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Re: [EMHL] ETC cyclic points

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  • Antreas Hatzipolakis
    Dear Nikos [ND] ... TI ROMIOSINI MIN TIN KLAIS ... I think that the following search in ETC would be interesting
    Message 1 of 12 , Jan 4, 2011
      Dear Nikos

      [ND]

      > We can see this in Greece because of wage cuts.
      > Our 1821 revolution comes to an end.
      > We'' ll set Greece on fire and
      > we'' ll spread the light of this fire in all European Union.
      > But don't be afraid.
      > Greece never sinks, is like an utricle in ocean.
      > Greece never dies, is always in intensive teatment unit.
      >
      TI ROMIOSINI MIN TIN KLAIS <http://www.youtube.com/watch?v=vZjtDRVJwdI>


      > Let us celebrate the beginning of the cycle of this year
      > by ETC cyclic points exploration.
      >
      >
      >
      I think that the following search in ETC would be interesting but I guess
      it is not easy!

      Let (A), (B), (C) be three circles defined with known points listed in ETC
      such that they do not have a common KNOWN point.

      Not like these circles, for example:

      1) 1 2 11 392
      2) 1 2 23 36
      3) 1 3 21 759

      Are there such circles which are concurrent?

      If there are, then the point of concurrence is a NEW interesting point !

      With my Best Wishes for the New Year

      Antreas


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    • Chris Van Tienhoven
      Dear Nikos and friends, I don t know if this set of 10 cyclic points is known: X(11) X(36) X(65) X(80) X(108) X(759) X(1354) X(1845) X(2588) X(2589) Best
      Message 2 of 12 , Jan 4, 2011
        Dear Nikos and friends,

        I don't know if this set of 10 cyclic points is known:
        X(11)
        X(36)
        X(65)
        X(80)
        X(108)
        X(759)
        X(1354)
        X(1845)
        X(2588)
        X(2589)

        Best Regards,

        Chris

        > Let us celebrate the beginning of the cycle of this year
        > by ETC cyclic points exploration.
        > Nikos Dergiades
      • Nikolaos Dergiades
        Dear Chris, very good case. I didn t know it. I had worked only with cases with X(1). Now with my computer seems that your points are indeed concyclic. The
        Message 3 of 12 , Jan 5, 2011
          Dear Chris,
          very good case.
          I didn't know it.
          I had worked only with cases with X(1).
          Now with my computer seems that your points
          are indeed concyclic.
          The center of this circle I think
          is the point with barycentrics
          ( (b-c)(abc(b+c-a)-(cc+aa-bb)(aa+bb-cc) : .. :..)
          not in ETC.
          Best regards
          Nikos Dergiades


          > Dear Nikos and friends,
          >
          > I don't know if this set of 10 cyclic points is known:
          > X(11)   
          > X(36)   
          > X(65)   
          > X(80)   
          > X(108)   
          > X(759)   
          > X(1354)   
          > X(1845)
          > X(2588)
          > X(2589)
          >
          > Best Regards,
          >
          > Chris
          >
          > > Let us celebrate the beginning of the cycle of this
          > year
          > > by ETC cyclic points exploration.
          > > Nikos Dergiades
          >
          >
          >
          >
          > ------------------------------------
          >
          > Yahoo! Groups Links
          >
          >
          >     Hyacinthos-fullfeatured@yahoogroups.com
          >
          >
          >
        • Nikolaos Dergiades
          Dear Antreas, ... Yes. It is not easy. ... Yes. There are. In order to reduce the time of calculation by the computer one of the three circles can be the
          Message 4 of 12 , Jan 5, 2011
            Dear Antreas,
            You wrote:
            > I think that the following search in ETC  would be
            > interesting but I guess
            > it is not easy!

            Yes. It is not easy.

            >
            > Let (A), (B), (C) be three circles defined with known
            > points listed in ETC
            > such that they do not  have a common KNOWN point.
            >
            > Not like these circles, for example:
            >
            > 1) 1 2 11 392
            > 2) 1 2 23 36
            > 3) 1 3 21 759
            >
            > Are there such circles which are concurrent?

            Yes. There are.
            In order to reduce the time of calculation by the
            computer one of the three circles can be the
            circumcircle (O).
            Hence one example is the circles
            (O) (11, 13, 14), (36, 65, 79)
            with common point P not in ETC and a second is
            (O), (11, 13, 14), (46, 57, 79)
            with the same common point P.
            Hence we have found 4 circles concurrent at P.
            (11, 13, 14) means the circle passing through
            the two Fermat points X(13), X(14)
            and the Feuerbach point X(11).

            >
            > If there are, then  the point of concurrence is a NEW
            > interesting point !

            I think that this point has not simple coordinates.

            Best regards
            Nikos
          • Barry Wolk
            ... I did this using complex numbers as coordinates, where the formula for M should be symmetric in the 4 complex variables (A1,A2,A3,A4), and have some
            Message 5 of 12 , Jan 9, 2011
              > Dear Jean-Pierre,
              >
              > I am quite thrilled.
              > As always when I find something beautiful.
              > You described a point M depending on the 4 points A1, A2,
              > A3, A4 of a quadrilateral. Here is the structure behind it.
              >
              > M = the inverse in the circumcircle of AiAjAk of the
              > isogonal conjugate of Al wrt AiAjAk,
              > where (i,j,k,l) is any permutation of (1,2,3,4).
              >
              > Strangely enough this is true for each permutation.
              > There should be more properties/references relating to this point.
              > Best regards,
              >
              > Chris van Tienhoven

              I did this using complex numbers as coordinates, where the formula for M should be symmetric in the 4 complex variables (A1,A2,A3,A4), and have some invariance properties. The answer can be expressed as follows:

              M is a quotient of two determinants. Each determinant is 4-by-4, and its rows correspond to the variables. In the numerator, the row for x is [1, x, x^2, x x#], where x# means the complex conjugate of x. In the denominator, the row for x is [1, x, x^2, x#].

              Can other ETC points be expressed in this way?
              --
              Barry Wolk
            • Francois Rideau
              Dear Jean-Pierre I add some circular properties of your drawing. There is a direct circular map f: A_1 -- B_1, A_2 -- B_2, A_3 -- B_3, A_4 -- B_4 and this
              Message 6 of 12 , Feb 25, 2011
                Dear Jean-Pierre
                I add some circular properties of your drawing.
                There is a direct circular map f: A_1 --> B_1, A_2 --> B_2, A_3 --> B_3, A_4
                --> B_4 and this map is,involutive with central point M that is to say f
                swaps M with the point at infinity.
                In the same way, there is a direct circular map f: A_1 --> O_1, A_2 --> O_2,
                A_3 --> O_3, A_4 --> O_4 and this map is,involutive with central point M.
                Here in France, we named these maps circular transpositions, Idon't know
                their english names.
                Friendly
                Francois


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