Can every curve be subdivided equichordally?
Niels Diepeveen helped me fill the gap in my incomplete proof, which I've deleted from the original question and moved here because it fits this question better.
We take the curve to be $c:[0,1]\to\mathbb R^m$ and assume that $c(0)\ne c(1)$ (otherwise a trivial solution exists). A subdivision into $n$ segments is determined by its vector of interval lengths, $s=(s_1,\dots,s_n)$ where $s_i=t_i-t_{i-1}$. The set of all valid $s$ forms the standard simplex $$\Delta^{n-1}=\left\{(s_1,\dots,s_n):\sum_{i=1}^n s_i=1, s_i\ge 0\text{ for all }i=1,\dots,n\right\},$$ which is an $(n-1)$-dimensional polytope embedded in $\mathbb R^n$. In fact, $\Delta^{n-1}$ lies in the nonnegative orthant $\mathbb R_+^n$ and its boundary lies in $\partial\mathbb R_+^n$: vertices lie on the coordinate axes, $1$-faces (edges) lie on the coordinate $2$-planes, and so on.
Consider the function $d:\Delta^{n-1}\to\mathbb R_+^n$ mapping the vector of interval lengths to the vector of chord lengths, $$d(s)=(\|c(t_1)-c(t_0)\|,\dots,\|c(t_n)-c(t_{n-1})\|),$$ where $t_i=\sum_{j=1}^i s_j$. This function is nonnegative ($d(s)_i\ge 0$), nondegenerate ($d(s)\ne0$ because $c(t_0)\ne c(t_n)$), and preserves zero coordinates ($d(s)_i=0$ if $s_i=0$). Zero coordinate preservation is the key property here: it means that while $d$ may transform $\Delta^{n-1}$ into an arbitrarily complicated, possibly self-intersecting $(n-1)$-dimensional surface, it cannot detach its boundary from the faces of $\partial\mathbb R_+^n$. Vertices still lie on the coordinate axes, edges become curves lying on the coordinate $2$-planes, and so on.
We want to prove that there exists an $s\in\Delta^{n-1}$ such that all the components of $d(s)$ are equal. Equivalently, we want to show that the surface $d(\Delta^{n-1})$ intersects the line $\{(a,\dots,a):a\in\mathbb R\}$.
From left to right: A curve $c([0,1])$, the corresponding deformed simplex $d(\Delta^{n-1})$ for $n=2$, $d(\Delta^{n-1})$ for $n=3$.
Rescaling $d(s)$ so that its components sum to $1$, we obtain the map $$\hat d(s) = \frac{d(s)}{\sum_{i=1}^n d(s)_i},$$ which is well-defined and continuous because $d(s)$ is never zero. It is easy to verify that $\hat d$ maps the simplex $\Delta^{n-1}$ to itself; further, zero coordinate preservation implies that $\hat d$ also maps each face of $\Delta^{n-1}$ to itself. It can be shown using Brouwer's fixed point theorem that such a mapping must be surjective. Therefore, there exists an $s\in\Delta^{n-1}$ such that $\hat d(s)=(\frac1n,\dots,\frac1n)\in\Delta^{n-1}$, which is equivalent to the desired result.
In fact, we have proved a slightly stronger property: For any vector of nonnegative chord length ratios $r=(r_1,\dots,r_n)$, we can find a subdivision $s$ such that $d(s)=ar$ for some $a\in\mathbb R$.
Here's an alternative approach that could work generally. One may be able to reason by induction on $n$ using a connectedness argument. One would need to establish connectedness of a certain level set, but this might be hard.
The proposition is quite obvious for $n=1$. Suppose we know the proposition to hold for $n\geq 1$, that is, suppose we know that for any (continuous, simple) curve $\gamma:[0,1]\to\Bbb R^d$ (I'll take $d=2$) there exist $0\leq t_1\leq\cdots\leq t_n\leq1$ such that, if we set $t_0=0$ and $t_{n+1}=1$, $$\forall k\in\lbrace 1,\dots,n\rbrace,\quad\|\gamma(t_{k+1})-\gamma(t_{k})\|=\|\gamma(t_{k})-\gamma(t_{k-1})\|$$ Note that since $\gamma(0)\neq\gamma(1)$ by non intersection, the $t_i$ must actually all be different, so that in actuality one has $0<t_1<\cdots< t_n<1$.
