Background
Let $c(t):[0,1]\mapsto\mathbb{R}^2$ be a curve with the following properties
- $c(t)$ is closed;
- $c(t)$ is symmetric with respect to the horizontal axis;
- $c(t)$ is contained in the rectangle $\mathcal{R}\triangleq[0,1]\times[-1,1]$;
- $c(t)$ interpolates the points $[0\,\,1]',[0\,\,-1]',[1\,\,0]$.
Now define the barycenter of $c(t)$ as \begin{equation*} \bar{c} \triangleq \frac{\int_0^1 c(t) \lVert\dot{c}(t)\rVert \text{ d}t}{\int_0^1 \lVert\dot{c}(t)\rVert \text{ d}t} \end{equation*} by symmetry, the barycenter $\bar{c}$ is located over some point of the horizontal axis, say \begin{equation*} \bar{c}=\left[\begin{array}{cc} \bar{x} & 0 \end{array}\right]' \end{equation*} for some $\bar{x}$. Clearly, since the curve is inside the rectangle $\mathcal{R}$, the value of $\bar{x}$ is in $[0,1]$. I'm wondering if we can say something more about $\bar{x}$. Consider the following limit curves shown in the picture. Let $\bar{x}_i$ be the horizontal component of the barycenter $\bar{c}_i$ associated to the curve $c_i$. We have \begin{equation*} \bar{x}_1 = 0 \qquad \bar{x}_2 = 1/3 \qquad \bar{x}_3 = 1/2 \end{equation*}
Questions
- Can we say that for any $c(t)$ with the previous properties hold $\bar{x}\in[0,1/2]$?
- Can we say that if $\bar{x} \geq 1/3$ then the curve $c(t)$ is convex, if $\bar{x}< 1/3$ then the curve is concave?