1964. A Construction For X(18569)

 Let A1B1C1 reflection triangle of X(3)-circumcenter on the side lines of ABC.

Tangents to circumcircle of A1B1C1 at A1, B1, C1 form a triangle A2B2C2.*

*Circumcenter of A2B2C2 lies on Euler line of ABC. It's X(18569).