Olympiad : The maximum number of pairs whose distance is $1$ when any pairwise distance is at least $1$

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 $A_3\le 11$.

Well, you are probably right. Here is a plan of a proof.

For convenience, let's build a graph. The set of vertices $V$ will consist of the given ten points. The set of edges $E$ will contain all pairs of vertices at distance $1$ from each other. We need to prove that $|E| < 20$. To do this it will be enough to prove that $$ \sum_{v \in V} \deg v < 40, $$ where $\deg v$ is the number of neighbours of vertex $v$. Let's rewrite this a little: $$ \sum_{v \in V} (\deg v - 4 ) < 0. $$ Let us call $\deg v - 4$ the "excess degree" of vertex $v$. We need to prove that the sum of all excess degrees is negative.

To do that, let's deal separately with positive and negative excess degrees.

Degrees 5 and 6. For positive excess degrees, it will suffice to show somehow that $$ \sum_{v \in V, \deg v > 4} (\deg v - 4) < 6. $$ I can't see a clever short way to prove this, but it can probably be done with some case analysis. First one can deal separately with the case when there is at least one vertex with degree $6$ (this is a very "rigid" situation). Then, if there are no vertices with degree $6$, it's enough to prove that there are at most 5 vertices with degree $5$, and that shouldn't be too hard.

Degrees 0, 1, 2, 3. For negative excess degrees, we can do this a bit more elegantly. Let us look at any vertex $v$ that is also a corner of the convex hull of set $V$. Let $\angle v$ be the value of this angle of the convex hull (considered as a polygon, not a set). It is easy to see that $$ \angle v \geqslant 60^\circ (\deg v - 1). $$ From this we know that every corner of the convex hull has negative excess degree. If we add together all these inequalities for all the corners of the convex hull, we get $$ 180^\circ (m-2) \geqslant 60^\circ \sum_{v\,\textrm{corner}} (\deg v - 1). $$ $v$ in this sum iterates over all the corners of the convex hull, and $m$ is the number of these corners. It's easy to see that the inequality is equivalent to this: $$ \sum_{v\,\textrm{corner}} (\deg v - 4) \leqslant -6. $$ Obviously, the bound will still hold if we extend the sum to all vertices with negative excess degrees: $$ \sum_{v \in V, \deg v < 4} (\deg v - 4) \leqslant -6. $$

If we combine this with the bound for positive excess, we will get exactly what we wanted.

So, the only weak link here is the bound for positive excess. What I described sounds like a doable thing, but it looks very clumsy. Maybe someone will come up with a way to avoid case analysis there.