A $10 \mathrm{~km}$ side square area has two parallel roads running along two of its sides. There are five observation stations on the territory. Connecting roads, consisting of segments parallel to the sides of the square, must be built so that it is possible to cycle from each station to each of the roads. (Cycling on the roads is not allowed.)

What is the shortest total road length that makes this possible, regardless of the placement of the observation stations? (See the solution to problem 3 of the 1972 Kürschák competition on pages 53-57 of the current volume of our journal.)
Let's say that we placed the map of the area under consideration in such a way that one of the highways is above and the other below, and let the five stations be $A, B, C, D, E$ from left to right. We plan three road systems. In the first, we lead a vertical 10 km road through $A$ and $D$, and from $B$ to the first, from $C$ and $E$ to the second, we lead horizontal road sections (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_6f200fdb421035b1b8ddg-1.jpg?height=353&width=399&top_left_y=344&top_left_x=844)

1st figure

In the second, the vertical roads run through $B$ and $E$, and we connect $A$ and $C$ to the first, and $D$ to the second with horizontal roads (2nd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_6f200fdb421035b1b8ddg-1.jpg?height=363&width=412&top_left_y=911&top_left_x=846)

2nd figure

In the third, there is only one vertical road, the one passing through $C$, to which we connect the other points with horizontal roads.

![](https://cdn.mathpix.com/cropped/2024_05_02_6f200fdb421035b1b8ddg-1.jpg?height=358&width=399&top_left_y=1442&top_left_x=844)

3rd figure

In the three road systems, there are five vertical roads in total, and their combined length is $50 \mathrm{~km}$. Project the points onto the lower highway and denote the projections by $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $D^{\prime}$, and $E^{\prime}$ (coincidence is of course possible). In the network formed by combining the three road systems, each of the segments $A^{\prime} B^{\prime}$, $B^{\prime} C^{\prime}$, $C^{\prime} D^{\prime}$, $D^{\prime} E^{\prime}$ is crossed by three horizontal roads, so the total length of the horizontal roads is three times the length of the $A^{\prime} E^{\prime}$ segment, which is at most $30 \mathrm{~km}$. The total length of the three road systems is therefore at most $80 \mathrm{~km}$, so the shortest of them is at most $26 \frac{2}{3} \mathrm{~km}$. A total road length of $26 \frac{2}{3} \mathrm{~km}$ is therefore always sufficient.

![](https://cdn.mathpix.com/cropped/2024_05_02_6f200fdb421035b1b8ddg-1.jpg?height=420&width=395&top_left_y=2224&top_left_x=846)

4th figure

We will show that with the arrangement in the 4th figure, a road network of $26 \frac{2}{3} \mathrm{~km}$ is also necessary. Based on any road network, group the points such that any points that can be reached from each other by bicycle are placed in the same group.

If this results in at least three groups, the total length is at least $30 \mathrm{~km}$, since at least 10 km of road is associated with each group, and these do not overlap.

If there are two groups, the total length of the vertical road segments associated with the road network is at least $20 \mathrm{~km}$. The total length of the horizontal road segments is at least the sum of the largest distances between the projections of the points in the groups. If the larger group has 4 points, this is at least as much as $A^{\prime} D^{\prime}$, which is $8 \frac{1}{3} \mathrm{~km}$, and if it has 3 points, it is at least $A^{\prime} B^{\prime} + C^{\prime} E = 6 \frac{2}{3} \mathrm{~km}$. The total length is at least $26 \frac{2}{3} \mathrm{~km}$ in both cases.

In road networks where the points form a single group, consider the roads from $A$ to $D$ and from $B$ to $E$ (or one of them if there are more). These necessarily intersect; denote by $P$ the point where, traveling from $A$ towards $D$, we first reach the $B E$ road, and by $Q$ the point where we finally leave it. If we reach $P$ from $B$ to $E$ before $Q$, then the road segments $A P B$ and $D Q E$ have no common points, so the total length of their vertical segments is at least $20 \mathrm{~km}$, to which at least $A E$ horizontal segments are added, and together they are at least $30 \mathrm{~km}$ (5th figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_6f200fdb421035b1b8ddg-2.jpg?height=393&width=353&top_left_y=931&top_left_x=867)

5th figure

If $P$ and $Q$ are the same, or if $P$ is after $Q$ on the $B E$ road, then the roads $A P E$ and $B Q D$ have no common horizontal segments, so the total length of the horizontal segments of the road network is at least $A E + B D = 16 \frac{2}{3} \mathrm{~km}$, which, together with at least 10 kilometers of vertical segments, is at least $26 \frac{2}{3} \mathrm{~km}$.

Note. The problem is related to the generalization of a problem from the 1972 Kürschák competition. Accordingly, at the time of the problem's proposal, the editorial referred to the following article: János Surányi: The solution of the problems of the 1972 Kürschák József mathematical student competition, K.M.L. 46 (1973) 49-57, specifically pages 53-57.

The above solution can be derived from one of the solutions of the cited Kürschák competition problem by replacing every consideration in the original solution that exploits the fact that there are exactly 4 observation stations with a consideration that ensures the desired result for 5 stations. It may be instructive to compare the two solutions from this perspective.