User:Group1-Origami/sandbox

Origami-Constructible Numbers
An Origami-Constructible Number is defined to be a real number r where two points of distance $$|r|$$apart can be constructed in a finite number of steps. This can be defined by the set $$\mathbb{F}_{0}=$$$$\{a\in\mathbb{R}|\exists {v_{1}},{v_{2}}\in O$$such that$$|a|=dist(v_{1},v_{2})\}$$. In addition, we can say that any element $$x \in\mathbb{R}_{\geq0}$$ is Origami-Constructible if and only if $$-x$$ is also Origami-Constructible. We can further define Origami-Constructible points as the set $$O_{0}=\cap\{O | (0,0),(0,1)\in O$$ and $$O$$ is closed under origami constructions$$\}$$, where $$(0,0),(0,1)$$ are points from the Cartesian coordinate system in the plane.

Defining an origami pair $$(P,L)$$, where $$P$$ is a set of points in $$\mathbb{R}^2$$and $$L$$ is a collection of lines in $$\mathbb{R}^2$$, is more complicated since we need to ensure that a set of conditions are satisfied. These conditions are:


 * 1) The point of intersection of any two non-parallel lines in $$L$$ is a point in $$P$$.
 * 2) Given any two distinct points in $$P$$, there is a line in $$L$$ going through them.
 * 3) Given any two distinct points in $$P$$, the perpendicular bisector of the line segment with given end points is a line in $$L$$.
 * 4) If $$L_{1}$$ and $$L_{2}$$ are lines in $$L$$, then the line which is equidistant from $$L_{1}$$ and $$L_{2}$$ is in $$L$$.
 * 5) If $$L_{1}$$ and $$L_{2}$$ are lines in $$L$$, then there exists a line $$L_{3}$$ is the mirror reflection of $$L_{2}$$ about $$L_{1}$$.

History
Origami spans over 1,000 years and is the art of Japanese paper folding. However, Origami was first referenced in China when paper was introduced in 200AD as a cheaper alternative to silk and was known as Zhezhi. It was brought to Japan by Monks in the 6th century and became popular from then on. The word "Origami" is from the Japanese language where "ori" means fold and "kami" (or "gami") means paper. Paper-folding was restricted to formal ceremonies and religious rituals since paper was expensive. Origami was linked with Mathematics when Akira Yoshizawa used it to help teach children geometry. He realised that it would help them understand geometrical problems, such as angles, lines and shapes, in a simpler form. Akira Yoshizawa continued to enjoy origami and developed a technique of wet folding which allowed for much more intricate designs to be created, these advances interested mathematicians who also recognised the link between origami and geometry.

After the link between origami and mathematics was found, Humiaki Huzita, a Japanese-Italian mathematician, founded six single-fold axioms in 1989. A further seventh axiom was discovered by Koshiro Hatori in 2003. Hence these seven axioms were to be known as the Huzita-Hatori axioms. However, it was later found that Jacques Justin had discovered these same seven axioms in 1986 but they were overlooked, so Huzita and Hatori had just rediscovered them. Huzita was born in Japan but emigrated to Italy to study nuclear physics at University of Padua.

Although the Huzita-Hatori axioms are an important aspect of origami geometry, especially in solving two of the Greek problems: Trisecting the Angle and Doubling the Cube, this would not be possible without the discovery of the Beloch fold. Margherita Piazzolla Beloch was able to use her Beloch fold and Beloch square to construct solutions to cubic equations. However, Beloch needed to use Eduard Lill's method of finding real roots of polynomials of any degree before she could consider solving cubic equations. It was in 1936 when Beloch was able to extend Lill's method to show that we can solve cubic equations using purely origami.

A Japanese Astrophysicist, Koryo Miura, developed an origami design where you can fold a flat surface, namely a piece of paper, into a tesselation of parallelograms and each part of the fold remains flat. This method of paper folding was given the name of the Miura Fold, after the founder, and has since been used for modern day developments. In 1995, Japan's Space Flyer Unit used the Miura Fold to deploy its solar panels. This choice of design method was possible since the Miura Fold is easily expanded and collapsed and hence the solar panels can be collapsed for the launch of the Space Unit and expanded once in space.

