Bororin P.A.

On the surface of a spherical planet, there are four continents separated from each other by an ocean. We will call a point in the ocean special if for it there are at least three nearest (equidistant from it) land points, and all are on different continents. What is the maximum number of special points that can exist on this planet?
If the continents represent the vertices of a regular tetrahedron inscribed in a sphere, then there are exactly four special points - these are the ends of the radii drawn from the center of the sphere through the centers of the faces of the tetrahedron.

Let's show that there cannot be five or more special points regardless of the arrangement of the four continents. Let \( A \) be a special point, \( r \) be the distance from \( A \) to the nearest points of land (at least three of which, by condition, lie on different continents). The ocean points that are closer to point \( A \) than \( r \) form a spherical "cap" \( H(A) \). Spherical arcs \( A X \) and \( B Y \), leading from different special points \( A \) and \( B \) to some of their nearest points of land \( X \) and \( Y \) respectively, can intersect only at their endpoints, i.e., in the case \( X = Y \). Indeed, let the arcs \( A X \) and \( B Y \) intersect at point \( O \) and \( O Y \leq O X \). Then \( A Y < A X \leq r \), which contradicts the definition of \( r \).

Problem 65257 Topics: [Examples and counterexamples. Constructions] [Dirichlet's principle (other). \(\quad\)]

Difficulty: 5 Classes: 10,11

Authors: Bogdanov I.I., Podlipskii O.K.

Given natural numbers \( a \) and \( b \), where \( a < b \), and a positive integer \( N \), we need to prove that there exists a positive integer \( K \) such that in a \( K \times K \) square grid, if we mark at least \( \alpha K^2 \) cells, where \( \alpha = \frac{1}{2a^2 - 2ab + b^2} \), then there is a \( N \times N \) subgrid containing at least \( \frac{N^2}{D} \) marked cells, where \( D = 2a^2 - 2ab + b^2 \).

To prove this, we use the following inequality:
\[
\frac{K(K-a-b)}{D} = \frac{(a+b)^2 N^2 (N^2 - 1)}{D}
\]

By dividing our \( K \times K \) square into \( (a+b)^2 N^2 \) smaller \( N \times N \) squares, we get that in one of them, there are more than
\[
\frac{N^2 - 1}{D}
\]
marked cells; hence, there are at least
\[
\frac{N^2 - 1}{D} + \frac{1}{D} = \frac{N^2}{D}
\]
marked cells, which is what we needed to prove.

## Answer

\[
\alpha = \frac{1}{2a^2 - 2ab + b^2}
\]