mathematical methods in origamiorigametry.net/tokyo/class1.pdf · credit: hisashi abe, 1980....

Mathematical Methods in Origami

University of Tokyo, Komaba Day 1 Slides, Dec. 16, 2015

Thomas C. Hull, Western New England University

Lecture notes and assignments available at: http://mars.wne.edu/~thull/tokyo/class2015.html

http://mars.wne.edu/~thull/tokyo/class2015.html

Overview of Origami-Math• origami geometric constructions

L 3

OC


• combinatorial geometry of origami

α1α2

α3

α4



• matrix models and rigid foldability




• computational complexity of origami





• folding manifoldsB(x, µ(x))

B(x, µ(x))

xx

(a) (b)

γv1 γv2

γv3γv4





• folding manifolds

• origami design






• origami design

• origami algebra






• origami design

• origami algebra

• modular origami

• applications

• curved folds

Folding a Equilateral Triangle

Folding a Regular Pentagon

method by David Chandler

A

B

If the side of the square = 2, then AB =

A

B

A

B

A

BA

BB

Folding a Regular Heptagon

method by Hull

(1)(2)

A

B(3)

P 1

P 2L 1

L 2(4)

P 1

P 2 L 3

(5)

L 3

OC

(6)

L 4O

C

(7)

O

C

(8)

OC

C 0

C 00

(9)

OC

C 0

C 00

(10)

C

C 0

C 00

(11)

O

C 00

(12)

E

F

Folding a Regular Heptagon

A+1/A2

A3

1/A

1

A=e 2πi/7A2

1/A2

1/A3

i

RA2+1/A2

2A3+1/A3

2

Origami angle trisection

L3 2θ 3

L1

θ

L1

L2

p1

p2

θ

L1

L2 L3

O

A

B C

D

Proof:

credit: Hisashi Abe, 1980

Straightedge and Compass basic operations

Given two points P1 and P2, we can draw the line P1P2.!Given a point P and a line segment of length r, we can draw a circle centered at P with radius r.!We can locate intersection points, if they exist, between lines and circles.

What are the Basic Operations of Origami?

Given two points P1 and P2, we can fold the crease line P1P2.!Given two points P1 and P2, we can make a crease that puts P1 onto P2.!Given two lines L1 and L2, we can make a crease that puts L1 onto L2.!and so on.

Folding a point to a line

pL

pL

p

Folding a point to a line

A different way...Given a family of curves in , its envelope consists of all points such that that for some ,!

The envelope is a single curve that is tangent to all the curves in the family.!In our case we have!

So becomes , or .!

Combining this with gives

and

F (x, y, t) = 0 R2

(x, y) t � RF (x, y, t) = 0

⇥

⇥tF (x, y, t) = 0

F (x, y, t) = t2 � 2xt + 2y � 1

2t� 2x = 0 x = t

�x2 + 2y � 1 = 0 y =12x2 +

12

⇥

⇥tF (x, y, t) = 0

t2 � 2xt + 2y � 1 = 0

or

a1

O

a0T

a3

a4

a5

a2

θ

θ

θ

θ

Lill’s geometric method for finding real roots of any polynomial:!anxn+an-1xn-1+...+a2x2+a1x+a0=0!

Start at O, go an, turn 90°,!go an-1, turn 90°, etc, ending at T.

(Lill, 1867)

Then shoot from O with an angle θ, bouncing off the walls at right angles, to hit T.!Then x = -tan θ is a root.

Lill’s Method

Why does Lill’s method work? !PnQn-1 /an = tan θ = -x!So PnQn-1 = -anx!

Pn-1Qn-2 /(an-1-PnQn-1) = -x!So Pn-1Qn-2 = -x(an-1 + anx)!Similarly,!Pn-2Qn-3 = -x(an-2+x(an-1+anx))!Continuing...!

a0 = P1T = - a1x - a2x2 - ... - an-1xn-1 - anxn !or, anxn + an-1xn-1 + ... + a2x2 + a1x + a0 = 0.! (Lill, 1867)

a1

O

a0T

an-1

an

a2

θ

θ

θ

θ

P n

Q n-1

P n-1 Q n-2 an-2

P1

Lill’s Method

Draw the turtle path!a3, a2, a1, a0.!

Draw D1 at dist a3 from and parallel to a2.!

Draw D2 at dist a0 from and parallel to a1.!

Then fold O onto D1 andT onto D2 at the same time.

(Beloch, 1936)

Beloch’s use of Lill to Solve Cubics

O

Ta2

a1

a0

a3 D1

D2Tψ

Oψ

XY

Origami can solve any cubic equationLet’s find the roots of the cubic z3 - 7z - 6. !

O

T

4 81 12-4

D2

D1


O

T

4 81 12-4

D2

D1

Here !tan θ = 1, !so!-tan θ = -1!is a root.


O

T

4 81 12-4

D2

D1

Here !tan θ = 2, !so!-tan θ = -2!is a root.


O

T

4 81 12-4

D2

D1

Here !tan θ = -3, !so!-tan θ = 3!is a root.


