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?

 A: 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 $\sin\theta+\cos\theta$.
A: I agree with the answer by @Jack D'Aurizio, but I just wanted to suggest a quicker way not involving too much algebra and no trigonometry:
Firstly we can establish, by consideration of equal tangents to the incircle, that if $P$ is the perimeter of the triangle and $r$ is the radius of the incircle, then $$P=1+1+r+r\Rightarrow r=\frac{P-2}{2}$$
We also have that $IG=r$, where $I(r,r)$ is the incentre and $G(\tfrac{a}{3},\tfrac{b}{3})$ is the centroid, and this gives us:  $$(\tfrac {a}{3}-r)^2+(\tfrac {b}{3}-r)^2=r^2$$ $$\Rightarrow \tfrac {1}{9}-\tfrac{2r}{3}(a+b)+r^2=0$$
But $P=1+a+b$, so this simplifies to $$1-6r(P-1)+9r^2=0$$
Finally we can substitute for $r$ and get $$1-6\big(\frac{P-2}{2}\big)(P-1)+\tfrac{9}{4}(P-2)^2=0$$ 
This simplifies very quickly  to become $$P^2=\tfrac{16}{3}$$ and hence the expected result.
