\section*{Problem 3 - 061213}

In a rectangular box with internal dimensions \(10 \mathrm{~cm}, 10 \mathrm{~cm}\) and \(1 \mathrm{~cm}\), identical spheres with a diameter of \(1 \mathrm{~cm}\) are to be placed. Someone claims that more than 105 of these spheres can fit into the box.

Determine whether this claim is correct!
}

In addition to the possibility of arranging 100 balls in 10 rows of 10 each, there is also the possibility of arranging 10 balls in one row, 9 balls in the next row, then 10 balls again, and so on. The distance \(d\) between the lines through the centers of the first and the second row of balls is \(d = r \cdot \sqrt{3}\), where \(r\) is the radius of the balls; because the distance between the lines through the centers is equal to the length of the height of a triangle whose vertices are the centers of three adjacent balls (see image). Such a triangle is equilateral and has a side length of \(2r\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1819.jpg?height=320&width=512&top_left_y=2193&top_left_x=772)

If we denote the number of rows of balls by \(n\), then for the distance \(h\) between the tangent planes in the described arrangement of the balls,

\[
h = 2r + (n-1) r \sqrt{3}
\]

For \(n = 11\), it follows (since \(r = \frac{1}{2}\)) that \(h < 10\).

Therefore, 11 rows can fit in the box. The balls can be arranged such that 6 rows contain 10 balls each and 5 rows contain 9 balls each, resulting in a total of 105 balls in the box. In the 'outer' rows, there are 10 balls each. If one of the rows with 9 balls is replaced by a row with 10 balls, the distance

\[
h = 2r + r(11-1) \sqrt{3} = 1 + 5 \sqrt{3}
\]

increases to

\[
h' = 6r + 8r \sqrt{3} = 3 + 4 \sqrt{3} < 3 + 4 \cdot 1.74 = 9.96
\]

Since \(h' < 10\), 106 balls can fit in the box. The claim is therefore correct. Adapted from \([5]\)