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 of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

• A. $\frac{3y}{2x(x+c)}$
• B. $\frac{3y}{2c(x+c)}$
• C. $\frac{3y}{4x(x+c)}$
• D. $\frac{3y}{4c(x+c)}$

Answer 1

The correct option is B $\frac{3y}{2c(x+c)}$

$a+b=xab=y{x}^2-c^2=y$
Using cosine rule,
$\cos C=\frac{a^2+b^2-c^2}{2ab}\Rightarrow \cos C=\frac{(a+b{)}^2-2ab-c^2}{2y}\Rightarrow \cos C=\frac{x^2-2y-c^2}{2y}=-\frac{1}{2}\Rightarrow C=\frac{2\pi }{3}$
Now,
$\frac{c}{\sin C}=2R\Rightarrow R=\frac{c}{\sqrt{3}}r=(s-c)\tan \frac{C}{2}\Rightarrow r=\sqrt{3}\left(\frac{a+b-c}{2}\right)=\frac{\sqrt{3}(x-c)}{2}$
Now,
$\frac{r}{R}=\frac{3(x-c)}{2c}=\frac{3y}{2c(x+c)}$