What if you have five towns at the corners of a regular pentagon:
which pattern would you choose here?

Suppose you are given three towns: what is your best strategy here?

How about six towns that are at the corners of two squares placed side by side?

Notes:

The most popular pattern students find for the four-town problem is usually a simple pair of crossed diagonals.
This uses 28.3 km of road. However, it can be improved upon, still keeping two lines of symmetry.
This uses 27.3 km of road: a saving of 1 million pounds perhaps?

This solution occurs where the angles at meeting points are 120°.
The solution is also demonstrated by soap bubbles,which by their nature take up the minimum answer each time.
The hexagon solution is easy once you have seen the square solution.

The pentagon solution is harder, but soap bubbles are helpful once again.

For a triangle, if it includes an angle that is 120° or greater,
then just take the two sides that include this angle.
Otherwise, there is a unique Steiner point in the triangle,
where three lines joining this point to the vertices of the triangle meet at 120° .

You can find this by constructing an equilateral triangle (say ABX) on one side of the triangle (AB), then drawing in its circumcircle.
Join up XC: where this line meets the circumcircle of ABX is the Steiner point.

The final problem has a pleasing answer: again, there is a kind of honeycomb solution,
but the honeycomb grid is this time on its side.