Origami and mathematics: why you are not just folding paper - - PowerPoint PPT Presentation

Origami and mathematics:

why you are not just folding paper Stefania Lisai MACS PhD Seminar 5th October 2018

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Pretty pictures to get your attention

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Pretty pictures to get your attention

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Pretty pictures to get your attention

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Pretty pictures to get your attention

Pretty pictures to get your attention

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Meaning and history

Origami comes from the Japanese words ori meaning "folding", and kami meaning "paper".

Meaning and history

Origami comes from the Japanese words ori meaning "folding", and kami meaning "paper". g Paper folding was known in Europe, Japan and China for long time. In 20th century, different traditions mixed up.

Meaning and history

Origami comes from the Japanese words ori meaning "folding", and kami meaning "paper". g Paper folding was known in Europe, Japan and China for long time. In 20th century, different traditions mixed up. g In 1986, Jacques Justin discovered axioms 1-6, but ignored.

Meaning and history

Origami comes from the Japanese words ori meaning "folding", and kami meaning "paper". g Paper folding was known in Europe, Japan and China for long time. In 20th century, different traditions mixed up. g In 1986, Jacques Justin discovered axioms 1-6, but ignored. g In 1989, the first International Meeting of Origami Science and Technology was held in Ferrara, Italy.

Meaning and history

Origami comes from the Japanese words ori meaning "folding", and kami meaning "paper". g Paper folding was known in Europe, Japan and China for long time. In 20th century, different traditions mixed up. g In 1986, Jacques Justin discovered axioms 1-6, but ignored. g In 1989, the first International Meeting of Origami Science and Technology was held in Ferrara, Italy. g In 1991, Humiaki Huzita rediscovered axioms 1-6 and got all the glory.

Meaning and history

Origami comes from the Japanese words ori meaning "folding", and kami meaning "paper". g Paper folding was known in Europe, Japan and China for long time. In 20th century, different traditions mixed up. g In 1986, Jacques Justin discovered axioms 1-6, but ignored. g In 1989, the first International Meeting of Origami Science and Technology was held in Ferrara, Italy. g In 1991, Humiaki Huzita rediscovered axioms 1-6 and got all the glory. g In 2001, Koshiro Hatori discovered axiom 7.

Compass and Straightedge construction

Basic constructions with compass and straightedge:

Compass and Straightedge construction

Basic constructions with compass and straightedge:

• 1. We can draw a line passing through 2 given points;

Compass and Straightedge construction

Basic constructions with compass and straightedge:

• 1. We can draw a line passing through 2 given points;
• 2. We can draw a circle passing through one point and centred in another;

Compass and Straightedge construction

Basic constructions with compass and straightedge:

• 1. We can draw a line passing through 2 given points;
• 2. We can draw a circle passing through one point and centred in another;
• 3. We can find a point in the intersection of 2 non-parallel lines;

Compass and Straightedge construction

Basic constructions with compass and straightedge:

• 1. We can draw a line passing through 2 given points;
• 2. We can draw a circle passing through one point and centred in another;
• 3. We can find a point in the intersection of 2 non-parallel lines;
• 4. We can find one point in the intersection of a line and a circle (if = ∅);

Compass and Straightedge construction

Basic constructions with compass and straightedge:

• 1. We can draw a line passing through 2 given points;
• 2. We can draw a circle passing through one point and centred in another;
• 3. We can find a point in the intersection of 2 non-parallel lines;
• 4. We can find one point in the intersection of a line and a circle (if = ∅);
• 5. We can find one point in the intersection of 2 given circle (if = ∅).

Compass and Straightedge construction

Basic constructions with compass and straightedge:

• 1. We can draw a line passing through 2 given points;
• 2. We can draw a circle passing through one point and centred in another;
• 3. We can find a point in the intersection of 2 non-parallel lines;
• 4. We can find one point in the intersection of a line and a circle (if = ∅);
• 5. We can find one point in the intersection of 2 given circle (if = ∅).

Using these constructions, we can do other super cool things: bisect angles, reflect points, draw perpendicular lines, find midpoint to segments, draw the line tangent to a circle in a certain point, etc...

Compass and Straightedge construction

Basic constructions with compass and straightedge:

• 1. We can draw a line passing through 2 given points;
• 2. We can draw a circle passing through one point and centred in another;
• 3. We can find a point in the intersection of 2 non-parallel lines;
• 4. We can find one point in the intersection of a line and a circle (if = ∅);
• 5. We can find one point in the intersection of 2 given circle (if = ∅).

