ADGEOM 2458

Dear friends,

I found the following:

Let ABC be triangle, A'B'C' its anticomplementary triangle, and A''B''C'' the circuncevian triangle of the infinite point of the Euler line of ABC, that is A''B''C'' are the second intersection points of the circumcircle and the parallel lines to Euler line through A,
B, C, respectively.

The conic ABCA'A'', with center the midpoint of BC is the locus of points P such the Euler lines of PBC and ABC are parallel.

The three conics ABCA'A'', ABCB'B'', and aBCC'C'' go through X1138.

Francisco Javier Garcia Capitan