Non-Congruent Tiling - 16
Adding a bit to an old and lengthy series, the latest instalment was this
Question: Is it possible to tile the plane with mutually non-congruent triangles with equal area and with the measure of one angle common? One can substitue area with perimeter in this question. The answer might depend on whether the angle is an irrational fraction of pi or not.
Note: I don't know the answer to even the basic question: whether a plane can be tiled with mutually non-congruent right triangles all with same area - even with no constraint on the perimeters.
This mathoverflow page also records this question: https://mathoverflow.net/questions/433795/tiling-with-non-congruent-triangles-all-of-which-have-an-equal-angle-and-equal-a
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Over the last week, 2 more posts have come up at mathoverflow on related matters:
1. https://mathoverflow.net/questions/433413/partition-of-polygons-into-strongly-acute-and-strongly-obtuse-triangles
2. https://mathoverflow.net/questions/433718/triangulation-of-polygons-with-all-triangles-having-a-common-angle
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(1) defines: obtuse triangles with the largest angle strictly above a given cutoff value as 'strongly obtuse' - the definition is parametrized by the cutoff value. Likewise, strongly acute triangles are acute triangles with the smallest angle strictly less than a specified cutoff.
General Question: Given an n-gon, how does one triangulate it into a finite number of strongly obtuse/ strongly acute pieces with specified cutoff - resulting in least number of pieces? And before that, how to decide whether such a triangulation exists?
Example: it seems impossible to cut a square into finitely many strongly obtuse triangles with cutoff 120.
Guess: Consider partitioning a given n-gon into strongly obtuse triangles with some cutoff greater than or equal to 120 degrees. It appears that to check if the n-gon can be so triangulated, we need to see only its triangulations using only segments connecting the vertices of the input n-gon - adding extra vertices for the triangulation does not help. And for any cutoff less than 120 degrees, any triangle and hence, any n-gon can be easily cut into strongly obtuse triangles with extra vertices.
(2) asks:Given an n-vertex polygonal region ("n-gon") and a real number theta less than pi, is it always possible to partition the n-gon into finitely many triangles such that each triangle has one angle equal to θ? Inumber of strongly obtuse/ strongly acute pieces with specified cutoff - resulting in least number of pieces? If the answer is "not always" (which appears to be the case), deciding whether such a triangulation exists for a given polygon and θ becomes another question. And in the cases where such a triangulation is possible, we could try to minimize the number of triangles.
Note: I don't know the answer to even the basic question: whether a plane can be tiled with mutually non-congruent right triangles all with same area - even with no constraint on the perimeters.
This mathoverflow page also records this question: https://mathoverflow.net/questions/433795/tiling-with-non-congruent-triangles-all-of-which-have-an-equal-angle-and-equal-a
-----------
Over the last week, 2 more posts have come up at mathoverflow on related matters:
1. https://mathoverflow.net/questions/433413/partition-of-polygons-into-strongly-acute-and-strongly-obtuse-triangles
2. https://mathoverflow.net/questions/433718/triangulation-of-polygons-with-all-triangles-having-a-common-angle
------------
(1) defines: obtuse triangles with the largest angle strictly above a given cutoff value as 'strongly obtuse' - the definition is parametrized by the cutoff value. Likewise, strongly acute triangles are acute triangles with the smallest angle strictly less than a specified cutoff.
General Question: Given an n-gon, how does one triangulate it into a finite number of strongly obtuse/ strongly acute pieces with specified cutoff - resulting in least number of pieces? And before that, how to decide whether such a triangulation exists?
Example: it seems impossible to cut a square into finitely many strongly obtuse triangles with cutoff 120.
Guess: Consider partitioning a given n-gon into strongly obtuse triangles with some cutoff greater than or equal to 120 degrees. It appears that to check if the n-gon can be so triangulated, we need to see only its triangulations using only segments connecting the vertices of the input n-gon - adding extra vertices for the triangulation does not help. And for any cutoff less than 120 degrees, any triangle and hence, any n-gon can be easily cut into strongly obtuse triangles with extra vertices.
(2) asks:Given an n-vertex polygonal region ("n-gon") and a real number theta less than pi, is it always possible to partition the n-gon into finitely many triangles such that each triangle has one angle equal to θ? Inumber of strongly obtuse/ strongly acute pieces with specified cutoff - resulting in least number of pieces? If the answer is "not always" (which appears to be the case), deciding whether such a triangulation exists for a given polygon and θ becomes another question. And in the cases where such a triangulation is possible, we could try to minimize the number of triangles.
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