Paul Erdos was a great mathematician who used to pose problems in mathematics periodicals (American Mathematical Monthly and others) with money prizes. I will do the same for a problem, but instead of money, the prize will be a classical book of geometry. It is a good coincidence that I have named the Point of the problem by the name of another great Paul: Pablo Picasso, and also today (18 January) is the birthday of my son Paul.
Let ABC be a triangle, A'B'C' the antipodal triangle of ABC (=circumcevian triangle of O) and D a point. Let A"B"C" be the circumcevian triangle of D with respect A'B'C' (ie A" is the other than A' intersection of A'D with the circumcircle etc)
N,N1,N2,N3 the NPC centers of the triangles ABC, A"BC, B"CA, C"AB.
The four centers N,N1,N2,N3 are concyclic.
The center of the circle is General Picasso Point [See HERE]
Terms of Proofs:
The proof must be Euclidean synthetic (ie with no use of algebra or coordinate geometry)
The first, who will post a proof to the list Hyacinthos, will win the book:
F.G.-M.: Exercices de Geometrie. Huitieme Edition.
The book has 1302 pages + a 36 pages supplement
From page 1130 to page 1259: Geometrie du triangle: