Four vehicles were traveling on the country road at constant speeds: a car, a motorcycle, a truck, and a bicycle. The car caught up with the truck at 12:00, met the cyclist at 14:00, and the motorcyclist at 16:00. The motorcyclist met the truck at 17:00 and caught up with the cyclist at 18:00.

At what time did the cyclist meet the truck?
I. solution. Let the speeds of the four vehicles be denoted by $v_{A}, v_{R}, v_{M}$, and $v_{K}$, respectively. (The indices are the initial letters of the corresponding vehicles.) According to the problem, the Car and the Robocar are traveling in the same direction, while the Motor and the Bicycle are traveling in the opposite direction. Let $d_{1}$ and $d_{2}$ be the distances of the Bicycle and the Motor from the Car at 12 o'clock, and let $12+t$ be the time of the meeting between the Bicycle and the Robocar (Fig. 1). According to the problem, $v_{A}>v_{R}$ and $d_{2}>d_{1}$.

# 1988-02-067-1.eps 

Fig. 1

Let's write down the equations that can be derived from the meetings of the individual vehicles. Car-Bicycle at $14^{h}$:

$$
d_{1}=2\left(v_{A}+v_{K}\right)
$$

Car-Motor at $16^{h}$:

$$
d_{2}=4\left(v_{A}+v_{M}\right)
$$

Robocar-Motor at $17^{h}$:

$$
d_{2}=5\left(v_{R}+v_{M}\right)
$$

Motor-Bicycle at $18^{h}$:

$$
d_{2}-d_{1}=6\left(v_{M}-v_{K}\right)
$$

Robocar-Bicycle at $(12+t)^{h}$:

$$
d_{1}=t\left(v_{R}+v_{K}\right)
$$

Subtracting twice (1) from (2), we get:

$$
d_{2}-2 d_{1}=4\left(v_{M}-v_{K}\right)
$$

which, according to (4), is exactly $\frac{2}{3}\left(d_{2}-d_{1}\right)$. From the obtained equation

$$
d_{2}-2 d_{1}=\frac{2}{3}\left(d_{2}-d_{1}\right)
$$

we get $d_{2}=4 d_{1}$. Based on this, from (3) and (4) we now get the following equations:

$$
\begin{gathered}
4 d_{1}=5\left(v_{R}+v_{M}\right) \\
\left(d_{2}-d_{1}\right): 3=d_{1}=2\left(v_{M}-v_{K}\right)
\end{gathered}
$$

If we subtract five times (4') from twice (3'), we get the equation

$$
3 d_{1}=10\left(v_{R}+v_{K}\right)
$$

from which

$$
d_{1}=\frac{10}{3}\left(v_{R}+v_{K}\right)
$$

Substituting this into (5), we get $t=\frac{10}{3}$, so the Bicycle and the Robocar met $\frac{10}{3}$ hours after 12 o'clock, that is, at 15:20.

II. solution. In the following graphical solution, we do not need to draw conclusions about the direction of movement of the two vehicles involved from the words "caught up with" or "met". We interpret both expressions as follows: the vehicles $X$ and $Y$ were at the same point on the road at time $T$, and the points $X_{T}$ and $Y_{T}$ representing their movements coincide on the graph.

A further simplification is achieved if we represent the events as if the Car, moving at a constant speed, were standing still. In this case, the movements of the other three vehicles are still represented by straight lines in the time-distance coordinate system, whose origin is the coinciding point $A_{12}=R_{12}$ (Fig. 2). Thus, the time axis represents the "movement" of the Car, and the points $K_{14}$ and $M_{16}$ on it represent the meeting of the Bicycle and the Motor with the Car. Thus, we know one point on the movement graph of each of the other three vehicles.

Let us choose the unit of distance so that the meeting of the Motor and the Bicycle at 18 o'clock occurs at a unit distance from the Car. This is certainly possible, since this distance cannot be zero, otherwise the four graphs would coincide, and the vehicles would move together. Whether this distance is positive or not depends solely on the direction in which we observe the movement of the other vehicles from the Car. Let us choose the direction of our observer so that this distance is positive. Thus, we get the point $M_{18} \equiv K_{18}$ on the 18 o'clock line.

Now the graphs of the other three vehicles are also determined. The graphs of the Motor and the Bicycle are obtained from the two points each ($M_{16}$ and $M_{18}$, and $K_{14}$ and $K_{18}$, respectively). Then the 17 o'clock line intersects the line $M_{16} M_{18}$ representing the movement of the Motor at the point $M_{17}=R_{17}$, and thus the points $R_{12}\left(\equiv A_{12}\right)$ and $R_{17}\left(\equiv M_{17}\right)$ give the graph of the Robocar.

The point $K_{T} \equiv R_{T}$, i.e., the intersection of the lines $K_{14} K_{18}$ and $R_{12} R_{17}$, represents the time and place of the meeting of the Bicycle and the Robocar. In the triangle $R_{12} M_{16} K_{18}$, $R_{12} R_{17}$ and $K_{18} K_{14}$ are both medians, so the point $K_{T}$ is the centroid of this triangle, and its abscissa, the sought value of $T$, is the arithmetic mean of the abscissas of the vertices, $T=(12+16+18): 3=15 \frac{1}{3}$.