Using these constructions, we can do other super cool things: bisect angles, reflect points, draw perpendicular lines, find midpoint to segments, draw the line tangent to a circle in a certain point, etc... We cannot solve the three classical problems of ancient Greek geometry using compass and straightedge!

Three geometric problems of antiquity

Doubling the cube: given a cube, find the edge of another cube which has double its volume, i.e. solve t3 = 2.

Three geometric problems of antiquity

Doubling the cube: given a cube, find the edge of another cube which has double its volume, i.e. solve t3 = 2. Squaring the circle: given a circle, find the edge of a square which has the same area as the circle, i.e. solve t2 = π.

Three geometric problems of antiquity

Doubling the cube: given a cube, find the edge of another cube which has double its volume, i.e. solve t3 = 2. Squaring the circle: given a circle, find the edge of a square which has the same area as the circle, i.e. solve t2 = π. Trisect the angles: given an angle, find another which is a third of it, i.e. solving t3 + 3at2 − 3t − a = 0 with a =

1 tan θ and t = tan

θ

3 − π 2

• .

Three geometric problems of antiquity

Doubling the cube: given a cube, find the edge of another cube which has double its volume, i.e. solve t3 = 2. Squaring the circle: given a circle, find the edge of a square which has the same area as the circle, i.e. solve t2 = π. Trisect the angles: given an angle, find another which is a third of it, i.e. solving t3 + 3at2 − 3t − a = 0 with a =

1 tan θ and t = tan

θ

3 − π 2

• .

You can do 2 of these 3 things with origami: guess which one is impossible?

Justin-Huzita-Hatori Axioms

Justin-Huzita-Hatori Axioms

• 1. There is a fold passing through 2 given points;

Justin-Huzita-Hatori Axioms

• 1. There is a fold passing through 2 given points;
• 2. There is a fold that places one point onto another;

Justin-Huzita-Hatori Axioms

• 1. There is a fold passing through 2 given points;
• 2. There is a fold that places one point onto another;
• 3. There is a fold that places one line onto another;

Justin-Huzita-Hatori Axioms

• 1. There is a fold passing through 2 given points;
• 2. There is a fold that places one point onto another;
• 3. There is a fold that places one line onto another;
• 4. There is a fold perpendicular to a given line and passing through a

given point;

Justin-Huzita-Hatori Axioms

• 1. There is a fold passing through 2 given points;
• 2. There is a fold that places one point onto another;
• 3. There is a fold that places one line onto another;
• 4. There is a fold perpendicular to a given line and passing through a

given point;

• 5. There is a fold through a given point that places another point onto a

given line;

Justin-Huzita-Hatori Axioms

• 1. There is a fold passing through 2 given points;
• 2. There is a fold that places one point onto another;
• 3. There is a fold that places one line onto another;
• 4. There is a fold perpendicular to a given line and passing through a

given point;

• 5. There is a fold through a given point that places another point onto a

given line;

• 6. There is a fold that places a given point onto a given line and another

point onto another line;

Justin-Huzita-Hatori Axioms

• 1. There is a fold passing through 2 given points;
• 2. There is a fold that places one point onto another;
• 3. There is a fold that places one line onto another;
• 4. There is a fold perpendicular to a given line and passing through a

given point;

• 5. There is a fold through a given point that places another point onto a

given line;

• 6. There is a fold that places a given point onto a given line and another

point onto another line;

• 7. There is a fold perpendicular to a given line that places a given point
• nto another line.

Folding in thirds

1 − x x y α β β α β α 1/2 1/2

x2 + 1 4 = (x − 1)2

• x = 3

8

• y

2 = 1 2

1 4

x = 1 3

Haga’s theorem

1 − x x y α β β α α β k 1 − k

x2 + k2 = (1 − x)2

• x = 1 − k2

2

• y

2 = k(1 − k) 1 − k2 = k 1 + k

Haga’s theorem

If we choose k = 1

N , for some N ∈ N, then

y 2 = 1/N 1 + 1/N = 1 N + 1, therefore starting from N = 2 we can obtain 1

n for any n > 2, hence any

rational m

n for 0 < m < n ∈ N.

