MATHEMATICS AND ORIGAMI

Mathematics and Origami

2

120

1

y= 30 + x → 2

120

1

30

3

130

y+ = + x

Which for y = 0 gives

30120 40

3

130

 =±









x=  −

What proves that the roots of one equation coincide with the cutting points of the curve repre-

senting that equation, and the x axis.

7.11 COMPLET EQUATION OF 3rd DEGREE (J. JUSTIN)
Let it be
t^3 +pt^2 +qt+r= 0 (1)

In Fig 1 we fix the points C(a,b) and D(c,d), from the coefficients of (1), as we ́ll see later.

Then we shall do simultaneously these foldings:

C → OY ; D → OX

The gradient of the normal to crease BF is the solution of (1), i.e.,

t = tg (DGX) = tg α

With the configuration and scale of Fig 1, there is only one solution for t: there is one only way

of folding. That ́s because the following conditions are fulfilled:

0

3

3 2

>

q−p

and 0

27

2 3 9 27

<

p − pq+ r

So equation (1), in this case, has one real solution and two conjugate imaginary. Let ́s discuss

the solution (Fig 2):

tgα

d

OG=c−

OA=b−atgα

() 











= + = −

tgα

2

2

1

2

1 d

xF c OG c

2

a

xB=

yF d

2

1

= ()() 2 tgα

2

1

2

1

yB= b+OA = b−a

C(a,b)

X

O G

A

B

F

Y

1

D(c,d)