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A question

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  • Jean-Louis Ayme
    Dear Hyacinthists, 6 points on a circle leads to 60 hexagones and 60 Pascal s lines. These 60 Pascal s lines intersect in 20 Steiner s points. What is the
    Message 1 of 10 , Jun 4, 2011
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      Dear Hyacinthists,
      6 points on a circle leads to 60 hexagones and 60 Pascal's lines.
      These 60 Pascal's lines intersect in 20 Steiner's points.
      What is the geometric rule when we sta

      [Non-text portions of this message have been removed]
    • Jean-Louis Ayme
      Dear Hyacinthists, 6 points on a circle leads to 60 hexagones and 60 Pascal s lines. These 60 Pascal s lines intersect in 20 Steiner s points. What is the
      Message 2 of 10 , Jun 4, 2011
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        Dear Hyacinthists,
        6 points on a circle leads to 60 hexagones and 60 Pascal's lines.
        These 60 Pascal's lines intersect in 20 Steiner's points.
        What is the geometric rule when we sta
        Dear Hyacinthists,
        6 points on a circle leads to 60 cyclic hexagons and 60 Pascal's lines.
        These 60 Pascal's lines intersect in 20 Steiner's points which are triple...
        Now, what is the geometric rule in order to obtain the Steiner's points?
        For example we start withe the cyclic hexagon ABCDEF. How do we obtain geometrically the two other hexagons? I don't wish to use the term of permutation...
        Any geometric ideas?
        Sincerely
        Jean-Louis

        [Non-text portions of this message have been removed]
      • Barry Wolk
        ... The 3 hexagons with concurrwent Pascal lines are ABCDEF, ADCFEB and AFCBED. You fix ACE and cyclically rotate BDF. Is this what you wanted? -- Barry Wolk
        Message 3 of 10 , Jun 6, 2011
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          > Dear Hyacinthists,
          > 6 points on a circle leads to 60 cyclic hexagons and 60 Pascal's lines.
          > These 60 Pascal's lines intersect in 20 Steiner's points which are triple...
          > Now, what is the geometric rule in order to obtain the Steiner's points?
          > For example we start withe the cyclic hexagon ABCDEF. How do we obtain
          > geometrically the two other hexagons? I don't wish to use the term of
          > permutation...
          > Any geometric ideas?
          > Sincerely
          > Jean-Louis

          The 3 hexagons with concurrwent Pascal lines are ABCDEF,
          ADCFEB and AFCBED. You fix ACE and cyclically rotate BDF.

          Is this what you wanted?
          --
          Barry Wolk
        • Jean-Louis Ayme
          Dear Hyacinthists, what is the nature of the symmedian point of a triangle relatively to his orthic triangle?   Sincerely Jean-Louis [Non-text portions of
          Message 4 of 10 , Jul 11, 2012
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            Dear Hyacinthists,
            what is the nature of the symmedian point of a triangle relatively to his orthic triangle?
             
            Sincerely
            Jean-Louis

            [Non-text portions of this message have been removed]
          • Francisco Javier
            From a quick calculation I find that K is the Mittenpunkt of orthic triangle when ABC is acute. Best regards, Francisco Javier.
            Message 5 of 10 , Jul 11, 2012
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              From a quick calculation I find that K is the Mittenpunkt of orthic triangle when ABC is acute.

              Best regards,

              Francisco Javier.

              --- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@...> wrote:
              >
              > Dear Hyacinthists,
              > what is the nature of the symmedian point of a triangle relatively to�his orthic triangle?
              > �
              > Sincerely
              > Jean-Louis
              >
              > [Non-text portions of this message have been removed]
              >
            • Catalin Barbu
              Dear Jean-Louis, The symmedian point of a triangle is the center of homology of triangle and median triangle of his orthic triangle. In attach you find the
              Message 6 of 10 , Jul 11, 2012
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                Dear Jean-Louis,


                The symmedian point of a triangle is the center of homology of triangle and median triangle of his orthic triangle. In attach you find the proof of this theorem.


                Best regards,

                Catalin



                ________________________________
                From: Jean-Louis Ayme <jeanlouisayme@...>
                To: "Hyacinthos@yahoogroups.com" <Hyacinthos@yahoogroups.com>
                Sent: Thursday, July 12, 2012 8:11 AM
                Subject: [EMHL] A question


                 
                Dear Hyacinthists,
                what is the nature of the symmedian point of a triangle relatively to his orthic triangle?
                 
                Sincerely
                Jean-Louis

                [Non-text portions of this message have been removed]




                [Non-text portions of this message have been removed]
              • Barry Wolk
                ... The orthic triangle of the ex-central triangle IaIbIc is the original ABC. So if the P-point of IaIbIc is the Q-point of ABC, then the P-point of ABC is
                Message 7 of 10 , Jul 13, 2012
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                  > From a quick calculation I find that K is the Mittenpunkt of orthic triangle
                  > when ABC is acute.
                  >
                  > Best regards,
                  >
                  > Francisco Javier.
                  >
                  > Jean-Louis Ayme wrote:
                  >>
                  >> Dear Hyacinthists,
                  >> what is the nature of the symmedian point of a triangle relatively to�his
                  > orthic triangle?
                  >>
                  >> Sincerely
                  >> Jean-Louis

                  The orthic triangle of the ex-central triangle IaIbIc is the original ABC. So if the P-point of IaIbIc is the Q-point of ABC, then the P-point of ABC is the Q-point of the orthic triangle of ABC. If P=symmedian, the above "quick calxculation" gives Q=Mitten..And if P=orthocenter then Q=incenter, which is geometrically ttrivial. A few other cases I tried didn't give anything recognizable.
                  --
                  Barry Wolk
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