RADICAL CENTER- EULER LINE

Let ABC be a triangle and A'B'C' the cevian triangle of P = H (orthic tr.)

Denote:

Ab,Ac = the reflections of A' in BB', CC', resp.

Bc,Ba = the reflections of B' in CC', AA', resp.

Ca,Cb = the reflections of C' in AA', BB', resp.

Na, Nb, Nc = The NPC centers of A'AbAc, B'BcBa, C'CaCb, resp.

O* = The circumcenter of NaNbNc. It is the NPC center of A'B'C'.

R* = the radical center of (Na), (Nb), (Nc)

The NPC circles (Na), (Nb), (Nc) concur on the NPC of A'B'C'. The point of concurrence, the R*, is the Poncelet point of H wrt A'B'C' and since the H of ABC is the I of A'B'C', the point is the Feuerbach point [1] of A'B'C' = the center of the Feuerbach hyperbola of A'B'C'.

The points R*, H, O* are collinear. The line is the INF-line of A'B'C'

Generalization:

Let P = point on the Euler line of ABC.

Conjecture:

The points R*, P, O* are collinear.

Locus:

P = a variable point.

Which is the locus of P such that R*,P,O* are collinear?

Euler Line + ???

Is the incenter I on the locus?

Note [1]:

The Feuerbach point of ABC:

Let ABC be a triangle and A'B'C' the cevian triangle of I. Denote:

Ab, Ac = the reflections of A in BB', CC', resp.

Bc, Ba = the reflectuons of B in CC', AA', resp.

Ca, Cb = the reflections of C in AA', BB', resp.

The NPCs of AAbAc, BBcBa, CCaCb (and ABC) concur at Feuerbach point of ABC.

Antreas P. Hatzipolakis, 16 May 2013