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RavenclawPrefect
RavenclawPrefect
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Question 2 can be answered in the affirmative, at least: there are many triangles with the multi-way rep-tile property.

Every triangle has a simple tiling of itself with $k^2$ copies (just take the affine image of the standard equilateral tiling); these are called quadratic tilings by Michael Beeson in his 2010 paper on tiling triangles by congruent triangular tiles. So for a triangle to have this property, it suffices to have any non-quadratic tiling of size $m$, since we can choose a larger quadratic tiling with some $n>m$ tiles: the quadratic tiling has no subtriangles that are not themselves tiled quadratically, as should be obvious from looking at which straight lines can be formed. (This condition is of course also necessary.)

So we can then consult the linked paper for a variety of triangles with non-quadratic tilings: this is true of the $(30^\circ,60^\circ,90^\circ)$ triangle and any right triangle whose legs are in a rational ratio, which I think is exhaustive though I haven't read the full paper in a while to confirm this (it focuses mostly on the realizable numbers of tiles, which is related but not quite identical to our question here).

If you want to revise your definition to the stricter condition that there are two self-tilings, neither of which contains any other nontrivial self-tiling, there are still triangular examples: consider the unique $3$-tiling of the $(30^\circ,60^\circ,90^\circ)$ triangle by itself and its $k=2$ quadratic tiling.

As an addendum, the question of convex hexagonal rep-tiles ought to be much easier to resolve, since there are only three classes of convex hexagon which can tile the plane, and I believe none of them can even tile a half-plane (which is a prerequisite to being a polygonal rep-tile, though the proof involves a somewhat messy compactness argument). You can at least use the fact that any rep-tile must have an angle of at most $90^\circ$ to narrow the scope of possible tiles a little. Combined with a negative answer to Question 1, this would effectively reduce the classification problem of Question 2 to the convex quadrilaterals, which seems like it might prove rather difficult - it might end up being easiest to just classify all convex quadrilateral rep-tiles and then work out which ones are multi-way.

Question 2 can be answered in the affirmative, at least: there are many triangles with the multi-way rep-tile property.

Every triangle has a simple tiling of itself with $k^2$ copies (just take the affine image of the standard equilateral tiling); these are called quadratic tilings by Michael Beeson in his 2010 paper on tiling triangles by congruent triangular tiles. So for a triangle to have this property, it suffices to have any non-quadratic tiling of size $m$, since we can choose a larger quadratic tiling with some $n>m$ tiles: the quadratic tiling has no subtriangles that are not themselves tiled quadratically, as should be obvious from looking at which straight lines can be formed. (This condition is of course also necessary.)

So we can then consult the linked paper for a variety of triangles with non-quadratic tilings: this is true of the $(30^\circ,60^\circ,90^\circ)$ triangle and any right triangle whose legs are in a rational ratio, which I think is exhaustive though I haven't read the full paper in a while to confirm this (it focuses mostly on the realizable numbers of tiles, which is related but not quite identical to our question here).

If you want to revise your definition to the stricter condition that there are two self-tilings, neither of which contains any other nontrivial self-tiling, there are still triangular examples: consider the unique $3$-tiling of the $(30^\circ,60^\circ,90^\circ)$ triangle by itself and its $k=2$ quadratic tiling.

As an addendum, the question of convex hexagonal rep-tiles ought to be much easier to resolve, since there are only three classes of convex hexagon which can tile the plane, and I believe none of them can even tile a half-plane (which is a prerequisite to being a rep-tile, though the proof involves a somewhat messy compactness argument). You can at least use the fact that any rep-tile must have an angle of at most $90^\circ$ to narrow the scope of possible tiles a little. Combined with a negative answer to Question 1, this would effectively reduce the classification problem of Question 2 to the convex quadrilaterals, which seems like it might prove rather difficult - it might end up being easiest to just classify all convex quadrilateral rep-tiles and then work out which ones are multi-way.

