59th Ukrainian National Mathematical Olympiad

Until the first meeting point, the first cyclist travelled the distance $S_1=v_1$, while the second cyclist travelled $S_2=v_2$. Therefore, the distance between towns $A$ and $B$ equals $v_1+v_2$. There are two possible cases.

Case 1. Before their second meeting, both cyclists reached their destination town, and turned around. Suppose their second meeting occurred at the distance $S_3$ from town $B$ and time $t_2$ later after the first meeting (fig. 22). Then, we get the equations Fig. 23 $$v_2+S_3=v_1{t}_2\text{ and }{v}_1+(v_1+v_2-S_3)=v_2{t}_2.$$ By adding these equations, we obtain $$v_2+S_3+2v_1+v_2-S_3=(v_2+v_1)t_2\Rightarrow t_2=2.$$

Case 2. Before their second meeting, the first cyclist reached destination town $B$, turned around and reached the second cyclist before he/she reached town $A$ (fig. 23). Suppose their second meeting occurred at the distance $S_3$ from town $B$ and time $t_2$ later after the first meeting (fig. 4). Then, we get the equations $$S_3-v_2=v_2{t}_2\text{ and }{v}_2+S_3=v_1{t}_2.$$ Taking the difference of these equations, we obtain $$v_2+S_3+v_2-S_3=(v_1-v_2)t_2\Rightarrow t_2=\frac{2v_2}{v_1-v_2}.$$

Now, the only thing left is to figure out which of the two cases occurs for which $v_1,v_2$. Case 1 happens when the first cyclist reaches $A$ later than the second cyclist, i.e. $\frac{v_2+v_2+v_1}{v_1}>\frac{v_1}{v_2}\Rightarrow 2v_2^2+v_1{v}_2>v_1^2$. Let us denote $x=\frac{v_1}{v_2}$, then it must satisfy the equation $$x^2-x-2<0\Rightarrow (x+1)(x-2)<0\Rightarrow x<2,\text{ i.e. }{v}_1<2v_2.$$