A question about curves in $\mathbb{R}^2$ I need to show this result:
Let $\alpha :I\rightarrow \mathbb{R}^2$ a smooth curve, where $I$ is a compact interval of the real line. If $\lVert \alpha (s) - \alpha (t) \rVert$ depends only on $|s-t|$, for all $s$, $t$ in $I$, then $\alpha$ must be a subset of a line or a circle.
I have tried calling $\lVert \alpha (s) - \alpha (t) \rVert=f(|s-t|)$, where $f$ is smooth but this led me nowhere. Also, I had a suggestion to fixing $t$ as $0$ and $\alpha(0)=0$ through reparametrization and a rigid movement, so I would have $\lVert\alpha(s) \rVert=f(|s|)$, $f$ is smooth; this lead to nothing also. I strongly believe I must show that the curvature of this function if constant, either $0$ (then it is a line) or it is a constant $\neq 0$ from where it is in a circle.
Can someone please give me a hint?
 A: Ok let me try it again.
First of all, suppose that $\alpha$ is parametrized by arc length.
Write $$\alpha(s)-\alpha(t)=\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2}+o(|s-t|)$$
Then
\begin{eqnarray}
 \|\alpha(s)-\alpha(t)\|^{2} &=& \|\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2}+o(|s-t|)\|^{2}      \nonumber \\
   &=& \|\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2}\|^{2}+2\langle\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2},o(|s-t|)\rangle \nonumber \\
   &&+\|o(|s-t|)\|^{2}  \nonumber \\
   &=& |s-t|^{2}+k(t)\frac{|s-t|^{4}}{4}+o(|s-t|)
\end{eqnarray}
So, as you can see, if $k(t)$ is not constant $\|\alpha(s)-\alpha(t)\|$ depends on $k(t)$. Hence the only possibilities are the circle or the line. But an straightforward calculation shows that for the circle and the line the statement is true.
A: You do not need differentiability. Below is a proof that works when $\alpha$ is only continuous.
We may assume that $I$ starts with $0$ : $I=[0,M]$ for some $M$. Suppose that the image of $\alpha$ is not contained in a line. Then there are $x_1<x_2<x_3$ such that $\alpha(x_1),\alpha(x_2),\alpha(x_3)$ are not collinear. Let $t\in ]0,M-x_3[$. Let now $x=x_4$ be an arbitrary number in $[0,M-t]$. Consider the two families of four points $A_i=\alpha(x_i) (1 \leq i \leq 4)$ and $B_i=\alpha(x_i+t) (1 \leq i \leq 4)$. They are isometric by hypothesis : $d(A_i,A_j)=d(B_i,B_j)$. 
By this older StackOverflow question, there is an isometry $\gamma(x,t)$ of the plane sending $A_i$ to $B_i$. Since $\gamma(t)$ is already uniquely determined by its image on $\alpha(x_1),\alpha(x_2),\alpha(x_3)$, we see that $\gamma(x,t)=\gamma(t)$ is in fact independent of $x$,  and we can explicitly write out the coefficients of the matrix of $\gamma(t)$ in terms of the coordinates of
$\alpha(x_1),\alpha(x_2),\alpha(x_3)$ and $\alpha(x_1+t),\alpha(x_2+t),\alpha(x_3+t)$. So $\gamma$ will be continuous if $\alpha$ is. Unicity also ensures that $\gamma$ is a homomorphism (where it is defined) : $\gamma(s+t)=\gamma(s)\gamma(t)$ when $s,t$ and $s+t$ are all in $I$. 
Now ${\sf det}(\gamma(t))=1$ or $-1$, and by continuity this determinant is always $1$ or always $-1$. Since $\gamma(0)$ is the identity, we see that $\gamma$ is always an (affine) rotation. We have $\gamma(s)\gamma(t)=\gamma(s+t)=\gamma(t)\gamma(s)$. We see that $\gamma(s)$ and $\gamma(t)$ commute, so they must share the same center point. We deduce that the center $\Omega$ of $\gamma(t)$ is independent of $t$.  Now for any $s<t$, $\alpha(t)=\gamma(t-s)\alpha(s)$, so $\alpha(t)$ and $\alpha(s)$ are located at the same distance from $\Omega$. So $\alpha$ walks on a circle of cnter $\Omega$, qed.
