- Mar 1, 2011
> Let ABC be a triangle P = (x:y:z) a point, A'B'C'

It's a cubic, 0=cyclic sum of

> the cevian triangle of a point Q = (u:v:w).

>

> Which is the locus of P such that the circles

> (APA'),(BPB'),(CPC') are coaxial?

(u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz>

It's a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz

> Specialcase: P,Q are isogonal conjugate points.

>

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> Antreas

Barry Wolk - << Previous post in topic Next post in topic >>