The 2289th practice's three participants are setting out on their journey again. This time, they also have to cover a distance of 84 kilometers on two bicycles. On foot, they can cover 5 kilometers, and on a bicycle, they can cover 20 kilometers per hour. They still cannot ride two people on one bicycle, but now a cyclist can tow an empty bicycle, and can still maintain a speed of 20 kilometers per hour while doing so.

How long will it take for all three of them to reach the finish line?
For the sake of clarity, let $S=84$, the length of the path. The total distance traveled by the three children is at least $3S$—it can be more if they don't just move forward—and if the total distance walked is at least $S$—which is always true if an empty bicycle cannot be towed—then there is a child who walks at least one-third of the path, and thus needs at least $\frac{1}{3} \cdot S: 5 + \frac{2}{3} \cdot S: 20 = S / 10 = 8.4$ hours to reach the destination. We saw in the solution to Exercise 2289 that this amount of time is already sufficient.

According to the above, the new condition—towing an empty bicycle—can only make it possible to reduce the time spent if the total distance walked is less than $S$. In such cases, there are parts of the path where no one walks. Let's now examine such travels, i.e., let the total distance walked be $\alpha S$, where $0 \leq \alpha < 1$. In this case, at least $(1-\alpha) S$ length of the path is traveled by all three on bicycles—let's call these parts of the path "bicycle segments."

Let's examine how much time the bicycles and the children individually spend on the journey. Although the problem only specifies the children reaching the destination, it is clear that the bicycles also reach the destination.

On the bicycle segments, therefore, the children always travel by bicycle at every point $B$. Since there are only two bicycles, at least two people are on one of them when passing through point $B$, which means that this bicycle must also travel back empty through $B$. Since it cannot move independently and no one walks here, the other bicycle must tow it back, which is now possible. After the bicycles finally reach the destination, the two bicycles must again pass through $B$, this time moving forward.

On the bicycle segments, therefore, both bicycles must travel at least three times, twice forward, and once backward. We also know that the bicycles travel on the walking segments, so individually they must travel at least $\alpha S + 3(1-\alpha) S = (3-2\alpha) \cdot S$. This requires at least $(3-2\alpha) S: 20$ hours.

Now let's try to estimate the total time spent by the children. Out of the total $3S$ length of the path, $\alpha \cdot S$ is walked, and the remaining $(3-\alpha) S$ is traveled by bicycle. We have already seen, however, that on the bicycle segments, one cyclist must also travel backward—towing the other, empty bicycle—and thus this segment is traveled forward twice.

The total distance traveled by the children by bicycle is therefore actually at least twice the "pure" forward travel on the bicycle segments, i.e., at least $(3-\alpha) S + 2(1-\alpha) S = (5-3\alpha) S$.

The total time spent by the children on the journey is therefore at least $\frac{\alpha S}{5} + \frac{(5-3\alpha) S}{20} = \frac{(5+\alpha) S}{20}$. Therefore, there is one of the three children who spends at least one-third of this time, $\frac{(5+\alpha) S}{60}$ hours, on the journey. The time required to reach the destination is thus at least $\frac{(3-2\alpha) S}{20}$ hours due to the bicycles, and at least $\frac{(5+\alpha) S}{60}$ hours due to the children.

# 1987-02-071-1.eps

## 1. Figure

Let's plot the lower bounds obtained as a function of $\alpha$ (Figure 1). Then the possible $(\alpha, t)$ pairs are located in the shaded area. The slopes of the two boundary lines are of opposite sign, so the "deepest" point of the area is the intersection of the boundary lines, whose $\alpha_0$ abscissa is the solution of the equation $\frac{(3-2\alpha) S}{20} = \frac{(5+\alpha) S}{60}$, which is $\alpha_0 = \frac{4}{7}$.

Thus, the children need at least $\frac{3-2\alpha_0}{20} S = \frac{5+\alpha_0}{60} S = \frac{13}{140} S = 7.8$ hours to reach the destination. This is less than the 8.4 hours obtained in Exercise 2289, but we need to show that it is also achievable.

From the above, it is clear that this can happen if the total distance walked is $\frac{4}{7} S$.

Mark the points $T_1$ and $T_2$ on the path $AB = S$ such that $AT_1 = T_2B = \frac{2}{7} S$ (Figure 2).

## 1987-02-071-2.eps

## 2. Figure

Let one child start walking from $A$, while the other two ride bicycles. At point $T_2$, one of the cyclists dismounts and walks toward $B$, while the other turns back, towing the empty bicycle. When they reach $T_1$, they have traveled a total of $AT_2 + T_2T_1 = \frac{8}{7} S$, and thus the third child, who started walking at a quarter of the speed, has traveled $\frac{2}{7} S = AT_1$ and just arrives there. They can then immediately mount the returned bicycle and both can cycle all the way to the destination. It is also clear that in this way, the walker left at $T_2$ and the one towing the bicycle will arrive at the end of the path at the same time, since after the separation, the former travels the remaining $T_2B = \frac{2}{7} S$ in exactly the same time as the latter tows the bicycle over the four times longer $T_2T_1 + T_1B = \frac{8}{7} S$ (In Figure 2, dashed and solid segments indicate the segments traveled on foot and by bicycle, respectively).

Under these conditions, therefore, all three arrive at the destination at the same time, and the time spent is clearly $\frac{2}{7} S: 5 + \frac{5}{7} S: 20 = \frac{13}{140} S = 7.8$ hours.

Thus, the three children can reach the destination in 7.8 hours, but not in less time.

Remarks. 1. It is clear that the above is not the only travel plan that ensures the children reach the destination in exactly 7.8 hours. Additional possibilities arise if we divide the path $AB$ into segments and organize the travel on each segment according to the above method.

2. It is easy to see that towing an empty bicycle makes it possible to reduce the time required for the journey if the ratio of the cyclist's and the walker's speeds is greater than 3. In this case, it is worth inserting bicycle segments, as they can travel faster even with the return trip than on foot.