## Tuesday, November 10, 2009

### Unsolved: The Perfect Cuboid Problem

In Tomorrow's Math, C. Stanley Ogilvy asks
Does there exist a rectangular parallelepiped (box) all of whose edges and face diagonals are of integral length, and the length of whose main diagonal is also an integer?
Put another way, can you construct a rectangular solid where all three edge lengths, all three face diagonals, and the long space diagonal have integer values (or prove that such a box cannot be constructed)?

This deceptively simple-sounding problem is known as the Perfect Cuboid Problem (among other names). It is closely related to the problem of finding an Euler Brick, which has many solutions, the smallest of which is a box with edges 125, 244, and 267.

No perfect cuboids were found during an exhaustive search of the integers up to 100 billion (1010). This lends evidence (but is far from a proof) to the suspicion that no perfect cuboids exist. One near miss is the cuboid that has edges with lengths 44, 117, and 240, and diagonals with lengths 125, 244, and 267.

Anonymous said...

L=3 b=4 h=12
this is a perfectcuboid

Bill the Lizard said...

sunny,
The diagonal formed by the L=3 and h=12 edges would not have an integer length.

3^2 + 12^2 = 153

√153 = 12.3693...

You can use the same method to check the diagonal formed by each of the other edges and the long diagonal that runs through interior space of the box.

Bill said...

a semi-perfect cuboid could be constructed... 3, 4, 12 w/integral main diagonal 13, perhaps the only one of its kind which shows that the lateral faces would ALWAYS have a conflict--- 3 or 4 with 12 will not produce a face diagonal; thus, using logic alone and not an exhaustive search proves that an Euler Brick is not possible; had all the face diagonals been present, then the main diagonal wouldn't exist either.
*qed Bill the Monitor Lizard... lol