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It is well-known that in the Euclidean plane two simple polygons have the same area if and only if they are equidecomposable, i.e., can be decomposed into congruent triangles.

Question. Is an analogous equidecomposability result true in the hyperbolic plane? And if yes, how such equidecomposability can be fulfilled?

In the Euclidean plane the equidecomposability is proved by reducing the problem to the equidecomposability of a triangle to a rectangle with one side of length 1. But the hyperbolic plane contains no rectangles, so this method does not work. Then, which method for establishing the equidecomposability does work in the hyperbolic plane?

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    $\begingroup$ The question seems to be answered here math.stackexchange.com/questions/2138904 $\endgroup$
    – Anton Petrunin
    Commented Dec 17, 2022 at 10:14
  • $\begingroup$ @AntonPetrunin Thank you very much for the link. It is very interesting. In fact, I am interested in the equidecomposablity for synthetic geometries. I have learned in the paper of Giovannini and Lassalle-Casanave (cited in my question) that such an equidecomposability theory does need the Archimedean Axiom (this was observed by Hilbert in his Grundlagen der Geometrie). So, my actual question is to what extent the classical theory of measuring the area via the equidecomposability hold for geometries satisfying just some of Tarski Axioms (namely, Segment Construction, Five-Segments and Pasch). $\endgroup$
    – Taras Banakh
    Commented Dec 17, 2022 at 10:55

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A proof is given in Example 8.10. of https://www.amazon.de/-/en/Johan-L-Dupont/dp/9810245084 . It is however not at all the elementary proof that you seem to be after.

There is an exact sequence $$H_1(O(2),St(S^1)^t)\to {\mathcal P}(\partial H^2)\to {\mathcal P}(\overline{H^2})\to {\mathcal P}(S^1)/\Sigma{\mathcal P}(S^0)\to 0,$$ where the first term has order at most $2$, the last nontrivial term is ${\mathbb R}/\pi{\mathbb Z}$, and the terms in between are the scissors congruence groups (${\mathcal P}(H^2)={\mathcal P}(\overline{H^2})$). Because the isometry group acts transitively on the set of ideal triangles, one has ${\mathcal P}(\partial H^2)={\mathbb Z}$ and one obtains that area provides an isomorphism ${\mathcal P}(\overline{H^2})\to{\mathbb R}$. I haven‘t seen an elementary proof.

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  • $\begingroup$ @Thank you for the answer. Indeed, it is too difficult for my mainly dydactic purposes, which was to develop the equidecomposability theory for measurement of the area in synthetic geomertries satisfying some of Tarski Axioms (namely, Segment Construction, Five-Segments and Pasch). $\endgroup$
    – Taras Banakh
    Commented Dec 17, 2022 at 10:58
  • $\begingroup$ But the cases of Euclidean and hyperbolic geometries differ substantially in treating the area so, maybe there is no unique equidecomposability theory for both geometries and then no such a theory for general geometries satisfying just some weak axioms? $\endgroup$
    – Taras Banakh
    Commented Dec 17, 2022 at 11:10

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