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 subarcs is a straight line segment with distinct endpoints? 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.

5$\begingroup$ My first instinct is there must be points where at least three copies of $C$ meet, and there one of them must have an "angle", so no unique tangent. I don't entirely trust my instincts on this one though. $\endgroup$ – Robin Chapman Jun 17 '10 at 19:56

3$\begingroup$ It's not clear from the question if cusps are allowed. $\endgroup$ – Andrey Rekalo Jun 17 '10 at 20:24
$C = \lbrace (x,y) sin\ x\le y\le 2\pi+sin\ x ,\ sin\ y\le x\le 2\pi+sin\ y\rbrace$.

2$\begingroup$ This looks like it has cusps: imgur.com/iNd7t.png $\endgroup$ – Steve Huntsman Jun 17 '10 at 19:52

2$\begingroup$ Are we allowing this sort of cusp or not? In some sense there is "only one tangent line" at this cusp. If this is not allowed, then pretty clearly no such tiling exists. $\endgroup$ – Tom Goodwillie Jun 17 '10 at 21:27

$\begingroup$ And can you do it with just one cusp? $\endgroup$ – Joel David Hamkins Jun 17 '10 at 22:35
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 eyeshaped figure, but where the ends of the eyes meet at angle 0. This will allow unique tangents at the "corners".
alt text http://www.freeimagehosting.net/uploads/1e7dbd258e.jpg
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.

$\begingroup$ Oh, I see Tom posted a very similar example. $\endgroup$ – Joel David Hamkins Jun 17 '10 at 20:22

6$\begingroup$ Robin Chapman's comment proves that you cannot do without cusps. (The fact that there is a point where at least 3 pieces meet is due to "Lebesgue covering dimension" of the plane, see Wikipedia article with this title.) For piecewise $C^1$ tiles, it is easy to see that one cusp is not enough too  the total variation of turn (i.e. the integral of curvature) of the smooth pieces must be zero because adjacent ones cancel each other. But the total turn is always $\pi$ for a loop with one cusp. There can be weird tiles  differentiable but not $C^1$  but I believe they can be smoothened. $\endgroup$ – Sergei Ivanov Jun 18 '10 at 11:43

$\begingroup$ Thanks for your answers which seem to imply that no such tiling exists, since I wish to rule out "cusps" and all other "tangent line ambiguities". Perhaps the following more precise statement will help to clarify what I have in mind. "At each point of J there exists one and only one straight line in E that intersects J in only that one point. $\endgroup$ – Garabed Gulbenkian Jun 19 '10 at 17:43

$\begingroup$ Garabed, that statement would seem to imply that C is convex, which you said you didn't want to insist upon. $\endgroup$ – Joel David Hamkins Jun 19 '10 at 19:53

$\begingroup$ Joel, you are right. I apologize and have to correct my previous comment. I think I just need to require that at each point of J, the curvature should be defined, continuous and finite. It may, of course, change sign. That should get rid of all cusps and weirdly behaving tangent lines. $\endgroup$ – Garabed Gulbenkian Jun 19 '10 at 22:31