Maximum number of aeroplanes landing at a single airport I am new to Stackexchange so hope this problem is already known and a simple proof to the answer exists.
Flatland is a plane extending infinitely in all directions. It has an infinite number of airfields no two of which are exactly the same distance apart. 
A training execise involves a single auroplane taking off from each airfield and flying to and landing at its nearest adjacent airport.
What is the maximum number of aeroplanes that may land at any single airfield?
I think the answer is 5 but have no definitive proof. Am I correct and is there a simple geometrical or trigonometrical proof please?
 A: Suppose the planes from B, C, D, E, F, and G all land at A. Suppose the distance from A to B is 1. Then C must lie on the A-side of the perpendicular bisector of AB (in order to land at A instead of at B), and it must be outside the circle of radius 1 centered at B (so B lands at A, and not at C). Draw a picture and you'll be able to see that this forces the angle CAB to exceed 60 degrees. 
More generally, if you stand at A and look at one of the airports and then turn until you face another airport, you must turn more than 60 degrees. But if you do this six times, you have turned over 360 degrees, which leads to a contradiction. So only five planes can land at A. 
A: I think that the value of $5$ is correct. This is not a complete proof, but I think that this idea can be supplemented to a full proof.
Assume we have a configuration of airports $X_1, \ldots, X_n$ whose airplanes will all fly to the same airport $M$ and we have numbered $X_1, \ldots, X_n$ in a clockwise fashion around $M$. We need to show that $n \le 5$.
For each $1 \le i \le n$ consider the bisector $b_i$ between $X_i$ and $M$ and denote by $P_i$ the corresponding open half-plane in which $M$ lies. Since the planes $X_1, \ldots, X_{i - 1}, X_{i + 1}, \ldots, X_n$ all fly towards $M$, we need that $X_j \in P_i$ for $j \ne i$ (otherwise the plane from $X_j$ flies towards $X_i$ or has the same distance from $X_i$ and $M$).
The intersection of all the $P_i$ forms a convex polygon. Since we have numbered $X_1, \ldots, X_n$ in a clockwise fashion, the lines $b_i$ and $b_{i + 1}$ are neighboring line segments of the polygon.
Now let's consider the angle between $b_i$ and $b_{i + 1}$. Since the plane from $X_i$ flies towards $M$ and not towards $X_{i + 1}$, we know that $X_{i + 1}$ lies outside of a circle $C_i$ of radius $\overline{MX_i}$ around $X_i$. But this angle attains its maximal value of $120^\circ$ if (and only if) $x_{i + 1}$ lies on the intersection between $C_i$ and $P_i$ (this seems intuitively clear, but you would need a rigorous proof). Since this is not possible, the angle must be strictly greater thatn $120^\circ$.
Since the sum of all angles in an $n$-gon is $(n - 2) \cdot 180^\circ$, we must have $(n - 2) \cdot 180^\circ < n \cdot 120^\circ$, which is equivalent to $n < 6$.
