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11297Re: A Problem From an Old Swedish Friend

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  • xpolakis
    Jun 1, 2005
      --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis" <xpolakis@o...> wrote:
      > The problem:
      > Consider a triangle ABC with three concurrent cevians. Let P be their point of
      > intersection, and D, E, F their intersection points with the sides BC, AC, AB,
      > respectively.
      > The question is, can I find such a point P that the triangles APE, CPD and BPF
      > have the same area as the other three, given that none of the cevians are
      > medians of the triangle?
      > I have proven the fact for AD [or BE or CF] being a median but the question is
      > whether there exist such a point P not on a median where the areas are equal as
      > described
      > I hope you understand the description of the problem =)
      > Best wishes
      > --
      > Olle the Greatest

      If DEF is the cevian triangle of P, then I think that the locus of P
      such that areas

      APE + BPF + CPD = APF + BPD + CPE

      is the quartic with barycentric equation

      x^3(y-z) + y^3(z-x) + z^3(x-y) = 0

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