MATHEMATICS AND ORIGAMI

Mathematics and Origami

Heptagon resultant angle:

Ang. EOF = Ang. EOP = () 90 51 , 38219 º

2

1

+POA =

Regular heptagon central angle: 51 , 428571 º

7

360

=

10.3.3 HEPTAGON: KNOT TYPE SOLUTION

Like all of this kind, it is a perfect one.

Recall the remark in Point 10.2.2 regarding docility and flattening of paper strip.

10.4 OCTAGON
It becomes a perfect geometrical construction as we ́ll see immediately. Again, square
side AC is one unit.
In ∆AOB:

1 2

2

1

2

1

tg ED

ED

OA

OC CB

OA

OB

A = −

−

=

−

= = (1)

and in ∆ABD:

ED

ED

ED

ED

DA

BD

A

−

=

−

= =

2

2

1

tg^2 (2)

Equalising (1) and (2), and making ED = k (octagon side), it is:

k

k

k

−

− =

2

1 2

which leads to the following quadratic equation in k:

2 k − 4 k+ 2 = 0

whose roots are:

2 + 1 y 2 − 1

(^12)
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