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## 15391Re: Cevian Trace Equal Area Points

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• Aug 1, 2007
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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