Generalization of the NK-Transform

1)   Bernard Gibert (- see the notes just above X(2081) in ETC -) described the transform, given for X=(u:v:w) (barycentrics) by

NK(X) = (a^2(c^2v^2+b^2w^2)u : b^2(a^2w^2+c^2u^2)v : c^2(b^2u^2+a^2v^2)w).

NK(X) is the pole of the line XX* with respect to the conic that passes through points A, B, C, X, and X* (X* denote the isogonal conjugate of X).


2)   The NK-Transform can be generalized to the NKG-transform, given for X=(p:q:r) and Y=(u:v:w) by

NKG(X,Y) = (p^2(r^2v^2+q^2w^2)u : q^2(p^2w^2+r^2u^2)v : r^2(q^2u^2+p^2v^2)w).

NKG(X,Y) is the pole of the line YY* with respect to the conic that passes through points A, B, C, Y, and Y* (Y*=X²×Y^•=(p^2vw:q^2wu:r^2uv) denote the barycentric product: (barycentric square of the X)×(isotomic conjugate of Y).
NK(P)=NKG(I,P), I the incenter of the triangle ABC.


3)   Another geometric property of NKG (X, Y):

Let X^a, X^b and X^c be the harmonic associates of X=(p:q:r), i.e,

X^a=(-p:q:r),    X^b=(p:-q:r)    and    X^c=(p:q:-r).

The triangle X^aX^bX^c is called the precevian triangle of X.

The polar lines of Y with respect to all the conics of the pencil of conics Φ, passing through X, X^a, X^b and X^c, are concurrent at Y*=X²×Y^• (which is the pole of Y in the pencil of conics).

The two fixed points of the involution on the line YY* generated by the pencil Φ are Y and Y*.
Let γ(Y) be the conic of the pencil Φ passing through X, X^a, X^b, X^c and Y, and let γ(Y*) be the conic of the pencil passing through X, X^a, X^b, X^c and Y*. The conics γ(Y) and γ(Y*) are tangent to YY*.

The eight tangents to the conics γ(Y) and γ(Y*) in X, X^a, X^b, X^c are tangent to a the same conic Γ, and the pole of the line YY* with respect to the conic Γ is NKG(X,Y).

In fact, the polar of Y respect to the conic Γ is the tangent in Y* to the circumconic passing through Y and Y*, and the polar of the Y* respect to Γ is the tangent in Y to the circumconic passing through Y and Y*.

Creado con GeoGebra     Angel Montesdeoca (02/08/2012)