In the 2nd problem of the first round of the 1965 National High School Mathematical Competition, 3 solutions can be read on pages 98-101 of the 31st volume (November 1965) of our journal. In each solution, a quadratic equation appeared, and we only considered the solution that matched the described motion of the cars. Now let's write down a motion scenario for the two cars that corresponds to the other solution of the mentioned equations.[^0]

[^0]:    ${ }^{1}$ We quote the problem in question:

    A truck left city $A$ for city $B$ at midnight, and a car left city $B$ for city $A$ (on the same route) at $t_{1}$ hours. They met at $t_{2}$ hours. The car arrived at its destination $r$ hours later than the truck.

    - After completing their tasks, they set off again and met again at $t_{3}$ hours, and finally arrived home at the same time. At what time did they arrive?
    - Numerical data: $t_{1}=2^{\mathrm{h}} 40^{\mathrm{m}}, t_{2}=4^{\mathrm{h}}, r=0^{\mathrm{h}} 40^{\mathrm{m}}, t_{3}=14^{\mathrm{h}}$.)

    - However, to fully understand the solution, it is necessary to study the quoted solutions.
The calculation of the arrival time $x$ at the moment of arrival was prepared in all three solutions by calculating an additional data point for the movements. This always resulted from a quadratic equation. In the first solution, we calculated the ratio of the speeds of the truck and the car $v / V$, in the second, the driving time $t$ of the truck (using the driving time $T$ of the car in the consideration), and in the third, the time $u$ that elapsed from the first meeting until the car arrived at $A$. The results - denoting the square root of the common discriminant $r^{2}+4 t_{2}\left(t_{2}-t_{1}\right)$ of the three equations by $d$ - are as follows:

$$
\begin{aligned}
& \text { I. } \frac{v}{V}=\frac{r \pm d}{2 t_{2}}, \quad \text { and from this } \quad x=t_{3}+t_{2}-\frac{t_{1}}{1+v / V} \\
& \text { II. } t=t_{2}+\frac{-r \pm d}{2}, \quad T=\frac{t\left(t_{2}-t_{1}\right)}{t-t_{2}}, \quad \text { and } \quad x=t_{3}+\frac{t\left(t_{2}-t_{1}\right)}{t-t_{1}} \text {; } \\
& \text { III. } u=\frac{r+d}{2}, \quad \text { and } \quad x=t_{3}+u\left(1-\frac{r}{t_{2}+u}\right)
\end{aligned}
$$

The results of the problem's accepted solution were given with the positive value of $d$: $v / V=2 / 3, t=6$ hours, $(T=4$ hours), $u=8 / 3$ hours, and from each, $x=16$ hours 24 minutes. However, with the negative value of $d$, we got $v / V=-1 / 2, t=1$ hour 20 minutes, $(T=-0$ hours 40 minutes), $u=-2$ hours, and in each case, $x=12$ hours 40 minutes, which we previously ignored. Now we need to interpret these latter results.

Let's assume that the movements continued after reaching the destination and were already in progress before the start. We interpret the first two data points of the problem as follows: the truck passed through $A$ at midnight, and the car passed through $B$ at $t_{1}$ hours. The negative value of the speed ratio can be interpreted in the usual way of speaking to mean that one of the movements is in the opposite direction to what the text suggests, i.e., they are moving in the same direction; therefore, instead of the head-on meetings in the original problem, it is more intuitive to say that one car overtakes the other. This is consistent with the fact that in the equation setup, differences should be used instead of sums. However, since changing the sign of the speed also reverses the direction of the distance (since $s=v t$), our formulas and equations remain valid.

A problem statement leading to the ignored results is as follows:

"A long route is approached by two buses, a slower one, $a$, and a faster one, $b$. The garage of $a$ is in the city $A$ on the route, and that of $b$ is in $B$. They record the passage of the buses at the stations and also when they are at the same point on the route. - One day, $a$ passed through $A$ at midnight, heading towards $B$. Some time later, the $B$ station informed $A$ of $a$'s passage. $r$ hours later, $b$ passed through $A$, at $t_{1}$ hours through $B$, and at $t_{2}$ hours it overtook $a$. In the opposite direction, $b$ overtook $a$ at $t_{3}$ hours. Interestingly, in this journey, $a$ passed its garage at the same time as $b$ passed its garage. At what time did this happen?"

Indeed, here $b$ plays the role of the car in the previous problem, and $-V$ takes the place of $V$ in the previous problem, so $v / V=1 / 2$, meaning the speed of bus $a$ is half that of bus $b$. Their travel times on the $A B$ segment are $t=1 \text{ hour } 20 \text{ minutes}$ and $T=40 \text{ minutes}$, respectively. Thus, bus $a$ passed through $B$ at $1 \text{ hour } 20 \text{ minutes}$. Therefore, bus $b$ passed through $A$ $r$ hours later, at $2 \text{ hours}$, and $T$ hours later, at $2 \text{ hours } 40 \text{ minutes}$, through $B$ (indeed at $t_{1}$ hours). Since $t_{1}=2t$, by this time bus $a$ had traveled $2 \cdot A B$, giving it a lead over $b$ equal to the $A B$ segment. Since the speed advantage of $b$ over $a$ is exactly the speed of $a$, $b$ will catch up to $a$ after another $t$ time, i.e., at $4 \text{ hours}$, as required by the problem. - On the return trip, they pass through $B$ and $A$ in the opposite direction, and at the same time, $b$ is at $B$ and $a$ is already at $A$, so the next overtaking can only occur after $A$. Now $b$ must catch up a lead of $A B$ length, which, according to the calculated $x=12 \text{ hours } 40 \text{ minutes}$, will end after another $t$ time, when the clock shows $14 \text{ hours}$, thus fulfilling the requirement of the problem.

We arrive at the same pair of movements by interpreting the results $t=1 \text{ hour } 20 \text{ minutes}$ and $u=-2 \text{ hours}$ from the second and third solutions. The first was ignored because $t < t_{2}$. If $a$ had already passed through $B$ at $1 \text{ hour } 20 \text{ minutes}$ and was only overtaken by $b$ at $4 \text{ hours}$, then $b$ must have been traveling in the opposite direction. Similarly, $u=-2 \text{ hours}$ means that the car $b$ passed through $A$ 2 hours earlier, at $2 \text{ hours}$, and only reached $B$ at $2 \text{ hours } 40 \text{ minutes}$, so it was traveling from $A$ towards $B$.