Minimum number of distances determined by $n$ points on a plane Set $S$ is the set of the distances between any two dots in a panel. Prove that given $202$ dots in a panel, the set will have at least $11$ members.
I tried using pigeonhole principle. Every three dots can form a equilateral triangle but adding a fourth dot will definitely produce another distance. But I don't know how to proceed.
 A: This concerns a famous result of P.Erdos, later generalized by others. We in fact have :

If $f(n)$ denotes the minimum number of distances determined by $n$ points on a plane, then $$
\sqrt{n - \frac 34} - \frac12 \leq f(n)
$$

Proof of the lower bound : The $n$ points can be enclosed in a convex shape $P$, called the convex hull of the points. This can be proved using linear algebra, for example, and can be shown in this case to be a convex polygon which has vertices made up from these points.
Let $P_1$ be any one of the points lying on $P$ as well. Consider the distances given by $P_1P_i, i=2,3,...,n$, and let $K = |\{|P_1P_i|, i=2,3,...,n\}|$.
If $N$ is the maximum number of times a distance occurs, we must have $KN \geq n-1$, obviously i.e. $f(n) \geq \frac{n-1}{N}$.
Furthermore, if a distance $r$ occurs $N$ times, then these points occuring $n$ times must lie in the same semicircle of radius $r$ around $P_1$, since these points lie on one side of $P_1$ because $P_1$ is on the convex hull.
It is easy to see that these points among themselves have at least $N-1$ different distances (draw a diagram and see this). Therefore $f(n) \geq N-1$.
Thus, for any $N$ we have $f(n) \geq \max\{N-1 , \frac{n-1}{N}\}$. It is easy to minimize this over all $n>N >0$ and obtain the expression given.
It is even easier to obtain a rough upper bound, but remarkably there are better lower bounds than the above. One of these is $Cn^{\frac 57}$ for a fixed constant $C>0$, which I believe is the best.
