The locus of the centroid of a triangle when one vertex moves on a circle Let $ABC$ be a triangle whose base $BC$ is fixed. What is the locus of the centroid if the vertex $A$ moves on a given circle?
A demonstration using GeoGebra shows that the locus is a circle whose radius does depend only on the radius of the give circle not on the size or position of the fixed base $BC$.
 A: Geometrical proof: I will use the term "barycenter" for "weighted average".
Let $D$ be the midpoint of line segment $[BC]$.
By partial barycentration, it is equivalent to look for the barycenter $W$ of variable point $A$ with weight $1$ and fixed point $D$ with weight $2$. Therefore point $W$ is always such that $\vec{DW}=\frac13 \vec{DA}$. Consequently, the locus of point $W$ is the image of the locus of $A$ by a homothety with center $D$ and ratio $1/3$ ; therefore $W$ describes a circle with radius a third of the radius of the circle described by $A$.
Analytical proof:
$$\begin{cases}x_G&=&\frac13(x_A+r \cos \theta+x_B+x_C)&=&x_G+\frac13(r \cos \theta)\\y_G&=&\frac13(y_A+r \sin \theta+y_B+y_C)&=&y_G+\frac13(r \sin \theta)\end{cases}$$
with $G=(x_G=\frac13(x_A+x_B+x_C),y_G=\frac13(x_A+x_B+x_C))$ is the center of mass of triangle $ABC$.
A: 
Say, Vertex $A$ moves on circle $X$ of radius $r$.
Then $OA = r$ and $G$, the centroid of $\triangle ABC$ , divides median $AM$ in the ratio $2:1$. Now take point $K$ on segment $OM$ such that $OK:KM = 2:1$. Given $M$ and $O$ are fixed, point $K$ is fixed.
Regardless of where point $A$ is on the circle $X$, segment $KG$ divides sides $AM$ and $OM$ in the same ratio of $2:1$ in $\triangle OAM$. So, it must be parallel to base $BC$ and $KG = \frac{OA}{3} = \frac{r}{3}$. So the distance of $G$ from $K$ is fixed as $A$ moves on the circle with given radius $r$.
Please also note there are two points on the circle when $A, M$ and $O$ are collinear and we have a degenerate triangle $\triangle OAM$. It still holds that $KG = \frac{r}{3}$.
So as vertex $A$ of $\triangle ABC$ moves on circle $X$ with radius $r$, its centroid $G$ moves on a circle with center as $K$ and radius as $\frac{r}{3}$. $K$ is the centroid of $\triangle OBC$.
