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Re: [EMHL] Looking for an explanation

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  • jpehrmfr
    Dear Bernard and Steve ... through ... I don t think so ... Darboux ... Yes, indeed. These three points are UVW in your Darboux page. many thanks. It is quite
    Message 1 of 6 , Apr 1, 2006
      Dear Bernard and Steve

      > >> [JP] consider the rectangular hyperbola through G, O, K
      > >> (symedian), X[110).
      > >> His center is the midpoint of GX[110] and the hyperbola goes
      through
      > >> the infinite points of the Jerabek hyperbola, X[154], X[354], X
      > >> [392], X
      > >> [1201],...
      > >> Consider a common point P (not X[110]) of the hyperbola with the
      > >> circumcircle;
      >
      > These 3 points P lie on the Thomson cubic.

      I don't think so

      > >> let M be the homothetic of P in (O, -3) and A'B'C' the
      > >> pedal triangle of M.
      > >> Then AA', BB', CC' are parallel.
      > >> Any explanation?

      > there's a close analogy with the bottom of my web page on the
      Darboux
      > cubic.
      >
      > http://perso.wanadoo.fr/bernard.gibert/Exemples/k004.html

      Yes, indeed. These three points are UVW in your Darboux page. many
      thanks.
      It is quite interesting to see that these points are the only points
      P in the plane such as, if A'B'C' is the pedal triangle of P, the
      lines AA', BB', CC' are parallel (of course they are parallel to the
      asymptots of Lucas - or Thomson -)
      Friendly. Jean-Pierre
    • Bernard Gibert
      Dear Jean-Pierre, ... THEY DO, dear Jean-Pierre ! Best regards Bernard [Non-text portions of this message have been removed]
      Message 2 of 6 , Apr 1, 2006
        Dear Jean-Pierre,

        >>>> [JP] consider the rectangular hyperbola through G, O, K
        >>>> (symedian), X[110).
        >>>> His center is the midpoint of GX[110] and the hyperbola goes
        > through
        >>>> the infinite points of the Jerabek hyperbola, X[154], X[354], X
        >>>> [392], X
        >>>> [1201],...
        >>>> Consider a common point P (not X[110]) of the hyperbola with the
        >>>> circumcircle;
        >>
        >> These 3 points P lie on the Thomson cubic.
        >
        > I don't think so

        THEY DO, dear Jean-Pierre !



        Best regards

        Bernard



        [Non-text portions of this message have been removed]
      • jpehrmfr
        Dear Bernard ... Of course, you re right - in fact I was thinking to your points UVW - Looking at your nice page about Darboux, I have an obvious explanation
        Message 3 of 6 , Apr 1, 2006
          Dear Bernard
          > THEY DO, dear Jean-Pierre !
          Of course, you're right - in fact I was thinking to your points UVW -
          Looking at your nice page about Darboux, I have an obvious explanation
          of the nice property of the pedal triangle of U,V,W.
          Suppose that A'B'C' is the pedal triangle of M; then BB' and CC' are
          parallel iff M lies on your hyperbola Ha.
          As U,V,W are the common points of Ha,Hb,Hc, for these points, AA',
          BB', CC' are parallel.
          Thus, there is nothing new in what I said.
          Friendly.Jean-Pierre
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