Filling the Plane with non-Congruent pieces

Question: Consider the set of all triangles with the same specified area and perimeter - there is an infinity of them (aleph_1 it seems) except in degenerate cases. Can a selection of unique triangles from among this set fill the plane without gaps or overlaps? Can this be done for all valid and non-degenerate area-perimeter combinations?

IOW, can the plane be filled by triangles of same area and perimeter with no two triangles congruent to each other? Note: A tiling of the plane by mutually congruent pieces contains only aleph_0 tiles.

(The question can be rephrased with convex quadrilaterals replacing triangles or even with the pieces allowed to be any convex polygon with the area and perimeter constraints and the mutually non-congruent nature obeyed.)

Alternatively, one thinks of working with only triangles but relaxing the perimeter constraint. It is seen that if the non congruent equal area triangles can have arbitrary perimeter, the plane can be filled with equal area non-congruent triangles. A possible method fills each of the 4 infinite quadrants given by the two axes separately. For the first quadrant, the first piece is a 'seed' right triangle T with arms on the +X and +Y axes, then triangles with same area but progressively larger perimeter are laid out with bases on each axis alternately to fill the quadrant; obviously, the perimeter of the triangles added increases without bound. To fill each other quadrant start with a seed right triangle with same area as T but with arms in a fundamentally different ratio.

That raises the issue:

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What if the perimeters of the non-congruent, plane-filling, equal area triangles ought to have a finite upper bound? If such layouts do exist, how could one minimize the ratio between the largest and smallest perimeter among the triangles used?

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One could just as well ask: What happens if we try to fill the plane with non-congruent triangles all of same *perimeter* but with area relaxed (for fixed perimeter, area has bounds)?

And what about similar questions in 3D?

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Note: if we need to fill the plane with non-congruent *quadrilaterals* of same area and bounded perimeter, it is pretty easy. Indeed, any convex quadrilateral tiles the plane (well known). Take a convex quadrilateral Q with all sides and angles unequal and tile the plane with copies of it. Now Q can be divided into 2 non-congruent quadrilaterals of equal area and *bounded* (both upper and lower) perimeter in infinitely many ways; so simply divide each tile Q differently into two quads each. In the same way, we can also tile the plane with non-congruent quads of same perimeter but bounded area.

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Update (Dec 31, 2014): Another question: Can one fill the plane with mutually non-congruent rectangles all of same area and *bounded perimeter* (and with mutually non-congruent rectangles all of same perimeter)?

Indeed, with a spiral arrangement of rectangles, one can achieve a filling with non-congruent rectangles of equal area and *unbounded perimeterr*. Note: Such spiral arrangments have been used elsewhere. For example, the Blanche dissection (http://mathworld.wolfram.com/BlanchesDissection.html) and in this preprint: http://arxiv.org/abs/1307.3472, in the section 'On rectangling the rectangles'.

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The story of the post will be continued in more posts....