3-5. What is the smallest number of points that can be chosen on a circle of length 1956 so that for each of these points there is exactly one chosen point at a distance of 1 and exactly one at a distance of 2 (distances are measured along the circumference)?
Solve 5. Answer: 1304.

Let \(A\) be one of the selected points, \(B\) and \(C\) be the selected points, located at distances 1 and 2 from it, respectively. The arrangement \(A B C\) is impossible, since in this case, for point \(B\), there are two selected points at a distance of 1. Therefore, the points are arranged in the following order: \(C \quad A B\) (or \(B A \quad C\)). Let \(D\) be the point located at a distance of 1 from \(C\). The arrangement \(C D A B\) is clearly impossible. Therefore, the arrangement is as follows: \(D C \quad A B\). Let \(E\) be the point located at a distance of 1 from \(B\). It is arranged as follows: \(D C \quad A B \quad E\). Continuing these considerations, we will see that the circle of length 1956 will be divided into \(1956 / 3 = 652\) arcs of length 3 (the endpoints of these arcs are the points \(A, C, E, \ldots\)). Each arc contains one point. In total, we get \(2 \cdot 652 = 1304\) points. All these points must be present.

\section*{8th Grade}