O

T

4 81 12-4

D2

D1

How to fold a regular heptagon

(1) (4)

(7)

(10)

(2) (3)

(5) (6) (8)

(9) (11) (12)

A

B

P1

P2L1

L2

P1

P2 L3

L3

OC

C

L4

C

O

OC

OC

C∂

C∂

C∂∂

C∂∂

CC∂

C∂∂O

C∂∂

E

F

A6=1/AA5=1/A2

A4=1/A3

A3

1

A=e2ºi /7A2

i

RA+1/A2

12º/7

Heptagon -> cubic equation

The equation

has as a solution

Therefore, solving this equation will give us

Task 2: Solve by origami Fold onto the x-axis!and onto the !y-axis at the same time.

Claim: When we do this fold, P1 gets folded to! on the x-axis.

tL1

-1

1

1-1

P2=(-1,-1/2)

P1=(0,1)

s

L2

tL1

-1

1

P2=(-1,-1/2)

P1=(0,1)

s

L2

-2x

x3

x2

-1

Task 2: Solve by origami Start with P1 = (0,1).!Make a “turtle diagram” from !P1 using the coefficients/2.!(Go, turn 90°, repeat.)

tL1

-1

1

P2

P1

s

L2

θ

θ

θ

-1

1

-2x

x3

x2

-1

Task 2: Solve by origami

Then we “shoot” from P1 by angle θ, bounce off the x2 and -2x lines, to hit P2.!

This gives us three similar triangles.!

Also note that the hypothenuse of the biggest triangle is the crease line.

How did we pick P1 and P2?

Task 2: Solve by origami

Notice how the triangle made by P1, t, and the origin verifies that !t = tan θ = x = 2cos(2π/7).

tL1

-1

1

P2

P1

s

L2

θ

θ

θ

-1

1

(3)

P1

P2 L1

L2

(4)

P1

P2 L3

How we fold this

Step 3 has P1=(0,1) being folded to the x-axis and !P2=(-1,-1/2) being folded to the y-axis.!

In step 4 we mark P1’s location to get t=2cos(2π/7).

Does it make sense now?

(1) (4)

(7)

(10)

(2) (3)

(5) (6) (8)

(9) (11) (12)

A

B

P1

P2L1

L2

P1

P2 L3

L3

OC

C

L4

C

O

OC

OC

C∂

C∂

C∂∂

C∂∂

CC∂

C∂∂O

C∂∂

E

F

2

2cos( )2π 7

The Algebraic PerspectiveThe set of constructible numbers under SE&C is the smallest subfield of (complex #s) that is closed under square roots.!or…! is SE&C constructible if and only if!for some . In other words, is algebraic over and the degree if its minimal polynomial over is a power of .!

Origami version:!Let be algebraic over , and let be the splitting field of the minimal polynomial of over . Then is origami constructible from our list of BOOs if and only if ! for some integers .

� � C

C

Q

[Q(�) : Q] = 2n

n � 0 �2

Q

� � CQ

Q L � Q��

[L : Q] = 2a3b a, b � 0

Folding Knots

Take a strip of paper and tie it in a knot.!What do you get?

Folding Knots

The knot forms a perfect pentagon!!

Can you make a hexagon knot?

But you can make a hexagon !from two strips of paper:!

And you can make a heptagon (7 sides) !knot from one:

Answer: NO!!!

Robert Lang’s Angle QuintisectionAngle

QuintisectionDesigned by Robert J. Lang

Copyright ©2004.

All Rights Reserved.1. Start with a long strip 1 unit high and 5–6 units long. Angle EAB

is the angle to be quintisected. Make a vertical crease about 1/3 unit

from the right side.

D

A B

CE

2. Fold line FG down to lie along edge AB.

D

A B

CE

G

F

3. Fold point F over to point A.

D

A

BC

E

GF

4. Fold and unfold.

D

A

E

B

G

C

F

5. Squash-fold on

the existing

creases.

D

A

E

B

G

C

F

5. Make a horizontal fold aligned with point C.

D

A

E

B

G

C

F

6. Fold point C to point A and unfold, making

a second longer horizontal crease.

D

A

E

B

G

C

F

7. Mountain-fold corner D behind. 7. Here’s where it all happens. Fold edge AE down along

crease AJ. At the same time, fold the left flap up so that

point F touches crease HI at the same point that edge AE

does and point C touches crease AJ. You will have to adjust

both folds to make all the alignments happen at once.

8. Here’s what it looks like folded. Yours may

not look exactly like this, depending on the angle

you used and the length of your strip. Unfold to

step 7.

9. Bisect angle EAJ.

10. Fold crease AK down to

AM and unfold.

11. Bisect angle LAM.

12. Angle EAM is now divided into fifths.

Angle quintisection is division of

an arbitrary angle into fifths. This

requires solution of an irreducible

quintic equation and thus is not

possible with the 7 Huzita-Hatori

axioms, each of which defines a

single fold by simultaneous

alignment of points and lines. By

permitting the simultaneous creation

of two or more folds that satisfy

various combinations of point/line

alignments, it is possible to solve

higher-order equations, as this

example illustrates.

