It's already solved. See here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2699668&sid=0ecfd9a96555096cbce4cb53167d61d4#p2699668

I liked this problem the most of the geos on the test This is much shorter, I think: Let $I$ be the incenter of $\triangle ABC$ and reflect $A_1$ over $BI_b$ to $A_2$, $B_1$ over $CI_c$ to $B_2$, $C_1$ over $AI_a$ to $C_2$. Now our condition is $A_1B_2=B_1C_2=C_1A_2$. Since the perpendicular bisector of each segment is an angle bisector of $ABC$, we see $(A_1B_1C_1A_2B_2C_2)$ has center $I$. It suffices to show $I_a,I_b,I_c$ lie on the same circle. Now $A_2I_b=I_bA_1$ and $\angle A_2C_1I_b=\angle I_bC_1A_1$, so $I_b$ lies on the circle, and similarly for the others, as desired. The weird length conditions and angle bisectors scream "reflections!"