We can arrange the numbers $1, 2, \ldots, 17, 18$ on the 18 nodes of Figure 18, placing one number in each node.

a) Show that in any arrangement, there exists a pair of adjacent (i.e., connected by a straight line) numbers whose difference is greater than 3.

b) Write the difference of the numbers at the endpoints of each connecting line on the line itself. Find an arrangement such that the smallest of the written differences is as large as possible. 1 1

![](https://cdn.mathpix.com/cropped/2024_05_02_ab5c3731209cd87d099bg-1.jpg?height=287&width=282&top_left_y=388&top_left_x=911)[^0]


[^0]:    ${ }^{1}$ Sketch the thought process for the search, highlight the result.
![](https://cdn.mathpix.com/cropped/2024_05_02_755a373737e285d97c6bg-1.jpg?height=271&width=247&top_left_y=122&top_left_x=923)

Figure 1

a) Let's call the line segments between two adjacent nodes in the figure as path segments, and specifically, those whose extension is also a path segment as ray path segments, and the rest as circular path segments. From any node in the figure, we can reach any other node by traversing at most 5 path segments, specifically 3 circular path segments and 2 ray path segments. Therefore, for example, from the node labeled 1 to the node labeled 17, on the shortest path, the increase of 17 - 1 = 16 units must include at least one path segment with an increase greater than 3, even if we end up at a larger number with each of the 5 path segments, since 16 > 5 * 3.
![](https://cdn.mathpix.com/cropped/2024_05_02_755a373737e285d97c6bg-1.jpg?height=462&width=1182&top_left_y=789&top_left_x=450)

Figure 2

b) Figure 2 shows three arrangements where the smallest difference (in absolute value) is 6. This was achieved in the I. and II. arrangements by dividing the numbers into "small" ones: 1, 2, ..., 9, and "large" ones: 10, 11, ..., 18, and writing large numbers next to small ones and vice versa. The III. arrangement achieved the goal without this additional requirement. In the I. and III. arrangements, however, the numbers on the outer and inner circular paths that are 5 path segments apart (i.e., the farthest apart) from each other, and also the numbers on the opposite points of the middle circular path, add up to 19. Both of these additional self-imposed requirements facilitated and accelerated the arrangement.

Gönczi István (Miskolc, Földes F. Gymnasium II. grade) Greskovics Klára (Budapest, Eötvös J. Gymnasium I. grade)

Remarks. 1. The statement can also be proven by the following local (pertaining to only a small part of the figure) consideration. Consider the neighbors of the number 1. If the number 1 is at one of the nodes of the middle circular path, it has 4 neighbors, so a difference of 4 already occurs if we choose the closest 2, 3, 4, and 5 for its neighbors. If the number 1 has only 3 neighbors, choosing 2, 3, and 4 for them—and only then—there is no difference greater than 3. Thus, the number 2 is not a neighbor of 3 and 4, and it must have two additional neighbors. If these are the smallest possible: 5 and 6, a difference of 6 - 2 = 4 still occurs.

Similarly, it can be shown that the smallest difference cannot be 8: Otherwise, the neighbors of 9 could only be 1, 17, and 18, and the neighbors of 10 could only be 1, 2, and 18, so neither 9 nor 10 would be on the middle circular path, and yet they would have two common neighbors: 1 and 18. This is impossible.

Bauer Katalin (Budapest, Berzsenyi D. Gymnasium II. grade)

2. Further remarks are contained in the article published on page 202 of this issue.