MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

9.10 TRISECTION OF ANY ANGLE (H. HUZITA)
Let AIB = 3α the angle to trisect.

In Point 7.14.5 this equation was solved:

t^3 +t^2 − 2 t− 1 = 0 (3)

Similarities and dissimilarities between (2) and (3):

• Both are complete equations of third degree.


• With negative discriminants (ignore in (2) the values of
  2

3

π

α=n , n being a natural

number; these exceptions will be studied below).

• Every coefficient of (3) can be expressed in the same unit.


• (2), on the contrary, requires two different types of unit to express its coefficients:
  the unit as such, like in (3), for the monomial terms of third and first degree, and the unit a
  for the terms of second degree and independent.


• Fortunately, coefficients using as unit 1 and a, alternate in (2): so, all the horizontal
  vectors will be measured with a type of unit and the vertical ones, with the other.
  Before proceeding, let ́s solve (2) for 3α equal to 90º and 180º respectively.
  Being a = tg 90º = ∞, let divide (2) by a:

− 3 t^2 + 1 = 0 ;

3

1

t= ; 30 º

3

1

α=Arctg =

As a = tg 180º = 0, (2) becomes:

t^3 − 3 t= 0 ; t^2 − 3 = 0 ; t= 3 ; α=Arctg 3 = 60 º

It ́s evident that these particular solutions are direct: 30

3

90

= ; 60

3

180

=

In Point 8.2.2.2 we already studied the obtention of angles of 30º and 60º by mere folding.

Let ́s go on with fig. 1. If we make IC = 1, we ́ll have:

DC = tg 3α = a

Therefore (fig. 2) we are able to draw the co-ordinate plane with origin at I (initial

point), abscissas measured in conventional units and ordinates measured with a as unit.

Looking at Fig.2 we have:

• First vector will be [I → (1,0)] (which gives the axle 2,Y)


• The second vector will be [(1,0) → (1,3a)]


• The third is [(1,3a) → (4,3a)]

As

α

α

α 2

1 tg

2 tg

tg 2

−

= , it is:

()

α

α α

α α α 2

3

1 3 tg

3 tg tg

tg 3 tg 2

−

−

= + =

Making tgα=t, we ́ll have:

2

3

1 3

3

tg 3

t

t t

−

−

α= which leads to the complete equation of third

degree:

t^3 − 3 tg 3 α×t^2 − 3 t+tg 3 α= 0 (1)

Making a=tg 3 α, (1) takes the form:

t^3 − 3 at^2 − 3 t+a= 0 (2)

B

1

I C A

2345

a

2a

3a

4a

D

1