Non-Congruent Tiling - an Ongoing Story
News Update: Dirk Frettloh has written a preprint on some interesting fresh work of his and consolidating whatever is known on the non-congruent equipartition problem here:
http://arxiv.org/abs/1603.09132
Recording some speculations and thoughts on similar lines:
1. It appears that a plane cannot be tiled by mutually non-congruent, equal area rectangles if the following constraint is applied: The thinnest of the rectangles has length less than twice its thickness. It seems this can be proved on similar lines to the proof that with near equilateral triangles all with same area and perimeter, the plane cannot be tiled (mentioned in an earlier post). One is not sure if relaxing the constraint partially (ie with some finite upper bound greater than 2 on the length to width ratio) will enable such a tiling.
2. Can the plane be tiled with non congruent right triangles all of the same area? (With equal area rectangles, one can have spiral arrangements; with right triangles, one is not sure)
3. Can the plane be tiled with non congruent right triangles all with the same hypotenuse (ie all those right triangles inscribable in the same semi circle)? Can one manage with non congruent triangles with same area and same longest side? Or (this one bit added on May 6th 2019) with non congruent triangles with same perimeter and same longest side?
4. Can the plane be tiled with equilateral triangles, all of different sizes? More generally, can a plane be tiled by non congruent but mutually similar triangles? These questions let go of the equal area requirement.
Note 1: As was mentioned in an earlier post (http://nandacumar.blogspot.in/2015/01/filling-plane-with-non-congruent-pieces.html), is is not very hard to tile the plane with squares all of different sizes. Method: a way to dissect a square into 21 squares all of different sizes is known. Do this to a unit square, then along side this dissected unit square, patch another unit square, then build outwards a Fibonacci spiral of squares of increasing size. Of course, this will need arbitrarily large squares. See the Wiki article on 'squaring the square' for details.
It looks very likely that *any* such tiling of the plane with unequal squares has to have either arbitrarily large or arbitrarily small squares - even if we use squares of irrational side. Not sure of this! At least as shown in the Wiki article mentioned above, we can have a lower bound on the size of the squares used.
Note 2: With equilateral triangles, such a tiling seems doable but unlike in the square case, a lower bound on the size of the tiles may not exist. Not sure!
5. CAn the plane be filled with rectangles none of which share either dimension? This appears doable with spiral arrangements starting with a Blanche dissection. But then, what if one wants some upper bound on the length to width ratio?
Shall update this page as answers come up...
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Update (Nov 29th 2017):
Christian Richter and Melchior Wirth (https://arxiv.org/pdf/1711.08903.pdf) have just proved (among plenty of other things) that the guess made in Note 2 above is valid: if one partitions the plane into equilateral triangles of different sizes, then arbitrarily small triangles necessarily occur.
A further thought: what if the tiles are to be *isosceles and non-congruent* (with various possibilities for area and perimeter)?
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Update (May 27th 2018)
Can the plane be tiled with triangles that are similar to one another but all of different sizes? Further we could insist on both upper and lower bounds on the sizes. Looks possible but not sure.
As usual, One can also ask the same question about quadrilaterals and so on and about higher dimensions. Curiously, if one goes to the hyperbolic plane, all similar triangles are congruent so the problem seems to collapse.
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Update (June 10th, 2020):
Here is a nice discussion on the question of tiling with non-congruent isosceles triangles (see above, the November 29th 1017 update), dating back to 2015!
https://mathoverflow.net/questions/221431/tiling-the-plane-with-incongruent-isosceles-triangles
Two constructions are illustrated both showing what appear to be tilings of the plane with pairwise non-congruent isosceles triangles with no upper bound on the size of the tiles. So, one naturally wonders if such a tiling with such an upper bound can be worked out. And of course, area and perimeter requirements on the tiles can be added!
Indeed, we can tile plane with non-congruent isosceles triangles with bounds only on edge lengths: perturb each vertex of an equilateral triangle tiling infinitesimally in a unique direction, to get a tiling with non-congruent acute triangles with every side unique and with both upper and lower bounds on max edge length.. Then, divide each triangle into isosceles triangles by joining its circumcenter with the vertices. That should be it.
