\section*{Problem 1 - 061231}

In the same plane, let \(n\) points \((n>2)\) be distributed such that for each of them, there is only one nearest neighbor among the others. For each of these \(n\) points, the vector starting from it and ending at its nearest neighbor, and only this vector, is drawn.

Determine the maximum number of these vectors that can end at the same one of the \(n\) points.
}

The maximum possible number is 5.

To demonstrate this, let's first assume there are 6 points \(P_{1}\) to \(P_{6}\), each of whose nearest point is a different point \(Q\).

W.L.O.G., let the points be labeled such that the vectors from \(P_{i}\) to \(Q\) are encountered in ascending order when moving in a mathematically positive orientation around \(Q\).

For simpler notation, let's additionally define \(P_{7} := P_{1}\). Thus, the 6 angles \(\angle P_{i} Q P_{i+1}\) together form a full angle at \(Q\). Consequently, at least one of these angles is at most \(60^{\circ}\), W.L.O.G., let this be \(\angle P_{1} Q P_{2}\). Then, one of the other interior angles of the triangle \(\triangle P_{1} Q P_{2}\) is at least as large, W.L.O.G., let this be \(\angle Q P_{1} P_{2}\).

Since in a triangle, the side opposite the larger interior angle is always longer, the segment \(\overline{P_{2} Q}\) is at least as long as the segment \(\overline{P_{1} P_{2}}\), meaning the point \(P_{1}\) is at most as far from \(P_{2}\) as \(Q\) is.

This, however, contradicts the assumption that the uniquely determined nearest point to \(P_{2}\) would be \(Q\). Therefore, at most 5 vectors can arrive at a point.

That this is indeed possible is shown by the configuration of a regular pentagon and its center. Then, the above consideration always leads to the result that the connecting segment between two "adjacent" vertices is always longer than the respective radii. (And the other diagonals are even longer.)