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A Conjugation Based On Crossdifference

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  • Quang Tuan Bui
    Dear All My Friends, 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
    Message 1 of 4 , Jul 14 3:57 AM
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      Dear All My Friends,

      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?

      Best regards,
      Bui Quang Tuan
    • Bernard Gibert
      Dear Tuan, ... This is similar to the Cundy-Parry transformations in http://pagesperso-orange.fr/bernard.gibert/Classes/cl037.html your Y is the intersection
      Message 2 of 4 , Jul 14 4:49 AM
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        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

        [Non-text portions of this message have been removed]
      • 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 3 of 4 , Jul 15 2:26 AM
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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 4 of 4 , Jul 15 11:18 PM
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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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