Mathematical_Recreations-Kraitchik-2e

108 Mathematical Recreations
The triangles will then be determined by the smes of the
arithmetical angles A and B as before.
14. Find a cuboid (rectangular parallelepiped) in which
the edges and the diagonals of the faces are all rational. This
is a more difficult problem. (See Fig. 16.)
We can take x = a(4b 2 - c 2 ), y = b(4a 2 - c 2 ), Z = 4abc,
where a, b, c are the sides of a rational right triangle:
a = m2 - n2, b = 2mn, c = m2 + n2. Then
These formulas do not give the complete solution. Here
are some that cannot be obtained from them:
x 85 187 195 231 275 855 1,105 1,155 1,155 ...
y 132 1,020 748 792 252 2,640 9,360 1,100 6,300 ...
z 720 1,584 6,336 160 240 832 35,904 1,008 6,688 ...
From any solution we can derive a new one by taking
u = yz, v = zx, w = xy.