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Re: Arnold theorem

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  • Darij Grinberg
    ... call them ABC and A B C ... ... and searched for another point collinear with the two orthologic centers P and P and the perspector Q. If triangle A B C
    Message 1 of 13 , Mar 5, 2003
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      I wrote:

      >> If two orthologic triangles
      call them ABC and A'B'C'
      >> are perspective, then the perspector is
      >> collinear with the two orthologic centers,
      >> and the join of the three points is
      >> orthogonal to their perspectrix!

      ... and searched for another point collinear with the two orthologic
      centers P and P' and the perspector Q.

      If triangle A'B'C' is inscribed in triangle ABC, then the Longchamps
      point L of triangle ABC is collinear with P, P' and Q.

      If triangle A'B'C' is inscribed in the circumcircle of triangle ABC,
      then (I conjecture that) the circumcenter O is collinear with P, P'
      and Q.

      THERE MUST BE A GENERALIZATION!
      ----

      Sincerely,
      Darij Grinberg
    • Barry Wolk
      ... Correct. ... I call a calculation trivial if I can do it without needing a computer to handle the polynomials. In many problems the polynomials tend to
      Message 2 of 13 , Mar 5, 2003
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        Darij Grinberg wrote:

        > I don't know of which computational verification is Barry Wolk

        > speaking, but I think he uses barycentric coordinates.

        Correct.

        > Perhaps a reason why
        > I had not found a trivial coordinate proof of the Arnold
        > theorem is that I am still too lazy to learn representing
        > perpendicular lines in barycentrics.

        I call a calculation "trivial" if I can do it without needing a
        computer to handle the polynomials. In many problems the
        polynomials tend to get ridiculously large.

        It is very easy to handle perpendicular lines. Take two points
        at infinity, P(x:y:z) and P'(x':y':z'), so x+y+z = x'+y'+z' = 0.
        Then lines through P are perpendicular to lines through P' iff
        x x' SA + y y' SB + z z' SC = 0

        So given P, we can find P' by letting x' = y SB - z SC, y' and
        z' similarly.

        And the line through H and P' has coordinates
        [x SA : y SB : z SC]
        --
        Barry Wolk


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      • Darij Grinberg
        Dear Barry Wolk, ... Many thanks... the next step would be (perhaps) generalizing to Lamoen s Pl-perpendicularity, but I think this is simply replacing SA, SB,
        Message 3 of 13 , Mar 5, 2003
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          Dear Barry Wolk,

          In Hyacinthos message #6661, you wrote:

          >> Darij Grinberg wrote:
          >>
          >> > I don't know of which computational verification is Barry Wolk
          >> > speaking, but I think he uses barycentric coordinates.
          >>
          >> Correct.
          >>
          >> > Perhaps a reason why
          >> > I had not found a trivial coordinate proof of the Arnold
          >> > theorem is that I am still too lazy to learn representing
          >> > perpendicular lines in barycentrics.
          >>
          >> I call a calculation "trivial" if I can do it without needing a
          >> computer to handle the polynomials. In many problems the
          >> polynomials tend to get ridiculously large.
          >>
          >> It is very easy to handle perpendicular lines. Take two points
          >> at infinity, P(x:y:z) and P'(x':y':z'), so x+y+z = x'+y'+z' = 0.
          >> Then lines through P are perpendicular to lines through P' iff
          >> x x' SA + y y' SB + z z' SC = 0
          >>
          >> So given P, we can find P' by letting x' = y SB - z SC, y' and
          >> z' similarly.
          >>
          >> And the line through H and P' has coordinates
          >> [x SA : y SB : z SC]
          >> --
          >> Barry Wolk

          Many thanks... the next step would be (perhaps) generalizing to
          Lamoen's Pl-perpendicularity, but I think this is simply replacing
          SA, SB, SC by some constants.

          Sincerely,
          Darij Grinberg
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