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21333Re: Happy New Year and Nik's locus for 2013

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  • yiuatfauedu
    Jan 1, 2013
    • 0 Attachment
      Dear Nik and Francisco Javier,

      ND: 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?

      PY: If A_0BC, B_0CA, C_0AB are the equilateral triangles
      constructed externally on the sides BC, CA, AB respectively,
      ... A*B*C* is perspective with ABC if and only if P [lies on]
      the isocubic K(Q,X(13)), where
      Q = ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
      has (6-9-13) search number 0.245203113787…


      I think we can go a little bit further and get a Euclidean constructible example. The point $Q$ can be regarded as the isogonal conjugate of the barycentric product of $X(15)$ (isodynamic point) and $X(17)$ (Napoleon point):
      Q = (X(15)*X(17))^*.
      It is an interior point of triangle $ABC$ if all angles are less than $120^\circ$.

      In this case, the barycentric square root $\sqrt Q$ is a point on the locus. With $P = \sqrt Q$, the perspector is the barycentric square root of
      (1/((S_A+S \sqrt 3)(S_A+S/\sqrt 3)) : ... : ...)
      = (1/(S_A+S \sqrt 3)^2 : ... : ...)*
      ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
      = (1/(S_A+S \sqrt 3)^2 : ... : ...)*Q,
      and is the barycentric product $X(17)*P$.

      Best regards
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