Problem 10.3. At exactly noon, a truck left the village and headed for the city, at the same time, a car left the city and headed for the village. If the truck had left 45 minutes earlier, they would have met 18 kilometers closer to the city. And if the car had left 20 minutes earlier, they would have met $k$ kilometers closer to the village. Find $k$.
Answer: 8.

Solution. We will express all distances in kilometers, time in hours, and speed in kilometers per hour. Let the distance between the village and the city be $S$, the speed of the truck be $x$, and the speed of the car be $y$. The time until the first meeting is $\frac{S}{x+y}$, so the distance from the village to the actual meeting point is $\frac{S x}{x+y}$.

In the first hypothetical scenario, the truck will travel $0.75 x$ kilometers in 45 minutes, meaning it will take another $\frac{S-0.75 x}{x+y}$ hours to reach the expected first meeting point, and the truck will travel another $\frac{(S-0.75 x) x}{x+y}$ kilometers. According to the problem, we know that

$$
18=0.75 x+\frac{(S-0.75 x) x}{x+y}-\frac{S x}{x+y}=0.75 \frac{x y}{x+y}
$$

from which we find $\frac{x y}{x+y}=18 \cdot \frac{4}{3}=24$.

In the second scenario, the car will travel $y / 3$ kilometers in 20 minutes, meaning it will take another $\frac{S-y / 3}{x+y}$ hours to reach the expected second meeting point, and the truck will travel $\frac{(S-y / 3) x}{x+y}$ kilometers during this time. The value we need to find is

$$
\frac{S x}{x+y}-\frac{(S-y / 3) x}{x+y}=\frac{x y / 3}{x+y}=24 / 3=8
$$

Another solution. Instead of the scenario "the truck left 45 minutes earlier," we will consider the scenario "the car left 45 minutes later" - it is clear that the meeting point will be the same in this case.

We will solve the problem graphically. Along the $O x$ axis, we will plot the coordinates of the vehicles such that the city corresponds to the coordinate 0 and the village to the coordinate $p>0$. Along the $O t$ axis, we will plot time (the common start of movement in the original scenario will correspond to $t=0$). We will represent the car's movement with a red line and the truck's movement with a blue line; in addition, we will mark two alternative graphs of the car's movement (with a start 45 minutes later and 20 minutes earlier). The meeting point in the original scenario (location and time) will be denoted by $A$, and in the alternative scenarios by $B$ and $C$:

![](https://cdn.mathpix.com/cropped/2024_05_06_d4986512fdb0330805a0g-34.jpg?height=493&width=711&top_left_y=551&top_left_x=371)

By Thales' theorem,

$$
\frac{45}{20}=\frac{A B}{A C}=\frac{\Delta x_{1}}{\Delta x_{2}}
$$

Since $\Delta x_{1}=18$ kilometers according to the problem, then $\Delta x_{2}=8$ kilometers.