Singapore Mathematical Olympiad (SMO)

Since the sum is $0$, there are positive and negative integers among the three. Without loss of generality, we assume that there is $1$ negative and two positive integers. (The other case is obtained by changing the signs of all the integers.) Then $c < 0$ since it has the smallest absolute value. Let $\gcd(a, b) = p$, $\gcd(b, c) = q$ and $\gcd(a, c) = r$. Then $\gcd(p, q) = \gcd(q, r) = \gcd(r, p) = 1$. Therefore $a = pr s$, $b = pq t$ and $c = -qr$ for some integers $s$, $t$, $u$ with $\gcd(s, t) = \gcd(s, u) = \gcd(t, u) = 1$. Substitute into the original equation and simplify, we have
$$qt + rs = \frac{pst}{u}.$$
Now, $u$ has no common factor with $s$ and $t$. Thus $u \mid p$. This means $u$ is a common divisor of $a$, $b$, $c$. By our assumption, $u = 1$. By symmetry, we also have $s = t = 1$. Therefore $q + r = p$ and $a = pr$, $b = pq$, $c = -qr$. Thus every triple $p$, $q$, $r$, pairwise coprime with $q + r = p$ uniquely determines the triple $a$, $b$, $c$. We have the following triples, with the product of the larger $2 \leq 50$:
$$(1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (1, 6, 7), (2, 3, 5), (2, 5, 7), (3, 4, 7), (3, 5, 8), (4, 5, 9).$$
Therefore, we have $20$ such triples.