Banach-Tarski nonparadox: Reassembling a ball into two balls with a total volume equal to the original volume 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?
 A: 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. 
