Let ABC be a triangle, and PaPbPc the pedal triangle of P.

The line IPa intersects the circumcircle of ABC at A1,A2

The line IPb intersects the circumcircle of ABC at B1,B2

The line IPc intersects the circumcircle of ABC at C1,C2

(where A1,B1,C1 lie on the positive side of the corresponding

triangle sides, and A2,B2,C2 on the negative side).

Which is the locus of P such that the triangles

ABC, A1B1C1; ABC, A2B2C2 are perspective?

In general: Instead of the incenter I, a fixed point Q = (f:g:h).

Antreas

PS: Sorry if this isn't "The first time that Antreas asked this

question,"