1. Petar and Toso came to visit Misha in city $A$. After some time, they decided to return to their city $B$ and organized the transport in the following way: Toso cycled from city $A$ to city $B$. Misha drove Petar towards city $B$ for some time and then left him to continue on foot, while he immediately turned back towards his city $A$. All of them left city $A$ at the same time and arrived in their respective cities at the same time. If Petar walks at $6 \mathrm{~km} / \mathrm{h}$ and Toso cycles at $22 \mathrm{~km} / \mathrm{h}$, at what speed did Misha drive the motor?
Solution. Let's denote by $s$ the distance from $A$ to $B$ and by $t$ the time they needed to reach their destinations. The speeds of the motor, bicycle, and walking, i.e., of Misha, Tosh, and Petar, let's denote them by $v_{M}, v_{T}$, and $v_{\Pi}$ respectively. Tosh was riding the bicycle from $A$ to $B$ the whole time, so,

$$
v_{T} \cdot t=s
$$

Misha, for the time $t$, drove a certain amount of time towards $B$ and then immediately returned to his city $A$. This means that he spent half of the time driving towards $B$, and in the second half, he returned to $A$. This means that Petar spent half of the time riding on the motor with Misha, and in the second half of the time, he continued walking to $B$. Since Petar traveled the distance $s$, it can be written as

$$
v_{M} \frac{t}{2}+v_{\Pi} \frac{t}{2}=s
$$

The right sides of (1) and (2) are equal, so,

$$
v_{T} \cdot t=v_{M} \frac{t}{2}+v_{\Pi} \frac{t}{2}
$$

from which we get $v_{M}=38$.