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21330Re: Happy New Year and Nik's locus for 2013

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  • yiuatfauedu
    Jan 1, 2013
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      Dear Nik and Antreas,

      Happy New Year to you and all friends of Hyacinthos.

      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?


      If A_0BC, B_0CA, C_0AB are the equilateral triangles constructed externally on the sides BC, CA, AB respectively, your A* is the midpoint of A'A_0. If P = (u:v:W) in homogeneous barycentric coordinates, then

      A* = (-(S_B+S_C)(v+w) : (S_C+S/Sqrt[3])v + (S_C+S Sqrt[3])w
      : (S_B+S/Sqrt[3])v + (S_B+S Sqrt[3])w),
      and similarly for B* and C*.

      A*B*C* is perspective with ABC if and only if

      Cyclic sum [(S_B+S/Sqrt[3])(S_C+S Sqrt[3]) uv^2
      - (S_B+S Sqrt[3])(S_C+S/Sqrt[3]) uw^2]= 0.

      Rewriting this equation as

      Cyclic sum u/(S_A+S/Sqrt[3])[(S_C+S Sqrt[3])/(S_C+S/Sqrt[3]) v^2
      - (S_B+S Sqrt[3])/(S_B+S/Sqrt[3])w^2] = 0 ,
      we identify the locus of P as 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…

      Best regards
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