Question 2 can be answered in the affirmative, at least: there are many triangles with the multi-way rep-tile property.

Every triangle has a simple tiling of itself with $k^2$ copies (just take the affine image of the standard equilateral tiling); these are called quadratic tilings by Michael Beeson in his 2010 paper on tiling triangles by congruent triangular tiles. So for a triangle to have this property, it suffices to have any non-quadratic tiling of size $m$, since we can choose a larger quadratic tiling with some $n>m$ tiles: the quadratic tiling has no subtriangles that are not themselves tiled quadratically, as should be obvious from looking at which straight lines can be formed. (This condition is of course also necessary.)

So we can then consult the linked paper for a variety of triangles with non-quadratic tilings: this is true of the $(30^\circ,60^\circ,90^\circ)$ triangle and any right triangle whose legs are in a rational ratio, which I think is exhaustive though I haven't read the full paper in a while to confirm this (it focuses mostly on the realizable numbers of tiles, which is related but not quite identical to our question here).

If you want to revise your definition to the stricter condition that there are two self-tilings, neither of which contains any other nontrivial self-tiling, there are still triangular examples: consider the unique $3$-tiling of the $(30^\circ,60^\circ,90^\circ)$ triangle by itself and its $k=2$ quadratic tiling.

As an addendum, the question of convex hexagonal rep-tiles ought to be much easier to resolve, since there are only three classes of convex hexagon which can tile the plane, and I believe none of them can even tile a half-plane (which is a prerequisite to being a polygonal rep-tile, though the proof involves a somewhat messy compactness argument). You can at least use the fact that any rep-tile must have an angle of at most $90^\circ$ to narrow the scope of possible tiles a little. Combined with a negative answer to Question 1, this would effectively reduce the classification problem of Question 2 to the convex quadrilaterals, which seems like it might prove rather difficult - it might end up being easiest to just classify all convex quadrilateral rep-tiles and then work out which ones are multi-way.

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RavenclawPrefect
RavenclawPrefect
  • 3.1k
  • 14
  • 24

Question 2 can be answered in the affirmative, at least: there are many triangles with the multi-way rep-tile property.

Every triangle has a simple tiling of itself with $k^2$ copies (just take the affine image of the standard equilateral tiling); these are called quadratic tilings by Michael Beeson in his 2010 paper on tiling triangles by congruent triangular tiles. So for a triangle to have this property, it suffices to have any non-quadratic tiling of size $m$, since we can choose a larger quadratic tiling with some $n>m$ tiles: the quadratic tiling has no subtriangles that are not themselves tiled quadratically, as should be obvious from looking at which straight lines can be formed. (This condition is of course also necessary.)

So we can then consult the linked paper for a variety of triangles with non-quadratic tilings: this is true of the $(30^\circ,60^\circ,90^\circ)$ triangle and any right triangle whose legs are in a rational ratio, which I think is exhaustive though I haven't read the full paper in a while to confirm this (it focuses mostly on the realizable numbers of tiles, which is related but not quite identical to our question here).

If you want to revise your definition to the stricter condition that there are two self-tilings, neither of which contains any other nontrivial self-tiling, there are still triangular examples: consider the unique $3$-tiling of the $(30^\circ,60^\circ,90^\circ)$ triangle by itself and its $k=2$ quadratic tiling.

As an addendum, the question of convex hexagonal rep-tiles ought to be much easier to resolve, since there are only three classes of convex hexagon which can tile the plane, and I believe none of them can even tile a half-plane (which is a prerequisite to being a rep-tile, though the proof involves a somewhat messy compactness argument). You can at least use the fact that any rep-tile must have an angle of at most $90^\circ$ to narrow the scope of possible tiles a little. Combined with a negative answer to Question 1, this would effectively reduce the classification problem of Question 2 to the convex quadrilaterals, which seems like it might prove rather difficult - it might end up being easiest to just classify all convex quadrilateral rep-tiles and then work out which ones are multi-way.