7.5. On a chessboard, the centers of some cells are marked in such a way that no triangle with marked vertices is a right triangle. What is the maximum number of points that could have been marked?
Answer: 14 points. Solution. Let's call a marked point vertical if there are no other marked points on its vertical line. Similarly, we define a horizontal point. Note that any marked point is either vertical or horizontal (or both at the same time). Indeed, if the marked point $A$ had another point $B$ on its vertical line and a point $C$ on its horizontal line, then the triangle $BAC$ would be a right triangle. It is not difficult to provide an example of 14 points satisfying the condition: mark the central points in the cells a2, a3,...,a8 and in the cells $\mathrm{b} 1, \mathrm{c} 1, \ldots, \mathrm{h} 1$ (i.e., in all cells of the first vertical and the first horizontal except the bottom-left cell). Obviously, all triangles with vertices at these marked points are obtuse. We will now show that there cannot be more than 14 points. If there are 8 vertical points, then they lie in eight different vertical lines, and therefore there are no other marked points. Similarly, if there are 8 horizontal points, then there cannot be a ninth marked point. Therefore, in the optimal example, there are no more than 7 horizontal and no more than 7 vertical marked points, and the total number of marked points is no more than 14.