Let C be a compact subset of the Euclidean plane E whose boundary is a Jordan curve J. If C tiles the plane, can J be such that it has a unique tangent line at each point and none of its sub-arcs is a straight line segment with distinct end-points? If so, can you give an example? J does not need to be convex and the tiling need not be regular. The only requirement is that the plane E be a countable union of congruent copies of C, no two of which have a common interior point.
$C = \lbrace (x,y)| sin\ x\le y\le 2\pi+sin\ x ,\ sin\ y\le x\le 2\pi+sin\ y\rbrace$.
I have a way to do it, but read my note at the bottom about why you may not find it to be valid.
Use an eye-shaped figure, but where the ends of the eyes meet at angle 0. This will allow unique tangents at the "corners".
The eyes will fit together in a standard brick pattern.
In the image, I used a sin function, but you can do it also using circle fragments. In this case, the tiling corresponds to a standard penny tiling of the plane, but apportioning the empty space to adjacent pennies in order to make the eye shapes.
Since my Jordan curves turn directly around in the opposite direction at those cusps, however, you may not consider this to be a valid example, since perhaps you regard this as two tangent lines at those points, pointing in opposite directions.
I think someone will show up and prove that you cannot do it without any cusps. I would like to know whether you can do it with only one cusp.