4. On each side of the square, 100 points are chosen, and from each chosen point, a segment perpendicular to the corresponding side of the square is drawn inside the square. It turns out that no two of the drawn segments lie on the same line. Mark all the intersection points of these segments. For what largest $k<200$ can it happen that exactly $k$ marked points lie on each drawn segment? (N. Avilov, I. Bogdanov)
Answer. For $k=150$. Solution. Estimation. Suppose there is an example with $k>150$. We will associate it with a $200 \times 200$ table, the rows of which correspond to horizontal segments (ordered from bottom to top), and the columns to vertical segments (ordered from left to right). In the cell of the table, there is a 1 if the corresponding segments intersect, and 0 if they do not. According to our assumption, in each row and each column, there are $200-k<50$ zeros. Note that among the rows from 51 to 150, there is at least one that starts and ends with 1 (otherwise, we would have at least 100 zeros in the first and last columns together). The corresponding horizontal segment $T$ intersects both the leftmost and the rightmost vertical segments. But there are zeros in this row, which means that some segment $X$ "falls short" of $T$ either above or below. Then in the column corresponding to segment $X$, either

![](https://cdn.mathpix.com/cropped/2024_05_06_2fbed22acb1306d1b27dg-1.jpg?height=340&width=346&top_left_y=2143&top_left_x=1569)
above, or below our row, there are only zeros, meaning there are at least 50 of them. Contradiction. Example. On the right in the figure. Each of the segments represents 50 parallel and equal segments of the example.