Croatian Mathematical Society Competitions

Let $a$ and $b$ be the lengths of legs, $c$ be the length of hypotenuse, and $r=4$ be the radius of the incircle in the given triangle. Expressing its area in two different ways, we get $$r\cdot \frac{a+b+c}{2}=\frac{ab}{2},$$ $$ab-4a-4b=4c.$$ Since $c^2=a^2+b^2$, it follows that $ab(ab-8a-8b+32)=0$. Considering that $a$ and $b$ are positive integers, we have $ab-8a-8b+32=0$, i.e. $(a-8)(b-8)=32$, hence both $a-8$ and $b-8$ must be divisors of $32$. Without loss of generality, we can assume $a\le b$, i.e. $a-8\le b-8$. Therefore, among $$\begin{array}{rlrl}a-8& =-32,& a-8& =1,\\ a-8& =-16,& a-8& =2,\\ a-8& =-8,& a-8& =4,\end{array}$$ we find that $(a,b)\in \{(9,40),(10,24),(12,16)\}$ satisfy the conditions of the problem. Finally, these are all solutions: $$(a,b)\in \{(9,40),(10,24),(12,16),(16,12),(24,10),(40,9)\}.$$