Convex polyhedra that can be folded from convex polygons This question is based on http://www.science.smith.edu/~jorourke/Papers/FoldingPP.pdf.
Therein is stated the theorem: Every convex polygon folds to an infinite number (a continuum) of noncongruent convex polyhedra.
Question: What could one say in the reverse direction: given any convex polyhedron, what could one say about the polygon(s) that could fold into it? If there are convex polyhedrons that cannot be folded from convex polygons, then, how does one characterize convex polyhedrons (such as for example, the regular tetrahedron) that can be folded out of convex polygons?
 A: This answers the easiest question posed (1), and addresses part of the more general question (2).

(1) "If there are convex polyhedrons that cannot be folded from convex polygons,..."

Yes, there is an abundance of such polyhedra. For example, let $P$ be a
cube. To unfold it to the plane, one must form surface cuts that combinatorially constitute a spanning tree of the vertices, often called a cut tree $T$.
(Any vertex not on $T$ will retain its 3D structure, and so obstruct
planar development.) A tree has at least two leaves. Let $v$ be
a vertex of $P$ at a leaf of $T$. Then the corresponding unfolding
forms an interior reflex angle of $3\pi/2$ locally about the image of $v$.
This is a local nonconvexity.

(2) "how does one characterize convex polyhedrons (...) that can be folded out of convex polygons?"

Generalizing the argument above leads to a necessary condition:
A convex polyhedron $P$ needs to have
at least two vertices of curvature $\ge \pi$
(and so incident face angles $\le \pi$)
in order to serve as leaves for a cut tree $T$
that unfolds to a convex polygon.
The cube has all vertices with curvature $\pi/2$,
so there is no "home" for the leaves of $T$.
This is, however, not a complete characterization.
