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Re: [EMHL] Happy New Year

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  • Nikolaos Dergiades
    Dear Antreas, if A B C is the pedal triangle and not the cevian then the locus is a cubic but I think complicated. The same holds if we take circumenters. I
    Message 1 of 14 , Jan 2, 2013
      Dear Antreas,
      if A'B'C' is the pedal triangle and not
      the cevian then the locus is a cubic
      but I think complicated. The same holds
      if we take circumenters.
      I think that the most interesting case is:

      If A'B'C' is the cevian triangle of a point P,
      A1A'B, A2CA'are similar isosceles triangles
      outside of ABC with base angle w,
      (the bases are opposite of A1 and A2)
      A* is the centroid of triangle A'A1A2
      and similarly define B*, C*
      then the triangles ABC, A*B*C* are perspective
      if and only if the locus of P
      is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
      and pole Q = ((S_A+5S_w)/(S_A+S_w):..:..).

      Best regards
      Nikos


      > Dear Nikos, Paul
      >
      > Happy New Year!!!
      >
      > 1. We can take A'B'C' as the pedal triangle of P and ask for
      > the locus.
      >
      > 2. We can take A*,B*,C* as triangle centers (Circumcenters,
      > Orthocenters
      > etc) of
      > the triangles A'A1A2, B'B1B2, C'C1C2 and ask of the loci of
      > P such taht
      > ABC,A*B*C* be perspective.
      >
      > I think that interesting is the case of the circumcenters.
      >
      > Season's Greetings
      >
      > Antreas
      >
      > On Tue, Jan 1, 2013 at 2:15 AM, Nikolaos Dergiades <ndergiades@...>wrote:
      >
      > > **
      > >
      > >
      > > Dear Antreas
      > > Happy New Year to you
      > > and all friends of Hyacinthos.
      > >
      > > A locus for the begining of 2013.
      > > If A'B'C' is the cevian triangle of P
      > > and A'BA1, CA'A2 are equilateral triangles
      > > and A* is the mid point of A1, A2 in the
      > > opposite side of BC ralative to A,
      > > and similarly define B*, C* what is the locus
      > > of P such that ABC, A*B*C* are perspective.
      > > Best regards
      > > Nikos Dergiades
      > >
      > >
      >
      >
      > [Non-text portions of this message have been removed]
      >
      >
      >
      > ------------------------------------
      >
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      >
      >
      >     Hyacinthos-fullfeatured@yahoogroups.com
      >
      >
    • Bernard Gibert
      Dear Nikos, Paul and friends, ... If A* is the homothetic of A under h(A , k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are
      Message 2 of 14 , Jan 2, 2013
        Dear Nikos, Paul and friends,

        > [ND] A little generalization:
        > If A'B'C' is the cevian triangle of a point P
        > and A"BC, B"CA, C"AB are similar isosceles triangles
        > outside of ABC with base agnle w then the triangles
        > ABC, A*B*C* where A* is the mid point of A'A" . . .
        > are perspective if and only if the locus of P
        > is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
        > and pole Q = ((S_A+3S_w)/(S_A+S_w):..:..).
        > The point S lies on the Kiepert hyperbola
        > and corresponds to A", B", C".
        > The point S'= ((1/(S_A+3S_w):..:..)) lies also
        > on the Kiepert hyperbola and corresponds to the
        > Centroids of triangles A"BC, B"CA, C"AB.
        > The point Q is the barycentric product of S
        > and the isotomic of S'.

        If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is pK(Ω, S) with

        pole Ω = (SA k + (2-k)Sw) / (SA + Sw) : : ,

        pivot S = 1 / (SA + Sw) : : , on the Kiepert hyperbola,

        isopivot : SA k + (2-k)Sw : : , on the line X2-X69.

        When k varies, the locus of Ω is a line through X2,

        When w varies, the locus of Ω is a nodal circum-cubic with node X2, the nodal tangents being X2-X6 and Euler line.

        This cubic is a member of the pencil generated by :
        the three medians (k = 1)
        Kiepert + Euler (k = 0)
        line X2-X6 + isotomic of Euler (k = 2).

