59th Ukrainian National Mathematical Olympiad

Let $M_1$ be the point of the ray $A_1{C}_1$ such that $M_1{C}_1=B_1{C}_1$, $N_1$ be the point of the ray $C_1{A}_1$ such that $N_1{A}_1=B_1{A}_1$ (Fig. 39). Then $\angle B_1{M}_{1}K=\frac{1}{2}\angle A_1{C}_1{B}_1=\angle KM{B}_1$, whence $K{M}_{1}M{B}_1$ is inscribed and $M$ is the projection of $M_1$ on the line $AC$. Analogously, $N$ is the projection of $N_1$. Then, the segment $MN$ is the projection $M_1{N}_1$ on the line $AC$, hence $$MN\le M_1{N}_1=B_1{C}_1+A_1{C}_1+B_1{A}_1,$$ which proves the problem statement.

Fig. 39