Robert J Lang, an American Physicist, is the author who showed that the Huzita-Hatori axioms are complete, that is, there are no other folds in origami construction. Lang is also an origami artist and considered one of the pioneers of modern maths-based origami. He was intrigued by the idea of a computer being able to design an origami model and so in 1990 he created a computer program called TreeMaker. Lang continued to develop this program and by 1998 it was at an advanced stage and was capable of constructing full crease patterns for a range of designs.

Huzita-Hatori Axioms
Euclid, a Greek Mathematician, is widely known for being the "founder of geometry" and wrote 'The Elements', one of the leading books in the history of mathematics. Within this book, Euclid stated five assumptions which he used to solve geometric constructions, these assumptions are now more commonly known as Euclid's Axioms. Although these axioms can be used to solve some of the most complex geometric problems, they were unable to provide a solution to two of the most famous Greek problems: Trisecting the Angle and Doubling the Cube.

Similar to Euclid's axioms devised for planar geometry, there are a set of axioms to describe origami geometry. These axioms were developed by mathematicians Humiaki Huzita and Koshiro Hatori and are called the Huzita-Hatori axioms.

The axioms are as follows:


 * O1) Given points $$p_{1}$$and $$p_{2}$$, we can fold a line that goes through both of them.
 * Group1-Origami Huzita-Hatori Axiom 1.png) Given points $$p_{1}$$and $$p_{2}$$, we can fold $$p_{1}$$onto $$p_{2}$$(i.e. find the perpendicular bisector of segment $$p_{1}p_{2}$$).
 * Group1-Origami Huzita-Hatori Axiom 2.png) Given two lines $$l_{1}$$and $$l_{2}$$, we can fold $$l_{1}$$onto $$l_{2}$$ (i.e. bisect the angle between them).
 * Group1-Origami Huzita-Hatori Axiom 3.png) Given a point $$p$$ and a line $$l$$, we can fold a line perpendicular to $$l$$ that goes through $$p$$.
 * Group1-Origami Huzita-Hatori Axiom 4.png) Given two points $$p_{1}$$and $$p_{2}$$and a line $$l$$, we can fold $$p_{1}$$onto $$l$$ with a line that goes through $$p_{2}$$.
 * Group1-Origami Huzita-Hatori Axiom 5.png) Given two points $$p_{1}$$and $$p_{2}$$ and two lines $$l_{1}$$and $$l_{2}$$, we can fold $$p_{1}$$onto $$l_{1}$$ and $$p_{2}$$onto $$l_{2}$$ with single line.
 * Group1-Origami Huzita-Hatori Axiom 6.png) Given one point $$p$$ and two lines $$l_{1}$$and $$l_{2}$$, there is a fold that places $$p$$ onto $$l_{1}$$and is perpendicular to $$l_{2}$$.
 * Group1-Origami Huzita-Hatori Axiom 7.png

Properties
Properties in origami geometry can now be solved using purely the Huzita-Hatori axioms.

Addition and Subtraction of lengths
It is possible to add two given lengths $$a$$ and $$b$$ using the Huzita-Hatori axioms. Axiom O3 is used to fold the first line onto the second, which in turn will place the line segment somewhere on that line. Now copying the line segment to a particular point on the line will allow for the other end of this line segment to lie in the preferred direction. Axiom O4 is now used to create a perpendicular fold through one of the endpoints of the line segment. Then we copy the line segment to that part of the line and unfold. This allows our desired point to be free and untouched by the line segment. Using Axiom O2, we can fold the near endpoint of the line segment to the desired point. We can now use Axiom O4 if our line segment is not in the desired direction, simply fold the line segment in the other direction using Axiom O4 through our desired point.

Similarly, to subtract two given lengths $$a$$ and $$b$$, we want to move the line segment on top of the other in order to find the difference in their lengths.

Multiplication and Division of lengths
It is possible to multiply two given lengths $$a$$ and $$b$$ using the Huzita-Hatori axioms. We begin by placing the two lengths so that they form an acute angle, this can be done using the same method described in the addition of two lengths. Now, we want one end of the unit length segment to lie at our angle vertex, and to do this, we need to copy the unit length segment onto the line which contains line segment $$b$$. Axiom O1 is then used to create a new line from the end of the line segment $$a$$ to the end of the unit line segment. Axiom O4 is now used, twice, to construct a parallel line which goes through the point lying at the end of the line segment $$b$$. The line segment which contains $$a$$ should now intersect this constructed parallel line. Now, from similar triangles, the length from our original vertex to this intersection is exactly $$ab$$.

