$A$ and $B$ are connected by a railway line; the train arriving in $B$ immediately returns to $A$. A railway also runs alongside the track. A cyclist started from $A$ at the same time as the train. Half an hour later, the cyclist was as far from $A$ as the train was from $B$. After the cyclist had traveled an additional 2 km, the positions of the two vehicles were again symmetrical with respect to the midpoint of the $A B$ segment. (The train had meanwhile turned around in $B$). A quarter of an hour later, the two vehicles met for the first time. - What is the distance between $A$ and $B$ and the speeds of the two vehicles? (We assume that the speeds are constant.)
I. solution. After the first half hour, the train still has as much distance left to $B$ as the motorbike covered during this time, so together they have covered the distance of $A B$. According to the third data point, until the meeting, the two vehicles together had traveled the distance of $A B$ back and forth. This took twice as much time, so the meeting occurred 1 hour after the start. The second symmetric situation occurred $1/4$ hour before this, i.e., $3/4$ hour after the start. The motorbike thus covered the said $2 \mathrm{~km}$ in the third quarter of an hour, so its speed was $8 \mathrm{~km}$ per hour.

According to this, the motorbike was $4 \mathrm{~km}$ from $A$ in the first symmetric situation, and $6 \mathrm{~km}$ from $A$ in the other. The train was the same distance from $B$ at these two times, and since it went to $B$ and turned back in between, it covered $10 \mathrm{~km}$ in the third quarter of an hour, so its speed was $40 \mathrm{~km}$ per hour.

Finally, the distance between $A$ and $B$, as the sum of the half-hour distances of the vehicles, is $24 \mathrm{~km}$.

Most contestants worked with a system of equations:

II. solution. Denoting the speed of the train and the motorbike (in $\mathrm{km} /$ hour units) by $v_{1}$ and $v_{2}$, respectively, the distance $A B$ (in km) by $s$, and the time required for the motorbike to cover $2 \mathrm{~km}$ (in hours) by $t$, the conditions can be written as:

$$
\begin{array}{cl}
v_{2} \cdot \frac{1}{2}=s-v_{1} \cdot \frac{1}{2}, & v_{2} t=2 \\
v_{2} \cdot\left(\frac{1}{2}+t\right)=v_{1} \cdot\left(\frac{1}{2}+t\right)-s, & \left(v_{1}+v_{2}\right)\left(\frac{3}{4}+t\right)=2 s
\end{array}
$$

Rearranging the first equation:

$$
v_{1}+v_{2}=2 s
$$

Substituting this into the last equation and simplifying by the non-zero $2 s$, and using the second equation:

$$
\frac{3}{4}+t=1, \quad t=\frac{1}{4}, \quad v_{2}=\frac{2}{t}=8
$$

Substituting these into the last two equations and rearranging:

$$
\frac{3}{4}\left(v_{1}-8\right)=s, \quad v_{1}+8=2 s
$$

and substituting $v_{1}$ from the second equation into the first:

$$
\frac{3}{4}(2 s-16)=s, \quad \text { from which } \quad s=24 \text { and } v_{1}=40 .
$$

Remark. The thought process up to the determination of $v_{2}$ is visibly the same as in the first solution.

III. solution. Measure the train's path (measured from $A$) and time (from the common start) on the axes of a Cartesian coordinate system (1st figure). The train's motion is then represented by a broken line $A D E$ rising straight to height $B$ and then descending at the same angle. The line $D E$ is the reflection of $A D$ in the line $b$ parallel to the time axis through $B$. This essentially determines the ratio of the units on the time axis and the distance axis, so only a single ray $A C s$ corresponds to the motorbike's motion. We consider the construction of this ray to be our task, and we will read the answers to our questions from the correct graph pair.

![](https://cdn.mathpix.com/cropped/2024_05_02_e51b714b91a73a69c815g-1.jpg?height=631&width=756&top_left_y=1994&top_left_x=674)

The symmetric positions of the vehicles on the track also appear in the graphs, the corresponding points of the graphs are reflections of each other in the line $k$ passing through the midpoint $K$ of segment $A B$ and parallel to the time axis. These are cut out from the train's graph by the reflection of $A C s$ in $k$, let these be $F^{\prime}, G^{\prime}$, then the reflections of $F, G, H$ in $k$ are points of $A C s$ with the same abscissa. On the other hand, the meeting point of $A C s$ and the segment $D E$ of the graph corresponds to $H$.

Let the projections of $F, G, H$ on the time axis be $S_{1}, S_{2}, T_{1}$, respectively, so the half hour that has passed until the first symmetric position corresponds to the segment $A S_{1}$ on the time axis, and the quarter hour from the second symmetric position to the meeting corresponds to the segment $S_{2} T_{1}$. Now $A S_{1}, S_{2} T_{1}$ are equal to the perpendicular height of the triangles $A B F$ and $G^{\prime} G H$ to the distance axis, and these two triangles are similar because the angles on the sides $A B, G^{\prime} G$ are equal in pairs due to the performed reflections. Therefore,

$$
G^{\prime} G: A B=S_{2} T_{1}: A S_{1}=\frac{1}{4}: \frac{1}{2}
$$

so $G^{\prime} G=A B / 2$, and $G^{\prime} S_{2}=S_{2} G=A B / 4$, where $S_{2}$ is the projection of $G^{\prime}$ on $b-$.

The reflection of $G^{\prime}$ in $b$ is on the line $A D$, and here the reflection $B C s^{\prime \prime}$ of $B C s^{\prime}$ in $b$ also passes. On the other hand, the double reflection of $B C s^{\prime \prime}$ is the translation of $A C s$ that takes $A$ to $B$. Therefore, $G G^{\prime \prime}=A B$, and $S_{2} G^{\prime \prime}=5 A B / 4=5 S_{2} G$, which determines $G^{\prime \prime}$ on $A D$ and with it $B C s^{\prime \prime}$, and thus $A C s$. According to these, the train's speed is five times that of the motorbike.

On the other hand, the reflection of $F^{\prime}$ in $b$ is the intersection of the lines $D E$ and $B C s^{\prime \prime}$, so $F F^{\prime \prime}=A B=2 G G^{\prime}$, and the similarity of the triangles $F F^{\prime \prime} H$ and $G G^{\prime} H$ implies $S_{1} T_{1}=2 \cdot S_{2} T_{1}$ and $S_{1} S_{2}=S_{2} T_{1}$, and the motorbike needed $1 / 4$ hour to cover the 2 km segment between the two symmetric positions. Therefore, the motorbike's speed is $8 \mathrm{~km} /$, and the train's is $40 \mathrm{~km} /$. Furthermore, $A T_{1}$ corresponds to 1 hour, during which the train and the motorbike together covered $48 \mathrm{~km}$, which is twice the distance of $A B$, so $A B=24 \mathrm{~km}$.

Remark. On the 1st figure, a variant of determining the train-motorbike graph pair is also visible, in which the intersection points $F, G$ are cut out from the reflection $B D_{1} E_{1}$ of $A D E$ (as if we also started a train from $B$ towards $A$). Then $B D_{1} \| D E, D_{1} E=B D=A D_{1}, A E=2 A D_{1}$, and from the similarity of triangles $A F D_{1} \triangle \sim A H E \triangle$ we get $A H=2 A F, A S_{1}=$ $A T_{1} / 2$. On the other hand, $A S_{1}=2 S_{2} T$ and from the similarity of triangles $B F F^{\prime} \triangle \sim G^{\prime} H H^{\prime} \