Triangles from prime-difference chords on a circle

Twenty points on a circle are labelled $1$ through $20$. A chord is drawn between two points exactly when the difference of their labels is prime. How many triangles have all three sides among these chords?

$72$. For labels $x>y>z$ we need $x-y$, $y-z$ and their sum $x-z$ all prime. A sum of two primes is an odd prime only if one of them is $2$, so the gaps are $\{2,p\}$ with $p$ and $p+2$ twin primes: $(3,5),(5,7),(11,13),(17,19)$. Counting both orders and all valid starting labels gives $72$.