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21335Re: [EMHL] Re: Happy New Year and Nik's locus for 2013

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  • Nikolaos Dergiades
    Jan 2, 2013
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      Dear Paul,
      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'.

      Best regards
      Nik



      > 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
      > Sincerely
      > Paul
      >
      >
      >
      >
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