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.
 A: Let the segments of tangents from the vertices to the incircle be $x,y,z$ (what is popularly known as the Ravi-substitution.
We have $(x+y), (y+z), (z+x)$ are all integers and so is $2(x+y+z)$ so $x+y+z = \frac{n}{2}, n \in \mathbb{N}$. Now its easy to see that $x,y,z$ are in the form $\frac{a}{2}, \frac{b}{2}, \frac{c}{2}$. Since $x+y, y+z, z+x \in \mathbb{N}, a,b,c$ have the same parity 
We are given that $r=\frac{xyz}{x+y+z} =1 $ i.e. $abc = 4(a+b+c)$ from which its clear that $a,b,c$ are all even $\Rightarrow x,y,z \in \mathbb{N}$.
So we are looking for solutions in $\mathbb{N}$ for $xyz = x+y+z$.
WLOG $x \le y \le z$. Hence $xyz = x+y+z \le 3z \Rightarrow xy \le 3$
Now taking the cases $(x,y) = (1,1), (1,2), (1,3)$ we see that $x=1, y=2, z=3$ is the only solution.
This gives the sides as $3,4,5$ which is a Pythagorean Triplet and hence sides of a right angled triangle.
A: 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 $3,4,5$. 
A: Let $a$, $b$ and $c$ be sides-lengths of the triangle.
Thus, $$1=\frac{2S}{a+b+c}$$ or
$$4(a+b+c)^2=(a+b+c)(a+b-c)(a+c-b)(b+c-a)$$ or
$$4(a+b+c)=(a+b-c)(a+c-b)(b+c-a).$$
Now, it's obvious that $a+b+c$ is an even number.
Thus, $a+b-c$, $a+c-b$ and $b+c-a$ are even positive numbers.
Let $a+b-c=2w$, $a+c-b=2v$ and $b+c-a=2u$.
Thus, we need to solve
$$u+v+w=uvw$$ or
$$\frac{1}{uv}+\frac{1}{uw}+\frac{1}{vw}=1.$$
Let $uv=z$, $uw=y$ and $vw=x$.
Hence, $x$, $y$ and $z$ are natural numbers such that
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.$$
This equation we can prove by the following way.
Let $x\leq y\leq z$.
Hence, $$1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq\frac{3}{x},$$
which gives $x\leq3$.
$x=1$ is impossible and $x=2$ and $x=3$ give the following solutions:
$(2,3,6)$, $(2,4.4)$ and $(3,3,3)$.
Last two are impossible and $(x,y,z)=(2,3,6)$ gives $(u,v,w)=(3,2,1)$,
which gives $(a,b,c)=(3,4,5)$ and since $4^2+3^2=5^2,$ we are done!
