Wikipedia:Reference desk/Archives/Mathematics/2014 October 15
Mathematics desk | ||
---|---|---|
< October 14 | << Sep | October | Nov >> | October 16 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
October 15
edituntilable shape with only right angle corners?
editLet 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)
- 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)
- 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)
- 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)
- 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)
- 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)
- 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)
@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)
- Did you consider the tiling [2] I linked? PrimeHunter (talk) 21:11, 15 October 2014 (UTC)
- 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)
- 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)
Is there some side set of side lengths that will not allow these to tile the plane?
- 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)
- 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)
- 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)
- Arg. I meant 8. I agree with you on 6 being an L shape.Naraht (talk) 13:48, 17 October 2014 (UTC)
- Is something missing in my first post since people continue to discuss this? PrimeHunter (talk) 15:33, 17 October 2014 (UTC)
- 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)
- Is something missing in my first post since people continue to discuss this? PrimeHunter (talk) 15:33, 17 October 2014 (UTC)
- Arg. I meant 8. I agree with you on 6 being an L shape.Naraht (talk) 13:48, 17 October 2014 (UTC)
- 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)