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

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  • jpehrmfr
    Dear Paul and Bernard ... centroids of ... anticevian ... aP is ... known, ... intersections of  ... They are the common points of the circumconic with center
    Message 1 of 8 , 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
    • jpehrmfr
      Dear Paul and Bernard ... the ... 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
      Message 2 of 8 , Feb 2, 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).

        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.
        Friendly. Jean-Pierre
      • Bernard Gibert
        Dear Jean-Pierre and Paul, ... 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
        Message 3 of 8 , Feb 2, 2005
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          Dear Jean-Pierre and Paul,

          > > > >  [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.
          > >
          > [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.

          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).

          Best regards

          Bernard



          [Non-text portions of this message have been removed]
        • Paul Yiu
          * 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
          Message 4 of 8 , Feb 2, 2005
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            * 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
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