6. Let $V$ be the set of vertices of a regular 21-gon. Given a non-empty subset $U$ of $V$, let $m(U)$ be the number of distinct lengths that occur between two distinct vertices in $U$. What is the maximum value of $\frac{m(U)}{|U|}$ as $U$ varies over all non-empty subsets of $V$ ?
Solution: To simplify notation, we will let $m$ be $m(U)$ and let $n$ be $|U|$. First note that there are 10 different diagonal-lengths in a regular 21-gon. Now consider the following set of 5 vertices.

Note that each of the 10 different diagonal-lengths appear (exactly once each). So for this set of 5 vertices we have $\frac{m}{n}=\frac{10}{5}=2$. We will now show that this is the maximum possible value for $\frac{m}{n}$. If $U$ is an arbitrary non-empty set of vertices, then there are two cases:
- Case 1: $n<5$. The total number of pairs of vertices in $U$ is given by $\frac{1}{2} n(n-1)$. Since $n-1<4$ this gives us the bound:
$$
m \leq \frac{n(n-1)}{2}<\frac{n \times 4}{2}=2 n .
$$

Thus $\frac{m}{n}<2$ in this case.
- Case 1: $n \geq 5$. The total number of distances in $U$ is at most 10 because there are only 10 different diagonal lengths in the 21-gon. Therefore
$$
\frac{m}{n} \leq \frac{10}{n} \leq \frac{10}{5}=2
$$
as required.
Remark: The construction given is unqiue up to rotations and reflections. I.e. all sets that achieve the value $\frac{m}{n}=2$ are congruent to the example given here.