<s>[INST] A triangle problem In a triangle, the sum of two sides is $x$ and the product of the same two sides is $y$. 
If $x^2 - c^2=y$ where c is the third side, then what is the ratio of the inradius to the circumradius of the triangle?
I guess I have found half of it: if the two sides of the triangle are $a$ and $b$, then $x=a+b$ and $y=ab$. Therefore
$x^2 - c^2 = y \Rightarrow (a+b)^2 - c^2 = ab \Rightarrow
            a^2 + b^2 +ab = c^2.$
But
   $a^2 + b^2 - 2ab\cos\theta=c^2$ (as c is the third side),
    therefore $\theta= 120^{\circ} = \frac{2\pi}{3}$.
OK, so what about the inradius and circumradius ?
 [/INST] $a+b=x$, $a.b=y$ and it is found $m(\widehat{ACB})=\frac{2\pi }{3}$ since $%
x^{2}-c^{2}=y$. Then area of the triangle is $S(ABC)=\frac{1}{2}ab\sin (%
\frac{2\pi }{3})=\frac{abc}{4R}=\frac{a+b+c}{2}r$, where $R$ and r are
circumradius and inradius respectively. Then you can easily calculate $R=%
\frac{c}{\sqrt{3}}$ and $r=\frac{\sqrt{3}y}{2(x+c)}$. Finally, we get $\frac{%
r}{R}=\frac{3y}{2c(x+c)}$.
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