Timeline for Tiling the plane with quadrilaterals that are mutually non-congruent and affine equivalent
Current License: CC BY-SA 4.0
10 events
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| May 24, 2023 at 17:02 | comment | added | Nandakumar R | From what I understand, you need to split each square by a slanting line into 2 right trapezoids that are affinely equivalent to each other. Further, since every trapezoid pair needs to be mutually affine equivalent, each trapezoid needs to have the same base ratio and that might constrain the slopes of the slant lines too much. Hope you can clarify this. A pic might help. | |
| May 23, 2023 at 19:44 | comment | added | Nick S | @NandakumarR See the edit, it is not a complete proof, but I think it works. | |
| May 23, 2023 at 19:44 | history | edited | Nick S | CC BY-SA 4.0 |
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| May 23, 2023 at 18:43 | comment | added | Nick S | The hard part seems to be getting affinely equivalent quadrilaterals. I think that the following approach works, needs to be treated with care, but start with a square tiling, and divide it into non-congruent trapezoid tiling. Now, inductively, for each trapezoid $ABCD$ with $AB \parallel CD$, pick points $E \in AD, F \in BC$ such that $EF \not\parallel AB$ and $ABEF, CDEF$ are affinely equivalent to all previous tiles. | |
| May 23, 2023 at 18:16 | comment | added | Nandakumar R | Kevin, if I understood you right,, we will get a tiling with mutually non-congruent parallelograms and they are all trapezoids. | |
| May 23, 2023 at 18:10 | comment | added | Kevin | @NandakumarR Can't you start with a tiling with parallelograms instead of squares, and do the same thing? | |
| May 23, 2023 at 17:43 | history | edited | Nick S | CC BY-SA 4.0 |
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| May 23, 2023 at 17:36 | history | edited | Nick S | CC BY-SA 4.0 |
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| May 23, 2023 at 17:30 | comment | added | Nandakumar R | thanks. this is certainly another solution but again involving trapezoids - rectangles are trapezoids. | |
| May 23, 2023 at 17:25 | history | answered | Nick S | CC BY-SA 4.0 |