Can every smooth space curve be realized as an origami curved crease? Many years ago, Ron Resch told me that he proved that every smooth simple space curve $C$ could be realized as a curved crease $\gamma$ in the interior of a piece of paper.
He never published this (as far as I know),
and perhaps never wrote it down formally. I would like
to know if some form of this claim has been, or can now be, proved.
By a curved crease in a piece of paper, I mean what
David Huffman
achieved with his remarkable curved-crease origami "sculptures":



Huffman proved1 that at any point $x$ on $\gamma$, the osculating plane of $\gamma$ at $x$ bisects the two planes tangent to the surface at $x$ from either side.
Aside from that, I do not know if further structural properties of $\gamma$ are known
and would help prove Resch's claim.

1David A. Huffman. Curvature and creases: A primer on paper.
IEEE Transactions on Computers, C-25(10): 1010–1019, Oct.1976.
 A: Note:  I'm revising my answer to make the argument/construction more transparent. In the previous version, I stated an existence result about flat surfaces, but didn't indicate a proof (because, at the time, I didn't remember a simple proof).  Now, having thought about it (or maybe just remembered the original construction), I can just provide the flat surface explicitly.
Here is a straightforward proof that works as long as the curvature of $\gamma$ is nonzero.  I don't know where it might be proved in the literature, but it's really just a standard `curves and surfaces' argument.
Suppose that one has an embedded, unit speed space curve $\gamma:I\to \mathbb{R}^3$ as well as a unit vector field $\nu:I\to S^2$ `that is normal along $\gamma$', i.e. $\gamma'(s)\cdot \nu(s) =0$ and also has the property that $\gamma'\cdot \nu'$ is non-vanishing.
Then the triple $(e_1,e_2,e_3) = (\gamma', \nu\times\gamma', \nu)$ is an orthonormal frame field along $\gamma$, so there are functions $a$, $b$, and $c$ on $I$ such that
$$
\gamma'' = a\,\nu + c\,\nu\times\gamma',\qquad
\nu' = - a\,\gamma' - b\,\nu\times\gamma',\qquad
(\nu\times\gamma')' = b\,\nu - c\,\gamma'.
$$
Of course, $a=-\gamma'\cdot\nu'$ is nonvanishing.  It is then not difficult to show that the mapping $X:I\times\mathbb{R}\to\mathbb{R}^3$ defined by
$$
X(s,t) = \gamma(s) + t\,\bigl(b(s)\,\gamma'(s) - a(s)\,\nu(s)\times\gamma'(s)\bigr)
$$
is a smooth embedding on an open neighborhood $U$ of $t=0$ and that the image is a ruled surface with vanishing Gauss curvature, i.e., it is locally isometric to the plane.  Moreover, calculation shows that, in this surface, oriented so that $\mathrm{d}s\wedge\mathrm{d}t$ is positive, the function $c$ is the geodesic curvature of $\gamma(I)$ in this surface $X(U)$. Finally,
although $(e_1,e_2,e_3)$ is not (usually) the Frenet frame of $\gamma$, one can compute the Frenet frame, and one finds that the curvature $\kappa$ and torsion $\tau$ of $\gamma$ as a space curve are given by the following formulae:
$$
\kappa = \sqrt{a^2 + c^2},\qquad
\tau = \frac{c\,a'-a\,c' + b\,(c^2-a^2)}{a^2+c^2}
$$
Now, suppose that one is only given a unit speed space curve $\gamma:I\to\mathbb{R}^3$ that has nonvanishing curvature $\kappa$
and torsion $\tau$ (which might or might not vanish).  Choosing a function $\theta:I\to\mathbb{R}$ such that $\sin\theta$ and $\cos 2\theta$ are nonvanishing, we can write down a solution of the above equations for $a$, $b$, and $c$ as
$$
a = \kappa\sin\theta,\quad b = (\tau-\theta')\sec2\theta,\quad
c = \kappa\cos\theta
$$
If $T=\gamma'$, $N$ and $B$ are the Frenet frame of $\gamma$, this solution corresponds to choosing $\nu_\theta = \sin\theta\, N + \cos\theta\, B$, as is easy to check.
Now, consider the two ruled surfaces $X$ that one gets by using $\nu_\theta$ and $\nu_{-\theta}$.  They both have the same geodesic curvature along $\gamma$, and so joining the $t\ge0$ part of the $\theta$-surface and the $t\le 0$ part of the $(-\theta)$ surface,
we get a flat surface that is creased along $\gamma$ at a normal angle of $2\theta$ (which is the angle between their normals along $\gamma$), but whose induced metric is flat in a neighborhood of $\gamma$.
Thus, we have a creased flat surface (i.e., a 'piece of paper') whose crease is the given space curve $\gamma$.
