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15388Re: Cevian Trace Equal Area Points

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  • Nikolaos Dergiades
    Jul 31, 2007
      Dear Francois
      thank you very much for explaining
      what is equicenter E and areal point Q.
      Another simple construction for Q comes from the
      following property:

      If A'B'C', A"B"C" are two inscribed triangles
      in ABC we construct the line La that passes
      through A and is parallel to the line that connects
      the mid points of B'C", B"C' and similarly
      construct the lines Lb, Lc. These three lines
      are concurrent at Q the areal point such that
      the signed area (QA'A") = (QB'B") = (QC'C").

      If in barycentrics the points are
      A' = (0 : p : 1-p)
      B' = (1-q : 0 : q)
      C' = (r : 1-r : 0)
      A" = (0 : p' : 1-p')
      B" = (1-q' : 0 : q')
      C" = (r' : 1-r' : 0)
      then the point Q is the isotomic conjugate
      of (p-p' : q-q' : r-r') and the signed area is
      (ABC).(p-p').(q-q').(r-r')/K where
      K = pq+qr+rp+p'q'+q'r'+r'p'+pp'+qq'+rr'-
      -(p+q+r)(p'+q'+r').

      Best regards
      Nikos Dergiades

      > Given any 2 inscribed triangles PaPbPc and UaUbUc of
      > the reference triangle
      > ABC, (cevian triangles are not needed!), there
      > exists (in general), a unique
      > point Q in the plane, such that:
      >
      > Area(QPaUa) = Area(QPbUb) = Area(QPcUc)
      >
      > In this formula , area means signed area of course!
      >
      > Now I give you a way (without proof for lack of
      > room) to construct this
      > point.
      > 1° Let E be the fixed point of the affine map f
      > sending PaPbPc on UaUbUc.
      > E is the equicenter, following Neuberg notation. Why
      > equicenter? Simply
      > because E has same barycentrics in PaPbPc and
      > UaUbUc.
      > There is a very known construction of E, due to
      > Sollertinski. I have already
      > talked about it in some previous posts.
      >
      > 2° Let g be the affine map sending PaPbPc on ABC,
      > then Q = g(E).
      > It is also very easy to have a construction of Q.
      >
      > This area equality is the reason why I have called Q
      > the areal center in
      > choo-choo theory.




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