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Re: [EMHL] Pedal triangles

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  • Francois Rideau
    Dear Nikos Excuse me for answering you so late. You are right of course provided that you are using oriented, (or signed ?), angles of lines. I give you two
    Message 1 of 37 , Nov 24, 2008
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      Dear Nikos
      Excuse me for answering you so late.
      You are right of course provided that you are using oriented, (or signed
      ?), angles of lines.
      I give you two constructions of this point but as I think they are too
      complex, that's why I ask for a more direct one.
      Triangles ABC and A'B'C' are given and are supposed to be not inversely
      similar for in this case the following constructions lead to nowhere.
      1° First construct a ABC-inscribed triangle <abc> homothetic with A'B'C' ,
      (very known construction) and then the point P is the common point of
      circles (bcA), (caB), (abC), (the Miquel point).
      2° First construct a ABC-circumscribed triangle <abc> homothetic with
      A'B'C', (straightforward!), then the common point P* of circles (aBC),
      (bCA), (cAB).
      P is the isogonal conjugate of P* wrt ABC.

      Friendly
      Francois
      PS
      Another way still more elaborate to recover P is the following:
      You look at the direct circular map <f> sending ABC onto A'B'C' and you look
      at the pôle P' of <f> such that f(P') = Infty, the point at infinity.
      Then P is the point inverse of P' wrt circle ABC.
      Or you can look at the indirect circular map <g> sending ABC onto A'B'C',
      then P is the pôle of <g> such that g(P) = Infty.
      So any simple construction of a pôle of a circular map must lead to a simple
      construction of P.


      [Non-text portions of this message have been removed]
    • Nikolaos Dergiades
      Dear Jeff, Yes. And the points A1, B1, C1 are found in this case as previously but by negative rotations. Have a Happy New Year. Best regards Nikos Dergiades
      Message 37 of 37 , Dec 30, 2008
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        Dear Jeff,
        Yes.
        And the points A1, B1, C1
        are found in this case as previously
        but by negative rotations.
        Have a Happy New Year.
        Best regards
        Nikos Dergiades

        > Dear Nikos and Francois,
        >
        > So, if instead of the triangle A'B'C', we use a
        > triangle inversely
        > similar to A'B'C', then we find the pedal
        > triangle of the point P*
        > inverse of P in the ABC circumcircle?
        >
        > Sincerely, Jeff
        >
        >
        > [ND]
        > > Dear Francois,
        > > I think you mean if we construct the point C1
        > > as we constructed the points A1, B1,
        > > if the reflections
        > > of C in A1B1,
        > > of A in B1C1
        > > of B in C1A1
        > > coincide?
        > > Yes.
        > >
        > > Best regards
        > > Nikos Dergiades
        > >
        > > > Dear Nikos
        > > > Thank you very much for your help.
        > > > Now, I will think about your construction and
        > compare it
        > > > with mines.
        > > > Have you already look at the configuration you
        > get when you
        > > > perform your
        > > > construction with other vertices?
        > > > Friendly
        > > > Francois
        > > >
        > > > 2008/11/24 Nikolaos Dergiades
        > <ndergiades@...>
        > > >
        > > > > Dear Francois,
        > > > > you are right.
        > > > > A simpler construction is the following:
        > > > >
        > > > > Let O be the circumcenter of ABC.
        > > > > The perpendicular from O to BC meets the
        > > > > rotation of line CO about C by the signed
        > angle
        > > > > B'A'C' at the point A1.
        > Similarly
        > > > > the perpendicular from O to CA meets the
        > > > > rotation of line AO about A by the signed
        > angle
        > > > > C'B'A' at the point B1.
        > > > > The reflection of C in A1B1 is the required
        > point P.
        > > > > Best regards
        > > > > Nikos Dergiades
        > > > >
        >
        >
        >
        > ------------------------------------
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