Similarly, to divide two given lengths $$a$$ and $$b$$, we use Axiom O1 to join the end of the line segment $$a$$ to the end of the line segment $$b$$. Again, we construct a parallel line which goes through the end of the unit line segment. The line segment which contains $$a$$ should now intersect this constructed parallel line and this intersection point is the end point of a line segment which has length $$a/b$$.

Square root of lengths
It is possible to find the square root of a length $$a$$ using the Huzita-Hatori axioms. We start by considering the parabola $$y=x^2$$ as this will help us to find the square root of our length. We copy our length $$a$$ onto the $$y$$-axis and using Axiom O4, twice, we can construct the horizontal line $$y=a$$. Next we can use Axiom O5 to make a fold through point $$(0,1/4)$$ which takes the endpoint to our horizontal line $$y=a$$. There are two possibilities where our image point can lie, on the horizontal line $$y=a$$, since it will lie on the parabola $$y=x^2$$ and the distance between the image point and our point $$(0,1/4)$$ will be $$n+1/4$$ (the same as the distance between the image point and the point $$(0,-1/4)$$). Hence, the distance from the two possible image points to the $$y$$-axis ($$x=0$$) is exactly $$\sqrt{a}$$.

Theorems

 * A point $$(a,b)\in$$ $$\mathbb{R}^{2}$$ is origami-constructible if and only if its coordinates $$a$$ and $$b$$ are origami-constructible elements of $$\mathbb{R}$$.
 * If two lengths $$a$$ and $$b$$ are given, one can construct the lengths $$a\pm b, ab, \frac{a}{b},a^{-1} $$ (if $$a\neq 0 $$) and $$\sqrt{a}$$ in origami (i.e. they are origami-constructible).
 * The set of origami constructible numbers is closed under taking square roots (i.e. given a length $$r$$, which is origami-constructible, $$\sqrt{r}$$ is origami-constructible).
 * The set of origami constructible numbers is closed under taking cube roots (i.e. given a length $$r$$, which is origami-constructible, $$\sqrt[3]{r} $$ is origami-constructible).
 * The collection of origami numbers, $$\mathbb{F}_{0}$$, is a field closed under the operation $$a\mapsto{\sqrt{1+a^2}}$$.
 * It is possible to construct a line parallel to a given line through any given point using origami.



Applications

 * Another application of the Miura-fold is a folded paper Lithium-Ion battery, created by scientists at Arizona State University. By using this design, the scientists were able to stack the paper-based Lithium-Ion battery 25 times and this resulted in the energy density being increased by 14 times.


 * Two researchers, Crooks and Liu, from the University of Texas developed an "origami Paper Analytical Device" in October 2011. It is in clinical stages of development but should be able to be used for detecting and hence diagnosing diseases such as Malaria and HIV.


 * Another application of origami in medicine is the development of an origami stent which can be used to get into problematic areas in the body and once located in the correct position it can be expanded. This was developed by Zhong You and Kaori Kuribayashi from University of Oxford in 2003. A self-deployable origami stent was developed in 2005.
 * In 2016, mechanical engineers at Brigham Young University began a project for law enforcement officers which utilised the Yoshimura folding pattern to transform flat layers of Kevlar into a curved shield. This origami-inspired shield weighs just under 15 kilograms yet it can stand up on its own whilst shielding two or three officers and stop bullets from a .44Magnum, which is one of the most powerful handguns.

Robert J. Lang
Robert J. Lang elevated the art form of origami by using his extensive knowledge, as an American Mathematical Society member, to practical applications in modern technology. Examples include a laser cutter which is used to score paper to assist in complicated folds and developing designs of the folding patterns that would be used by a manufacturer of a German airbag. Moreover, Lang also worked with a team of fellow scientists to solve the problem of creating a space telescope that would have a lens measuring 100 m, which became known as the Eyeglass telescope, such that it could fold compactly and be placed into a small rocket without causing damage to it or compromising its quality and purpose.

In addition, Lang created a computer program, TreeMaker, which uses descriptions of numbers, lengths and the connectedness of the flaps to design origami figures. Lang stated that "TreeMaker could solve crease patterns that I couldn't construct by pencil and paper". In order to create your origami base, you first need to create a stick figure of your base where each stick will be represented by a flap. You then place relevant constraints on these flaps in order for TreeMaker to compute the crease pattern.

