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)
3
4
5
6