HYACINTHOS 6352

Floor en Lyanne van Lamoen wrote:

>
> Dear Darij and Jean-Pierre,
>
> > > [DG]:
> > > > > Prof. Clark Kimberling has updated ETC, as we have seen. But
> > there
> > > > > are still some points whose trilinears we don't know -
> > especially
> > > > I
> > > > > with my knowledge on trilinears which doesn't go further than
> > line
> > > > > equations. I am interested in the trilinears of the Schröder
> > > > point.
> > >
> > > [JPE]:
> > > > Your point is trilinear x = (b-c)^2 + a(b+c-2a)
> > > > Friendli. Jean-Pierre
> >
> > [FVL]
> > > It seems that there are more points P satisfying the
> > > condition that if A'B'C' is the pedal triangle of P, then the
> > > circumcircles of AA'P, BB'P and CC'P meet in a second point. The
> > > circumcenter seems to be one of these points. Perhaps the locus is
> > the
> > > Stammler hyperbola? I am now running out of time, later I will
> > invest
> > > more time.
>
> [JPE]:
> > This locus is the Darboux cubic. Friendly. Jean-Pierre

Now we may try to find some second intersections for given P:

We look for second points of intersection of the three circles (AA'P),
(BB'P), (CC'P) in barycentrics:

P=I gives a(2a^2-ab-ac-(b-c)^2):: see above.

P=O gives a^2(2a^2-b^2-c^2):: = X(187) = inverse of K in circumcircle.

P=L=X20 gives (-2a^2+b^2+c^2)/(-a^2+b^2+c^2):: .
Is this one in ETC?

P=X40=reflection of I through O gives a*(-2a+b+c)/(-a+b+c):: .
Is this one in ETC?

I don't think that P=H gives a well defined point (that is why I first
rejected the locus to be Darboux...).

Kind regards,
Sincerely,
Floor van Lamoen