Take a map and connect each town shown on it to its nearest (as the crow flies) neighbor by means of a straight line segment.

The maximum number of cities that can be nearest to the same city is
6.  The UNIQUE configuration is the 6 cities are at the corners of
regular hexagon centered around the other city.  But in this
configuration the distances are not distinct.  Since this configuration
is the unique one for 6, the most possible using distinct distances is
5.

5 is in fact obtainable as follows.  Draw a regular pentagon centered
around city X and initially place the other cities at the corners.  Let
me call the other cities in clockwise order, A,B,C,D,E.  Leaving corner
city A fixed, it is possible to push each of the remaining cities along
the radial lines to the center so that each of these cities nearest is
the central city and such that the pairwise distances are distinct.
One such method: Let d = the max. distance B can be pushed along the
radial line towards X so that both A and B's nearest neighbor is X. (d
is the distance from the corner to the point at the intersection of the
radial line with the circle centered at A with radius AX).

Let e be anything such that 0 < e < d/4.  Push city B e units towards
X; push C 2*e units towards X; push D 3*e units towards X; and push E
4*e units towards X.  The nearest point from each of A,B,C,D,E is X.
(note that E has been pushed < d units towards X as required.)

-- Glenn C. Rhoads <rhoads@paul.rutgers.edu>

[ Editor: the same puzzle, posted in 1994 as "Panic in Cucumberland": ]

> There are 100 airports in Cucumberland. All the distances between
> them are different. There is exactly one airplane on every airport.
> Suddenly, all the planes start and move to the closest airport.
> What maximal number of the airplanes can arrive to one airport?

-- Pertsel Vladimir <voldemar@wiscon.weizmann.ac.il>

Say that the destination airport is at the origin, and the the plane at
(1,0) goes to the origin.  There cannot be any other airport within the
circle of radius 1 centered at (1,0), and any plane whose x-coordinate
is greater than 1/2 will go to (1,0) in preference to the origin.  Any
plane whose path to the origin makes an angle of pi/3 or less with the
x-axis must either have x-coordinate of at least 1/2 or else be inside
the forbidden circle (since the chord from the origin to
(1/2,sqrt(3)/2) is inside the circle makes an angle of pi/3 with the
x-axis).

Thus, if planes A and B both go to airport O, the angle AOB must be
more than pi/3.  The sum of the angles between consecutive pairs must
be 2*pi, so there are at most five pairs.

It is possible for five planes to fly to the origin; put them near the
vertices of a regular pentagon centered at the origin.  (They must be
near, not at, the vertices, so that distances aren't equal.)

-- David Grabiner <grabiner@math.harvard.edu>