64*. Can 1965 points be placed in a square with a side of 1 so that any rectangle of area $1 / 200$ with sides parallel to the sides of the square contains at least one of these points?
64. Answer: it is possible.

In the square $0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1$ and on the line $y=1 / 2$, uniformly place $c_{0}=200$ points: ( $\left.k / 201 ; 1 / 2\right), k=1,2, \ldots$ $\ldots, 200$. Then on each of the lines $y=1 / 4$ and $y=3 / 4$, place $c_{1}=100$ points ( $\left.k / 101 ; 1 / 4\right),(k / 101 ; 3 / 4), k=1,2, \ldots$ $\ldots, 100$. Repeating this process, for $m=2,3, \ldots, 7$ and on each line $y=(2 l-1) 2^{-m-1}, 1 \leqslant l \leqslant 2^{m}$, place $c_{m}$ points $\left(\frac{k}{c_{m}+1} ; \frac{2 l-1}{2^{m j+1}}\right)$, where $c_{m}=\left[200 \cdot 2^{-m}\right]$ (when $m=7$, 128 corresponding lines will have one point each). In total, the number of points placed is

$$
\begin{aligned}
& \sum_{m=0}^{7} 2^{m} c_{m}=200+2 \cdot 100+4 \\
& \cdot 50+8 \cdot 25+16 \cdot 12+32 \cdot 6+ \\
& \quad+64 \cdot 3+128=1704 \text { points. }
\end{aligned}
$$

This construction is illustrated in Fig. 45.

It is clear that no rectangle of area $1 / 200$ can fit between the placed points. If it intersects the line $y=1 / 2$, then its base is not

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-128.jpg?height=381&width=411&top_left_y=817&top_left_x=635)

Fig. 45 greater than $1 / 201$; if not, then it is located entirely above or below this line. In this case, if it intersects the line $y=3 / 4$ or $y=1 / 4$, its height is not more than $1 / 2$, and its base is not more than $1 / 101$, and so on. If the rectangle is entirely located between lines of the form $y=$ $=n / 256$ ( $n=1,2, \ldots, 255$ ), then its height is not more than $1 / 256$.