### Going Hyperbolic

Let me state the broad question first:

Specific examples:

1. Is it possible to tile the hyperbolic plane with equilateral triangles all of different sizes? In the Euclidean setting, the answer to this is that there will necessarily be arbitrarily small triangles in such a tiling (as has recently been proved by experts).

Note: On the Hyperbolic plane, we can have equilateral triangles (all angles equal, all sides equal) with every vertex angle value less than 60 degrees and there is a unique equilateral triangle with a given vertex angle.

2. Can one tile the hyperbolic plane with triangles all of same area and perimeter but mutually non-congruent (in Euclidean case, this question has recently been given a negative answer by Kupavski, Pach and Tardos)?

On the Hyperbolic plane, one needs to check if triangles with same area (same angle sum) and perimeter can be at all non-congruent and if so by how much.

Remark: The Hyperbolic plane has lots more room than Euclidean (for example, infinitely many ways to tile with regular polygons as opposed to just 3 in Euclidean) but does this also imply that the answer to question 2 might be "possible"? (By the same token, in elliptic geometry, things just possible in Euclidean might become impossible).

3. I have no idea if the the basic fair partition question - and the complete answer thereof recently found by Akopyan, Avvakumov and Karasev - goes over 'uneventfully' to hyperbolic.

Or the old congruent partition question. A preliminary finding there was: "A quadrilateral with all angles irrational fractions of pi but adding up to 2*pi cannot be broken into any number of mutually congruent pieces"

4. Or for that matter, if the question of 'fair and non-congruent tiling of the plane' (tiling the plane with convex regions of equal area and perimeter, but all mutually non-congruent and with

At a more fundamental level, I am yet to understand well how perimeter behaves on the Hyperbolic plane.

Update (May 27th 2018):

Here is another question that was recently put up as an addition to the two year old post 'Non Congruent Tilings - an Ongoing Story' (http://nandacumar.blogspot.in/2016/06/non-congruent-tiling-ongoing-story.html): Can the (Euclidean) plane be tiled with triangles that are

However, on the Euclidean plane itself, one could ask if there is any qualitative difference from the case of

Update (June 9th 2018):

Adding one more question:

Find a set of rectangles all of same perimeter (let us call such rectangles isoperimetric rectangles) but with different areas which together form a neat big rectangle. Note that we are on the Euclidean plane now.

This question was attempted at https://arxiv.org/ftp/arxiv/papers/1307/1307.3472.pdf and one ended up at:

One wonders how going to hyperbolic geometry would change things.

Note: A rectangle on the hyperbolic plane can be viewed as a quadrilateral with all angles equal but less than 90 degrees and opposite sides of equal length. But more than 4 rectangles need to be put around each internal vertex in the layout and that *might* mean that making the overall layout rectangular would be harder than in Euclidean case and perhaps impossible.

Aside: The 'rectangling the rectangle' problem with rational sides for tiles (http://cs.smith.edu/~jorourke/TOPP/P78.html#Problem.78) seems still open even on the Euclidean plane.

Updates/Corrections shall follow as and when I find out more from experts....

*do some of the questions that were asked on these pages over the years have nontrivially altered answers if one goes to hyperbolic geometry from euclidean?*Specific examples:

1. Is it possible to tile the hyperbolic plane with equilateral triangles all of different sizes? In the Euclidean setting, the answer to this is that there will necessarily be arbitrarily small triangles in such a tiling (as has recently been proved by experts).

Note: On the Hyperbolic plane, we can have equilateral triangles (all angles equal, all sides equal) with every vertex angle value less than 60 degrees and there is a unique equilateral triangle with a given vertex angle.

2. Can one tile the hyperbolic plane with triangles all of same area and perimeter but mutually non-congruent (in Euclidean case, this question has recently been given a negative answer by Kupavski, Pach and Tardos)?

On the Hyperbolic plane, one needs to check if triangles with same area (same angle sum) and perimeter can be at all non-congruent and if so by how much.

Remark: The Hyperbolic plane has lots more room than Euclidean (for example, infinitely many ways to tile with regular polygons as opposed to just 3 in Euclidean) but does this also imply that the answer to question 2 might be "possible"? (By the same token, in elliptic geometry, things just possible in Euclidean might become impossible).

3. I have no idea if the the basic fair partition question - and the complete answer thereof recently found by Akopyan, Avvakumov and Karasev - goes over 'uneventfully' to hyperbolic.

Or the old congruent partition question. A preliminary finding there was: "A quadrilateral with all angles irrational fractions of pi but adding up to 2*pi cannot be broken into any number of mutually congruent pieces"

4. Or for that matter, if the question of 'fair and non-congruent tiling of the plane' (tiling the plane with convex regions of equal area and perimeter, but all mutually non-congruent and with

**no restrictions of their number of sides**) becomes easier on the Hyperbolic plane - of course, I don't know even its Euclidean answer yet.At a more fundamental level, I am yet to understand well how perimeter behaves on the Hyperbolic plane.

Update (May 27th 2018):

Here is another question that was recently put up as an addition to the two year old post 'Non Congruent Tilings - an Ongoing Story' (http://nandacumar.blogspot.in/2016/06/non-congruent-tiling-ongoing-story.html): Can the (Euclidean) plane be tiled with triangles that are

**similar to one another but all of different sizes**? This question is irrelevant in Hyperbolic plane because similar triangles are automatically congruent there.However, on the Euclidean plane itself, one could ask if there is any qualitative difference from the case of

**equilateral triangles with all having different sizes**and the case of**mutually similar but unequal triangles**(the mutually unequal equilateral triangle case has been done as mentioned above). These questions can be generalized to quadrilaterals from triangles and to higher dimensions.Update (June 9th 2018):

Adding one more question:

Find a set of rectangles all of same perimeter (let us call such rectangles isoperimetric rectangles) but with different areas which together form a neat big rectangle. Note that we are on the Euclidean plane now.

This question was attempted at https://arxiv.org/ftp/arxiv/papers/1307/1307.3472.pdf and one ended up at:

*Conjectures:*

It appears that no such rectangular layouts can be made with less than 7 isoperimetric tiles (we mean any rectangular layout, not only spirally arranged ones). There may be no layouts whe re (1) the big rectangle is a square (at least when the dimensions of the tiles are rational) or (2) with the number of isoperimetric tiles arbitrarily large (indeed, we know of no rectangular layout formed with more than 10 isoperimetric tiles). We do not yet know of any solution where the ratios between dimensions of the tiles are irrational.It appears that no such rectangular layouts can be made with less than 7 isoperimetric tiles (we mean any rectangular layout, not only spirally arranged ones). There may be no layouts whe re (1) the big rectangle is a square (at least when the dimensions of the tiles are rational) or (2) with the number of isoperimetric tiles arbitrarily large (indeed, we know of no rectangular layout formed with more than 10 isoperimetric tiles). We do not yet know of any solution where the ratios between dimensions of the tiles are irrational.

One wonders how going to hyperbolic geometry would change things.

Note: A rectangle on the hyperbolic plane can be viewed as a quadrilateral with all angles equal but less than 90 degrees and opposite sides of equal length. But more than 4 rectangles need to be put around each internal vertex in the layout and that *might* mean that making the overall layout rectangular would be harder than in Euclidean case and perhaps impossible.

Aside: The 'rectangling the rectangle' problem with rational sides for tiles (http://cs.smith.edu/~jorourke/TOPP/P78.html#Problem.78) seems still open even on the Euclidean plane.

Updates/Corrections shall follow as and when I find out more from experts....