Let P = (x:y:z) be a point on the Euler line of ABC.
P1, P2, P3 = the P-points of the triangles A'B"C", B'C"A", C'A"B", resp. ie P1 = (x:y:z) wrt A'B"C", P2 = (x:y:z) wrt B'C"A", P3 = (x:y:z) wrt C'A"B". Or alternatively, since the triangles share the same circumcenter, which is the NPC center N of ABC: NP1/NH1 = NP2 / NH2 = NP3 / NH3 = t (where H1,H2,H3 are the orthocenters of the triangles A'B"C", B'C"A", C'A"B")
P'1, P'2, P'3 = the P-points of the triangles A"B'C', B"C'A', C"A'B', resp.
The circumcenter of P1P2P3 is N, the common circumcenter of the triangles.
The triangles P1P2P3, P'1P'2P'3 are perspective and inscribed in a r. hyperbola with center the perspector of the triangles.
As P moves on the Euler Line (or t varies) which is the locus of the perspector?
Antreas P. Hatzipolakis, 13 May 2013