Problem A. 559. (March 2012)
A. 559. The incircle of triangle ABC is k. The circle kA touches k and the segments AB and AC at at A', AB and AC respectively. The circles kB, kC and the points B', C' are defined analogously. The second intersection point of the circles A'B'AB and A'C'AC, other than A', is K. The line A'K meets k at R, other than A'. Prove that R lies on the radical axis of the circles kB and kC.
(Kolmogorov's Cup, 2011; a problem by F. Ivlev)
Deadline expired on 10 April 2012.