75th Romanian Mathematical Olympiad

a) Since $c\mid a+b+2c$ and $c\mid 2c$, $c\mid a+b$. Now $a<b<c$ implies $a+b<2c$, hence $a+b=c$. Then $(a+b,c)=c$.

b) From $b\mid a+b+2c$ and (a), $b\mid 3a+3b$. Since $b\mid 3b$, $b\mid 3a$. From $3a<3b$ follows $3a\in \{b,2b\}$.

Case I: $3a=b$. Then $a=n$, $b=3n$, $c=4n$, whence $abc=12n^3$. The values of $n$ so that $2^2\cdot 3\cdot n^3=k^2$, $k<1000$ are: $n=3$, implying $abc=18^2$; $n=3\cdot 2^2$, implying $abc=144^2$; $n=3^3$, implying $abc=486^2$.

Case II: $3a=2b$. Then $a=2p$, $b=3p$, $c=5p$. So $abc=30p^3$, yielding $p=30$ and $abc=900^2$.

In conclusion, $k\in \{18,144,486,900\}$.