Kornél found two rings of the same size, and the side of each ring is divided into 36 equal parts. On both rings, 18 parts are painted yellow, and 18 are painted green. Can the rings be placed on top of each other so that the dividing lines on the side of the rings align, and the overlapping side parts are of the same color in at least 18 cases?
Solution. Let's fix one of the disks and rotate the other around it. There are 36 positions where the dividing lines coincide. If we look at a yellow part of the lower disk, during the rotation, there will be 18 positions where yellow aligns above. This is true for each of the 18 yellow parts, so during the full rotation, the number of yellow-yellow alignments will be $18^{2}$. Clearly, the number of green-green alignments is the same. Therefore, if we count the number of color alignments in all 36 positions, we get a total of $2 \cdot 18^{2}$ alignments. The average number of color alignments is thus 18, and there must be a position where at least 18 alignments occur.

In general, suppose each disk has $n_{1}$ parts of color $a_{1}$, $n_{2}$ parts of color $a_{2}$, and generally $n_{i}$ parts of color $a_{i}$ ($i=1,2, \ldots, m$). In this case, there is a position where at least $[\mathrm{n} / \mathrm{m}]$ color alignments occur, where $m$ is the number of colors, and $n$ is the number of equal parts on each disk, $n=n_{1}+n_{2}+\cdots+n_{m}$. If we fix one disk and rotate the other over it, then a part of color $a_{i}$ will align with a part of the same color $n_{i}$ times, and thus the total number of color alignments is $n_{1}^{2}+n_{2}^{2}+\cdots+n_{m}^{2}$. According to the inequality between the arithmetic and quadratic means, this is at least $\left(n_{1}+n_{2}+\ldots n_{m}\right)^{2} / m=n^{2} / m$. Since the full rotation creates $n$ different positions, the average number of color alignments is $n / m$, so the maximum number is also at least $n / m$. Since the number of alignments is an integer, in the position where the most alignments occur, this number is at least $[n / m]$.

Hajnal József (Bp., I. István Gimn., III. o. t.)