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Ukrainian National Mathematical Olympiad

Geometry Difficulty 6.8 National Olympiad Prove it Ukraine

A square of size n×nn \times n is given. Some of its 1×11 \times 1 cells are marked. It turned out that there is no convex quadrilateral with vertices at these marked points. For each natural number n≥3n \ge 3, find the largest value of mm for which this is possible.
A quadrilateral is called convex if both of its diagonals lie inside the quadrilateral.

Figure 1
Fig. 6

Solution

We mark two points in the corner cells of the left column, as well as all the points in some non-edge row. Then we have n+2n + 2 marked points, none of which form a vertex of a convex quadrilateral (Fig. 6).

We will show by contradiction that it is not possible to mark more than n+2n + 2 points. Suppose at least n+3n + 3 cell centers are marked. It is clear that if there are two rows, each with at least 2 marked points, then by taking exactly 2 points from each of these two rows, we obtain a convex quadrilateral. Otherwise, in at least the (n−1)(n-1)-th row, no more than 1 point is marked. Then there must be a row in which at least 4 points are marked. Similarly, there must be a column in which at least 4 points are marked. Let this row and column intersect at point AA. It is easy to see that there are at least 2 points in this row that lie on one side of point AA, and similar 2 points can be found for the column. These 4 points form a convex quadrilateral.

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