Given three points A, B and C,
find the point P such that angle(PAB) = angle(PBC) = angle(PCA). Express tan(PAB) in terms of A, B and C. 

The trilinear coordinates of P are the 3 values (PF, PG, PE) or any 3 multiples of them by a same factor. Therefore we have (PF, PG, PE) = (PB, PC, PA) since PA = (A1,A)*cos(A1,A,P) = (A1,A)*sin(P,A,B) = CA/sin(CAB)*sin(PAB) therefore (PB, PC, PA) = (AB/sin(ABC), BC/sin(BCA), CA/sin(CAB)) Simplifying notation, The trilinears of P are (c/sin(B), a/sin(C), b/sin(A)) And for its isogonal conjugate (b/sin(C), c/sin(A), a/sin(B)) Remarkably simple ... with a twist.