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Cevian triangles similar to ABC

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  • Jean-Pierre.EHRMANN
    Dear Hyacinthists, If ABC is obtusangle and D the line isotomic conjugate of the circumcircle, D intersects the circumcircle in two points V and W ( W is the
    Message 1 of 32 , May 31, 2000
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      Dear Hyacinthists,
      If ABC is obtusangle and D the line isotomic conjugate of the circumcircle,
      D intersects the circumcircle in two points V and W ( W is the isotomic
      conjugate of V and the reflection of V w.r.t. the Euler line).
      V and W have a very nice property : their Cevian triangles are directly
      similar to ABC (the two centers of similitude are the common points of the
      circumcircle and the nine-points circle ); more over their Cevian triangles
      are congruent.

      Has any one some references about those points V and W?
      I think they should have a lot of other nice properties.

      Thank you for your answers.
      Friendly from France. Jean-Pierre
    • John Conway
      ... It seems to me that the conic through the traces of P and Q deserves a name! Let s call it [P,Q] for the moment, so that your theorem asserts that
      Message 32 of 32 , Jun 16, 2000
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        On Fri, 16 Jun 2000, Paul Yiu wrote [edited]:
        >
        > Let P be a fixed point, and L a given line with tripole L*.
        > Then P/L is the conic through the traces of P and L*.

        It seems to me that "the conic through the traces of P and Q"
        deserves a name! Let's call it [P,Q] for the moment, so that
        your theorem asserts that P/Q* = [P,Q], which is symmetric in
        P and Q, so you've proved that P/Q* = Q/P* - wonderful!

        This has, I think, some notational consequences. It's natural to
        use 1/P for the tripolar, in view of the coordinate form:

        tripolar of the point (X:Y:Z) is the line (1/X:1/Y:1/Z|).

        In this notation, your result becomes

        P/(1/Q) = Q/(1/P) is a symmetric "product" [P,Q]

        which might almost be expected! I wonder if there's any more
        "algebra" of this type?

        > Corollary:
        >
        > Let P be a fixed point. Then P/P* is the inscribed conic with
        > Brianchon point P. The center of the conic is the inferior of the
        > isotomic conjugate of P.

        I think your "Brianchon point" of an inconic is the same as my
        "perspector" of it? I used to use various names for these notable
        invariant points of inconics and circumconics until I noticed only
        last year the important fact that ABC is in perspective with its
        dual triangle with respect to ANY conic, not just in- and circum- ones,
        and so switched to the term "perspector" for all cases.

        Regards, John Conway
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