ADGEOM422

  Let ABC be a triangle, P a point, K the symmedian point, K*P= cevapoint of K and P, gP = isogonal conjugate of P, agP = anticomplement of gP, (K*P) x agP= barycentric product of (K*P) and agP.

(notations in Bernard Gibert:
http://bernard.gibert.pagesperso-orange.fr/files/isocubics.html )

The point (K*P) x agP is on the line at infinity if and only if P is on the quartic of the equation barycentric:

-b^2 c^4 x^3 y - a^2 c^4 x^2 y^2 - b^2 c^4 x^2 y^2 + c^6 x^2 y^2 -
a^2 c^4 x y^3 - b^4 c^2 x^3 z - a^2 b^2 c^2 x^2 y z -
a^2 b^2 c^2 x y^2 z - a^4 c^2 y^3 z - a^2 b^4 x^2 z^2 + b^6 x^2 z^2 -
b^4 c^2 x^2 z^2 - a^2 b^2 c^2 x y z^2 + a^6 y^2 z^2 -
a^4 b^2 y^2 z^2 - a^4 c^2 y^2 z^2 - a^2 b^4 x z^3 - a^4 b^2 y z^3=0.


This quartic passes through the vertices of the reference triangle and the tangential triangle.

Angel Montesdeoca, 2 Agosto 2013, Creado con GeoGebra