Related Fields
If $$K=\{x\in\mathbb{R}|$$ $$x$$ ruler and compass constructible$$\}$$, the set of constructible numbers, and $$O=\{x\in\mathbb{R}|$$$$x$$ origami constructible$$\}$$, the set of origami-constructible numbers, then we have that $$\mathbb{Q}\subset{K}\subset{O}\subset\mathbb{A}\subset\mathbb{R}$$, where $$\mathbb{Q}$$ is the set of rational numbers, $$\mathbb{A}$$ is the set of algebraic numbers and $$\mathbb{R}$$ is the set of real numbers. This chain of containment is depicted in the diagram on the right.

Constructible Numbers
Given a unit length, a real number $$a$$ is constructible if and only if a line segment of length $$|a|$$can be constructed from the unit length in a finite number of steps using only a ruler (i.e. a straight unmarked edge) and pair of compasses. It is clear that 0 and 1 are constructible and it can be shown that a real number $$a$$ is constructible if and only if $$-a$$ is constructible.

Classical constructions

 * 1) Perpendicular bisector of a line segment.
 * 2) Bisector of an angle.
 * 3) Given a line L and a point P not on L, construct the perpendicular line from P to L.
 * 4) Given a line L and a point P not on L, construct the line through P parallel to L.

Properties

 * 1) The set of constructible numbers is a field.
 * 2) If $$a$$ is a constructible real number and $$a>0$$ then $$\sqrt{a}$$ is a constructible real number.
 * 3) An angle $$\theta$$ can be constructed by a ruler and compass if and only if $$\cos(\theta)$$ is a constructible number.

Relation to Origami Numbers
Origami-Constructible Numbers are a powerful field of mathematics and are somewhat more powerful than Constructible Numbers. All numbers that are constructible using a ruler and compass are also constructible using origami. In addition, steps and processes which are allowed in ruler and compass constructions can also be performed by origami. The Huzita-Hatori Axioms describe origami geometry but Axioms 1-5 can also be performed using only ruler and compass constructions.

Constructions
In 1837 it was proved by Pierre Laurent Wantzel, a French Mathematician, that Doubling the Cube and Trisecting the Angle was impossible using only straightedge and compass construction. However, when Margherita Piazzolla Beloch showed that Origami can be used to solve cubic equations (and hence quartic equations) it became known that it was also possible to solve two of the classical Greek problems that the straightedge and compass constructions could not. With the help of Beloch's findings and Axiom 6 of the Huzita-Hatori axioms, it is possible to trisect angles, double cubes, and otherwise solve cube roots using origami.

Doubling the Cube
This simply means finding the real root to the cubic equation $$x^3-2=0$$, and hence constructing the cube root of 2, $$\sqrt[3]{2}$$.

Steps taken to solve the problem of Doubling the Cube using Origami:


 * 1) Divide a square piece of paper into 3 equal parts, where each crease is parallel to $$BC$$.
 * 2) We denote the vertex on the lower line (of the two lines constructed in step 1) on the edge $$CD$$, by point $$p_{1}$$.
 * 3) Now fold vertex $$C$$ onto the edge $$AB$$ and fold point $$p_{1}$$onto the upper line (of the two lines constructed in step 1).
 * 4) Denote the length $$AC_{1}$$ by $$X$$ and the length $$C_{1}B$$ by $$Y$$. The ratio of lengths of $$X$$ to $$Y$$ is $$\sqrt[3]{2}$$.



Trisecting the Angle
Steps taken to solve the problem of Angle Trisection when the angle is acute using Tsune Abe's method:


 * 1) Mark the angle you wish to trisect on the corner of the square. In the diagram below, the chosen angle is $$\alpha$$.
 * 2) Let the bottom fold any new crease which is parallel to the edge $$BC$$.
 * 3) Now fold the edge $$BC$$ upwards so that it meets the crease $$EF$$ (constructed in step 2). Then unfold $$BC$$ from $$EF$$.
 * 4) Fold the corner $$B$$ upwards so that the following are satisfied:
 * 5) *Point $$E$$ lies on the line $$BP$$
 * 6) *Corner $$B$$ lies on the line $$GH$$
 * 7) Crease along the existing crease through the point $$G$$ (constructed in step 4), making sure to crease through all the layers. Then unfold this crease.
 * 8) Extend the crease $$J$$ from the point it meets the line $$GH$$ back to the corner $$B$$ by folding along this crease.
 * 9) Bring the edge $$BC$$ to the fold $$BJ$$ and crease. Then unfold this crease.
 * 10) The angle $$\alpha$$ has now been trisected.



