Olympiad : The maximum number of pairs whose distance is $1$ when any pairwise distance is at least $1$ Consider the ten points in the plane. Each distance between any two points is larger than or equal to $1$. Then what is the maximum number of pairs whose distance is $1$? 
 I guess that the answer is $19$: Consider the two regular hexagons placed so that the center of each is a corner of the other.
My Trial : As like Dan, I will use the same notations. 
Consider the graph such that there exist ten vertices and edges between two vertices has length 1. 
Step 1 : The graph is connected. 
Proof : If not, we push a component to another component closely. So we can draw another edge. This completes the proof. 
Step 2 : Note that any such connected graph has the equality : 
  $v-e + f=1$ where $v$, $e$ and $f$ are the numbers of vertices, edges, and faces respectively. 
So $10- e + f =1$ This implies that the maximum of $f$ gives the maximume of $e$. 
Consider the graph whose figure is the two regular hexagons placed so that the center of each is a corner of the other.
Note that any vertex in the graph has degree $\geq 3$. This implies that the maximum of $f$ occurs. 

The pairs give you a planar graph (the shortest intervals cannot intersect). Assume we have 20 pairs. Then, denoting the total number of faces by $A$, we get $10+A-20=2$ by Euler, so $A=12$. However, we also have $2\cdot 20\ge 3A_3+4\cdot(12-A_3-1)+Q$ where $A_3$ is the number of triangles and $Q$ is the number of boundary edges for the unbounded face. Thus, we must have $Q\le A_3-4$, $A_3\le 11$. On the other hand, $Q$ unit intervals can bound at most $\frac 1{\pi\sqrt 3}Q^2$ triangles, so $(A_3-4)^2\ge \pi\sqrt 3 A_3>5A_3$ which is impossible for $A_3\le 11$. 
