Equilaterally triangulated surfaces with prescribed boundary There is a problem in Richard Kenyon's list (Wayback Machine) which I would like to post here, because although I have thought about it from time to time, I have not been able to make the slightest progress on it:

Question: "Given a closed polygonal path $p$ in $R^3$ composed of unit segments, is there an immersed polygonal surface
whose faces are equilateral triangles of edge length $1$, spanning $p$?"

The condition that the configuration is an "immersed" surface means that the whole configuration may be parametrized by a continuous locally one-to-one mapping from a compact connected 2-dimensional manifold with boundary into $R^3$. In other words at most two triangles may share an edge, and triangles sharing a vertex form an embedded topological disk, but two triangles which do not share an edge or a vertex may intersect.
I cannot answer the question even for the case where $p$ is a quadrilateral. Has there been any partial results for this problem?
 A: arXiv:2208.05076
Another solution, along the lines of Ian Agol's comment.
We consider the natural map $\delta$ from the moduli space $P$ of polygonal surfaces of given combinatorial type with boundary (up to ambient isometries) with fixed metric $\rho$ to the Kapovich-Millson-Klyachko-.. space $L$ of polygons in $\mathbb R^3$.
It turns out that indeed $\delta(P)$ is an isotropic subset of $L$ (it is generally a semi-alebraic subset containing an open dense isotropic manifold). In particular, the image of $\delta$ has measure $0$ in $L$. But there are only countably many polygonal surfaces whose faces are equilateral triangles of edge length 1. Thus the set of all spannable polygonal paths is a countable union of sets of measure $0$ in $L$.
N.B. This answers the question only for orientable surfaces, thus we solve a weaker problem (on the other hand we don't require a surface to be immersed). It is likely that for a non-orientable surface the isotropic property of $\delta$ doesn't hold (see Remark 4.4 and Conjecture 4.5 in our paper) and the argument doesn't work out.
A: (It is intended as an extended comment and thinking out loud rather than an answer)
That seems that the case when $p$ is a quadrilateral is exactly the bottleneck, the general case will hopefully follow from it by an induction argument.
Without loss of generality we can prove the statement for paths of even length (glue a triangle to any edge of an odd-length path and span the resulting even-length path).
Definition: The waist $w(p)$ of a closed polygonal path $p$ of lenght $l$ is the shortest line segment connecting two vertices of $p$ that separate $p$ into pieces of equal lenght ($=\frac{l}{2}$)
Statement: $|w(p)|\le C\cdot l$, where $C$ is the ratio of the diameter of the regular polygon with $l$ edges of length 1 to its perimeter $l$.
In other words, a waist of a path of length $l$ cannot be longer than a diagonal of the regular $l$-gon. In particular it is quite short for large $l$, shorter than $\frac{l}{2}$.
We can turn the waist into a polygonal path (with unit segments) of length $\lceil{|w|}\rceil$ (smallest integer greater than $|w|$). Starting from $l=6$, when $C=\frac{1}{3}$ and a poligonized waist (let's call it $W$) has lenght at worst $2$ -- everything is more or less fine: span the path consisting of the first half of  $p$ and $W$ (its length at worst $5$ -- 5 is not even, of course, but at least less than 6 :). For $l=8$ it's surely curable) and in the same manner span the second half of $p$ and $W$ and take the union of these two spanning surfaces, doing an induction step.
For $l=4$ that does't work, however, since $W$ can have length $2$.
Even for a transcendentally angled flat rhombus it is not clear to me whether one can span it. If no, the answer can be probably given by some sort of extension of fields argument (in any construction I tried to think of the transcendence of the angle survived in some form preventing to complete it). In the positive direction, there is the industry of flexible polyhedra, which potentially could provide an example of a polyhedra with movable quadrilateral living on it that could be a hope for a construction, but I could not readily find in the literature ones with equilateral triangle sides with enough flexibility to do something like we want.
A: We resolve Kenyon's problem in this paper.  We discuss in Section 5 a number of further conjectures and open problems.
