International Mathematical Olympiad

Assume $A_1{A}_2=d$, and $AB=a$. Construct an equilateral triangle $A_0{B}_0{C}_0$ with side of length $a-d$. Points $A^{\prime },B^{\prime },C^{\prime }$ are chosen on the sides of triangle $A_0{B}_0{C}_0$ such that $A^{\prime }{C}_0=A_{2}C$, $B^{\prime }{A}_0=B_{2}A$, and $C^{\prime }{B}_0=C_{2}B$.

Therefore, $$A^{\prime }{B}_0=a-d-A^{\prime }{C}_0=BC-A_1{A}_2-A_{2}C=B{A}_1$$

$$B^{\prime }{C}_0=B_{1}C,C^{\prime }{A}_0=C_{1}A.$$ Since $$\angle B_{1}C{A}_2=\angle B^{\prime }{C}_0{A}^{\prime },\angle B_{2}A{C}_1=\angle B^{\prime }{A}_0{C}^{\prime },\angle C_{2}B{A}_1=\angle C^{\prime }{B}_0{A}^{\prime },$$ thus $$△C{B}_1{A}_2\cong △C_0{B}^{\prime }{A}^{\prime },△A{B}_2{C}_1\cong △A_0{B}^{\prime }{C}^{\prime },△C_{2}B{A}_1\cong △B_0{C}^{\prime }{A}^{\prime },$$ which implies that $B^{\prime }{C}^{\prime }=C^{\prime }{A}^{\prime }=A^{\prime }{B}^{\prime }=d$ and triangle $A^{\prime }{B}^{\prime }{C}^{\prime }$ is equilateral. So $$\angle A{B}_2{C}_1=\angle A_0{B}^{\prime }{C}^{\prime }=180^{\circ }-\angle C^{\prime }{B}^{\prime }{A}^{\prime }-\angle A^{\prime }{B}^{\prime }{C}_0=120^{\circ }-\angle A^{\prime }{B}^{\prime }{C}_0,\angle C_1{B}_2{B}_1=\angle B_1{A}_2{A}_1.$$ In view of $B_2{C}_1=B_1{B}_2=A_2{B}_1=A_1{A}_2=d$, triangles $C_1{B}_2{B}_1$ and $B_1{B}_2{A}_1$ are congruent, implying that $B_1{C}_1=A_1{B}_1$. Together with $C_1{C}_2=A_1{C}_2$, we show that $C_2{B}_1$ is the perpendicular bisector of $A_1{C}_1$ and $C_2{B}_1$ is the height of triangle $A_1{B}_1{C}_1$ on side $A_1{C}_1$. Similarly, $C_1{A}_2$ and $A_1{B}_2$ are the altitudes of triangle $A_1{B}_1{C}_1$ to the sides $A_1{B}_1$ and $B_1{C}_1$ respectively.

Therefore the lines $A_1{B}_2$, $B_1{C}_2$ and $C_1{A}_2$ are concurrent.