# What's wrong with this solution of Tarski's circle-squaring problem?

Tarski's circle-squaring problem asks whether it is possible to cut up a circle into a finite number of pieces and reassemble it into a square of the same area. Note that this is different from the problem of squaring the circle (which is about compass and straightedge constructions) and also from the Banach-Tarski paradox (which involves a change of volume). Laczkovich showed in 1990 that the answer was yes, and proved the existence of a solution with about $10^{50}$ pieces. The proof was nonconstructive.

What is the error in the following proof that the answer to the problem is no, contrary to Laczkovich? At a given stage in the process of cutting and reassembly, let $u$ be the total length of all convex edges that are circular arcs whose radius equals the radius $r$ of the original circle, and similarly let $v$ be the total length of all such concave edges. Initially, $u=2\pi r$ and $v=0$. At the end of the construction, we would have to have $u=v=0$. But $u-v$ is conserved by both cutting and reassembly, so this is impossible.

[EDIT] Edited to remove mistake about Banach-Tarski.

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According to the Wikipedia article you link to, the pieces in Laczkovich's decomposition are non-measurable and do not have Jordan curve boundary, so it is impossible to define the length of their edges. – Rahul Feb 21 '12 at 0:56
well, the solutions uses the axiom of choice and non-measurable pieces, so $u$ and $v$ are meaningless in the decomposition. – mercio Feb 21 '12 at 0:56
@RahulNarain: If you make this an answer, I'll accept it. – Ben Crowell Feb 21 '12 at 0:59
The Banach-Tarski paradox also involves a decomposition into only finitely many subsets. – Pete L. Clark Feb 21 '12 at 1:27
@Ben Crewell: The reason is very similar to the reason that the Banach-Tarski "paradox" is not a paradox. The decomposition of a ball of radius $1$ into a finite number of pieces that are then reassembled to make two balls of radius $1$ violates our notion of volume. But the sets are non-measurable and do not have a volume. With circle-squaring, it has not been proved that some pieces are non-measurable, but before Laczkovich, there were a number of papers showing the pieces could not, in various ways, be "nice." – André Nicolas Feb 21 '12 at 1:32