Let's solve the following problem graphically.

A passenger train traveling from $A$ to $B$ at a speed of $30 \mathrm{~km} /$ hour is usually overtaken by a fast train traveling from $A$ to $B$ at a speed of $60 \mathrm{~km} /$ hour upon reaching $B$. However, due to a technical malfunction, the passenger train continues its journey at half its speed after traveling $2 / 3$ of the distance, and as a result, the fast train overtakes it $271 / 9 \mathrm{~km}$ before $B$. What is the distance between $A$ and $B$?
a) We choose the point that represents the departure of the passenger train $S z$ from $A$ as the origin $O$ of the conventional time-distance $(t, s)$ coordinate system. (In other words: time is measured from the departure of the train, and distances are measured from $A$.)

![](https://cdn.mathpix.com/cropped/2024_05_02_ed2a9bd0a440faa853c4g-1.jpg?height=770&width=895&top_left_y=314&top_left_x=623)

The graph of the planned motion of $S z$ can be drawn as the line segment $O C$, leaving the units on the axes undetermined for now. Here, the abscissa of $C$ is the travel time $T$, and its ordinate is the distance $A B = d$. Similarly, the graph of the motion of the express train $G y$ is the segment $D C$, where $D$ is the midpoint of $O T$, since the speed of $G y$ is twice that of $S z$, so it needs $T / 2$ time to cover the distance $d$, and thus it must depart at the time $T / 2$ to arrive at $B$.

From the graph of $S z$, only the segment $O H$ has been realized, which is below the line $h$ parallel to the $t$-axis through the second third point $E$ of the segment $A B$. From the point $H$ of the malfunction, the half-speed of $S z$ is represented by the segment $H F$, whose slope is half that of $H C$, i.e., the slope of $O C$. This can be constructed, for example, by bisecting the segment $G C$ above $h$ with the point $J$ and drawing the line $H J$ to the intersection with $B C$. (The dashed lines on the diagram outline another possibility.)

The meeting point of $S z$ and $G y$ corresponds to the common point $K$ of the graphs $O H F$ and $D C$. Projecting this point onto the $s$-axis parallel to $t$ gives the corresponding point on the track $A B$, the point $U$ of the meeting. Thus, the segment $U B$ corresponds to $271 / 9 = 244 / 9 \text{ km}$. Based on this, we can approximately determine the distance $d$ by measuring the segments $A B$ and $U B$, from the ratio

$$
d: 244 / 9 = A B: U B
$$

In our diagram, $A B = 60 \text{ mm}$ and $U B = 13.5 \text{ mm}$, so $d \approx 121 \text{ km}$. (The accuracy of the result depends on the size of the diagram and the accuracy of the measurements.)

b) We can also obtain an exact result for $d$, for example, from the observation that $K$ is the centroid of the triangle $H G C$, so the height $E U$ above $H G$ is $1/3$ of the height from $C$ to $E B$, hence $U B = E B \cdot 2 / 3$, and since $E B = A B / 3$, we have $U B = d \cdot 2 / 9$. From this, $d = 122 \text{ km}$. Indeed, according to our construction, $H J$ is a median of the triangle $H G C$, and similarly, the segment $C D$ inside the triangle is also a median, since $C D$ bisects $A T$, and thus it also bisects $H G$, as this is the $1/3$ scaled image of $A T$ from the center $C$.

Horváth Mária (Hódmezővásárhely, Liszt F. Music and General School, 7th grade) Mészáros Imre (Budapest, Piarist High School, 2nd grade)