**A**
Yes.

Label the vertices A,B,C,D,A',B',C',D', so that we want to show ABCD and A'B'C'D' are opposite squares. It is given that

The problem and my solution first appeared in postings to sci.math.research in mid-November, 1993. Bruce Reznick wrote:The same method shows more generally that the eight vertices of any rectangular box are rigidified by the twelve sides and four body diagonals of the box. These are special configurations: in general it takes 3(8-2)=18 distances to rigidify 8 points in 3-space, but here 16 distances suffice, even in dimensions higher then 3.The following question has been submitted to “School Science and Mathematics” by Eugene Levine and Harry Ruderman. Harry Ruderman wrote me about it and has given permission for me to post it on Internet.The original sci.math.research thread can be located via this Google Groups search.

Exercise (see these rec.puzzles threads from earlier that year): use the same Ptolemy ptrick to prove that a regular hexagon is rigidified in space by its six sides and three long diagonals (even though generically 9 distances would be barely enough to rigidify six points in the plane). Generalize.