Centroid of a Triangle on a inscribed circle $AB$ is the hypotenuse of the right $\Delta ABC$ and $AB = 1$.
Given that the centroid of the triangle $G$ lies on the incircle of $\Delta ABC$, what is the perimeter of the triangle?


Assume that $C=(0,0),B=(a,0),A=(0,b)$ and $c=\sqrt{a^2+b^2}=AB$. We have:
$$ G = \left(\frac{a}{3},\frac{b}{3}\right)  \tag{1}$$
and the inradius is given by 
$$ r = \frac{2\Delta}{a+b+c} = \frac{ab}{a+b+c},\tag{2}$$
so the equation of the incircle is given by:
$$ \left(x-\frac{ab}{a+b+c}\right)^2 + \left(y-\frac{ab}{a+b+c}\right)^2 = \left(\frac{ab}{a+b+c}\right)^2\tag{3} $$
and since $G$ lies on the incircle, by setting $a=\sin\theta,b=\cos\theta$ we get:
$$ 5-3\sin\theta-3\cos\theta-3\cos\theta\sin\theta\tag{4} $$
or:
$$ (1+\sin\theta)(1+\cos\theta)=\frac{8}{3} \tag{5}$$
hence:
$$ a+b+c = \sin\theta+\cos\theta+1 = \color{red}{\frac{4}{\sqrt{3}}}\tag{6}$$
since $(4)$ can be solved as a quadratic equation in $\sin\theta+\cos\theta$.
