Mathematics and OrigamiThose 6 rectangles are centered on the faces of a cube as can be seen in Fig. 4. Each pair
of opposite rectangles is homothetic and determines two planes parallel to the sides l in Fig. 4
right, Point 18.6.1; l is the pentagon ́s side of the dodecahedron in Fig. 2.
The distance between those planes is the side AB of the cube in present Fig. 4. We can
figure it out from Fig. 5 that is an enlarged detail of said Fig. 4 right.
= =
= − × − (^18.^6.^1 )
2AB D 2 FDcos 90 verPunto
εl 2 , 3416407 l
2116 , 565052 , 618034 2 0 , 1624599 cos (^90) =
= − × −
After what we have seen, it is evident the coexistence in the macle of a hexahedron and
a pentagonal-dodecahedron.
Before drafting the pyritohedron ́s folding diagram (Fig. 7), we shall dig out on how the
wedge of Fig. 3 is constructed; see enlarged Fig. 6:
CD = l = side of the pentagonal-dodecahedron.
Ang. ECD = Ang. ECF = Ang. FCD = 108º
CE = CF = DG = AC (Point 18.6.1) = 0,2763932 l
EG parallel to CD
Now we can draw Fig. 8 that is the folding diagram for a hemi-pyritohedron; we require
two of them to be mounted in opposition. As each produces 4 rectangles and we only need a
total of 6, eventually we must discard 2 of those rectangles.
Previous to Fig. 8 we have drawn Fig. 7 which is one hemi-dodecahedron ́s folding dia-
gram including the lines needed to transform the dodecahedron into a pyritohedron. In Fig. 8
have disappeared the lines of Fig. 7 not needed anymore. Note the required pleat fold of Fig. 8.
l
W
l
V Z
D
5
A
B