## 11297Re: A Problem From an Old Swedish Friend

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• 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

APH
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