A triangle problem

Question

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 ?

Answer 1

$a+b=x$, $a.b=y$ 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 $\frac{% r}{R}=\frac{3y}{2c(x+c)}$.

Answer 2

We have equations for the inradius $r$ and the circumradius $R$, namely $r=\dfrac{\triangle}{s}$, and $R=\dfrac{abc}{4rs}$ where $\triangle$ denotes the area of the triangle and $s$ denotes the semiperimeter. Hence we have $$\frac{r}{R}=\frac{4r^2s}{abc}=\frac{4\triangle^2}{sabc}$$ We have $c=\sqrt{x^2-y}$, $s=\dfrac{1}{2}(a+b+c)=\dfrac{1}{2}(x+\sqrt{x^2-y})$, and $\triangle=\dfrac{1}{2}ab\sin\theta=\dfrac{\sqrt{3}}{4}y$. Plugging everything in, we get

$$\frac{r}{R}=\frac{3y^2}{4(\frac{1}{2}(x+\sqrt{x^2-y})y(\sqrt{x^2-y}))} =\frac{3y}{2(x\sqrt{x^2-y}+x^2-y)}$$