Affine Geometry - 3 questions
1. Can any convex polygon C
be partitioned into some finite number m
of quadrilaterals that are mutually affine-equivalent? If the answer is "yes", how does one do it for a given n
-gon efficiently, minimizing m
?
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2. Question: Can the plane be tiled with convex quadrilaterals that are (1) mutually non-congruent in a Euclidean sense and (2) mutually affine-equivalent?
Remark: Every trapezoid is affine equivalent to every other trapezoid with same base ratio (ratio between lengths of mutually parallel pair of sides). So one can find keep tiling the plane in parallel strips of same width; indeed, each strip can be tiled with mutually non-congruent trapezoids that all have same height and parallel sides that only need to be some constant ratio. So, the question is whether there is any other solution - ideally, a solution not involving trapezoids?
Another question: Can the plane be tiled with pentagons that are mutually non-congruent and projectively equivalent? (indeed, any two convex quadrilaterals are projectively equivalent so we go to pentagons).
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A possibly related qn:
Question: Is it possible to tile the plane with triangles that are (1) mutually similar, (2) pairwise non-congruent and (3)non-right? No other constraints.
Note 1: Reg requirement 3 above: since any right triangle can be cut into 2 pieces similar to itself and non-congruent, we can start with any right triangle and grow a plane-tiling layout outwards with progressively scaled copies of itself.
Note 2: If the question has a "yes" answer, one could consider a more constrained version of the question is got by demanding the tiles to have bounded size. Even if the basic shape of the triangle is a right triangle, this constrained version could be asked.
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These have been posted at overflow here , here and here
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2. Question: Can the plane be tiled with convex quadrilaterals that are (1) mutually non-congruent in a Euclidean sense and (2) mutually affine-equivalent?
Remark: Every trapezoid is affine equivalent to every other trapezoid with same base ratio (ratio between lengths of mutually parallel pair of sides). So one can find keep tiling the plane in parallel strips of same width; indeed, each strip can be tiled with mutually non-congruent trapezoids that all have same height and parallel sides that only need to be some constant ratio. So, the question is whether there is any other solution - ideally, a solution not involving trapezoids?
Another question: Can the plane be tiled with pentagons that are mutually non-congruent and projectively equivalent? (indeed, any two convex quadrilaterals are projectively equivalent so we go to pentagons).
---------
A possibly related qn:
Question: Is it possible to tile the plane with triangles that are (1) mutually similar, (2) pairwise non-congruent and (3)non-right? No other constraints.
Note 1: Reg requirement 3 above: since any right triangle can be cut into 2 pieces similar to itself and non-congruent, we can start with any right triangle and grow a plane-tiling layout outwards with progressively scaled copies of itself.
Note 2: If the question has a "yes" answer, one could consider a more constrained version of the question is got by demanding the tiles to have bounded size. Even if the basic shape of the triangle is a right triangle, this constrained version could be asked.
-----------
These have been posted at overflow here , here and here
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