I'm sure someone will pipe up with a more exact answer (likely using permutation and/or combination in their specific mathematical senses), but I believe it's because summing medals is algebraic, and with the given weights there's no way for the sum of any combination of non-identical medals that equals any other. The values of 𝛑 and 𝛑² (and the patient, unacknowledged workhorse 1) provide this. Looking at a simple case where s is the number of silvers and b the number of bronzes, there's no value of b that can equal any value of s·𝛑. You could use another triple of weights, such as (𝛑, e, 1) as well.
Medals are awarded in integer quantities, and you don't want j·g+k·s+l·b=0 to be possible for integer j,k,l (except when all three medal counts are 0).
This is actually a slightly stronger condition than is needed, because no country can earn negative medals and there are a finite number of medals available, so you could actually have the weights for eg. gold and silver medals differ by a rational factor as long as the denominator was large enough.
