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15391Re: Cevian Trace Equal Area Points

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  • Nikolaos Dergiades
    Aug 1, 2007
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      Dear friends,
      the point Q is on the line at infinity if the
      point (p-p' : q-q' : r-r') is on the
      Steiner circumellipse.
      Hence the denominator K can be written another way
      K = xy+yz+zx where x=p-p', y=q-q', z=r-r'.
      or
      area = (ABC)/[1/(p-p')+1/(q-q')+1/(r-r')]
      = (ABC)/(BC/A"A' + CA/B"B' + AB/C"C')

      Best regards
      Nikos Dergiades

      > 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




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