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Re: [EMHL] curve related to Darboux quintic

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  • rhutson2
    Thanks Bernard! I forgot to include X(1824) in list of centers on the curve. It appears this curve is the Zosma transform of K034. Is this correct? What is
    Message 1 of 6 , Mar 21, 2013
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      Thanks Bernard!

      I forgot to include X(1824) in list of centers on the curve. It appears this curve is the Zosma transform of K034. Is this correct? What is the isoconjugation of this curve?

      Randy

      --- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@...> wrote:
      >
      > Dear Randy,
      >
      > > What curve is the inverse-in-polar-circle of the anticomplement of the
      > > Darboux quintic, Q071
      > > <http://bernard.gibert.pagesperso-orange.fr/curves/q071.html> ? It
      > > passes through X(4), X(25), X(51), X(1827), X(1828), X(1843), and the
      > > crosssum of X(3) and X(20). I find no matches on Bernard's site.
      >
      > your curve is the pivotal cubic with pivot X(1843), isopivot X(4) with respect to the orthic triangle.
      >
      > the mapping M -> X(4) x ctM (barycentric product) where ctM is the center of the inconic with perspector M transforms the Thomson cubic into this cubic.
      >
      > Best regards
      >
      > Bernard
      >
      > [Non-text portions of this message have been removed]
      >
    • Bernard Gibert
      Dear Randy, ... I don t think so. It should be K184 = pK(X76, X76). Best regards Bernard [Non-text portions of this message have been removed]
      Message 2 of 6 , Mar 22, 2013
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        Dear Randy,

        > I forgot to include X(1824) in list of centers on the curve. It appears this curve is the Zosma transform of K034. Is this correct?


        I don't think so. It should be K184 = pK(X76, X76).

        Best regards

        Bernard

        [Non-text portions of this message have been removed]
      • rhutson2
        Dear Bernard, Are you sure? The Zosma transforms of each of the listed centers on K034 lie on this cubic: Zosma transform of X(1) = X(4) Zosma transform of
        Message 3 of 6 , Mar 22, 2013
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          Dear Bernard,

          Are you sure? The Zosma transforms of each of the listed centers on K034 lie on this cubic:

          Zosma transform of X(1) = X(4)
          Zosma transform of X(2) = X(1824)
          Zosma transform of X(7) = X(1827)
          Zosma transform of X(8) = X(1828)
          Zosma transform of X(63) = X(25)
          Zosma transform of X(75) = X(1843)
          Zosma transform of X(92) = X(51)
          Zosma transform of X(280) = inverse-in-polar-circle of anticomplement of 'Darboux quintic point of X(3345)'
          Zosma transform of X(347) = inverse-in-polar-circle of anticomplement of 'Darboux quintic point of X(1490)'

          By 'Darboux quintic point of P', I mean, for P on the Darboux cubic, the common point, other than P, of the circles PAPa, PBPb, PCPc, where PaPbPc is the pedal triangle of P (as described at Q071).

          Best regards,
          Randy


          --- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@...> wrote:
          >
          > Dear Randy,
          >
          > > I forgot to include X(1824) in list of centers on the curve. It appears this curve is the Zosma transform of K034. Is this correct?
          >
          >
          > I don't think so. It should be K184 = pK(X76, X76).
          >
          > Best regards
          >
          > Bernard
          >
          > [Non-text portions of this message have been removed]
          >
        • Bernard Gibert
          Dear Randy, ... You re right and I owe you an apology. I made a stupid mistake using the trilinear macro of Zosma instead of the barycentric macro !!!
          Message 4 of 6 , Mar 22, 2013
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            Dear Randy,

            > Are you sure? The Zosma transforms of each of the listed centers on K034 lie on this cubic:

            You're right and I owe you an apology.

            I made a stupid mistake using the trilinear macro of Zosma instead of the barycentric macro !!!

            Nevertheless, the mapping M -> X4 x ctM is still valid.

            Sorry about that

            Best regards

            Bernard

            [Non-text portions of this message have been removed]
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