is a x + b y + c z = 0. {\displaystyle ax+by+cz=0.}
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 2 A sin ( B − C ) + y sin 2 B sin ( C − A ) + z sin 2 C sin ( A − B ) = 0. {\displaystyle 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 X647.
The Nagel line of △ABC 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 x a ( b − c ) + y b ( c − a ) + z c ( a − b ) = 0. {\displaystyle xa(b-c)+yb(c-a)+zc(a-b)=0.} This is the central line associated with the triangle center X649.
The Brocard axis of △ABC 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. {\displaystyle x\sin(B-C)+y\sin(C-A)+z\sin(A-B)=0.} This is the central line associated with the triangle center X523.
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