<s>[INST] Proving right angle triangle A triangle $A$$B$$C$ is integer sided and has inradius $1$.prove that it is right angled. Please help.give some hints please.
 [/INST] My idea :
$r = \Delta/s = 1 \implies \Delta = s \implies \Delta^2 = s^2 \implies (s-a)(s-b)(s-c) = s \implies (b+c-a)(c+a-b)(a+b-c) = 4(a+b+c) $.
If all $a,b,c$ are even or odd , then the parity doesn't satisfy. Thus  that means $a+b+c$ is even. So $s$ is an integer. Thus by the above argument, we have $ s-a|s,s-b|s,s-c|s $. Now let $r_a,r_b,r_c$ be the exradii of the triangle $ABC$. We have our formula $r_a= \Delta/s-a = s/s-a $. Thus $r_a,r_b,r_c$ are integers. Now also we have the equation : 
$\frac{1}{r_a}  + \frac{1}{r_b} + \frac{1}{r_c}  = \frac{1}{r} \implies\frac{1}{r_a}  + \frac{1}{r_b} + \frac{1}{r_c} = 1 $
 and this has only one solution namely $ 2,3,6$ and hence you get the values of sides as $3,4,5$. 
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