Can a million standard ping pong balls with a diameter of $4 \mathrm{~cm}$ fit into a box measuring $200 \mathrm{~cm} \times 164 \mathrm{~cm} \times 146 \mathrm{~cm}$?
Place the box on one of its $200 \times 164 \mathrm{~cm}$ sides and place the ping-pong balls - which we will consider as spheres from now on - such that they touch each other and the bottom of the box (1st figure, top view). This way, we can place $200: 4=50$ spheres in one row and $164: 4=41$ rows in one layer; thus, a total of $50 \times 41=2050$ can be placed. If we continued this "packing" method, then - since $146: 4=36.5$ - at most $2050 \times 36=73800<100000$ balls would fit in the box.

Let's find a more efficient placement. Place the balls of the second layer such that each one fits into the depression created by the four balls below it. In such a layer, $49 \times 40=1960$ balls can be placed.

The centers of the four lower balls form a square $O_{1} O_{2} O_{3} O_{4}$, and the center of the fifth upper ball $O_{5}$ together with them forms a square pyramid (2nd figure). The distance from the apex of the pyramid to the base can be determined using the Pythagorean theorem:

$$
O O_{5}=\sqrt{4^{2}-(2 \sqrt{2})^{2}}=2 \sqrt{2}=d
$$

This is the height by which the centers of the balls in the second layer rise above the plane defined by the centers of the balls below them.

The distance between two even (or odd) layers - or the distance defined by the centers of the balls placed in the same manner - is $2 d=4 \sqrt{2}$. The centers of the balls in the lowest and highest layers are $2-2 \mathrm{~cm}$ away from the bottom and top of the box, respectively. Subtract this from 146, and in the remaining space, $\left[\frac{142}{4 \sqrt{2}}\right]=25$ even or odd layers fit. Thus, the total number of balls that can be placed is: $25(2050+1960)+2050=102300$. Therefore, 100000 ping-pong balls fit in the box.

András Katona (Debrecen, Fazekas M. Gymnasium, 9th grade)

Note. The illustration for the solution can be seen on the first inside cover.