Can we characterize the following kinds of plane coverings? (Open-ended, but provide some description more "useful" than the constraints given.) For a more answerable question, is there an effective method for continuing any initial finite placement of line segments (or proving it impossible)? Do some initial placements force periodic coverings? I have a more specific question/conjecture at the end, but it requires explanation.

We must cover the plane with line segments such that each is parallel to two other line segments on opposite sides at unit distance. Each line segment must touch one or two others at its endpoints, and the angle between them must be 120 degrees. Moreover, the channels between segments must not have any branching or dead-ends (unit hexagons are allowed).

Several obvious coverings satisfy the above. E.g., a hexagonal grid with unit spacing works, as does a banded pattern of "zig-zag" lines, or a series of concentric hexagons going out to infinity. There are less obvious coverings, such as:

In case the above verbal description is inadequate, I think it is equivalent to describe the problem in terms of requirements on local neighborhoods of individual points, illustrated in the following figure:

Local neighborhoods may be rotated as needed.

(1) Every point is on a line segment with parallel segments at unit distance.

(2) Segments can only terminate if they meet one or two other segments at a 120 degree angle.

(3) The channels between segments cannot branch or terminate in a dead end (unit hexagons are allowed). Specifically, the neighborhoods shown with a red dotted boundary cannot appear anywhere in the covering.

Conjecture: Apart from concentrically nested hexagons, only one size of hexagon can be formed in such a covering. That is, if a covering contains a hexagon in which three segments meet at a corner, then there are no larger hexagons in the covering, and any smaller ones are concentrically nested in these. [Update: false as stated, but something similar may hold.] The example above satisfies this. Moreover, if unequal-length hexagon sides are coincident, then these form a 60 degree angle, which is forbidden, so I think the placement of one constrains the entire covering. Can this be proved or is there a counterexample?

An observation that may help is that the plane can be tessellated into regions of parallel line segments.

Assuming these regions are always parallelograms (as above, but not proved), then it may be reducible to a more conventional tiling problem. It looks like three segments join when the 120 degree corners meet and two segments join at adjacent sides. Unit hexagons occur where 60 degree corners meet.

(Note: I don't know the answer. I am actively working on this, but if it's solved, I'll gladly read that.)

channels. $\endgroup$ – Joseph O'Rourke Jun 15 '17 at 15:35