        ----------------------------------------------------------

        The locus of X is another pK(Ω', S') with

        pole Ω' = 1 / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

        pivot S' = (SA k + Sw) / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

        isopivot : 1 / (SA k + Sw) : : , on the Kiepert hyperbola.


        Best regards and Happy New Year to you all.

        Bernard

        [Non-text portions of this message have been removed]
      • Paul Yiu
        Dear Nik, Bernard, and Francisco Javier, Thank you for these generalizations. FJ, I did not realize that the locus contains G! Best regards Sincerely Paul
        Message 3 of 14 , Jan 2, 2013
          Dear Nik, Bernard, and Francisco Javier,

          Thank you for these generalizations. FJ, I did not realize that the locus contains G!

          Best regards
          Sincerely
          Paul
          ________________________________________
          From: Hyacinthos@yahoogroups.com [Hyacinthos@yahoogroups.com] on behalf of Bernard Gibert [bg42@...]
          Sent: Wednesday, January 02, 2013 8:00 AM
          To: Hyacinthos@yahoogroups.com
          Subject: Re: [EMHL] Re: Happy New Year and Nik's locus for 2013

          Dear Nikos, Paul and friends,

          > [ND] A little generalization:
          > If A'B'C' is the cevian triangle of a point P
          > and A"BC, B"CA, C"AB are similar isosceles triangles
          > outside of ABC with base agnle w then the triangles
          > ABC, A*B*C* where A* is the mid point of A'A" . . .
          > are perspective if and only if the locus of P
          > is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
          > and pole Q = ((S_A+3S_w)/(S_A+S_w):..:..).
          > The point S lies on the Kiepert hyperbola
          > and corresponds to A", B", C".
          > The point S'= ((1/(S_A+3S_w):..:..)) lies also
          > on the Kiepert hyperbola and corresponds to the
          > Centroids of triangles A"BC, B"CA, C"AB.
          > The point Q is the barycentric product of S
          > and the isotomic of S'.

          If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is pK(Ω, S) with

          pole Ω = (SA k + (2-k)Sw) / (SA + Sw) : : ,

          pivot S = 1 / (SA + Sw) : : , on the Kiepert hyperbola,

          isopivot : SA k + (2-k)Sw : : , on the line X2-X69.

          When k varies, the locus of Ω is a line through X2,

          When w varies, the locus of Ω is a nodal circum-cubic with node X2, the nodal tangents being X2-X6 and Euler line.

          This cubic is a member of the pencil generated by :
          the three medians (k = 1)
          Kiepert + Euler (k = 0)
          line X2-X6 + isotomic of Euler (k = 2).

          ----------------------------------------------------------

          The locus of X is another pK(Ω', S') with

          pole Ω' = 1 / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

          pivot S' = (SA k + Sw) / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

          isopivot : 1 / (SA k + Sw) : : , on the Kiepert hyperbola.


          Best regards and Happy New Year to you all.

          Bernard

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



          ------------------------------------

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        • Ricardo Barroso
          Happy New year 2013 . From Sevilla  Ricardo http://personal.us.es/rbarroso/trianguloscabri/ [Non-text portions of this message have been removed]
          Message 4 of 14 , Jan 2, 2013
            Happy New year 2013 .

            From Sevilla 

            Ricardo

            http://personal.us.es/rbarroso/trianguloscabri/

            [Non-text portions of this message have been removed]
          • Bernard Gibert
            Dear all, If A B C is the pedal triangle of P (instead of cevian triangle of P), and A , B , C as before, If A* is the homothetic of A under h(A , k), B*
            Message 5 of 14 , Jan 2, 2013
              Dear all,

              If A'B'C' is the pedal triangle of P (instead of cevian triangle of P), and A", B", C" as before,

              If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is a central cubic passing through X3 with asymptotes parallel to the altitudes of ABC and concurring on the Brocard axis, very similar to the Darboux cubic which is obtained when k = 0.

              The locus of the perspector X is pK(X2, T) where

              T = (-a^4+b^4+c^4)k-4 SA Sw : : , on the line X4-X69.

              If k = 0, T = X69 and this is the Lucas cubic.

              Best regards

              Bernard

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