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19878Re: [EMHL] A generalization...

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  • Barry Wolk
    Mar 1, 2011
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      > Let ABC be a triangle P = (x:y:z) a point, A'B'C'
      > the cevian triangle of a point Q = (u:v:w).
      >
      > Which is the locus of P such that the circles
      > (APA'),(BPB'),(CPC') are coaxial?

      It's a cubic, 0=cyclic sum of
      (u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz
      >
      > Specialcase: P,Q are isogonal conjugate points.

      It's a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz

      >
      > Antreas

      --
      Barry Wolk
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