. Proof, Assume towards contradiction that some r ? S 1 ? SEC 2 This implies, according to C1, that ?x ? S 2 : x ? SEC 1 . But |S 2 | ? n/2 and there must be at least one robot of S 1 in SEC 1 . Hence this also implies |SEC 1 | > n

. Proof, Assume that some r ? S 2 ? SEC 1 . Observe that according to C2 all robots x that are in S 1

. Hence, the robots of SEC 1 are those in S 1 and robots of SEC 2 are those of S 2 . The SEC of S 2 is the SEC of T ?1 . Hence, the center of SEC 1 is the Weber point. According to the definition of B, the robots at SEC 1 form a regular polygon. Hence, there is some symmetry maintained between T ?1 and T . It contradicts the fact

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