Dear Fred, Darij and Antreas

> Dear Jean-Pierre, Antreas and Darij

> > [FL]

> > > Let ABC be a triangle, P a point, and (Oa, Ob, Oc)

> > > the Circumcenters of the triangles PBC, PCA, PAB, resp.

> > >

> > > The locus of P such that the perpendiculars

> > > from A,B,C to POa, POb, POc are concurrent is:

> > > infinity line*circumcircle*KjP cubic.

> > [JPE]

> > Consider three lines La, Lb, Lc going respectively through A, B,

C

> > and such as (La,POa) = (Lb,POb) = (Lc,POc) = phi where (L1,L2)

is

> > the measure modulo Pi of the oriented angle of lines L1, L2.

> > Then La, Lb, Lc concur if and only if

> > (AP,BC)+(BP,CA)+(CP,AB) = Pi/2-phi

> > Thus, we get the MacCay cubic for phi = 0, the Kjp cubic for phi

=

> > Pi/2 and, for any theta, a member of the pencil generated by Kjp

and

> > MacCay.

> > Friendly. Jean-Pierre

> [FL]

> That's fine.

> A link between nK0 and pK.

>

> I have the pencil MacCay -4s tan(phi) Kjp.

>

> This cubic is an isocubic iff phi = 0 or phi = Pi/2

>

> Remark: MacCay and Kjp are isogonal cubic but this cubic is not

> (if phi non 0 or Pi/2).

>

> Question (probably difficult) to JPE: Locus of perspector?

This is the same cubic with -phi instead of phi - ie the isogonal

conjugate of the locus of P -.

Of course, in the case of Mac Cay or Kjp, the locus of the common

point of La, Lb, Lc and the locus of P are the same ones.

Friendly. Jean-Pierre