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October 15

untilable shape with only right angle corners?

Let A be the set of all polygons with *only* right angle corners. (Everything that isn't a rectangle will be convex and all with have an even number of corners. Let B be the subset of A where the polygons can not tile the plane. What is the smallest number of corners that a polygon in B can have? I know that with 10 corners, I can make an untilable (consider the heptile with the positions 1,2,3,4,6,7 & 8 in a 3x3 and then shrink tile 6 a titch). But I don't know if 6 or 8Naraht (talk) 14:23, 15 October 2014 (UTC)[reply]

Are you talking about a convex equiangular polygon that has 10 right angle corners? Is that even possible? "For an equiangular n-gon each angle is 180° - (360°)/n ; this is the equiangular polygon theorem." I don't understand what you mean about a "heptile" either, but maybe that's my fault :) SemanticMantis (talk) 17:12, 15 October 2014 (UTC)[reply]

It's about a rectilinear polygon. The mentioned "heptile" must mean the heptomino with a hole. A rectilinear hexagon can only have an L-like shape and that can always tile the plane like in [1] regardless of the side lenghts. A rectilinear octagon cannot in general. Consider for example a U pentomino where the gap is made too narrow to fit another part. PrimeHunter (talk) 17:41, 15 October 2014 (UTC)[reply]

Ok, that explains my "heptile" confusion, but I'm still at a loss on the convexity claim. I think you're right that the left picture is mostly what the OP describes, but is it convex under any standard definition of convexity? It's certainly not a convex polygon... SemanticMantis (talk) 18:19, 15 October 2014 (UTC)[reply]

Right, the poster meant concave. A rectangle is convex and all rectilinear polygons with more than four sides are concave. PrimeHunter (talk) 19:13, 15 October 2014 (UTC)[reply]

@Naraht:, are you restricting yourself to polyominoes and allowing both rotations and translations? If not:

If you allow only translations: a simple L-shape polyomino (with 6 corners) won't tile the plane.
Even allowing rotations, if you don't restrict yourself to polyominoes: I believe, a L-shaped polygon with appropriately chosen (ie, non-commensurate) sides won't tile the plane.

Abecedare (talk) 19:54, 15 October 2014 (UTC)[reply]

Did you consider the tiling [2] I linked? PrimeHunter (talk) 21:11, 15 October 2014 (UTC)[reply]

Perhaps I'm misunderstanding you but as far as I can see any L shape with right angles will tile the plane with just translations no matter what length the sides are.Dmcq (talk) 21:14, 15 October 2014 (UTC)[reply]

You are obviously right. Sorry for the brain-freeze (even I can't replicate what I was thinking). Abecedare (talk) 21:26, 15 October 2014 (UTC)[reply]

Is there some side set of side lengths that will not allow these to tile the plane?
Ok, I think we've mostly gotten the question figured out, and some partial answers. As I now understand the problem, the question is "is there a rectilinear, 8-cornered polygon that cannot tile the plane. We know that there is at least 1 10-cornered that will not, and it seems that any 6-cornered one will. SemanticMantis (talk) 15:30, 16 October 2014 (UTC)[reply]

Candidate: take the 'T' tetromino at right. Assume it is made of unit squares. Remove the descender square, and replace it by a rectangle of size 1/3 X 1/4. I don't think that will tile, and it remains an 8-angled rectilinear polygon. SemanticMantis (talk) 17:36, 16 October 2014 (UTC)[reply]

Original poster here. Yes, sorry about the convex/concave confusion, you are right. And I think the proposed candidate by @SemanticMantis: is very good - a rectangle with a tiny rectangular "tumor" on the side (not at an end) will have 6 corners and not tile. Thank You!Naraht (talk) 00:56, 17 October 2014 (UTC)[reply]

If the "tumor" is not at one end of the side then surely that is 8 corners, not 6 ? Seems to me that any rectilinear 6-cornered polygon must be an L-shape and so can tile the plane as Dmcq points out above (but I could be wrong). Gandalf61 (talk) 11:50, 17 October 2014 (UTC)[reply]

Arg. I meant 8. I agree with you on 6 being an L shape.Naraht (talk) 13:48, 17 October 2014 (UTC)[reply]

Is something missing in my first post since people continue to discuss this? PrimeHunter (talk) 15:33, 17 October 2014 (UTC)[reply]

No, I think your post nailed it. Everything after is just clearing up confusions, typos and terminology. Gandalf61 (talk) 16:21, 17 October 2014 (UTC)[reply]