Geometric problems of antiquity: double the cube

P Q

Geometric problems of antiquity: double the cube

P Q x y

The wanted value is given by the ratio y

x =

3

√ 2. Doubling the cube is equivalent to solving the equation t3 − 2 = 0.

Geometric problems of antiquity: trisect the angle

Q θ P

Geometric problems of antiquity: trisect the angle

θ P A B Q

The angle PAB is θ

3.

Trisecting the angle is equivalent to solving the equation t3 + 3at2 − 3t − a = 0 with a =

1 tan θ and t = tan

θ

3 − π 2

• .

Geometric problems of antiquity: square the circle

Unfortunately, π is still transcendental, even in the origami world. This problem is proved to be impossible in the folding paper theory.

Solving the cubic equation t3 + at2 + bt + c = 0

K L M

Q Q′ P P′ slope(M)

P = (a, 1), Q = (c, b), L = {x = −c}, K = {y = −1}. The slope of M satisfies the equation.

Solving the cubic equation t3 + at2 + bt + c = 0

ψ φ K L M

Q Q′ P P′ R S 1 slope(M)

P = (a, 1), Q = (c, b), L = {x = −c}, K = {y = −1}. If a = 1.5, b = 1.5, c = 0.5, then t = slope(M) = −1.5 satisfies t3 + at2 + bt + c = 0.

Solving cubic equations

Want to solve t3 + at2 + bt + c = 0. φ = {4y = (x − a)2}, ψ = {4cx = (y − b)2}, M = {y = tx + u}. M is tangent to φ at R, then u = −t2 − at, M is tangent to ψ at S, then u = b + c

t .

M is the crease that folds P onto K and Q onto L.

Applications in real world

g Solar panels and mirrors for space;

Applications in real world

g Solar panels and mirrors for space; g Air bags;

Applications in real world

g Solar panels and mirrors for space; g Air bags; g Heart stents (Oxford);

Applications in real world

g Solar panels and mirrors for space; g Air bags; g Heart stents (Oxford); g Self-folding robots (Harvard, MIT);

Applications to real world

g Solar panels and mirrors for space; g Air bags; g Heart stents (Oxford); g Self-folding robots (Harvard, MIT); g Decorations for your flat.

References

[Lan08] Robert Lang. The math and magic of origami. ❤tt♣s✿✴✴✇✇✇✳②♦✉t✉❜❡✳❝♦♠✴✇❛t❝❤❄✈❂◆❨❑❝❖❋◗❈❡♥♦, 2008. [Lan10] Robert Lang. Origami and geometric constructions. ❤tt♣✿✴✴✇❤✐t❡♠②t❤✳❝♦♠✴s✐t❡s✴❞❡❢❛✉❧t✴❢✐❧❡s✴❞♦✇♥❧♦❛❞s✴❖r✐❣❛♠✐✴❖r✐❣❛♠✐✪ ✷✵❚❤❡♦r②✴❘♦❜❡rt✪✷✵❏✳✪✷✵▲❛♥❣✪✷✵✲✪✷✵❖r✐❣❛♠✐✪✷✵❈♦♥str✉❝t✐♦♥s✳♣❞❢, 2010. [Mai14] Douglas Main. From robots to retinas: 9 amazing origami applications. ❤tt♣s✿✴✴✇✇✇✳♣♦♣s❝✐✳❝♦♠✴❛rt✐❝❧❡✴s❝✐❡♥❝❡✴ r♦❜♦ts✲r❡t✐♥❛s✲✾✲❛♠❛③✐♥❣✲♦r✐❣❛♠✐✲❛♣♣❧✐❝❛t✐♦♥s★♣❛❣❡✲✹, 2014. [Tho15] Rachel Thomas. Folding fractions. ❤tt♣s✿✴✴♣❧✉s✳♠❛t❤s✳♦r❣✴❝♦♥t❡♥t✴❢♦❧❞✐♥❣✲♥✉♠❜❡rs, 2015. [Wik17] Wikipedia. Mathematics of paper folding — wikipedia, the free encyclopedia. ✧❤tt♣s✿✴✴❡♥✳✇✐❦✐♣❡❞✐❛✳♦r❣✴✇✴✐♥❞❡①✳♣❤♣❄t✐t❧❡❂▼❛t❤❡♠❛t✐❝s❴♦❢❴♣❛♣❡r❴❢♦❧❞✐♥❣✫ ♦❧❞✐❞❂✽✵✼✽✷✼✽✷✽✧, 2017.