On a circle with a circumference of 15 units, we select a finite set of points such that from any point in this set, measured along the circumference, there is exactly one point in the set at a distance of 2 units and exactly one point at a distance of 3 units. Is it possible for the number of points in the set not to be a multiple of 10?

[^0]:    ${ }^{1}$ Our question is a slight modification of the following problem. "On a circle with a circumference of 15 units, we select a finite set of points such that, measured along the circumference, from any point in this set, there is exactly one point in the set at a distance of 1 unit and exactly one point at a distance of 2 units. Prove that the number of points in the set is a multiple of 10." This problem appeared in the following publication: D. O. Skljarskij-N. N. Chencov-I. M. Yaglom: Selected Problems and Theorems in Elementary Mathematics, Part II (Volume 1). Tankönyvkiadó, Budapest, 1972. Problem 42.
A single suitable example is enough to answer the question: yes, it is possible. Draw a regular 15-sided polygon in the circle, mark one of its vertices as \(A_{1}\), one of its neighbors as \(A_{2}\), and then the subsequent vertices in this direction as \(A_{3}\), ..., \(A_{15}\). In this direction, the distance between any two vertices measured along the arc is as many units as the difference in their indices \(\overparen{A_{i} A_{j}}=j-i\) (if \(j>i\)), and in the opposite direction, it is the complement to 15: \(15-(j-i)=(15+i)-j\), always a positive integer.

![](https://cdn.mathpix.com/cropped/2024_05_02_ea6a42e4485a5fcfa0ebg-1.jpg?height=599&width=531&top_left_y=405&top_left_x=775)

Figure 1

Choose the first point of our set to be \(A_{1}\), and then sequentially those vertices that we find by advancing 2 and 3 units of arc alternately in the direction. These "additional" points are suitable because one of their nearest (closest) selected points is 2 units away in one direction, and 3 units away in the other direction, while the next neighbor is \(2+3=5\) units away in both directions.

Since the arc length covered with this \((2+3)\) step pattern is 3 times (an integer multiple) the circumference of our circle, after 3 step patterns we return exactly to \(A_{1}\), and the second part of the requirement is also met for \(A_{1}\), and the number of selected points is \(3 \cdot 2=6\), not a multiple of 10. - Therefore, the answer to the question is indeed yes. (In Figure 1, the line connecting \(A_{1} A_{3} A_{6} A_{8} A_{11} A_{13} A_{1}\) is not essential, it only aims to highlight better.)

Remarks. 1. Human weakness, to give a hasty, exaggerated answer to questions that can be quickly reviewed. The submitted solutions often lack the statement that from no point is there more than one point in the set 2 or 3 units away. (However, this omission was overlooked in the grading by the editorial board.)

2. From vertex \(A_{6}\), we can return to \(A_{1}\) more quickly by moving in the opposite direction through \(A_{4}\) (dashed line in Figure 1). However, this already cannot omit the missing proof, as it cannot be said that any further point is only 2 or 3 units away from the set, since \(A_{3} A_{4}=1\).
3. In the source of the problem given at the time of its release (where 1 and 2 units are used instead of 2 and 3 units), we return to the starting point with only \(15:(1+2)=5\) "mixed pairs" of steps, and thus we mark \(5 \cdot 2\) points. The recently mentioned return is not possible, as in the path \(A_{1} A_{2} A_{4} A_{3}\), \(A_{3}\) would also be one unit away from \(A_{2}\) (Figure 2). This is the basis for the statement there.

![](https://cdn.mathpix.com/cropped/2024_05_02_ea6a42e4485a5fcfa0ebg-1.jpg?height=735&width=246&top_left_y=1858&top_left_x=926)

4. The hasty conclusion mentioned in Remark 1 is even more pronounced in such a statement: "the set can only consist of 6 (or only 4) points" (this can also be found scattered in the submitted solutions). Here, the preparatory 15-sided polygon has already risen to the challenge of further selections. By combining the marked \(B_{i}\) and \(C_{j}\) points in Figure 3, we have exactly 10 points. By rotating the preparatory 15-sided polygon by an arc smaller than one unit, we can mark any number of additional 4 or 6 element subsets for our set.

![](https://cdn.mathpix.com/cropped/2024_05_02_ea6a42e4485a5fcfa0ebg-2.jpg?height=802&width=746&top_left_y=282&top_left_x=668)