Problem
Suppose that we have a P$3$ Penrose tiling, using rhombi of smallest angles $36^\circ$ and $72^\circ$:
Which convex shapes can be found as a union of rhombi in this tiling?
The answer appears to be $8$, as shown below:
The regular decagon here can be formed in another way, but both have the same union so I am counting them once.
(As a remark, these eight convex polygons must show up in any Penrose tiling, because all Penrose tilings contain every legal finite patch infinitely often within them.)
Note that there are many convex polygons that could be formed from the Penrose rhombs, but it seems that only those pictured above can be extended to a tiling that respects the matching rules.
Classification
We can represent any convex polygon formed from Penrose rhombs by describing its side lengths along each of ten possible directions ($0^\circ, 36^\circ, \ldots, 324^\circ$), since there are only $5$ orientations of undirected edges in a Penrose tiling. It is also easy to see (by considering what happens when we repeatedly cross from one edge of a rhombus to the opposite parallel edge) that opposite edges in such a polygon must be of the same length. Subject to these restrictions, it appears that the permissible cyclic orders of sides in such a polygon fall into one of two classes, if we normalize the rhombs to have unit edge length:
All side lengths are either $0$ or $1$ (i.e. we have some kind of unit equilateral quadrilateral, hexagon, octagon, or decagon).
All side lengths are either $1$ or $2$, and no two sides of length $2$ touch.
The above two categories exactly describe the eight convex polygons I have found. However, it is not apparent to me why this should be the case; why can't we have a polygon with sides in cyclic order $(2,1,2,0,0,2,1,2,0,0)$, for instance? Many such polygons can be tiled with the rhombi, just in ways that turn out to be incompatible with the Penrose matching rules.
Proof strategies
It seems straightforward, if tedious, to classify all possible ways that one can legally arrange tiles along the border of a convex curve and discard those which prove incompatible with the P$3$ matching rules until all cases have been exhausted. However, I am hopeful that a more enlightening proof exists, given all the nice properties that Penrose tilings are known to have and the apparent simplicity of the resulting classification.
One avenue of attack might be to show several local rules about Penrose tiles that together imply this classification. For instance (I believe both of these are in fact true):
Three parallel edges cannot meet end-to-end (so a polygon cannot have edges of length $\ge3$).
No vertex in the tiling is the endpoint of two different "spokes" of two end-to-end parallel edges (so edges of length $2$ cannot be adjacent).
Some other properties?
Put together, these might be able to show that no polygons other than the ones pictured are possible, but this sort of approach alone would not show existence for the cases that are possible; given that these configurations are inevitable in any tiling, it would be nice to see a natural proof that they must appear which does not manually construct each one and show that it fits into a valid Penrose tiling.