Angle


Copyright ©2004.




D

A B

CE


D

A B

CE

G

F


D

A

BC

E

GF

4. Fold and unfold.

D

A

E

B

G

C

F

5. Squash-fold on

the existing

creases.

D

A

E

B

G

C

F


D

A

E

B

G

C

F



D

A

E

B

G

C

F









step 7.



AM and unfold.

















Angle


Copyright ©2004.




D

A B

CE


D

A B

CE

G

F


D

A

BC

E

GF

4. Fold and unfold.

D

A

E

B

G

C

F

5. Squash-fold on

the existing

creases.

D

A

E

B

G

C

F


D

A

E

B

G

C

F



D

A

E

B

G

C

F









step 7.



AM and unfold.

















Angle


Copyright ©2004.




D

A B

CE


D

A B

CE

G

F


D

A

BC

E

GF

4. Fold and unfold.

D

A

E

B

G

C

F

5. Squash-fold on

the existing

creases.

D

A

E

B

G

C

F


D

A

E

B

G

C

F



D

A

E

B

G

C

F









step 7.



AM and unfold.

















Angle


Copyright ©2004.




D

A B

CE


D

A B

CE

G

F


D

A

BC

E

GF

4. Fold and unfold.

D

A

E

B

G

C

F

5. Squash-fold on

the existing

creases.

D

A

E

B

G

C

F


D

A

E

B

G

C

F



D

A

E

B

G

C

F









step 7.



AM and unfold.

















Angle


Copyright ©2004.




D

A B

CE


D

A B

CE

G

F


D

A

BC

E

GF

4. Fold and unfold.

D

A

E

B

G

C

F

5. Squash-fold on

the existing

creases.

D

A

E

B

G

C

F


D

A

E

B

G

C

F



D

A

E

B

G

C

F









step 7.



AM and unfold.

















Angle


Copyright ©2004.




D

A B

CE


D

A B

CE

G

F


D

A

BC

E

GF

4. Fold and unfold.

D

A

E

B

G

C

F

5. Squash-fold on

the existing

creases.

D

A

E

B

G

C

F


D

A

E

B

G

C

F



D

A

E

B

G

C

F









step 7.



AM and unfold.

















Robert Lang’s Angle QuintisectionAngle


Copyright ©2004.






4. Fold and unfold. 5. Squash-fold on

the existing

creases.

5. Make a horizontal fold aligned with point C. 6. Fold point C to point A and unfold, making


7. Mountain-fold corner D behind.

D

A

E

B

G

C

F

7. Here’s where it all happens. Fold edge AE down along





D

A

E

B

G

C

F

H I

J




step 7.

D

A

I

J

E


D

A

E

B

G

C

F

I

J

K


AM and unfold.

D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

J

K

L

M

N















Angle


Copyright ©2004.







the existing

creases.




D

A

E

B

G

C

F






D

A

E

B

G

C

F

H I

J




step 7.

D

A

I

J

E


D

A

E

B

G

C

F

I

J

K


AM and unfold.

D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

J

K

L

M

N















Angle


Copyright ©2004.







the existing

creases.




D

A

E

B

G

C

F






D

A

E

B

G

C

F

H I

J




step 7.

D

A

I

J

E


D

A

E

B

G

C

F

I

J

K


AM and unfold.

D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

J

K

L

M

N















Angle


Copyright ©2004.







the existing

creases.




D

A

E

B

G

C

F






D

A

E

B

G

C

F

H I

J




step 7.

D

A

I

J

E


D

A

E

B

G

C

F

I

J

K


AM and unfold.

D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

J

K

L

M

N















Angle


Copyright ©2004.







the existing

creases.




D

A

E

B

G

C

F






D

A

E

B

G

C

F

H I

J




step 7.

D

A

I

J

E


D

A

E

B

G

C

F

I

J

K


AM and unfold.

D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

J

K

L

M

N















Angle


Copyright ©2004.







the existing

creases.




D

A

E

B

G

C

F






D

A

E

B

G

C

F

H I

J




step 7.

D

A

I

J

E


D

A

E

B

G

C

F

I

J

K


AM and unfold.

D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

J

K

L

M

N















Angle


Copyright ©2004.







the existing

creases.




D

A

E

B

G

C

F






D

A

E

B

G

C

F

H I

J




step 7.

D

A

I

J

E


D

A

E

B

G

C

F

I

J

K


AM and unfold.

D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

I

J

K

L

M


D

A

E

B

G

C

F

J

K

L

M

N















Robert Lang’s Angle Quintisection

Draw C’F’ to be the image of CF under the folding.

Origami Constructions TheoremHow to solve quintics with a 3-fold and Lill’s Method:

Origami Constructions TheoremTheorem: (Alperin and Lang, 2009)

Every polynomial equation of degree n with real solutions can be solved by an (n-2)-fold.

mathematical methods in origamiorigametry.net/tokyo/class1.pdf · credit: hisashi abe, 1980....

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