\section*{Problem 4 - 321024}

Determine whether it is possible to inscribe more than \(64\) circles, each with a diameter of \(1 \mathrm{~cm}\), in a square with a side length of \(8 \mathrm{~cm}\), such that no two circles overlap and no point of any circle lies outside the square!
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Yes, this is possible. We provide a configuration with 68 such circles below:

To do this, we set up a coordinate system in the plane of the square such that one of its corners lies at the coordinate origin, the edges emanating from this corner lie on the positive coordinate axes, and it has an edge length of 16. (One unit of length in the coordinate system corresponds to \(5 \mathrm{~mm}\).) Furthermore, let \(h:=\frac{7}{4}\).

We now choose two types of points:

\[
\begin{aligned}
M_{1} & :=\{(1+2 i ; 1+2 j \cdot h) \mid 0 \leq i \leq 7 \wedge 0 \leq j \leq 4\} \\
M_{2} & :=\{(2+2 i ; 1+(2 j+1) \cdot h) \mid 0 \leq i \leq 6 \wedge 0 \leq j \leq 3\}
\end{aligned}
\]

Here, \(M=M_{1} \cup M_{2}\) is the set of the centers of the selected circles. First, we establish \(\left|M_{1}\right|=8 \cdot 5=40\) and \(\left|M_{2}\right|=7 \cdot 4=28\), so \(|M|=68\), since the points in \(M_{1}\) have odd \(x\)-coordinates and those in \(M_{2}\) have even \(x\)-coordinates, making the two sets disjoint.

Furthermore, the \(x\)-coordinates of all points in \(M\) are at least 1 and at most \(1+14=15\), and the same applies to the \(y\)-coordinates due to \(1+8 \cdot h=1+14=15\), ensuring that no circle with a radius of 1 around any of these centers contains points outside the square.

One can imagine these centers arranged in individual rows of the same \(y\)-coordinate: At the bottom, a row of eight centers on the line \(y=1\), then a second row of seven centers, "offset in the middle" relative to the first row, on the line \(y=1+h\), and so on, alternating accordingly.

Two centers in the same row clearly have a distance of at least 2, so the circles with a radius of 1 around them do not overlap.

Two adjacent centers in a row form, with the "center between them" in an adjacent row, an isosceles triangle with a base of length 2 and a height of \(h\). Thus, the two legs have a length of \(\sqrt{1+h^{2}}=\sqrt{1+\frac{49}{16}}=\sqrt{\frac{65}{16}}>\sqrt{\frac{64}{16}}=\sqrt{4}=2\), according to the Pythagorean theorem, so no two circles with a radius of 1 around these centers overlap. The same applies even more to centers that are further apart.

Thus, no two circles with a radius of 1 around the centers specified in \(M\) overlap, and none of these 68 circles contain points outside the square, so this arrangement of more than 64 circles satisfies the statement of the problem.