Amazing Triangle!!

Let the sides be $a-1$ , $a$ and $a+1$ , let $r$ be the inradius, $R$ the circumradius and $A$ the area.

By Heron's formula:

$A=\dfrac{\sqrt{(a-1+a+a+1)(a-1+a-a-1)(a-1-a+a+1)(-a+1+a+a+1)}}{4}$

$A=\dfrac{a\sqrt{3(a^2-4)}}{4}$

The well-known formulas tell us that:

$r=\dfrac{2A}{a-1+a+a+1}=\dfrac{\sqrt{3(a^2-4)}}{6}$

But $r=4$ , so:

$4=\dfrac{\sqrt{3(a^2-4)}}{6}$

Solving for $a$ we get $a=14$ .

Also, the formula that relates $r$ and $R$ is:

$rR=\dfrac{(a-1)(a)(a+1)}{2(a-1+a+a+1)}$

$4R=\dfrac{13 \times 14 \times 15}{2(42)}$

$R=\dfrac{65}{8} = \boxed{8.125}$