A five rhombic problem              Torsten Sillke, FRA, 1998-09-01

    R     R'
   / \   / \
  /   \ /   \
 A     X     A'
  \   / \   /
   \ /   \ /
    B     B'
   / \   / \
  /   \ /   \
 C     Y     C'
  \   / \   /
   \ /   \ /
    S     S'

The smalles angle phi of each rhombic has 60 degree.
Fold this figure up that the two segments RX, R'X as well as SY, S'Y match.
(Folding is allowed along the edges only.)

   Determine angle(ABC).

Show that this angle is equal to the smaller rhombic angle
of a rhombic-dodecahedron.
Show that this angle is equal to the center-angle of a regular
tetrahedron. (The angle under which the center sees two vertices
of the tetrahedron.) 

Can explain this coincidence?


angle(ABC) = arccos(1/3) = 70.5288 degree

If we increase phi then angle(ABC) well decrease.
For which phi will be phi = angle(ABC)?



Replace the rhombics by two equilateral triangles.

           R     R'
          / \   / \
         /   \ /   \
        A-----X-----A'
       / \   / \   / \
      /   \ /   \ /   \
     D-----B-----B'----D'
          / \   / \
         /   \ /   \
        C-----Y-----C'
         \   / \   /
          \ /   \ /
           S     S'

Now fold up that R=R', S=S', C=D, and C'=D'.
The points A, A', C, C', R, and S shall be in a plane.