'Circumcircular conjugation' of X(1) with respect to X(3)
Circumcircular conjugate is defined here:
https://math.stackexchange.com/questions/3786640/new-conjugation
https://www.geogebra.org/m/c8xfvkyc
Let's find the circumcircular conjugate of the incenter of the triangle ABC with respect to the circumcenter of ABC.
I is the incenter, R - circumcenter of the triangle ABC
B1 is a point lying on the circumcircle of ABC where it intersects with the circumcircle of the triangle IRB
A1 is a point lying on the circumcircle of ABC where it intersects with the circumcircle of the triangle IRA
C1 is a point lying on the circumcircle of ABC where it intersects with the circumcircle of the triangle IRC
AA1, BB1, CC1, IR meet at the point I1. This Point I1 is the circumcircular conjugate of I with respect to R.
Also Oa,Ob,Oc (3 centers of circles ) are collinear. IE=ID IF=IG. GB1 is parallel to A1E.
3 lines drawn through points (A1,Oa), (B1,Ob), (C1,Oc) intersect at some point X.
X is always lying on the circumcircle of the triangle ABC. It must be a classic Triangle center. (or it even might be a new one/lesser known one)
Finally, the line that is drawn through the points (Oa,Ob,Oc) is perpendicular to the line that goes through the points (R, I, I1) and RJ=JI. J is probably another lesser known triangle center.