Darboux cubic

On Fri, 18 Aug 2000, Lang wrote:

> About inflection points:

> 1) O is an inflection point of the Darboux cubic, with inflectional tangent

> equal to OK (K = Lemoine point).

  Oh yes; of course I knew THAT one, which is obvious from the symmetry. I believe I proved once that the other 8 require the solution of an octic that's irreducible in general.

> 2) The isogonal inverse of OK is Kiepert hyperbola, so the cubic of Darboux

> is tangent to this hyperbola with triple contact

     [at H, of course]

 the other 3 points of intersection are A, B and C.

    This is a nice remark!

> E. Turriere proves the following property of the Darboux:

> "La cubique de Darboux est une solution du probleme suivant:

> Determiner une cubique circonscrite au triangle ABC ayant le centre O du

> cercle circonscrit comme point d'inflexion et du type x^2(m y-n z) + y^2(n

> z-l x) + z^2(l x - m y) = 0. " (normal coordinates)

    that last phrase just says "is an isogonal cubic", of course, so

this just says that the Darboux one is the only isogonal cubic for

which  O  is an inflection point.  But more generally, it's the only

such cubic through  O,  since an isogonal cubic is determined by any

point on it that's not a vertex or incenter of the triangle.

> 4) Thank to John for his explanations.

    You're very welcome!

   John Conway