--- 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