Loading ...
Sorry, an error occurred while loading the content.

Re: [EMHL] Pedal triangles

Expand Messages
  • Nikolaos Dergiades
    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
    Message 1 of 37 , Nov 24, 2008
    • 0 Attachment
      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

      > 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]
      >
      >
      > ------------------------------------
      >
      > Yahoo! Groups Links
      >
      >
      >


      ___________________________________________________________
      Χρησιμοποιείτε Yahoo!;
      Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
      διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
      μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr
    • 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
      • 0 Attachment
        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
        > > > >
        >
        >
        >
        > ------------------------------------
        >
        > Yahoo! Groups Links
        >
        >
        >


        ___________________________________________________________
        Χρησιμοποιείτε Yahoo!;
        Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
        διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
        μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr
      Your message has been successfully submitted and would be delivered to recipients shortly.