62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour

Make the following transformations: $$mn={\left(\frac{m+n}{2}\right)}^2-{\left(\frac{m-n}{2}\right)}^2=45^2-{\left(\frac{m-(90-m)}{2}\right)}^2=2025-(m-45{)}^2\to \max .$$ So the max will be achieved, when the value of the expression $(m-45{)}^2$ is minimized, but here we have to not forget that $m,n$ are coprime.

$m=45$, then $n=45$ and numbers $m,n$ aren't coprime.

$m=46$, then $n=44$ and numbers $m,n$ aren't coprime.

$m=47$, then $n=43$ and numbers $m,n$ are coprime, as required. So the largest value of the product is $2025-(47-45{)}^2=2021$.