The Banach-Tarski paradox states that a ball can be partitioned into finitely many pieces which can be rotated and translated into two balls identical to the original one.

But can a ball be partitioned into finitely many pieces which can be rotated and translated to form two balls with non-zero radiuses such that the sum of their volumes is equal to the original volume? If yes, can this be done without choice?

  • $\begingroup$ Banach's trick was to dissect the unit ball into carefully constructed sets which are not measurable. $\endgroup$ – Maksim Mar 22 '19 at 23:15
  • $\begingroup$ So... you want to cut the ball into two balls such that the sum of their volumes is the original one? Why should this be a serious problem? $\endgroup$ – Asaf Karagila Mar 22 '19 at 23:26
  • $\begingroup$ No, you dissect a ball in a finite number of sets and only by means of translation and rotation you move these sets in a way that in the end you have 2 balls, both having the same volume as the original one. That is serious. And called a paradox. $\endgroup$ – Maksim Mar 22 '19 at 23:32
  • $\begingroup$ @Asaf It is not immediately clear to me that this is possible with measurable pieces. Is it? $\endgroup$ – Andrés E. Caicedo Mar 23 '19 at 0:33

This is a special case of a problem posed by Tarski and Wagon who asked if for any pair of measurable subsets $A, B\subset R^n$ of the same (positive) volume, there is a subdivision of $A, B$ into measurable subsets $A_1,...,A_k$, $B_1,...,B_k$ such that $A_i$ is congruent (via translations) to $B_i$, $i=1,...,k$. This question has an affirmative answer (assuming that $A, B$ are bounded and their boundaries satisfy a certain technical condition) given in

M. Laczkovich, Decomposition of sets with small boundary. J. London Math. Soc. (2) 46 (1992), no. 1, 58–64.

The technical condition is satisfied for instance by subsets with smooth boundary like in your question when $A$ is a ball and $B$ is the union of two balls.

Laczkovich's proof was nonconstructive and used AC (axiom of choice), but it was improved in

A. Marks, S. T. Unger, Borel circle squaring. Ann. of Math. (2) 186 (2017), no. 2, 581–605.

who gave a constructive proof which does not rely upon AC.

Edit: I just noticed that a very similar question was asked earlier here.

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  • $\begingroup$ Thank you! I keep meaning to read tye Marks-Unger paper in detail. I guess this is a good excuse to do it now. $\endgroup$ – Andrés E. Caicedo Mar 23 '19 at 17:23
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    $\begingroup$ @Ha! I had forgotten that previous question. I'll update it later tonight. $\endgroup$ – Andrés E. Caicedo Mar 24 '19 at 0:27

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