MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

A version of the latter case consists in cutting the triangle out of the square:

8.2.2.2 MAXIMUM EQUILATERAL TRIANGLE (four solutions)
Solution 1

To produce the crease AB by taking corner E over the perpendicular bisector of the horizontal

sides of square: we get B and H. Do the same simmetrically to diagonal through A: we get C.

Triangle ABC is the solution.

AE=AH= 1

2

2 1

1

sen = =

AH

α ; α= 30 º ; ang. BAC = 60º

1. 0352762 1

cos 15

1

AB= = >AE=

A

B C

A

B C

1

Fold 1 obtains point A:

C → perpendicular bisector of

BC; B → B

B

A

B C

A A

✂

B

A

C

H

E B F B

A

C

= =

=

=