Seven vertices are chosen in each of two congruent regular 16-gons. Prove that these polygons can be placed one atop another in such a way that at least four chosen vertices of one polygon coincide with some of the chosen vertices of the other one.
If possible, please provide a detailed explanation of the solution.
There are 16 ways (if no flip is allowed) to place one polygon on top of the other. For each way count the number of matching pairs. This gives numbers $a_1, \ldots, a_{16}$. Now each chosen position from one polygon will match exactly seven times with some position of the other. Therefore $\sum_k a_k = 7\cdot 7=49$. This means that $a_k \geq 4$ for at least one index $k$ since $16\cdot 3$ is only $48$.
An explanation of the numbers in the problem.
For two random (independent, uniform) subsets of size $k$ from an $n$ element set, the expected size of their intersection is $\frac{k^2}{n}$. Here $k=7$ and $n=16$, and the average intersection size is $49/16$, which is slightly more than $3$, so that above-average intersections have $4$ or more elements.
In this question, you have rotations that can be used to adjust the relative position of the two sets, and problem is essentially to show that this repositioning can be done in a way that the intersection is larger ($\geq$) than average. Of course you can arrange for it to be larger than the average over all rotated placements, but the point is to make it larger than the average over all random placements. The calculations of "average" in its two meanings are almost identical, and the answers are equal.
Thus, to force intersection size of at least $s$ one needs $k^2 \geq (s-1)n+1$ with the most interesting cases happening when there is equality.