However, it looks like further constraints such as equality of area/perimeter would make the problem harder
A further variant: to tile the plane with mutually non-congruent acute isosceles triangles ( without or with area/perimeter constraints),
Recording some speculations and thoughts on similar lines:
1. It appears that a plane cannot be tiled by mutually non-congruent, equal area rectangles if the following constraint is applied: The thinnest of the rectangles has length less than twice its thickness. It seems this can be proved on similar lines to the proof that with near equilateral triangles all with same area and perimeter, the plane cannot be tiled (mentioned in an earlier post). One is not sure if relaxing the constraint partially (ie with some finite upper bound greater than 2 on the length to width ratio) will enable such a tiling.
2. Can the plane be tiled with non congruent right triangles all of the same area? (With equal area rectangles, one can have spiral arrangements; with right triangles, one is not sure)
3. Can the plane be tiled with non congruent right triangles all with the same hypotenuse (ie all those right triangles inscribable in the same semi circle)? Can one manage with non congruent triangles with same area and same longest side? Or (this one bit added on May 6th 2019) with non congruent triangles with same perimeter and same longest side?
4. Can the plane be tiled with equilateral triangles, all of different sizes? More generally, can a plane be tiled by non congruent but mutually similar triangles? These questions let go of the equal area requirement.
Note 1: As was mentioned in an earlier post (http://nandacumar.blogspot.in/2015/01/filling-plane-with-non-congruent-pieces.html), is is not very hard to tile the plane with squares all of different sizes. Method: a way to dissect a square into 21 squares all of different sizes is known. Do this to a unit square, then along side this dissected unit square, patch another unit square, then build outwards a Fibonacci spiral of squares of increasing size. Of course, this will need arbitrarily large squares. See the Wiki article on 'squaring the square' for details.
It looks very likely that *any* such tiling of the plane with unequal squares has to have either arbitrarily large or arbitrarily small squares - even if we use squares of irrational side. Not sure of this! At least as shown in the Wiki article mentioned above, we can have a lower bound on the size of the squares used.
Note 2: With equilateral triangles, such a tiling seems doable but unlike in the square case, a lower bound on the size of the tiles may not exist. Not sure!
5. CAn the plane be filled with rectangles none of which share either dimension? This appears doable with spiral arrangements starting with a Blanche dissection. But then, what if one wants some upper bound on the length to width ratio?
Shall update this page as answers come up...
------------
Update (Nov 29th 2017):
Christian Richter and Melchior Wirth (https://arxiv.org/pdf/1711.08903.pdf) have just proved (among plenty of other things) that the guess made in Note 2 above is valid: if one partitions the plane into equilateral triangles of different sizes, then arbitrarily small triangles necessarily occur.
A further thought: what if the tiles are to be *isosceles and non-congruent* (with various possibilities for area and perimeter)?
-----------
Update (May 27th 2018)
Can the plane be tiled with triangles that are similar to one another but all of different sizes? Further we could insist on both upper and lower bounds on the sizes. Looks possible but not sure.
As usual, One can also ask the same question about quadrilaterals and so on and about higher dimensions. Curiously, if one goes to the hyperbolic plane, all similar triangles are congruent so the problem seems to collapse.
-----------
Update (June 10th, 2020):
Here is a nice discussion on the question of tiling with non-congruent isosceles triangles (see above, the November 29th 1017 update), dating back to 2015!
https://mathoverflow.net/questions/221431/tiling-the-plane-with-incongruent-isosceles-triangles
Two constructions are illustrated both showing what appear to be tilings of the plane with pairwise non-congruent isosceles triangles with no upper bound on the size of the tiles. So, one naturally wonders if such a tiling with such an upper bound can be worked out. And of course, area and perimeter requirements on the tiles can be added!
Indeed, we can tile plane with non-congruent isosceles triangles with bounds only on edge lengths: perturb each vertex of an equilateral triangle tiling infinitesimally in a unique direction, to get a tiling with non-congruent acute triangles with every side unique and with both upper and lower bounds on max edge length.. Then, divide each triangle into isosceles triangles by joining its circumcenter with the vertices. That should be it.
However, it looks like further constraints such as equality of area/perimeter would make the problem harder
A further variant: to tile the plane with mutually non-congruent acute isosceles triangles ( without or with area/perimeter constraints),
posted by R.Nandakumar @ 11:37 AM
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