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11044Re: [EMHL] Steiner circumconic, X(1962)...

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  • jpehrmfr
    Feb 1, 2005
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      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?
      > >
      > > [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 the
      trilinear polar of i.h.i(q) where q = aP/G, i = isotomic
      conjugation, h = homothecy (G,1/4).
      Friendly. Jean-Pierre
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