Cubic Equations
In 1936 Margharita Piazolla Beloch, an Italian Mathematician, found that Origami constructions can be used to solve general cubic equations. Hence, it can be concluded that Origami constructions are more powerful than straightedge and compass constructions. It must be remembered that Beloch's discovery would not have been possible without Eduard Lill, an Austrian Engineer, and his discovery of Lill's Method.

The Beloch Origami Fold is an axiom or folding move considered to set Origami apart from straightedge and compass constructions.

Lill's Method
This method is used to find a real root, if one exists, of a given polynomial $$f(x)=a_nx^n+...+a_1x+a_0$$which has real coefficients. Lill's method is graphical whereby based on the coefficients of $$f(x)$$, you create a path in the plane.

The Beloch Fold
Given two points $$p_{1}$$and $$p_{2}$$and two lines $$l_{1}$$and $$l_{2}$$ we can, whenever possible, make a single fold that places $$p_{1}$$onto $$l_{1}$$and $$p_{2}$$onto $$l_{2}$$ simultaneously.

The Beloch Square
Given two points $$p_{1}$$and $$p_{2}$$ and two lines $$l_{1}$$and $$l_{2}$$ in the plane, construct a square $$WXYZ$$ with two adjacent corners $$X$$ and $$Y$$ lying on $$l_{1}$$and $$l_{2}$$, respectively, and the sides $$WX$$ and $$YZ$$, or their extensions, passing through $$p_{1}$$and $$p_{2}$$, respectively.

Squaring the circle
Squaring the circle is the problem of constructing a square of equal area to the unit circle, which has radius of length 1 and therefore area equal to $$\pi$$. Then in order for the area of the square to be $$\pi$$, each side must have length $$\sqrt{\pi}$$. Hence one needs to prove that is it impossible to construct a line segment of length $$\sqrt{\pi}$$ in order to prove that squaring the circle is impossible.

The Greeks did know that the problem was impossible, but they were unable to prove so. In 1761, it was proved by John Heinrich Lambert that $$\pi$$ is irrational, and it was proposed as conjectures that $$e$$ and $$\pi$$ were both transcendental numbers in his 1768 paper. Ferdinand von Lindemann, a German mathematician, published his first complete proof in 1882 for the transcendence of $$\pi$$, but in 1880, Lindemann proved that $$\pi$$ is transcendental, using Euler's identity ($$e^{i\pi}+1=0$$), and the fact that $$e$$ is transcendental. This meant that $$\pi$$ could not be the root of any non-zero polynomial which has rational coefficients and that $$\pi$$ is not constructible, so it is impossible to construct a line segment of length $$\pi$$ or $$\sqrt{\pi}$$, and hence squaring the circle is impossible using straightedge and compass construction.

Since $$\pi$$ is still a transcendental number in the set of origami-constructible numbers, then the problem of squaring the circle is also impossible using origami.

Approximations
Even though one is unable to construct a line segment of length exactly equal to $$\pi$$, the following are two examples of very close approximations:


 * $$\pi\approx\surd(40/3-2\surd3)\approx3.14153$$
 * $$\pi\approx355/113=3+16/113=3+4^2/(7^2+8^2)\approx3.1415929$$



Rolling a circle
Besides using straightedge and compass construction, there are other methods that does allow the problem of squaring the circle to be possible, and we will discuss one of these called "rolling a circle". If a unit circle is rolled halfway along a horizontal, straight line, beginning at point $$A$$ and ending up at point $$B$$, then because $$AB$$ is equal to half of the circumference of the unit circle which is $$2\pi$$ then $$AB$$ $$=\frac{1}{2}\times2\pi$$ $$=\pi$$. Next, allow segment $$AB$$ to be extended to the point $$C$$ so that we have $$AC=AB+1=\pi+1$$, and construct a semicircle with diameter $$AC$$. Then let $$D$$ be the point of intersection of the semicircle with a line drawn through $$B$$ that is perpendicular to $$AB$$. We see that $$AB\times BC=BD^2$$ together with $$AB=\pi$$ and $$BC=1$$ implies that $$BD=\sqrt{\pi}$$. Hence if one constructs a square with $$BD$$ as one of the sides, then the area of this square is exactly equal to the area of a unit circle.