Airport problem mathematical circles There are several airports in a country, and it is known that distances  between them are all different. An airplane takes off from each airport and flies to the closest airport. Prove that there will be no more than five of the planes at each airport after they all have landed
 A: This problem is similar to a challenge problem I came across a few years back$\,-\,$not sure where, and the solution I present below is just a replay from memory of the published solution.

Suppose that after all the planes have landed, at least $6$ of the planes have landed at airport $B$.

Our goal is to derive a contradiction.

Assume the airports are points of the $xy$-plane, with $B$ as the origin, and an arbitrary direction chosen for the positive $x$-axis.

Let $A_1,...,A_6$ be the respective airports from which each of the $6$ planes originated, labeled so that $\theta_1 \le \cdots \le \theta_6$, where $\theta_i$ is the least nonnegative angle between the ray $BA_i$ and the positive $x$-axis.

Since $B$ is the closest neighboring airport to each of $A_1,...,A_6$, it follows that the angles $\theta_1,...,\theta_6$ are distinct, hence
$\theta_1 < \cdots < \theta_6$.

For $1 \le i \le 6$, let $i' = i+1$ if $i < 6$, and $i' = 1$ if $i=6$.

For each $i$, let $t_i$ be the least positive angle between the rays $BA_i$ and $BA_{i'}$.

Then $t_1 + ... + t_6 = 360^\circ$, hence $t_i \le 60^\circ$, for some $i$. 

Fixing $i$ with $t_i \le 60^\circ$, consider triangle $ABC$, where $A = A_i$ and $C = A_{i'}$.

By the closest neighbor condition (and since all distances are distinct), side $BC$ must be strictly the largest side of triangle $ABC$, hence $\angle ABC > 60^\circ$, and thus, $t_i > 60^\circ$, contradiction.

This completes the proof.
