Problem
GEO-B3-M05-P022 Center of a Tucker Circle
Let \(K\) be the Lemoine point and \(O\) the circumcenter of triangle \(ABC\). Triangle \(A'B'C'\) is obtained from \(ABC\) by a homothety centered at \(K\). Extensions of the sides of \(A'B'\), \(B'C'\), \(C'A'\) meet the sides of \(ABC\) in six points lying on a Tucker circle. Prove that the center of this circle lies on line \(KO\).
C. Hint 1. Tucker circles are best viewed as images of one circle under a homothety-like variation.
D. Hint 2. Use that the construction direction is controlled by the Lemoine point.
Under the homothety centered at \(K\), the sides \(A'B'\), \(B'C'\), \(C'A'\) remain parallel to the sides of \(ABC\). The six side intersections form a configuration in which corresponding segments are antiparallel. Hence they lie on a Tucker circle.
As the homothety ratio varies, this circle continuously changes into a Lemoine circle. The centers of all these circles keep the same direction of variation: it is the line joining the circumcenter \(O\) to the center of the symmedian structure \(K\). In coordinate form, the equation of a Tucker circle is a linear combination of the circumcircle equation and the circle equation associated with parallels through \(K\); the centers of such circles lie on the line joining the original centers, namely \(KO\).
This is a preview-level problem; a coordinate proof is an acceptable main route.