To double triacontahedron

From the table of space angles we read that a space angle of the prolate rhombohedron is [Graphics:Images/part2_gr_1.gif] of the full space angle, so 20 rhombohedra form a hexecontahedron. We can then put 12 icosahedra to get a 120-hedron. Filing the niches with 20 oblate rhombohedra we get doube triacontahedron.

[Graphics:Images/part2_gr_2.gif]

[Graphics:Images/part2_gr_3.gif]

From double triacontahedron to fourfolded triacontahedron

But we could put 12 triacontaherda on the hexecontahedron. Then we fill niches at points of 2-fold symmetry with 2 prolate rhombohedra and then put one prolate rhombohedron at each point of 3-fold symmetry.

[Graphics:Images/part2_gr_4.gif]

[Graphics:Images/part2_gr_5.gif]

[Graphics:Images/part2_gr_6.gif]

At this stage we put 3 dodecahedra around each point of 3-fold symmetry and then fill niches at points of 5-fold symmetry with two layers of prolate rhombohedra.

[Graphics:Images/part2_gr_7.gif]

[Graphics:Images/part2_gr_8.gif]

[Graphics:Images/part2_gr_9.gif]

We then put two dodecahedra at points of 2-fold symmetry, a oblate rhombohedron at each point of 3-fold symmetry and an icosahedron at each point of 5-fold symmetry.

[Graphics:Images/part2_gr_10.gif]

[Graphics:Images/part2_gr_11.gif]

Then we fill niches with layers of oblate rhombohedra and get a fourfold triacontahedron.

[Graphics:Images/part2_gr_12.gif]

[Graphics:Images/part2_gr_13.gif]

[Graphics:Images/part2_gr_14.gif]


Converted by Mathematica      May 14, 2001