Picasso Point
by RSM, Jan 19, 2012, 2:38 am
Problem:-
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.
Source:-
http://anthrakitis.blogspot.com/2012/01/paul-prize.html
Proof:-
Suppose,
are the orthocenters of 
.Note that, those nine-point centers are the midpoints of
. So its enough to prove that 
is cyclic. Now note that
is a parallelogram. So 
and 
and similar for others.Suppose,
are the diametrically opposite points of 
wrt 
. Note that 
are concurrent at the reflection of 
on 
.Now suppose,
are the foot of the perpendiculars from to . Note that 
and 
. Similar for others.So the configuration
is homothetic to 
.Clearly
lie on the circle with diameter 
where 
is reflection of on .So
is cyclic. So done.
May 2012
April 2012