Puzzle 10: Solution

Q     A unit cube has twelve pairs of vertices at distance 1 (the sides) and four pairs at distance 31/2 (the body diagonals). Must every of eight points with the same configuration of twelve unit distances and 4 distances of 31/2 be a unit cube? (NB it is not required that the remaining twelve distances equal 21/2.)

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

AB=BC=CD=DA = AA'=BB'=CC'=DD' = A'B'=B'C'=C'D'=D'A' = 1
AC' = BD' = CA' = DB' = 31/2.
Apply Ptolemy's Inequality to ABCD and A'B'C'D' to find
AC · BD ≤ AB · CD + AD · BC = 1 · 1 + 1 · 1 = 2,
and likewise A'C' · B'D' ≤ 2, with equality if and only if ABCD (respectively A'B'C'D') is a plane cyclic quadrilateral, and thus a unit square. On the other hand, the same inequality applied to ACA'C' and BDB'D' yields
AC · A'C' ≥ AC' · A'C - AA' · CC' = 3 - 1 = 2
and likewise BD · B'D' ≥ 2. Now consider
T = AC · BD · A'C' · B'D'.
Multiplying our first pair of inequalities yields T≤4, while multiplying the second pair yields T≥4. Hence equality holds throughout. In particular, ABCD and A'B'C'D' are both unit squares. Likewise the same is true of ABA'B', BCB'C', CDC'D' and DAD'A'. Therefore ABCDA'B'C'D' is indeed a cube, Q.E.D.

The problem and my solution first appeared in postings to sci.math.research in mid-November, 1993. Bruce Reznick wrote: 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.
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.

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.