RMO 1990 question

Question

Prove that the inradius of a right angled triangle having integer sides is also integral

I tried it and got something like

$r=\frac {(a.b)}{(a+b+c)}$

How to proceed after this.

Answer 1

First, note that $$r=\frac{a+b-c}{2}$$ Also, the Pythogorian theorem says $$a^2+b^2=c^2$$ $\textbf{Case 1: }$ If both $a$ and $b$ are even numbers then so is $c$, which implies that $a+b-c$ is also even and $r=\frac{a+b-c}{2}$ is integer.

$\textbf{Case 2: }$ If both $a$ and $b$ are odd numbers then $c$ is even, which implies that $a+b-c$ is also even and $r=\frac{a+b-c}{2}$ is integer.

$\textbf{Case 3: }$ If $a$ is an even and $b$ is an odd number then $c$ is odd, which implies that $a+b-c$ is even and $r=\frac{a+b-c}{2}$ is integer.

$\textbf{Case 4: }$ If $a$ is an odd and $b$ is an even number then $c$ is odd, which implies that $a+b-c$ is even and $r=\frac{a+b-c}{2}$ is integer.