MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

9.7 A SQUARE IN THREE EQUAL PARTS (Corollary P)

This is a consequence of the fact that a / x is biunivocal (see Point 5).

The first thing is to fold BD in such a way that:

• lower side BD lies on A, mid-point of BC.


• simultaneously, D must lie on the upper side of the square.

If the square has one unit side, the result is that

3

1

FE=. Justification:

• The two angles α are congruent because their sides are perpendicular.


• In ∆ACF:
  2 CF

1

tgα= ; In ∆FED: =FE

2

tg

α

• 
  

2

1 tg

2

2 tg

tg

2 α

α

α

−

= ; 2

1

2

2

1

FE

FE

CF −

= ;

()()()FE FE

FE

FE − +

=

− 1 1

2

21

1

• 1 +FE= 4 FE ;
  3

1

FE=

9.8 A SQUARE BY TRISECTING ITS DIAGONALS

The solution is also applicable to a rectangle.

• To get A and B, the midpoints of respective sides.


• C and D trisect the diagonal in equal parts:
  
  
  • Triangles EFB and AGH are congruent (equal and parallel legs). So EB and AH are
    parallel.
  
  
  • In the pencil of rays FG-FH, FB = BH. Hence FC = CD (Thales theorem).
  
  
  • In the pencil of rays GE-GF, GA = AE. Hence CD = DG (Thales theorem)
  
  
  • That is: FC = CD = DG =
    3
  
  
  
  

1

FG

• Obtained C and D, folding of last square determines EI = IJ = JG (Thales theorem
  applied to rays GE-GF)

B D

A

=

=

C C

A

F E

A

C F E

2

D

F

C C

C

F F

AAE E

B B

D

D D

G

H

I J G