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Suppose I take a union of nonoverlapping $1-1-\sqrt{2}$ triangles in the plane:

enter image description here

The same shape can be tiled in another way, by flipping two triangles joined together into a square:

enter image description here

In general, are these the only such decompositions possible? That is, given two planar configurations of finitely many nonoverlapping $1-1-\sqrt{2}$ triangles whose unions are the same, are they necessarily related by some number of "square flipping" operations?

In the infinite case, this is not true, as seen by the following tilings of an infinite quadrant:

enter image description here

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    $\begingroup$ If the answer is yes (as in, only square flipping can keep it the same), I would guess the proof has something do with the irrationality of $\sqrt{2}$. $\endgroup$ Dec 21, 2020 at 0:01

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While it feels a little like cheating, the following configurations seem to provide a counter-example to the stated conjecture:

Octagon with octagonal hole, formed by eight triangles The same octagon, reflected along y-axis

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    $\begingroup$ Looks OK, unless OP wants to stipulate union must be simply connected. $\endgroup$
    – brainjam
    Dec 22, 2020 at 0:12
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    $\begingroup$ Ooh, thanks! Barring any simply-connected or more surprising counterexamples discovered, the bounty’s yours. $\endgroup$ Dec 22, 2020 at 2:05

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