4. From the town "U $" to the town "A x" at $10^{00}$ AM, Ivan set off on his bicycle. After traveling two-thirds of the way, he passed the town "Ox," from which at that moment Peter set off on foot towards the town "U x". At the moment Ivan arrived in the town "A x", Nikolai set off from there in the opposite direction on his bicycle and arrived in the town "U x" by the morning of the same day. At what distance from the town "A x" did Nikolai catch up with Peter, if Peter arrived in the town "U x" at $12^{00}$ AM of the same day, and the speed of each participant was constant, with the distance between the towns "U x" and "A x" being 10 km in total. (7 points)
Answer: 6 km.

## Solution:

We will solve the problem using a graphical-geometric method. Let's represent Ivan's movement as segment $K L$, Nikolai's movement as segment $L M$, and Petr's movement as segment $N P$ in a coordinate system $(t ; s)$, where $t$ is time in hours and $s$ is distance in kilometers from point A (Fig.1). Let $Q$ be the intersection point of $L M$ and $N P$. According to the problem, $M K=2$ and $P M=1$. Draw $M G \| N Q, G \in K L$, then by Thales' theorem we have

$$
N G: G K=P M: M K=1: 2 .
$$

Thus, if $N G=2 x$, then $G K=4 x$, and $L N=3 x$. Therefore, by Thales' theorem again, we have

$$
L Q: Q M=L N: N G=3 x: 2 x=3: 2
$$

Thus, the required distance is $\frac{3}{5} \cdot 10=6$ (km.)

Note that when finding the ratio $L Q: Q M$, we can use Menelaus' theorem. Points $N, Q$, and $P$ lie on the same line, so $\frac{K N}{N L} \cdot \frac{L Q}{Q M} \cdot \frac{M P}{P K}=1$ or $\frac{2}{1} \cdot \frac{L Q}{Q M} \cdot \frac{1}{3}=1$. Therefore, $\frac{L Q}{Q M}=$ $\frac{3}{2}$.