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Re: [EMHL] Looking for an explanation

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  • Bernard Gibert
    Dear Jean-Pierre and Steve ... These 3 points P lie on the Thomson cubic. We ve already seen this hyperbola but where ? ... the Darboux cubic ... that s right.
    Message 1 of 6 , Mar 31, 2006
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      Dear Jean-Pierre 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.

      We've already seen this hyperbola but where ?


      >> 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?
      >
      >
      > [SS] So the pedal triangle of M is a Cevian triangle of an infinite
      > point.
      > This puts it on one of the famous cubics (I cannot remember which),

      the Darboux cubic

      > and the infinite point must be an infinite point on that cubic. If so
      > there must be 2 other points that serve as M.

      that's right. they are 3 common points of the Darboux cubic and C(O,3R).

      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



      Best regards

      Bernard



      [Non-text portions of this message have been removed]
    • 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 2 of 6 , Apr 1, 2006
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        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 3 of 6 , Apr 1, 2006
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          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 4 of 6 , Apr 1, 2006
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            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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