Mathematics Education: ORIGAMI

ORIGAMI（折り紙）

GEOMETRY　

YOSHIHIRO ASANO

　　　　　　　　　　　　　　　　（JICA VOLUNTEER）

Introduction

In this course , I share the presence of regular polyhedron with you at the same time while introducing the Japanese traditional culture “ORIGAMI” . “ORIGAMI” is the Japanese paper craft . But I'm using some method other than “ORIGAMI” in somewhere . I prove first the existence of five regular polyhedra , but this proof is not intuitive and boring . Everyone can intuitively understand this to make five regular polyhedra with “ORIGAMI” . Let’s try !!

Why there are only 5 regular polyhedra ?

(If you are interested, please try to read .)

Euler characteristic . f + v － e = 2

f , v and e is the number of face ,vertices and edges of a polyhedron respectively .

(This proof is put on at the end of this paper .)

1 , The face is equilateral triangle and three faces are gathered into one vertex .

3f = 2e 3v = 2e and f + v － e = 2

We solve this simultaneous equation .

f = 4

So , in this case this regular polyhedron is regular tetrahedron .

2, The face is equilateral triangle and four faces are gathered into one vertex .

3f = 2e 4v = 2e and f + v － e = 2

We solve this simultaneous equation .

f = 8

So , in this case this regular polyhedron is regular octahedron

3, The face is equilateral triangle and five faces are gathered into one vertex .

3f = 2e 5v = 2e and f + v － e = 2

We solve this simultaneous equation .

f = 20

So , in this case this regular polyhedron is regular icosahedron .

When the face is an equilateral triangle , it is impossible more than five faces gather into one vertex. We see this in the experiment .

4, The face is square and three faces are gathered into one vertex .

4f = 2e 3v = 2e and f + v － e = 2

We solve this simultaneous equation .

f = 6

So , in this case this regular polyhedron is regular hexahedron .

When the face is a square , it is impossible more than three faces gather into one vertex.

5, The face is regular pentagon and three faces are gathered into one vertex .

5f = 2e 3v = 2e and f + v － e = 2

We solve this simultaneous equation .

f = 12

So , in this case this regular polyhedron is regular dodecahedron .

Why other types of polyhedron does not exist ?

The face is n-regular polygon .

r is the number of faces that are gathered into one vertex .

Case A　If n = 3 ( triangle) and r ≧ 6 .

f + v － e = 2 , 3f = 2e , rv = 2e

We solve this simultaneous equation .

(2/r －1/3)e = 2

From r ≧ 6 , 2/r ≦ 1/3 . So 2/r －1/3 ≦ 0

e is negative . This is the contradiction !

There for r < 6

Case B　If n = 4 (square) and r ≧ 4 .

f + v － e = 2 , 4f = 2e , rv = 2e

We solve this simultaneous equation .

(2/r －1/2)e = 2

From r ≧ 4 , 2/r ≦ 1/2 . So 2/r －1/2 ≦ 0

e is negative . This is the contradiction !

There for r < 4

Case C　If n = 5 (pentagon) and r ≧ 4 .

f + v － e = 2 , 5f = 2e , rv = 2e

We solve this simultaneous equation .

(2/r －3/5)e = 2

From r ≧ 4 , 2/r ≦ 1/2 ≦ 3/5 . So 2/r －3/5 ≦ 0

e is negative . This is the contradiction !

There for r < 4

Case D　If n ≧6 and r ≧ 3 .

f + v － e = 2 , nf = 2e , rv = 2e

We solve this simultaneous equation .

(2/r + 2/n －1)e = 2

From n ≧ 6 , 2/n ≦ 1/3 and 2/r ≦ 2/3 .

So 2/r + 2/n －1 ≦ 0

e is negative . This is the contradiction !

There for , there is no regular polyhedra in this case .

Proof is over

Proof of Euler characteristic .

We think the case was removed the removable edge from some polyhedron like as following drawing .

We can consider the relationship between vertices , edges and faces of the shape are the same as polyhedron C. f, v, and e are respectively the number of faces, vertices and edges of the original polyhedron respectively . F, V, and E are respectively the number of faces, vertices and edges of the new polyhedron respectively .

Case1 Case2