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Θέμα: Re: [EMHL] Re: Involutive transfor mation

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  • Nikolaos Dergiades
    ... I don t think I understood your problem, so I ask the following: The reflection about a line that passes through the center of the given circle is
    Message 1 of 10 , Jul 26 10:12 AM
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      Dear Francois, you wrote:
      > > given a
      > conic <Gamma>, what are the
      > > (algebraic) involutive maps (m <--> m') such that
      > m and m' are conjugate wrt
      > > <Gamma>?
      > > If <Gamma> is a circle, inversion wrt <Gamma> is
      > such a map but what are
      > > the others?

      I don't think I understood your problem, so I
      ask the following:
      The reflection about a line that passes through
      the center of the given circle <Gamma> is such a map?
      Best regards
      Nikos Dergiades



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    • Jeff
      Dear Francois, ... I like the setup but I don t know what is the equalizer conic of ABC, with barycentric equation x^2 + y^2 + z^2 = 0. Why is this conic
      Message 2 of 10 , Jul 27 9:03 PM
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        Dear Francois,

        >> the situation is very tricky but maybe
        >> this problem is not formulated well enough?

        I like the setup but I don't know what is the "equalizer conic of ABC,
        with barycentric equation x^2 + y^2 + z^2 = 0." Why is this conic
        called 'equalizer'?

        Sincerely, Jeff
      • Jeff
        ... Dear Francois, Sorry, I just now saw the reference in msg 14109. Sincerely, Jeff
        Message 3 of 10 , Jul 30 6:27 PM
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          --- In Hyacinthos@yahoogroups.com, "Jeff" <cu1101@...> wrote:
          >
          > Dear Francois,
          >
          > >> the situation is very tricky but maybe
          > >> this problem is not formulated well enough?
          >
          > I like the setup but I don't know what is the "equalizer conic of
          > ABC, with barycentric equation x^2 + y^2 + z^2 = 0." Why is this
          > conic called 'equalizer'?
          >
          > Sincerely, Jeff
          >

          Dear Francois,
          Sorry, I just now saw the reference in msg 14109.
          Sincerely, Jeff
        • Francois Rideau
          Dear Jeff Yes, I think this conic is the equalizer, following Steve notations. Of course this conic has no real points but it does not matter! The point m with
          Message 4 of 10 , Jul 30 7:25 PM
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            Dear Jeff
            Yes, I think this conic is the equalizer, following Steve notations.
            Of course this conic has no real points but it does not matter!
            The point m with barycentrics (a:b:c) has for polar line wrt this conic the
            line with barycentric equation:
            a.x + b.y + c.z = 0
            But if you don't like this point of view, you can also use duality wrt the
            quadratic form:
            (x,y,z) -->x^2 + y^2 + z^2
            Friendly
            Francois



            On 7/28/07, Jeff <cu1101@...> wrote:
            >
            > Dear Francois,
            >
            > >> the situation is very tricky but maybe
            > >> this problem is not formulated well enough?
            >
            > I like the setup but I don't know what is the "equalizer conic of ABC,
            > with barycentric equation x^2 + y^2 + z^2 = 0." Why is this conic
            > called 'equalizer'?
            >
            > Sincerely, Jeff
            >
            >
            >


            [Non-text portions of this message have been removed]
          • Francois Rideau
            Dear Nikos Nope! Look at the circle x^2 + y^2 = 1 and at the symmetry: (x,y) -- (x, -y) wrt the x-axis The points (a,b) and (a, -b) are in general never
            Message 5 of 10 , Jul 30 7:35 PM
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              Dear Nikos
              Nope!
              Look at the circle x^2 + y^2 = 1 and at the symmetry: (x,y) --> (x, -y) wrt
              the x-axis
              The points (a,b) and (a, -b) are in general never conjugate wrt this this
              circle.
              I want to find all involutive transformations (x,y) --> f(x,y) defined on
              some open set of the projective (resp conformal) plane such that points
              (x,y) and f(x,y) are always conjugate wrt some conic (resp circle)
              Friendly
              Francois

              On 7/26/07, Nikolaos Dergiades <ndergiades@...> wrote:
              >
              > Dear Francois, you wrote:
              > > > given a
              > > conic <Gamma>, what are the
              > > > (algebraic) involutive maps (m <--> m') such that
              > > m and m' are conjugate wrt
              > > > <Gamma>?
              > > > If <Gamma> is a circle, inversion wrt <Gamma> is
              > > such a map but what are
              > > > the others?
              >
              > I don't think I understood your problem, so I
              > ask the following:
              > The reflection about a line that passes through
              > the center of the given circle <Gamma> is such a map?
              > Best regards
              > Nikos Dergiades
              >
              >
              >
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