Given positive integers $a, b, c$ satisfying $a<b<c$, find all triples $(a, b, c)$ that satisfy the equation
$$
\frac{(a, b)+(b, c)+(c, a)}{a+b+c}=\frac{1}{2}
$$

where $(x, y)$ denotes the greatest common divisor of the positive integers $x$ and $y$.
Let $\operatorname{gcd}(a, b, c)=d$, and let $a=d a_{1}, b=d b_{1}, c=d c_{1}$, then $\operatorname{gcd}\left(a_{1}, b_{1}, c_{1}\right)=1$.
(1) If $a_{1}=1$, then we have
$$
\left.\frac{1+\left(b_{1}, c_{1}\right)+1}{1+b_{1}+c_{1}}=\frac{1}{2} \Rightarrow 3+2\left(b_{1}, c_{1}\right)=b_{1}+c_{1} \Rightarrow\left(b_{1}, c_{1}\right) \right\rvert\, 3,
$$

Thus, $\left(b_{1}, c_{1}\right)=1$ or 3. When $\left(b_{1}, c_{1}\right)=1$, $b_{1}=2, c_{1}=3$; when $\left(b_{1}, c_{1}\right)=3$, $b_{1}=3, c_{1}=6$. Therefore, $(a, b, c)=(d, 2 d, 3 d)$ or $(d, 3 d, 6 d)$.
(2) If $a_{1} \geqslant 2$, let $\left(a_{1}, b_{1}\right)=p,\left(b_{1}, c_{1}\right)=q,\left(c_{1}, a_{1}\right)=r$, then $p, q, r$ are pairwise coprime. We can set $a_{1}=p r a_{2}, b_{1}=q p b_{2}, c_{1}=q r c_{2}$, thus
$$
\begin{array}{l}
\frac{(a, b)+(b, c)+(c, a)}{a+b+c}=\frac{p+q+r}{p r a_{2}+p q b_{2}+q r c_{2}}=\frac{1}{2} \\
\Rightarrow 2(p+q+r)=p r a_{2}+p q b_{2}+q r c_{2} .
\end{array}
$$

Assume $\min \left\{a_{2}, b_{2}, c_{2}\right\}=a_{2} \geqslant 2$, then $p+q+r \geqslant p r+p q+q r$. Clearly, at least one of $p, q, r$ is not less than 2, otherwise $a_{1}, b_{1}, c_{1}$ are pairwise coprime, in which case
$$
\frac{(a, b)+(b, c)+(c, a)}{a+b+c}=\frac{3}{a_{1}+b_{1}+c_{1}} \leqslant \frac{3}{8}<\frac{1}{2} .
$$

Assume $p \geqslant 2$, then $p+q+r \geqslant 2 r+p+q$, thus, $r \leqslant 0$, which is a contradiction. Therefore, $a_{1} \geqslant 2$ does not hold, i.e., $a_{1}$ can only be 1.

In conclusion, all the triples $(a, b, c)=(d, 2 d, 3 d)$ or $(d, 3 d, 6 d)$ (where $d$ is any positive integer).