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Suppose $A,B$ are polygons of equal area. By the Wallace-Bolyai-Gerwien theorem, $A$ and $B$ are equidissectable: we can make finitely many straight-line cuts in $A$ and rearrange the resulting pieces via rigid motions to get a copy of $B$ (allowing pieces to overlap at their boundaries).

Define the cost of an equidissection of $A$ to $B$ to be the sum of the lengths of the cuts involved. Note that an equidissection with many cuts can still have a small cost, so this doesn't necessarily align with the more natural idea of minimizing the number of cuts involved.

Question: Does a minimal-cost equidissection always exist?

Since the cost of a given dissection is a positive real number, there is no obvious problem with always being able to find a slightly cheaper equidissection. However, I have no idea how to produce a counterexample. More embarrassingly, I don't even see whether minimal-cost equidissections exist in simple cases - e.g. a square to an equal-area equilateral triangle.

I would also be interested in the analogous question for other geometries (e.g. for the hyperbolic plane, or for polyhedra in Euclidean space of equal volume and Dehn invariant with cut surface area in place of cut length), but I'm primarily interested in polygons in the Euclidean plane.

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    $\begingroup$ I think this needs a lemma that if there is a shorter dissection, then there is a shorter dissection without increasing the number of cut points or edges. A compactness argument would probably be enough after that. $\endgroup$
    – user44143
    Commented Jan 4, 2023 at 14:26
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    $\begingroup$ I don't know how to justify it clearly, but it seems that for the example of equidissection in the wikipedia article you provided, every side of pieces that are on the inside of the square end up outside of the triangle. So it seems an equidissection is at least the perimeter of the triangle, ie. the one provided is optimal. $\endgroup$
    – Qise
    Commented Jan 13, 2023 at 12:03
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    $\begingroup$ @MattF. That lemma is just false. Take a square with two (or twenty) Pinocchio noses on the top side and one big ear on the right side. Now take the same square with noses and the ear on the top and bottom sides respectively. You can relocate the ear in one cut but it is long, so it will be more beneficial to relocate all the noses and rotate the square, but it requires many cuts and it seems that with one cut allowed relocating the ear is optimal. $\endgroup$
    – fedja
    Commented Jan 15, 2023 at 17:56
  • $\begingroup$ "see e.g. here" is supposed to be a link, but it isn't one. $\endgroup$
    – Gerry Myerson
    Commented Jan 26 at 22:18
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    $\begingroup$ @GerryMyerson Whoops! Since I have no idea what I had in mind, I've removed that (and fixed a typo). $\endgroup$
    – Noah Schweber
    Commented Jan 26 at 22:53

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