A cylindrical, closable container has a diameter of 82 units and a length of 225 units. How many spheres with a diameter of 38 units can be placed inside, and then the container can be closed?
We set the cylinder's axis vertically so that the opening is at the top. The diameter of $2R = 82$ units is clearly the internal diameter of the container. When 3 spheres fit exactly into the container's neck, such that their centers $O_{1}, O_{2}, O_{3}$ lie in a horizontal plane $S_{1}$, and they are arranged so that each pair touches each other, the sides of the triangle $O_{1} O_{2} O_{3}$ are equal to the diameter $2r$ of the spheres. The radius $r_{1}$ of the circumcircle $k_{1}$ of this triangle is $\frac{2}{3}$ of the triangle's height $r \sqrt{3}$. The radius $R^{*}$ of the smallest circle concentric with this, which encloses the three mutually touching great circles of the spheres in the plane $S_{1}$, is $r_{1} + r$, i.e.,

\[
R^{*} = r \left( \frac{2}{3} \sqrt{3} + 1 \right) = \frac{r}{\sqrt{3}} (2 + \sqrt{3}) = r \cdot 2.1547
\]

and for $r = 19$, $R^{*} = 40.94$ units, which is smaller than the container's radius $R = 41$. Considering the slight (less than $0.2\%$) deviation, we will henceforth consider the cylinder's diameter to be $2R^{*}$.

Clearly, another layer of 3 spheres can be placed above the existing layer, such that the lines connecting the centers of the upper and lower spheres, $O_{4} O_{1}, O_{5} O_{2}, O_{6} O_{3}$, are vertical, and the plane $O_{4} O_{5} O_{6}$ is $2r$ units above $S_{1}$, since each new sphere touches one of the lower ones. However, if we rotate the new layer until the center $O_{4}$ of the middle sphere touches the center $O_{2}$ of the lower middle sphere, the plane $S_{2}$ of the new centers will be deeper, and more spheres can likely fit into the cylinder. In this case, the projection $O_{4}^{*}$ of $O_{4}$ onto $S_{1}$ will lie on the perpendicular bisector of the segment $O_{1} O_{2}$, since the projections of the segments $O_{4} O_{1} = O_{4} O_{2} = 2r$ are equal, and $O_{4}^{*}$ is $r$ units from the wall of the container, i.e., on the circle $k_{1}$. Thus, $O_{4}^{*} O_{2} = r_{1}$, and the height difference between the planes $S_{1}$ and $S_{2}$ is given by the right triangle $O_{4} O_{2} O_{4}^{*}$:

\[
d = \sqrt{(2r)^{2} - r_{1}^{2}} = r \sqrt{\frac{8}{3}} ( = 1.6330 r = 31.027 \text{ units } ).
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_287cb38b4ad375890270g-1.jpg?height=631&width=639&top_left_y=1056&top_left_x=732)

By placing the subsequent layers in the same manner, the plane $S_{n}$ of the centers of the 3 spheres in the topmost (the $n$-th) layer can be at most $m - r$ units high, where $m$ is the (internal) height of the cylinder, otherwise the container cannot be closed. Thus, the number of layers following the first layer is the integer part of the ratio $(m - 2r) : d$:

\[
n - 1 = \left[ \frac{m - 2r}{d} \right] = \left[ \frac{187}{31.027} \right] = 6
\]

so at least 21 spheres can be placed in the container.

Remarks. 1. We do not get a better arrangement if we loosen the first three spheres so that they touch the wall of the cylinder. In this case, the layers will indeed be lower, but (as can be verified by some calculations) only seven layers will fit into the cylinder. Of course, if the cylinder were very tall, this small modification would also be worth considering.

2. We could also place the spheres such that between two layers of 3 spheres each, we place one sphere that touches all 6 spheres. However, this would require increasing the distance between the layers so much that ultimately fewer spheres would fit into the cylinder.

3. Our problem "How many spheres can be placed?" was interpreted in our solution as: "Place as many of the given spheres as possible in the cylinder." According to this, it would be an acceptable solution if we did not rotate the second layer to fit into the gaps of the first layer and placed only 15 spheres. Of course, according to the requirement to place "as many as possible," this solution would be considered of lesser value than the above solution. The fact that we suspect that more than 21 spheres cannot be placed in the cylinder does not significantly increase the value of our solution. A significant addition would be a proof of this, but this would exceed the scope of our paper, and we did not expect it from our contestants. (Here we note that we do not consider it correct to always clearly define what we consider a complete solution to each problem. We also want to educate our solvers to decide for themselves what they need to do in various situations: we feel that this brings their thinking closer to practical, everyday problems.)