In a triangle the sum of two sides is x and the product of the same two sides is y. If ${\mathrm{x}}^2 – {\mathrm{c}}^2 = \mathrm{y}$, where c is the third side of the triangle, then the ratio of the in radius to the circum-radius of the triangle is

Suppose 

a + b = x, ab = y

and ${\mathrm{x}}^2 – {\mathrm{c}}^2 = \mathrm{y}$

$\begin{array}{ll}\Rightarrow & (a+b{)}^2-c^2=ab\\ \Rightarrow & a^2+b^2-c^2=-ab\\ \Rightarrow & \cos C=\frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}\\ \Rightarrow & C=\frac{2\pi }{3}\end{array}$

We know that

$R=\frac{c}{2\sin C}=\frac{c}{2\sin \left(\frac{2\pi }{3}\right)}=\frac{c}{\sqrt{3}}$

Also, $\begin{array}{l}r=(s-c)\tan \left(\frac{C}{2}\right)\\ =\left(\frac{a+b+c}{2}-c\right)\tan \left(\frac{C}{2}\right)\\ =\frac{1}{2}(x-c)\sqrt{3}\end{array}$

Now, $\begin{array}{r}\frac{r}{R}=\frac{(x-c)\sqrt{3}}{2c/\sqrt{3}}=\frac{3(x-c)}{2c}\\ =\frac{3\left(x^2-c^2\right)}{2c(x+c)}=\frac{3y}{2c(x+c)}.\end{array}$