Question: Can a shape which is curved everywhere tile the plane? If so, find at least one. If not, why not?
Answer: No. A shape which is curved everywhere cannot tile the plane. If some shape tiles the plane, then two copies of that shape must touch. Two curved shapes can only touch at a point. The space between the two touching shapes creates a region that can only be filled by a cusp.
Question: Can a shape with just 1 cusp tile the plane?
You might stumble upon the following idea, as I did, and be convinced the answer is yes.
However, making the pieces, and attempting to fit them together, reveals the problem. No matter how you arrange the pieces, you will eventually come to a region which cannot be filled. Below are two such regions, highlighted in red and green. One piece added to the red region, or two pieces added to the green region, will cause a hole. Notice that for any of these tiles to be surrounded, they need to be in contact with 5 consecutive cusps, and this is not accomplished for the pieces within the red and green regions.
This tiling was first found to fail by Lietzmann, who proved this by showing the Heesch number of this tile is 1.
Question: Can a shape with just 2 cusps tile the plane?
Answer: Yes. If you graph \(\sin(x),\) \(\sin(x + \pi) + 2,) and \(\sin(x) + 4,\) you will clearly see a shape with exactly 2 cusps that tiles the plane.
Every shape which can tile the plane by translation only could be defined by a linear combination of any two non-parallel vectors. Still, if I give you some shape that is known to tile the plane, and exactly two vectors, can you always tile the plane? In other words, are there any shapes that would require 3 or more translation vectors, or do all tilings sort of "fan out" from a single shape? If only 2 are necessary, what's the proof? And if not, what's a counterexample? This raises another somewhat related question: Which shapes can tile the plane by translation only (no rotation)?
If you have some shape which can tile the plane by translation only, is it always possible to create a tiling such that no vertex lies midway on some edge? Here's an example of vertices lying midway on edges:
But this can be made into a tiling where no vertex lies on an edge by shifting every-other row. If this is always the case, then the number of translation vectors necessary to tile a plane must be finite. More specifically, the number of vectors must be n choose 2 or fewer, where n is the number of vertices. In addition, we only need the number of nonparallel vectors, so often many of these vectors can be disregarded right away. For example, a rectangle would have only two nonparallel vectors, while 4 choose 2 = 6.