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Question: Can the plane be tiled with convex quadrilaterals that are (1) mutually non-congruent in a Euclidean sense and (2) mutually affine-equivalent?

Remark: Every trapezoid is affine equivalent to every other trapezoid with same base ratio (ratio between lengths of mutually parallel pair of sides). So one can find keep tiling the plane in parallel strips of same width; indeed, each strip can be tiled with mutually non-congruent trapezoids that all have same height and parallel sides that only need to be some constant ratio. So, the question is whether there is any other solution - ideally, a solution not involving trapezoids?

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I think the answer is trivially yes.

Tile the plane with squares of side 1.

Then split each square into two rectangles of sides $1, \alpha$ and $1, 1-\alpha$, of course with pairwise different $\alpha$ for different squares.

P.S. I think you can do the following. Start with a square tiling. By drawing a different slant line in each square, you can create a tiling with right trapezoids of pairwise distinct bases and angles.

Now, enumerate these trapezoids, $A_nB_nC_nD_n$ with $A_nB_n \parallel C_nD_n$.

Now, pick $E_1 \in A_1D_1$ and $F_1 \in B_1C_1$ such that $E_1F_1 \not\parallel A_1B_1$ and $A_1B_1E_1F_1, E_1F_1C_1D_1$ are afinelly equivalent.

Next, for each $N$ pick, $\phi : A_1B_1C_1D_1 \to A_nB_nC_nD_n$ an affine equivalence. Let $E_nF_n = \phi(E_1F_1)$. Split $A_nB_nC_nD_n$ into $A_nB_nF_nE_n$ and $E_nF_nC_nD_n$.

All the quadrilaterals are not trapezoids, and they are affinely equivalent. Now, making the right choices of distinct bases and angles in the first step, I think you can make sure all these quadrialterals are non-congruent (they all have a $90^\circ$ angle, and you have control on an edge to this angle and the angle at the other end of this edges).

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    $\begingroup$ thanks. this is certainly another solution but again involving trapezoids - rectangles are trapezoids. $\endgroup$
    – Nandakumar R
    Commented May 23, 2023 at 17:30
  • $\begingroup$ @NandakumarR Can't you start with a tiling with parallelograms instead of squares, and do the same thing? $\endgroup$
    – Kevin
    Commented May 23, 2023 at 18:10
  • $\begingroup$ Kevin, if I understood you right,, we will get a tiling with mutually non-congruent parallelograms and they are all trapezoids. $\endgroup$
    – Nandakumar R
    Commented May 23, 2023 at 18:16
  • $\begingroup$ The hard part seems to be getting affinely equivalent quadrilaterals. I think that the following approach works, needs to be treated with care, but start with a square tiling, and divide it into non-congruent trapezoid tiling. Now, inductively, for each trapezoid $ABCD$ with $AB \parallel CD$, pick points $E \in AD, F \in BC$ such that $EF \not\parallel AB$ and $ABEF, CDEF$ are affinely equivalent to all previous tiles. $\endgroup$
    – Nick S
    Commented May 23, 2023 at 18:43
  • $\begingroup$ @NandakumarR See the edit, it is not a complete proof, but I think it works. $\endgroup$
    – Nick S
    Commented May 23, 2023 at 19:44

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