The Locus of the Centroid of a Variable Triangle For the variable Triangle $\Delta ABC$ with fixed vertex at $C(1,2)$ and $A,\,B$ having co-ordinates $(\cos t, \sin t)$, $(\sin t, -\cos t)$, find the locus of its centroid.
 A: The parametric equation of the locus curve for centroid is
$$\begin{align}
\vec G &= \frac{1}{3} (\vec A + \vec B + \vec C)\\
&= \frac{1}{3}[(1+\cos t+\sin t){\bf{i}}+(2+\sin t -\cos t){\bf{j}}] \\
&\equiv x_G {\bf{i}} + y_G {\bf{j}}
\end{align}$$
In fact, this locus is a circle and the details of it is given in the answer by marwalix. It's Cartesian equation is
$$(x_G-{1\over 3})^2+(y_G-{2\over 3})^2={2\over 9}$$
So it is a circle centered at $(\frac{1}{3},\frac{2}{3})$ with the radius $R=\frac{\sqrt{2}}{3}$.
This animation helps to visualize the locus better. As you can see, if one of the points $A$ or $B$ lies at the intersection of the two circles then all the points lie on a line and the centroid is $A$ or $B$ itself!

A: The. Centroid $G$ of ${A,B,C}$ is such that
$$\vec{GA}+\vec{GB}+\vec{GC}=\vec{0}$$
Using Chasles identity with the origin $O$ one gets
$$\vec{GO}+\vec{OA}+\vec{GO}+\vec{OB}+\vec{GO}+\vec{OB}=\vec{0}$$
Using $\vec{GO}=-\vec{OG}$ one gets
$$\vec{OG}={\vec{OA}+\vec{OB}+\vec{OC}\over 3}$$
And this translates in coordinates
$$\vec{OG}={\left(1+\cos{t}+\sin{t}\right)\over 3}\vec{i}+{\left(2+\sin{t}-\cos{t}\right)\over 3}\vec{j}$$
One can check that 
$$(x_G-{1\over 3})^2+(y_G-{2\over 3})^2={\cos^2{t}+2\cos{t}\sin{t}+\sin^2{t}+\sin^2{t}-2\sin{t}\cos{t}+\cos^2{t}\over 9}$$
And this simplifies to
$$(x_G-{1\over 3})^2+(y_G-{2\over 3})^2={2\over 9}$$
And this is the equation of a circle centred in $(1/3,2/3)$ with radius ${\sqrt{2}\over 3}$
