\section*{Task 4 - 331024}

Determine whether it is possible to fit more than 64 metal spheres with a diameter of \(1 \mathrm{~cm}\) inside a cubic box that has an inner dimension of \(4 \mathrm{~cm}\) for length, width, and height, such that none of the spheres protrude over the edge!
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This is possible. We give below a distribution with 66 balls. To this end, let all lengths be measured in the unit cm, so that we only give the numerical values.

Along one edge of the base, four balls can be placed side by side such that two adjacent ones touch each other and the two outer ones touch the respective edge of the box. Four such rows can be placed on the base, forming a square grid of \(4 \times 4\) balls. The centers of these balls all lie in a plane parallel to the base at a distance of \(\frac{1}{2}\) from it.

On this layer of 16 balls, a second layer can now be placed such that each ball of the second layer rolls into the "gap" formed by four balls of the first layer that form a " \(2 \times 2\)-square" and takes a position as close to the base as possible (meaning the distance of the center of the ball from the base).

Since the \(4 \times 4\) grid of balls in the first layer results in exactly \(3 \cdot 3\) such gaps, a total of 9 balls can be placed in the second layer.

The centers of four balls of such a " \(2 \times 2\)-square" in the first layer and the center of the ball in the corresponding gap in the second layer form a square pyramid, all of whose edges have a length of 1. By symmetry, the foot of the perpendicular from the apex of this pyramid to its base lies at the center of the square, thus forming a right triangle between a vertex of the square, its center, and the apex of the pyramid. Therefore, the length of the height \(h\) of this pyramid can be calculated using the Pythagorean theorem as

\[
h=\sqrt{1^{2}-\left(\frac{\sqrt{2}}{2}\right)^{2}}=\frac{\sqrt{2}}{2}
\]

In the same way, a third layer, which is identical to the first, can be placed on the second layer, a fourth layer, which is identical to the second, and a fifth layer, which is again identical to the first.

By construction, \(16+9+16+9+16=66\) balls have been placed in the box in this way, with none of them protruding beyond the base or any of the four side faces of the box. It remains to show that the balls of the fifth layer do not protrude beyond the top face of the box.

To this end, we note that by construction, the centers of the balls in a layer always lie in a plane parallel to the plane of the centers of the balls in the layer below, and these two planes have a distance of \(h\).

Thus, the centers of the balls in the fifth layer lie in a plane parallel to the base of the box and at a distance of \(\frac{1}{2}+4 \cdot h=\frac{1}{2}+2 \sqrt{2}=\frac{1}{2}+\sqrt{8}<\frac{1}{2}+3=4-\frac{1}{2}\) from it, and thus have a distance of more than \(\frac{1}{2}\) from the top face, so that no ball of the fifth layer (and thus none of the underlying ones) protrudes beyond the top face.

Thus, it is possible to accommodate more than 64 balls in the box according to the problem statement, \(\square\).