Now let $c:[0,1]\to\Bbb R^d$ be a (continuous, simple) curve. We shall write vectors $x\in\Bbb R^{m}$ as $x_\bullet=(x_1,\dots,x_m)$ et us define $$T_{n+1}=\lbrace t_\bullet\in\Bbb R^{n+1}\mid 0\leq t_1\leq\cdots\leq t_n\leq t_{n+1}\leq1\rbrace$$ (again, we write $t_0=0$ and $t_{n+2}=1$) and set, for $\theta\in[0,1]$, $$T_{n+1}^\theta=\lbrace t_\bullet\in T_{n+1}\mid t_{n+1}=\theta\rbrace\simeq\begin{cases}* & \text{if }\theta=0\\[3mm]\theta\cdot T_n & \text{if }\theta>0 \end{cases}$$ Let us consider the function $\varphi:T_{n+1}\to\Bbb R_+$ defined by the formula $$\varphi(t_\bullet)=\sum_{k=1}^{n}\Big[\|c(t_{k+1})-c(t_{k})\|^2-\|c(t_{k})-c(t_{k-1})\|^2\Big]^2$$ Note that the sum only goes to $t_{n+1}$. Let us set $Z=\varphi^{-1}(0)$. This is the set of all equichordal subdivisions that start at $c(0)$, but may end before $c(1)$, that is, we drop the requirement that the last chord from $c(t_{n+1}$ to $c(1)$ have the same length as the other chords.
What can we say about $Z$ ?
- $Z$ is a closed subset of $T_{n+1}$,
- for every $\theta\in[0,1]$, the set $Z\cap T_{n+1}^\theta$ is non empty,
- actually, for every $\theta\in(0,1]$, the set $Z\cap T_{n+1}^\theta$ is included in the interior of $T_{n+1}^\theta$ (where all inequalities are strict).
The second point follows from induction : for every $\theta\in[0,1]$, the (continuous, simple if $\theta>0$, constant if $\theta=0$) curve $$c^\theta:[0,1]\to\Bbb R^d,\;t\mapsto c(\theta t)$$ has an equichordal subdivision with $n$ nodes.
Question. Is $Z$ connected ?
Suppose it were, then consider the functions $\Delta^L,\Delta^R:T_{n+1}\to\Bbb R_+$ defined by $$\Delta^L(t_\bullet)=\|c(t_1)-c(0)\|^2,\quad \Delta^R(t_\bullet)=\|c(1)-c(t_{n+1})\|^2$$ Then at the point $t^0_\bullet=(0,0,\dots,0)\in Z$, $$\Delta^L(t_\bullet^0)=0\quad\text{and}\quad\Delta^R(t_\bullet^0)=\|c(1)-c(0)\|^2>0$$ while for every $t_\bullet^1\in Z\cap T_{n+1}^1\;(\neq\emptyset)$, $$\Delta^L(t_\bullet^1)>0\quad\text{and}\quad\Delta^R(t_\bullet^1)=\|c(1)-c(1)\|^2=0$$
Thus, the continuous function $$\Delta^L-\Delta^R$$ switches sign on $Z$. If $Z$ were connected, then there would exist a point where the two functions coïncide, and such a point is exactly an equichordal subdivision.
Here are a few illustrations of this in the case $n=1$ (i.e. $n+1=2$).
Consider the curve $c$ drawn in the plane as follows:
The following represents $Z$ in this particular case : there can be either one, two or three equichordal subdivisions of $c$ with prescribed endpoint and one node, and this is how these possibilities are distributed :
The triangle represents $T_2$ in the plane, the horizontal lines are the $T_2^\theta$ for four values of $0<\theta<\theta'<1$. One sees in this example that the level set $Z$ is path connected. It even looks like a manifold ($c$ is rather smooth).