User:MA783-NS476/sandbox

Origami Numbers

The classical Greek problems

There were three classical problems that the ancient Greeks (600BC to 400AD) tried to solve by ‘ruler and compass’ constructions. These problems form the basis of what this module is trying to solve.

1. Can we square the circle? If we have a circle of radius 1 unit (and hence of area $\pi$), is it possible to construct a square of the same area and thus to construct the length ${\sqrt {\pi }}$?
2. Can we double the cube? If we have a cube of edge length 1 unit (and hence volume 1), is it possible to construct a cube of volume twice that of the original cube and thus to construct the length ${\sqrt[{3}]{2}}$?
3. Can we trisect an arbitrary angle? Suppose that we can construct a triangle (so the length of each side is a constructible real number) in which one of the angles in the triangle is $\theta$. Is it always possible to construct a triangle (so each side is again a constructible real number) with one of the angles being $\theta \over 3$?

Throughout the module, we have examined the relation between field extensions and irreducible polynomials as well as ruler and compass constructions. Through this, we have understood that each of these questions cannot be solved using ruler and compass constructions. So, to help us answer these questions, we shall consider origami constructible numbers but first, let’s consider what we mean by origami. Origami simply involves taking a square piece of paper and folding the paper to form numerous three-dimensional shapes, which produces constructible numbers. This was used as an essential tool in the development of modern algebra.

Note, a real number $r$ is origami-constructible if one can construct, in a finite number of steps, two points which are a distance of $|r|$ apart. Since the set of real numbers associated with lengths in $\mathbb {R}$ that are produced by origami constructions together with their negatives are origami-constructible elements of $\mathbb {R}$, for the remainder of this project, we will not distinguish between the terms origami-constructible lengths and origami-constructible real numbers.

Looking at the lecture notes, we have found that a real number $\alpha$ is constructible if, in a finite number of steps, we can construct a line segment of length $|\alpha |$ (from the unit length) using only a ruler and compass.

We will study this in more depth throughout this project including the Huzita-Hatori axioms and any theorems that may aid our understanding of how these numbers can be used to solve the three classical Greek problems.

Huzita-Hatori axioms

There are seven axioms, referred to as the Huzita- Hatori axioms, that helps aid our understanding regarding the rules of origami constructions which are listed below:

Axiom 1

Given two distinct points $p_{1}$ and $p_{2}$, there is a distinct fold that maps through them. This is illustrated by figure 1.

Figure 1: Axiom 1

Axiom 2 Given two distinct points $p_{1}$ and $p_{2}$, there is a distinct fold that maps $p_{1}$ onto $p_{2}$. This is illustrated by figure 2.      

Figure 2: Axiom 2

From the figure, we can see that this is the same as constructing the perpendicular bisector of the line segment $p_{1}p_{2}$.

Axiom 3

Given two lines $l_{1}$ and $l_{2}$, there is a fold that maps $l_{1}$ onto $l_{2}$. This is illustrated by figure 3.

Figure 3: Axiom 3

From this figure, we can see that this is the same as constructing the bisector of an angle between the two lines.

Axiom 4

Given a line $l_{1}$ and a point $p_{1}$, there is a distinct fold that goes through $p_{1}$ and perpendicular to $l_{1}$. This is illustrated by figure 4.

Figure 4: Axiom 4

From this figure, we can see that this is the same as constructing a perpendicular to $l_{1}$ that goes through $p_{1}$.

Axiom 5

Given a line $l_{1}$ and points $p_{1}$and $p_{2}$, there is a fold mapping $p_{1}$onto $l_{1}$that goes through the point $p_{2}$. This is illustrated by figure 5.

Figure 5: Axiom 5

From this figure, we can see that this is the same as constructing the intersection between a circle and line $l_{1}$, and so having 0,1 or 2 solutions.

Axiom 6

Given two lines $l_{1}$ and $l_{2}$ and two points $p_{1}$ and $p_{2}$, there is a fold mapping $p_{1}$ onto $l_{1}$ and $p_{2}$ onto $l_{2}$. This is illustrated by figure 6.

Figure 6: Axiom 6

Figure 7: An application of axiom 6

From figure 7, we can see that axiom 6 is the same as constructing a line that is tangent to two parabolas simultaneously where we refer to each point as a focus and each line as a directrix. Thus, the foci of the two parabolas are at $p_{1}$ and $p_{2}$ respectively and the directrices are given by $l_{1}$ and $l_{2}$ .

This axiom enables the construction of cube roots, solving problem 2 which we will discuss in more detail later on.

Axiom 7

Given two lines $l_{1}$ and $l_{2}$ and a point $p$, there is a fold mapping point $p$ onto the line $l_{1}$ and is perpendicular to $l_{2}$. This is illustrated by figure 8.         

Figure 8: Axiom 7

A possible eighth axiom?

In 2017, Jorge Lucero claimed there were in fact eight axioms, the seven axioms stated above and an additional operation; given a line $l_{1}$, there is a fold along this line.

