For any piece wise smooth, simple closed curve $\gamma$ in the Euclidean plane $E^2$ and fix a point $G$ inside the area circled by $\gamma$.


**Show**: There exists three points $A,B$ and $C$ on the $\gamma$, such that $G$ become the barycenter of the triangle $\Delta ABC$ and $\Delta ABC$ locates inside the area circled by $\gamma$.


**Remark**: 


1. It is not my idea, I just saw it on Weibo <http://weibo.com/1648021814/EaRJiy1au?from=page_1005051648021814_profile&wvr=6&mod=weibotime&type=comment#_rnd1475372956164>
, I think it is quite interesting , so I pose it.
2. Maybe there is theorem to admit it or a counter-example in computational geometry I do not know.



Triangle inside the Closed Curve
This is true for convex curves, but false in general for non-convex ones.


Proof for convex curves: First find two points $A, D \in \gamma$ such that $G$ is the midpoint of $AD$ (by continuity argument or by intersecting $\gamma$ with its symmetral wrt $G$). Then take a point $E$ on $AD$ one third on the way from $D$ to $A$. For this point, too, there are $B, C \in \Gamma$ such that $E$ is the midpoint of $BC$. Now since $E$ is the barycenter of $B$ and $C$, and $G$ is the barycenter of $A$ weighted by 1 and $E$ weighted by 2, $G$ is the barycenter of $A$, $B$, $C$.


A counterexample for non-convex curves, if we require the triangle $ABC$ to lie inside $\gamma$: take a half-circle of a very small radius in the lower half-plane, centered at $(0,0)$. Take also a half-circle of a big radius in the upper half-plane, centered at $(1,0)$. Extend the ends of the big circle into the lower half-plane and join them to the endpoints of the small circle. Now, $(0,0)$ is not a barycenter of three points on $\gamma$ visible from it. Indeed, we need to take at least one point on the big half-circle. If we take one, it will "pull" too far from the origin. If we take two or three, they will also "pull away" because the big half-circle is "unbalanced".


