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21337Re: [EMHL] Re: Happy New Year and Nik's locus for 2013

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  • Bernard Gibert
    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]
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