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Re: [EMHL] Re: Involutive transformation

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  • Francois Rideau
    Dear Jeff Of course, I don t know the answer! But in case of a circle, inversion have infinitely many fixed points when isoconjugation have only 4 fixed points
    Message 1 of 10 , Jul 25, 2007
      Dear Jeff
      Of course, I don't know the answer!
      But in case of a circle, inversion have infinitely many fixed points when
      isoconjugation have only 4 fixed points real or not, inversion is defined on
      the extended conformal plane when isoconjugation is defined on the extended
      projective plane minus 3 points, so the situation is very tricky but maybe
      this problem is not formulated well enough?

      On 7/25/07, Francois Rideau <francois.rideau@...> wrote:
      >
      > Dear Jeff
      > When you look at some isoconjugation (m <--> m') with 4 real fixed points,
      > for example isotomic conjugation or isogonal conjugation, then m and m' are
      > conjugate wrt every conic on these 4 fixed points, that is to say conics in
      > a linear pencil;
      > That's why I ask the converse problem: 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?
      > Friendly
      > Francois
      >
      >
      > On 7/25/07, Jeff <cu1101@...> wrote:
      > >
      > > Dear Francois,
      > >
      > > I am attempting to tackle this problem with a direct proof. But, my
      > > methods are long and laborious. It seems that the tangent line at
      > > <p> works for all lines (even those not tangent to the incircle.)
      > >
      > > Am I wrong?
      > >
      > > Sincerely, Jeff
      > >
      > >
      > > > Dear friends
      > > > I give you an example of such an involutive map that I have found
      > > > while doing some chowchows.
      > > > I start with an equilateral triangle ABC with <Gamma> as
      > > > circumcircle and <gamma> as incircle.
      > > > I choose a point <p> on the incircle <gamma> of which the tangent
      > > > at <gamma> cuts side BC in <a>, side CA in <b> and side AB in <c>.
      > > >
      > > > Now given any point M in the plane, I give the construction of the
      > > > point M'= f(M), depending on the choice of <p> on the incircle
      > > > <gamma>.
      > > >
      > > > Line AM cuts again circumcircle <Gamma> in U.
      > > > Line aU cuts again circumcircle <Gamma> in U'.
      > > >
      > > > In the same way, we get the points V' and W'.
      > > >
      > > > Line BM cuts again circumcircle <Gamma> in V.
      > > > Line bV cuts again circumcircle <Gamma> in V'.
      > > >
      > > > Line CM cuts again circumcircle <Gamma> in W.
      > > > Line cW cuts again circumcircle <Gamma> in W'.
      > > >
      > > > Then lines AU', BV' and CW' are on point M'.
      > > >
      > > > Now it is easy to see that this construction is involutive.
      > > > Moreover M and M' are conjugate wrt the equalizer conic of ABC,
      > > > with barycentric equation: x² + y² + z² = 0.
      > > > That is to say M' is on the dual line of M wrt ABC and of course M
      > > > is also on the dual line of M' wrt ABC.
      > >
      > > > Friendly
      > > > Francois
      > > >
      > > > PS
      > > > 1° Equilateral triangles are not needed. Of course, all this stuff
      > > > is affine (as the chowchow theory), replacing ABC by any triangle
      > > > and circles by both Steiner ellipses.
      > > > But construction is easier with circles than with ellipses.
      > > > 2° I guess M and M' are isoconjugate in some isoconjugation.
      > > >
      > >
      > >
      > >
      >
      >


      [Non-text portions of this message have been removed]
    • 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 2 of 10 , Jul 26, 2007
        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 3 of 10 , Jul 27, 2007
          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 4 of 10 , Jul 30, 2007
            --- 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 5 of 10 , Jul 30, 2007
              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 6 of 10 , Jul 30, 2007
                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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