MATHEMATICS AND ORIGAMI

Mathematics and Origami

• Start in I (IO = 1) a series of coefficient vectors according to these criteria:
  IO = 1: coefficient of x^2.
  OA = 2: absolute value of the x coefficient; at right angle with IO; clockwise direction be-
  cause from 1st to 2nd term there is no sign change.
  AF = 3: absolute value of the independent term; at right angle with OA; anticlockwise di-
  rection because passing from the 2nd to the 3rd term there is sign change.
  Finally we get F which is the end point of the three successive vectors.


• To fold
  I → x ́ ; F → F

As can be seen, there are two solutions:
OX 1 = 1 ; OX 2 = -3
Justification:
∆IOX 1 ; ∆FAX 1 as well as ∆IOX 2 ; FAX 2 are similar, so:

AF

AX

OX

IO 1

1

= ;

AF

AX

OX

IO 2

2

=

To assign a value to these segments, we have to bear in mind:

• Independent variable x is to be given the correspondent sign in the Cartesian plane.


• Give to the rest of segments the absolute value they have in equation (1), because its sign
  was already taken into account when clockwise or anticlockwise direction was assigned:
  these segments (the equation coefficients) have not the dimension of the independent vari-
  able though they appear overlapped with it in the Cartesian plane.

Then it follows:

3

1 1 2

1

+

=

x

x

;

3

1 2 2

2

− −

=

−

x

x

that in both cases leads to the same result (x 1 ; x 2 been taking by x):
x^2 + 2x – 3 = 0

Yet we ́ll see another example to settle sign attribution.
Let the quadratic equation with roots x 1 = -1; x 2 = -3
(x+1) (x+3) = 0 ; x^2 + 4x + 3 = 0 (2)

• Absolute values:
  IO = 1: coefficient of x^2
  OA= 4: coefficient of x
  AF= 3: independent term


• Vectors ́ sequence: IOAF (clockwise all the time because in (2) there is not signs change).


• Fold: I → x ́; F → F


• Roots come out to be: Ox 1 = -1; Ox 2 = -3
  Similarity of ∆IOx 1 ~ ∆FAx 1 ; ∆IOx 2 ~ ∆Ax 2 F give:

AF

Ax

Ox

IO 1

1

= ;

AF

Ax

Ox

IO 2

2

=