If the distances between every pair of airports is distinct, prove that more than $5$ planes can't land at any airport 
The distances between any two airports of a country are distinct. From every airport one airplane takes off and lands at the closest airport. Prove that there doesn't exist an airport where more than 5 planes will land.

I proved it in the following way:
In order for a plane to take off from a certain airport and land at another, it means that the other airport is the closest to it at a distance $r$. Hence it has a circle with radius r with center the airport which it took off from and with a point on its circumference, the airport at which it landed. If there are more than $5$ points which land at a certain airport, then you get a polygon $A_1A_2A_3...A_n$ with a point inside it $O$. If $n\gt 5$ then we have that at least one $\angle OA_iA_{i+1}\lt60^o$ which is indeed a contradiction.
This is what I managed to do for this question, but I am not completely certain I am completely correct. Could you please verify my solution and explain any other nice approaches and also explain to me why my last statement that $\angle OA_iA_{i+1}\lt60^o$ indeed is a contradiction as I worked it out only intuitively?
 A: Your proof is pretty much complete. Here are the gory details.
The only facts you need about triangles are:

*

*For any two vertices $A$ and $B$, $m(\angle A)\ge m(\angle B)$ implies the side opposite $A$ is at least as long as the side opposite $B$. This follows from the law of sines.


*$m(\angle A)+m(\angle B)+m(\angle C)=180^\circ$
Suppose planes at airports $A$ and $B$ both fly to the airport at $C$. I claim this implies $m(\angle ACB)>60^\circ$.
Proof: Since $A$ flew to $C$ instead of $B$, the distance $AC$ must be strictly less than $AB$, implying $m(\angle B)<m(\angle C)$. Similarly, since $B$ flew to $C$ instead of $A$, it must be true that $m(\angle A)<m(\angle C)$. Finally, if we had $m(\angle C)\le 60^\circ$, we would have $$180^\circ=m(\angle A)+m(\angle B)+m(\angle C)<60^\circ+60^\circ+60^\circ=180^\circ,$$ a contradiction. We conclude $m(\angle C)>60^\circ$.
Now, suppose that planes at $A_1,\dots,A_n$ all fly to point $O$, where $A_i$ are listed in clockwise order around $O$. On the one hand,  $m(\angle A_1OA_2)+m(\angle A_2OA_3)+\dots+m(\angle A_nOA_1)=360^\circ$, since the angle form a complete circle. On the other hand, we just proved that each $m(\angle A_iOA_{i+1})>60$. This implies that $n\cdot 60<360$, so that $n<6$.
A: Here are airports:

We're considering the red (at the center).  Take a nearest neighbor (blue) airport a distant $R$ from the center, whose plane must land at red.  Take the "next" blue.  We want it to also land at red but not at the other blue.  Thus it a distance $R$ from red but must be as close to the first blue as possible (so we can pack as many other airports as possible).  Assume for the moment we're allowed to put it a distance $R$ from the first blue.  (We can't, but continue reading...)  We can continue like this to get the hexagonal packing in the figure.
BUT....
the problem states the airports distances must all be different.  Thus the angle from the red center to at least two "neighboring" blues must be larger than $\pi/3$ ($60^\circ$).  We thus cannot pack all six blue neighboring airports.
Of course we can pack $5$.
QED.
