MATHEMATICS AND ORIGAMI

Mathematics and Origami

Solution 4

Solution 5

9.5 A SQUARE IN THREE EQUAL PARTS (two approximate solutions)

Solution 1

Side BC of square = 1

∆ABC, equilateral; ang DCE = 30º ; ang DBC = 45º ; ang BDC = 105º

sen 30 sen 105

DB BC

= ; 0. 517638

sen 105

sen 30

DB= =

BE=DBcos 45 = 0. 3660254 ≠ 0. 333 ...

Practical folding process is:

• To get point D.

A

D

B 1/ 3 E 2 / 3 C

1

2

3

4

(^56)
DE EC
4
3
2
1
= =
3
2
EC= ;
3
1
BE=
B
E
F D G C
P
O
1 / 3
By folding, find the centers O (big square) and P
(square OBCD).
Triangles FEC and FPG are similar.
As FG FC
4
3
= , it will be:
PG GC EC
4
3
4
1
= = = , therefore
3
1
EC=
A
E D G
F
C
H
I
1
2
3
4
Begin with a square of side one unit.
Triangles CFD and DGH, are congruent.
Triangles ADE and DGH, are similar.
AsAE ED
3
1
= , it will be
12
1
4
1
3
1
3
1
GH = DG= × =
3
1
12
4
4
1
12
1
4
1
CI=CF+FI=GH+ = + = =
Therefore it ́s enough to fold the square horizontally
over C and then to fold its upper side also over C.