- Dear Paul and Bernard
> > [PY] Given a point P, can you find a point Q so that the

centroids of

> > the cevian triangle of Q is the same as the centroid of

anticevian

> > triangle of P?

aP is

> >

> > [BG]: There are at most three such points Q.

> > one of them is aP/G (cevian quotient or Ceva conjugate) where

> > the anticomplement of P.

known,

> > *** If there are at most three such points and one of them is

> > it seems that the other two can be described as the

intersections of

> > a conic and a line.

They are the common points of the circumconic with center P and the

trilinear polar of i.h.i(q) where q = aP/G, i = isotomic

conjugation, h = homothecy (G,1/4).

Friendly. Jean-Pierre - Dear Paul and Bernard

> > > [PY] Given a point P, can you find a point Q so that the

the

> centroids of

> > > the cevian triangle of Q is the same as the centroid of

> anticevian

> > > triangle of P?

> > >

> > > [BG]: There are at most three such points Q.

> > > one of them is aP/G (cevian quotient or Ceva conjugate) where

> aP is

> > > the anticomplement of P.

>

> > > *** If there are at most three such points and one of them is

> known,

> > > it seems that the other two can be described as the

> intersections of

> > > a conic and a line.

>

> They are the common points of the circumconic with center P and

> trilinear polar of i.h.i(q) where q = aP/G, i = isotomic

I just realize that the line above is homothetic of the trilinear

> conjugation, h = homothecy (G,1/4).

polar of q = aP/G in (G,4); this is, of course, an easier

characterization.

Friendly. Jean-Pierre - Dear Jean-Pierre and Paul,

> > > > [PY] Given a point P, can you find a point Q so that the

I have found the same thing.

> > centroids of

> > > > the cevian triangle of Q is the same as the centroid of

> > anticevian

> > > > triangle of P?

> > > >

> > > > [BG]: There are at most three such points Q.

> > > > one of them is aP/G (cevian quotient or Ceva conjugate) where

> > aP is

> > > > the anticomplement of P.

> >

> > > > *** If there are at most three such points and one of them is

> > known,

> > > > it seems that the other two can be described as the

> > intersections of

> > > > a conic and a line.

> >

> [JP] They are the common points of the circumconic with center P and

> the

> trilinear polar of i.h.i(q) where q = aP/G, i = isotomic

> conjugation, h = homothecy (G,1/4).

>

> I just realize that the line above is homothetic of the trilinear

> polar of q = aP/G in (G,4); this is, of course, an easier

> characterization.

the 3 points also lie on three (easy to draw) conics :

one of them passes through B, C, Ga (vertex of anticomplementary

triangle) and the reflections of A in the intersections of AB, AC with

the homothetic of BC under h(P,-1/2).

Best regards

Bernard

[Non-text portions of this message have been removed] - * was Steiner circumconic and X(1962)

Dear Jean-Pierre and Bernard,

[PY] Given a point P, can you find a point Q so that the centroids of

the cevian triangle of Q is the same as the centroid of anticevian

triangle of P?

[BG]: There are at most three such points Q. one of them is aP/G

(cevian quotient or Ceva conjugate) where aP is the anticomplement of P.

[PY]: If there are at most three such points and one of them is

known, it seems that the other two can be described as the

intersections of a conic and a line.

[JP] They are the common points of the circumconic with center P and

the ... homothetic of the trilinear polar of q = aP/G in (G,4).

[BG]: I have found the same thing. the 3 points also lie on three

(easy to draw) conics :

one of them passes through B, C, Ga (vertex of anticomplementary

triangle) and the reflections of A in the intersections of AB, AC with

the homothetic of BC under h(P,-1/2).

*** Thank you very much for these wonderful results.

Best regards

Sincerely

Paul