I made a presentation on flexible polyforms at the last Gathering for Gardner, but there were some polyform types that I didn’t get to, since I hadn’t yet come up with any good problems for them. One odd sort of polyform, which I am fancifully calling a constellation, can be obtained from configurations of points on the plane. We can consider two sets of points on the plane to be distinct if the pattern of collinearity among the points is different. Because every pair of points defines a line, the lines with only two points are, in a sense, not interesting; only the lines with three or more points need to be considered when determining whether two constellations differ. It seems reasonable to consider the order of points on a line to be significant; this gives us three different 5-point constellations with a pair of three point lines that meet at a point. There are 7 5-point constellations in all. Here’s the first tiling puzzle solution I found for them:
One rule for constellation tiling puzzles that I like is to disallow any point from one constellation from falling directly between two points in another constellation. This keeps the constellations more compact, and adds a little challenge to the puzzle. I like to get as much symmetry as possible in one of these flexible polyform tilings, so I decided to try for one with 7-fold symmetry. This was a little harder, but eventually I found the following tiling:
April 2018: Edited to update the image to a proper solution. Thanks to Bryce Herdt for noticing that the old solution was incorrect.
Where can we go from here? If I’ve counted right, there are 21 6-constellations. Of these, 7 can be formed by adding one independent point (a point on no line of 3) to each of the 5-constellations. The full set seems a little too big to solve by hand, but if we exclude the ones with independent points, a puzzle with the remaining 14 seems more manageable. (We may also want to exclude the 6-constellation with two separate lines of three points. With that one excluded, the remaining 13 6-constellations all can be formed from connected groups of lines with 3 or more points.)
The 14 6-constellations with no independent points.
Problem #42: Find a tiling of 6-constellations with 6-fold dihedral symmetry. Either the set of 13 or the set of 14 will do. Even more symmetry is even better.
“…I like… to disallow any point from one constellation from falling directly between two points in another constellation.”
You only mean when the two points in the second constellation are part of a line in that constellation, right? I ask because there are some points in the second figure that seem to be collinear but are not connected with lines.
Three points which might violate that rule are the two rightmost red points and the central green point. Three other points which certainly don’t violate the rule but look collinear are the lowest green and dark blue points and the top red point.
I intended that all lines with three or more points would be drawn in the diagrams. If the points you mention are in fact collinear, (which they certainly seem to be) I made a mistake. I’ll have to go back to the drawing board to look for a solution on that figure.