Lucero believed that this eighth axiom is required in the application of origami, even though it does not form a new line, as one is allowed to fold the layer of paper along a marked line on the layer directly below. Thus, this eighth axiom completes the other seven axioms. However, it is not seen as an axiom.

Theorems on origami constructible numbers

Before we examine theorems on origami constructions, let us first recall a few key theorems on real constructible numbers.

Note, from the lecture notes, we know that the set of constructible real numbers is a field. Also, this field is closed under addition, multiplication, inverses and taking the square roots of constructible real numbers. For example, using a ruler and compass, all the elements in the field $\mathbb {Q}$ can be constructed. Moreover, if r is a constructible real number, then $[\mathbb {Q} (r):\mathbb {Q} ]=2^{k}$ for some integer k $\geq$ 0. As a result, it is clear that straightedge and compass constructions can find solutions to polynomials of degree $2^{k}$. So, doubling the cube and trisecting an angle cannot be solved using this method as they require constructing lengths which are solutions to cubic equations.

Now, through application of the axioms above, we can derive the following theorems on origami constructions.

Theorem: A point $(a,b)$ $\in$ $\mathbb {R} ^{2}$ is origami constructible if and only if $a$ and $b$ are origami-constructible elements of $\mathbb {R}$.

Theorem: Let r ∈ $\mathbb {R}$. Then r ∈ O if and only if there is a finite sequence of fields $\mathbb {Q} =F_{0}\subset F_{1}\subset ...\subset F_{n}\subset \mathbb {R}$ such that $r\in F_{n}$ and $[F_{i}:F_{i-1}]=2$ or $3$ for each $1\leq i\leq n$.

Following on from this, we obtain that if r ∈ O then $[\mathbb {Q} (r):\mathbb {Q} ]=2^{a}3^{b}$ where a,b $\geq$ 0 are integers. More specifically, the set of origami constructible numbers is closed under taking cube roots.

Thus, we can see that two of the Greek problems, doubling the cube and trisecting an angle, can be solved by origami constructions. We will go into further detail on this later on. But first, let us observe the role of origami in solving the cubic equation.

Application of axioms 1 - 6 in finding solutions to the cubic equation

Using axioms 1 - 6, we can construct a real solution with real coefficients in the field O to a cubic equation. We can see this below when we consider the following parabolas.

Let $(y-{1 \over 2}a)^{2}=2bx$ and $y=\left({1 \over 2}x\right)^{2}$ be two parabolas, and let $(x_{0},y_{0})$ and $(x_{1},y_{1})$ be the points in which the common tangent intersects these two parabolas respectively. Suppose the gradient of the simultaneous tangent (see figure 7) is $\mu$.

If we differentiate this, we get that $2\displaystyle \left(y-{1 \over 2}a\right)\left({dy \over dx}\right)=2b$ and ${dy \over dx}=x$.

This would imply that ${dy \over dx}={2b \over 2y-a}$ . Now if we let $\mu ={2b \over 2y_{0}-a}=x_{1}$ , we get that $y_{1}={1 \over 2}(\mu )^{2}$ .

We also know that $(y_{0}-{1 \over 2}a)^{2}=2bx_{0}$ , which tells us that $x_{0}={(y_{0}-{1 \over 2}a)^{2} \over 2b}={({b \over \mu })^{2} \over 2b}={b \over 2\mu ^{2}}$ .

From this we get that $\mu ={y_{1}-y_{0} \over x_{1}-x_{0}}={{\mu ^{2} \over 2}-{a \over 2}-{b \over \mu } \over \mu -{b \over 2\mu ^{2}}}$ . This implies that $\mu ^{2}-{b \over 2\mu }={\mu ^{2} \over 2}-{a \over 2}-{b \over \mu }$.

Simplifying this, we get that $\mu ^{3}+a\mu +b=0$. From this, we can see that; given any two constructible real numbers $a$ and $b$, any cubic equation has constructible solutions.

Now that we have all the necessary theory, we can proceed determining whether the three classical Greek problems can be solved using origami.

Problem 1: Can we square the circle?

During the course of this module, we have learnt that we cannot 'square the circle' using ruler and compass constructions, i.e. we cannot construct a square with an area of $\pi$.

This is shown in Theorem 6.10, by proving that ${\sqrt {\pi }}$ is not a real constructible number; the proof by contradiction is shown below.

Proof by contradiction:

We need to construct a square with length ${\sqrt {\pi }}$. Assume ${\sqrt {\pi }}$ is a constructible real number. Then,  ${\sqrt {\pi }}\times {\sqrt {\pi }}=\pi$ is also a constructible real number using Lemma 6.2 and so $\pi$ is algebraic over $\mathbb {Q}$ by Theorem 6.9. This is a contradiction; from our previous knowledge we are aware that $\pi$ is transcendental over $\mathbb {Q}$. Thus, we cannot construct ${\sqrt {\pi }}$ and so we cannot square the circle.

