Problem 6

Use directed angles modulo $180^\circ$. The circumcenter result in Step 2 gives $B_1A=B_1C=B_1B_2$ and $C_1A=C_1B=C_1C_2$. Hence $A,C,B_2$ lie on the circle centered at $B_1$, while $A,B,C_2$ lie on the circle centered at $C_1$. Since $C_2$ lies on $AB_1$ and $B_2$ lies on $AC_1$, we calculate $\measuredangle B_2B_1C_2=180^\circ-\measuredangle AB_1B_2=180^\circ-2\measuredangle ACB_2=180^\circ-2(60^\circ-\alpha)=60^\circ+2\alpha$. Similarly, $\measuredangle B_2C_1C_2=180^\circ-\measuredangle AC_1C_2=180^\circ-2\measuredangle ABC_2=180^\circ-2(60^\circ-\alpha)=60^\circ+2\alpha$. Thus $B_2,C_1,B_1,C_2$ are concyclic; call their circle $\omega_a$. Cyclically define $\omega_b=(C_2A_1C_1A_2)$ and $\omega_c=(A_2B_1A_1B_2)$. The same angle computations prove these incidences, not merely by a picture. Finally, $A_2,C_1,B_2,A_1,C_2,B_1$ occur in this order on a strictly convex hexagon: each $A_i$ lies beyond the two adjacent points on the corresponding cevians because $0<\alpha,\beta,\gamma<30^\circ$. If the three auxiliary circles were not pairwise distinct, all six vertices would be concyclic; but a convex cyclic hexagon would satisfy $360^\circ=\angle C_2A_1B_2+\angle B_2C_1A_2+\angle A_2B_1C_2=\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ$, impossible. Hence they are pairwise distinct.