At most, how many circular sheets with a diameter of $5 \mathrm{~cm}$ can be cut from a rectangular sheet measuring $9 \times 100 \mathrm{~cm}$?
I. Solution. Let's assume that our disk is 1 mm thick, and we have already cut out many 5 cm diameter circular disks from such material, and now our task is to fit as many of these as possible into a $100 \times 9 \times 0.1 \, \text{cm}$ sized box (made, for example, of transparent material), through its $9 \times 0.1 \, \text{cm}$ sized side opening. This is obviously an equivalent problem to the original one. - Let's place the box vertically so that the opening is the lid, thus it is natural to say: the force of gravity ensures that the disks placed one after another slide to the deepest and most compact position.

The center $O_{1}$ of the first disk $K_{1}$ will be $2.5 \, \text{cm}$ above the bottom. The center $O_{2}$ of the next disk $K_{2}$, due to $K_{1}$, will be somewhere on the $5 \, \text{cm}$ radius circle around $O_{1}$, but higher than $O_{1}$ - because two disks cannot fit such that their centers are at the same height - and then $O_{2}$ will be the deepest if $K_{2}$ pushes $K_{1}$ against one side wall (I), while it itself leans against the other side wall (II). In this case, the horizontal projection of the $O_{1} O_{2}$ central line is $9 - 2 \cdot 2.5 = 4 \, \text{cm}$, and the height difference between $O_{1}$ and $O_{2}$ is $3 \, \text{cm}$ according to the Pythagorean theorem, so the highest point of $K_{1}$ is $5 \, \text{cm}$ and that of $K_{2}$ is $8 \, \text{cm}$ above the bottom.

It is easy to show that the center $O_{i}$ of any disk $K_{i}$ is at least $2.5 \, \text{cm}$ away from the closer side wall, and thus the centers $O_{i}$ are inside or on the boundary of the strip in the middle cross-section of the box, which is bounded by lines parallel to the long sides and $2.5 \, \text{cm}$ away from them; the distance between these lines is $4 \, \text{cm}$.

When inserting $K_{3}$, it cannot touch $K_{1}$, because for that it would also have to touch $K_{2}$, but the smallest width of the equilateral triangle $O_{1} O_{2} O_{3}$ (the distance between the parallel lines that can contain the triangle) is $2.5 \sqrt{3} = 4.33 \, \text{cm} > 4 \, \text{cm}$. Thus, the deepest position of $K_{3}$ is determined by $K_{2}$ and the I. side wall in the same way as the position of $K_{2}$ was determined by $K_{1}$ and the II. side wall.

Continuing, after inserting $n$ disks and placing them in the deepest position, the highest point of the disk system will be $5 + (n - 1) \cdot 3 = 3n + 2 \, \text{cm}$ above the bottom. $n$ can only be increased as long as this does not exceed 100: $3n + 2 \leq 100$, and from this, the maximum value of $n$ is 32. It is visible that this is also achievable.

Remark. The above solution is intuitive, it translated the "maximum number of" question into a practical consideration. The following elegant solution uses only mathematical tools, and is instructive in handling inequalities.

II. Solution. a) Let $S$ be any disk arrangement, that is, let $S$ be the set of circles placed on the given rectangle without overlapping, and let $S_{0}$ be the set of the centers of the circles in $S$. Place our disk in a coordinate system so that one of its vertices is at the origin, the sides starting from it lie on the positive half-axes of the coordinate axes, and the $100 \, \text{cm}$ side lies on the $x$-axis. For the coordinates $(x, y)$ of the centers in $S_{0}$,

\[
\begin{gathered}
2.5 \leq x \leq 97.5 \\
2.5 \leq y \leq 6.5
\end{gathered}
\]

holds, since the circles in $S$ are within the rectangle; furthermore, for any two arbitrary centers $\left(x^{\prime}, y^{\prime}\right), \left(x^{\prime \prime}, y^{\prime \prime}\right)$ in $S_{0}$,

\[
\left(x^{\prime}-x^{\prime \prime}\right)^{2}+\left(y^{\prime}-y^{\prime \prime}\right)^{2} \geq 25
\]

holds, since the circles in $S$ do not overlap.

The difference in the ordinates of any two arbitrary points in $S_{0}$, based on (2), is at most $6.5 - 2.5 = 4$, that is,

\[
\left|y^{\prime}-y^{\prime \prime}\right| \leq 4
\]

From this and (3), it follows that

\[
\begin{gathered}
\left(x^{\prime}-x^{\prime \prime}\right)^{2} \geq 25-16=9 \\
\left|x^{\prime}-x^{\prime \prime}\right| \geq 3
\end{gathered}
\]

that is, the difference in the abscissas of any two arbitrary points in $S_{0}$ is at least 3. According to this, the abscissas of the points in $S_{0}$ are different, and if we order them by size, and let $x_{i}$ be the $i$-th abscissa, then

\[
x_{i+1}-x_{i} \geq 3, \quad(i=1,2, \ldots, n-1)
\]

where $n$ is the number of circles in $S$. Adding the inequalities (6) for the indices $i=1,2, \ldots, n-1$, we get:

\[
x_{n}-x_{1}=\left(x_{n}-x_{n-1}\right)+\left(x_{n-1}-x_{n-2}\right)+\ldots+\left(x_{2}-x_{1}\right) \geq 3(n-1)
\]

However, according to (1), $x_{n}-x_{1} \leq 95$, so

\[
3(n-1) \leq 95
\]

that is, $n \leq 32$. According to this, we can cut out at most 32 circular disks from the sheet.
b) We show that this many can indeed be cut out. Let

\[
\begin{aligned}
& x_{i}=2.5+3(i-1), \\
& y_{i}=\left\{\begin{array}{l}
2.5 \text { if } i \text { is odd } \\
6.5 \text { if } i \text { is even, }
\end{array}\right.
\end{aligned}
\]

( $i=1,2, \ldots, 32$), then for these points $\left(x_{i}, y_{i}\right)$, (1), (2), and (3) hold, so the circles with centers at $\left(x_{i}, y_{i}\right)$ can be cut out from the rectangle.