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Re: [EMHL] Re: Happy New Year and Nik's locus for 2013

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  • 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 1 of 14 , Jan 2, 2013
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      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 2 of 14 , Jan 2, 2013
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        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 3 of 14 , Jan 2, 2013
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          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 4 of 14 , Jan 2, 2013
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            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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