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Re: [EMHL] A Conjugation Based On Crossdifference

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  • Quang Tuan Bui
    Dear Bernard, Thank you very much! You are right! I see following fact but not sure if is it true: If pK is any pK cubic and U is any point of pK then we can
    Message 1 of 4 , Jul 15, 2010
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      Dear Bernard,
      Thank you very much! You are right!
      I see following fact but not sure if is it true:

      If pK is any pK cubic and U is any point of pK then we can always find unique P on pK such that fPU(pK)=pK

      Best regard,
      Bui Quang Tuan

      --- On Wed, 7/14/10, Bernard Gibert <bg42@...> wrote:

      > From: Bernard Gibert <bg42@...>
      > Subject: Re: [EMHL] A Conjugation Based On Crossdifference
      > To: Hyacinthos@yahoogroups.com
      > Date: Wednesday, July 14, 2010, 6:49 PM
      > Dear Tuan,
      >
      > > We use following notations:
      > > Y*Z = barycentric product of Y and Z
      > > Crd(Y,Z) = Crossdifference of Y and Z
      > > 1/Z = isotomic conjugate of Z
      > > Y/Z = Y*(1/Z)
      > >
      > > Given two fixed points with barycentrics P = (p : q :
      > r) and U = (u : v : w). X is variable point with
      > barycentrics X = (x : y : z). We construct point Y as:
      > > Y = P*Crd(P,X)/Crd(U,X) and denote the transform as Y
      > = fPU(X)
      > >
      > > Barycentrics of Y = fPU(X) as following:
      > >
      > > p*(q*z - r*y)/(v*z - w*y) : q*(r*x - p*z)/(w*x - u*z)
      > : r*(p*y - q*x)/(u*y - v*x)
      > >
      > > The transform Y = fPU(X) is conjugation: fPU(fPU(X)) =
      > X
      > >
      > > Are there somethings interesting from this
      > conjugation?
      >
      > This is similar to the Cundy-Parry transformations in
      >
      > http://pagesperso-orange.fr/bernard.gibert/Classes/cl037.html
      >
      > your Y is the intersection of the lines UX  and PX*
      > where X* is the homologue of X in the isoconjugation that
      > swaps P and U.
      >
      > Best regards
      >
      > Bernard
      >
    • Bernard Gibert
      Dear Tuan, ... fPU(pK) is in general a quartic unless pK also contains P. fPU globally fixes pK(P x U, P) and pK(P x U, U) just like the cubics K003 and K004
      Message 2 of 4 , Jul 15, 2010
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        Dear Tuan,

        > If pK is any pK cubic and U is any point of pK then we can always find unique P on pK such that fPU(pK)=pK

        fPU(pK) is in general a quartic unless pK also contains P.

        fPU globally fixes pK(P x U, P) and pK(P x U, U) just like the cubics K003 and K004 in CL037.

        Best regards

        Bernard

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