My solution for Matt Parker's Maths Puzzles no.9
Considering all the refections and rotacions of the starting triangle, we can represent the starting three numbers such that
a ≤ a + b ≤ a + b + c,
a, b, c ∈ Z.
Grouping starting numbers like this, iterating the procedure, the sum of numbers will tend to
2 · GCF (b, c) ,
where GCF() gives the greatest common factor.