In this case, (with the hindsight of knowing the answer), it would probably be easier to enumerate the angles, as they were all rational multiples of pi (with t

In this case, (with the hindsight of knowing the answer), it would probably be easier to enumerate the angles, as they were all rational multiples of pi (with the biggest denominator being 12). Still, that gives a naive 5^144 combinations of angles to test. If you reduce this to proper, reduced fractions this comes down to 5^47 [1]. You can probably improve on this further by requiring that the angles form a (convex) polygon, but I suspect that still leaves you with to many to be practical. [1] http://oeis.org/A005728 http://oeis.org/A005728