# Dissections of golden rhombohedra

Izidor Hafner

Anja Komatar

Trzaska 25, 1000 Ljubljana, Slovenia

e-mail: izidor.hafner@fe.uni-lj.si

"A rhombohedron is a parallelepiped bounded by six congruent rhombs. It has two opposite vertices at which the three face-angles are equal; it is said to be acute or obtuse according to the nature of these angles. A golden rhombohedron has faces whose diagonals are in the golden ratio t : 1 " [1, pg. 161]. The volumes of A_{6} (the acute rhombohedron) and O_{6} (the obtuse rhombohedron ) are also in the ratio t : 1.
"The Fibonacci sequence 1, t, 1+t, 1+2t, 2+3t, 3+5t, 5+8t, 8+13t, is equal to geometric sequence 1, t, t^{2}, t^{3}, t^{4}, t^{5}, t^{6}, t^{7}"[2, pg. 94].
If we take the obtuse rhombohedron of volume 1, then the obtuse rhombohedron whose edges are t as long, has volume t^{3}=1+2t. So there exists a dissection of tO_{6} to one O_{6} and two A_{6}.
On the other hand the volume of tA_{6} is t^{4}=2+3t, so there exists a dissection of the solid to two O_{6} and three A_{6}.
But t^{4}=t^{2}+t^{3}=t^{3}+1+t, so there is a dissection of tA_{6} to tO_{6}, O_{6} and A_{6}.
Finaly we could dissect O_{6} to (1/t)O_{6} and two (1/t)A_{6}.
In this paper we shall illustrate the dissection of tA_{6} to tO_{6}, A_{6}, (1/t)O_{6} and two (1/t)A_{6}. (Or similar disection of t^{2}A_{6} to t^{2}O_{6}, tA_{6}, O_{6} and two A_{6}. Observe: 8+13t = 5+8t + 3+5t = (5+8t) + (2+3t) + (2t) + 1.
There is a three piece dissection of the red parallelepiped to A_{6}. There is a four piece dissection of the pink parallelepiped to tO_{6}. The blue and the gray parallelepipeds are (1/t)A_{6}. The pink and the cyan parallelepipeds are similar, so there exists a four piece dissection of the cyan parallelepiped to (1/t)O_{6}. Use the below nets to make a model. Since the pieces of the dissection of cyan parallelepiped are too small, it is given as a whole.

The visual proof that the volumes of A_{6} and O_{6} are in ratio t:1. The rhombic dodecahedron of the second kind can be dissected to two A_{6} and two O_{6}. The area of the rhombus is d1d2/2. The hight of A_{6} is d2/2, and the hight of O_{6} is d1/2. So the ratio of volumes is d2 : d1 = t .

To make a model of the combination of A_{6} and tO_{6} use the nets below. Observe that A_{6} has the same hight as tO_{6}. So, the ratio of hights of A_{6} and O_{6} is (ones again) t .
A dissection of two A_{6} and one O_{6} to tO_{6} given below will be explained in another paper.

References

[1] W.W. Rouse Ball, H.S.M. Coxeter, Mathematical recreations and essays, Thirteen Edition, Dover Pub., 1987.

[2] M. Gardner, AHA! Gotcha, Paradoxes to Puzzle and Delight (Slovenian eddition), 1992

[3] I. Hafner, T. Zitko, Introduction to golden rhombic polyhedra, http://symmetry-us.com/Journals/hafner2/IntrodRhombic.html