MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

cube as well as a beautiful illumination effect. The process simplicity is evident from the fold-

ing diagram of Fig. 3 that in turn produces Fig. 4.

To overcome the eventual virtual ambiguity of Fig. 4, the best is to get that figure done

and look from top, down to concave vertex O. The observer should take Fig. 4 holding by

means of his thumb and forefinger, the overhanging triangle B. It is also required that the inte-

rior faces of the trihedral be homogeneously lit and, at least, one of them would show a certain

light contrast with respect to the others.

Under these circumstances, the experiment is simple: To look intensively at O while

closing one eye. Concentrate in that vision till the moment you see that vertex O as a convex

one (it ́s easy): by so doing, the trihedral will become a convex cube. If at that moment the ob-

server moves his head sideways, he will see how the solid tips clearly around vertex A. The

cube appears like a lantern, recalling that of Goya ́s picture “May 3 shootings”. Besides, its

movement seems to be a matter of magic since the observer is conscious of the fact that he is –

holding tight the solid.

18.8.4 CUBE half (or double) the volume of other.

The Greek already knew that this problem could not be solved by means of a ruler and

compasses. At present we shall see three different origami based solutions: the first one is an

approximation but includes an exact version. The other two are exact and have to do with mat-

ters already dealt with before.

SOLUTION 1

It has the peculiarity that no one of both cubes is constructed, but the difference solid; a

rectangle is the base for the folding diagram.

Its inexactness is a consequence of the fact that the rectangles sides are divided into

three and four equal parts respectively for the sake of simplicity (see how Fig. 1 is designed in

Point 18.8.1). By so doing we obtain side L of the great cube. Being l = 3 / 4 L the side of the

small cube, the thickness of the difference polyhedron is L / 4.

If the volume of greater cube is double of the smaller, we ́ll have:

L^3 = 2 l^3 ; L=^32 ×l= 1 , 259921 l (wanted relation)

L l 1 , 333333 l

3

4

= = (obtained relation)

the error: 1,333333^3 = 2,3703702 ≠ 2 (indicator of double)

Fig. 1 is the folding diagram for the preceding conditions showing some diagonals and

the orthogonal segments L, l, L / 4 (or 3 / 4 L).

3

4

O O

A

B