Filling the plane with non-Congruent Pieces - II

This post continues the theme of the last post:

Just came to know of the "Heterogenous Tiling Conjecture" by Golomb (made famous by Martin Gardner, says Wiki). It states: "The plane can be tiled using squares of integer sides with all squares being of different size and *all integers appearing as sides of the squares used*". This was proved in 2008 by Henle and Henle. Obviously, the sides being integers implies no bounds on the sizes of squares. One can further ask: "Can one fill the plane with squares of sides being *real numbers* in the range [a,b]?". And what if instead of real values, we restrict the square sides to be *rationals* in [a,b] where a and b are non-zero?

Remark: Regarding filling the plane with non-congruent triangles( last post). There may not quite be aleph_1 triangles of specified area and perimeter as was guessed in the last post (I gathered this from a mail from Prof. Stan Wagon). But the number of such triangles will still be more than aleph_0, the number of finite area triangles needed to fill the plane.



Update (January 15th, 2015)

Some progress on the basic question: "Is it possible to tile the plane with non-congruent triangles all of same area and perimeter?"

The Answer: Not always. Example: For a given area, if we choose the perimeter to be slightly more than that of the equilateral triangle (which is a degenerate case), infinitely many non-congruent triangles can be made. But we see that with such NEAR-EQUILATERAL triangles (sides are nearly of equal length), the plane CANNOT be tiled - pairs of adjacent triangles always end up sharing a side, which would cause congruence if both area and perimeter are equal for all tiles. For {area, perimeter} values that yield triangles far from equilateral, one is not sure.

Update (January 23rd, 2015)

The basic version of the problem (Tiling the plane with non-congruent triangles with equal area and unbounded perimeter) has been listed at Prof. Stan Wagon's 'Problem of the Week': See here. A detailed solution has alwo been linked there. As reported there, the bounded perimeter case of the problem is under active attack....

A brief remark: If the tiles are allowed to be non-convex, it is not difficult to show a set of non-convex hexagons all mutually non-congruent and all of the same area and perimeter which fill the plane - same for n-gons for any n greater than 6.

Updates to follow...