MATHEMATICS AND ORIGAMI

Mathematics and Origami

That Fig. 4 gives:

b^2 =ac ; c^2 =bd

b^4 =a^2 bd ; b^3 =a^2 d= 2 a^3 ; 2

3

 =











a

b

; =^32 = 1 , 2

a

b

59921

b^4 =a^2 c^2

Said Fig. 4 shows as well how to get l, the small ube side: to take L (greater ube side)
along axle Y, and then producing a parallel to the hypotenuse a,b as indicated.
The only problem left is how to transport a segment by means of origami. The process

was already suggested in Points 1.1 and 1.2. Fig.6 shows the piece of paper p in which segment
AB to be transported, has being produced. Then cut another paper p ́ and fold it to (Fig. 5). Af-
terwards, make Fig. 5 to coincide with Fig. 6 producing in p ́ a normal fold in front of B.
Fig. 7 shows how p ́ incorporates segment AB ready for transport (e.g. to Fig. 3 of SO-
LUTION 1)

SOLUTION 3

The Greek were already mentioned in connection with this problem. We can add
now that Hipocrates of Chios (430 B.C.) proved that its solution is associated to the intersection
of two parabolas.
In Point 1.2.4 we saw how a parabola is produced by paperfolding. Now we shall jump
straight away to the wanted solution.
As before, let l be the side of a given cube and L the wanted side of another cube whose
volume is double of the former ́s.
Let the equations of two parabolas
x^2 =ly
y^2 = 2 lx

From Fig. 1 it is evident that both pass through the origin O. L is given by the abscissa
of the other point of intersection.
Solving the former system, we have:

lx

l

x

2 2

4

= ; x^3 = 2 l^3 ; x=^32 l ; x=L

p ́

p

A B

p ́

A B

5 6 7