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?


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.
