Let ABC be a triangle and A'B'C' the cevian triangle of I.
Denote:
Ab,Ac = the reflections of A' in CC',BB', resp. (lying on AC, AB, resp.)
Similarly Bc,Ba = the reflections of B' in AA',CC', resp. and Ca,Cb = the reflections of C' in BB',AA', resp.
A1B1C1 = the antipedal triangle of I wrt triangle A'AbAc
Similarly A2B2C2 = the antipedal triangle of I wrt triangle B'BcBa and A3B3C3 = the antipedal triangle of I wrt triangle C'CaCb,
Oa,Ob,Oc = the circumcenters of the triangles A1B1C1, A2B2C2, A3B3C3, resp.
The circumcenter O of ABC and the circumcenters Oa,Ob,Oc of A1B1C1,A2B2C2,A3B3C3, resp. are concyclic.
Which point is the center X of the circle?
Antreas P. Hatzipolakis, 4 June 2014
X is X(5495)
Peter Moses 5 June 2014
Δεν υπάρχουν σχόλια:
Δημοσίευση σχολίου