FINAL ROUND

Answer: 1. Let $\angle BMA=2x$, then $\angle AMC=180^{\circ }-2x$. Since $M{B}_1$ is the bisector of the angle $BMA$, we have $\angle BM{B}_1=\angle B_{1}MA=x$. Similarly, $\angle AM{C}_1=\angle C_{1}MC=90^{\circ }-x$. The triangle $BM{B}_1$ is a right-angled triangle, so $\angle B_{1}BM=90^{\circ }-\angle B_{1}MB=90^{\circ }-x$. The right-angled triangles $BM{B}_1$ and $MC{C}_1$ are equal ($BM=MC$, $\angle B_{1}BM=\angle C_{1}MC=90^{\circ }-x$). So $B{B}_1=M{C}_1$. Since $\angle B_{1}BM=\angle C_{1}MC$, we have $B{B}_1\parallel M{C}_1$. Therefore $BM{C}_1{B}_1$ is a parallelogram. Since $B_1{C}_1\parallel BM$, we have $\angle BM{B}_1=\angle A_1{B}_{1}M=\angle B_{1}M{A}_1$. So, the triangle $B_{1}M{A}_1$ is an isosceles triangle: $M{A}_1=A_1{B}_1$. Similarly, the triangle $M{C}_1{A}_1$ is an isosceles triangle: $M{A}_1=C_1{A}_1$. Therefore, we have $B_1{A}_1/A_1{C}_1=M{A}_1/M{A}_1=1$.