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K016 and another Tucker cubic

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  • Darij Grinberg
    Let ABC be a triangle, and A B C be the cevian triangle of any point P with respect to triangle ABC. Then: (1) The (signed) areas of triangles ABC and A B C
    Message 1 of 1 , Jan 25, 2004
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      Let ABC be a triangle, and A'B'C' be the cevian triangle
      of any point P with respect to triangle ABC.

      Then:

      (1) The (signed) areas of triangles ABC and A'B'C' are
      equal if and only if the point P lies on the cubic
      with barycentric equation

      x^2 * (y+z) + y^2 * (z+x) + z^2 * (x+y) = 0.

      This is Bernard Gibert's cubic K016.

      (2) The (signed) areas of triangles ABC and A'B'C' are
      oppositely equal if and only if the point P lies
      on the cubic with barycentric equation

      x^2 * (y+z) + y^2 * (z+x) + z^2 * (x+y) + 4xyz = 0,

      or, equivalently,

      xyz + (yz + zx + xy)(x + y + z) = 0.

      Is the latter one known?

      Darij Grinberg
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