The original puzzle was solved using Pythagorean triples. It’s impossible to find a Pythagorean triple where none of the numbers is divisible by 3.

Instead we can use the cosine rule, which is a kind of generalised version of Pythagoras theorem.

a^2 + b^2 - 2ab*cosC = c^2, where C is the angle opposite the side c. (When C is 90, cosC is zero and that whole term disappears).

Clearly if a b and c are integers, 2ab*cosC needs to be too. cosC can be anything from 1 to -1. We don’t care if C is rational (in whatever units: degrees, radians, etc), only that cosC is rational.

We need two adjacent central angles to add to 180 degrees. It turns out, conveniently for our purposes, that if C and C’ add to 180 degrees, then cosC and cosC’ add to zero.

So now we just need to find a convenient rational value for cosC, for example 1/13, then find a triple a,b,c that satisfies

a^2 + b^2 - 2ab/13 = c^2

none of which is a multiple of 3, for example 13,20,23

and other triples that satisfy

a^2 + b^2 + 2ab/13 = c’^2

with again, no multiples of 3, for example 10,13,17 and 13,40,43

Making a double size copy of 13,20,23 and also doubling 10,13,17 gives us four triangles of alternating central angles, which fit together as shown:

I thank Philip Morris Jones for this solution.

There is one more solution with a total mileage of less than 400 miles. It is based on a value of cosC of 1/5, and triples of 4,5,7 and 7,10,11, with roads leading from the centre with lengths of 50,40,28 and 35, and around the outside of 70,44,49 and 55, giving a total mileage of 371.