PICASSO POINT.
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)
Denote:
N1,N2,N3 the NPC centers of the triangles A"BC, B"CA, C"AB, resp.
The four NPC centers N, N1,N2,N3 are concyclic.
See Picasso Point Generalized.
Call the circle (N,N1,N2,N3) as Picasso Circle with respect ABC and D and let P0 be its center (ie the center of the Picasso circle with respect ABC and D)
Now, omit the vertex A and replace it with A" and denote Pa = the center of the Picasso circle with respect A"BC and D.
Similarly denote Pb = the center of the Picasso circle with respect AB"C and Pc = the center of the Picasso circle with respect ABC" and D.
Are the following true?
1. The four circles (P0), (Pa), (Pb), (Pc) are concurrent at a point Pp.
2. The four centers P0, Pa, Pb, Pc and the point D lie on a circle with center Pp.
APH, 5 February 2012
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1. It is true. The point Pp is the reflection of O on the midpoint of DN, that is DONPp is a parallelogram, and DPp is always parallel to Euler line.
2. The four centers P0, Pa, Pb, Pc, BUT NOT THE POINT D lie on a circle with center Pp. The radius of this circle is HALF the distance between D and O.
Francisco Javier García Capitán
6 February 2012