This leaves us with the following question: can we square the circle using origami construction?

Unfortunately, this problem still remains unsolved. However, by rounding $\pi$ to a certain decimal place, we are able to provide approximate solutions.

In 1914, an Indian mathematician, Srinivasa Ramanujan gave a ruler-and-compass construction. This was equivalent to taking the approximate value of $\pi$ to be ${\biggl (}9^{2}+{19^{2} \over 22}{\biggr )}^{\frac {1}{2}}$, giving eight decimal places of  $\pi$.

Problem 2: Can we double the cube?

During the course of this module, we have learnt that we cannot 'double the cube' i.e. we cannot construct a cube with a length of ${\sqrt[{3}]{2}}$ because it is not a constructible real number.

Proof by contradiction:

Assume $\alpha ={\sqrt[{3}]{2}}$ is a constructible real number. Then $\alpha$ has a minimal polynomial $x^{3}-2$ over $\mathbb {Q}$. This is irreducible due to Eisenstein’s Criterion. So, $[\mathbb {Q} (\alpha ):\mathbb {Q} ]=3$ but $3\neq 2^{k}$, so a contradiction. Therefore, ${\sqrt[{3}]{2}}$ is not constructible and so we cannot 'double the cube'.

Now, we ask the following question: can the cube be doubled using origami construction?

We can answer in the affirmative of this by showing that ${\sqrt[{3}]{2}}$ is constructible using origami.

Firstly, fold a square piece of paper into thirds as illustrated in figure 9a.

Figure 9a

Define $l_{1}$ to be the line of the left-hand side of the square piece of paper and $l_{2}$ as the top one-third crease line. Let $P_{1}$ represent the point on the bottom right-hand corner of the square piece of paper and let $P_{2}$ represent the point of intersection of the right hand-side of the square piece of paper with the bottom one-third crease line.

Now, applying Axiom 6 to lines $l_{1}$ and $l_{2}$ and points $P_{1}$ and $P_{2}$, we form a fold line, let’s call this $m$ which maps $P_{1}$ onto $l_{1}$ and $P_{2}$ onto $l_{2}$ as illustrated in Figure 9b.

Figure 9b

The position in which $P_{1}$ intersects line $l_{1}$ divides the edge by ${\sqrt[{3}]{2}}$ to 1 and is given by point $P'_{1}$. This can be seen in Figure 9c.

Figure 9c

Let $x=B\cdot P_{1}$ and $y=A\cdot D$

$\Rightarrow AB=x+1$, $D\cdot P_{1}=x+1-y$, $P_{1}\cdot P_{2}={(x+1) \over 3}$, $C\cdot P_{1}=x-{(x+1) \over 3}={(2x-1) \over 3}$.

Using Pythagorean Theorem on $\bigtriangleup$ $ADP_{1}$, we obtain

$(x+1-y)^{2}=1+y^{2}$ which implies $y={x^{2}+2x \over 2x+2}$ (*).

Also, $m\angle P_{1}CP_{2}$ $=m\angle DAP_{1}=90^{\circ }$ and

$m\angle CP_{1}P_{2}+m\angle CP_{2}P_{1}=90^{\circ }$,

$m\angle ADP_{1}+m\angle AP_{1}D=90^{\circ }$,

$m\angle CP_{1}P_{2}+m\angle AP_{1}D=90^{\circ }$.

Thus, $m\angle CP_{1}P_{2}=m\angle ADP_{1}$ and $m\angle CP_{2}P_{1}=m\angle AP_{1}D$ and so $\bigtriangleup P_{1}CP_{2}\backsim \bigtriangleup DAP_{1}$.

Hence, ${DA \over DP_{1}}={CP_{1} \over P_{1}P_{2}}$

$\Rightarrow {y \over x+1-y}={{2x-1 \over 3} \over {x+1 \over 3}}={2x-1 \over x+1}$.

Substituting (*) into the above, we obtain

${{x^{2}+2x \over 2x+2} \over x+1-{x^{2}+2x \over 2x+2}}={{x^{2}+2x \over 2x+2} \over {x^{2}+2x+2 \over 2x+2}}={x^{2}+2x \over x^{2}+2x+2}={2x-1 \over x+1}$

$\Rightarrow (x^{2}+2x)(x+1)=(2x-1)(x^{2}+2x+2)$

$\Rightarrow x^{3}+3x^{2}+2x=2x^{3}+3x^{2}+2x-2$

$\Rightarrow x^{3}=2$

$\Rightarrow x={\sqrt[{3}]{2}}$.

Hence, ${\sqrt[{3}]{2}}$ is constructible using origami.

Problem 3: Can we trisect the angle?

We have learnt during the course of this module that using ruler and compass constructions, not all angles can be trisected. For example, $\pi \over 9$ cannot be constructed. We shall prove this using a contradiction.

Proof by contradiction:

Suppose $\pi \over 9$ can be constructed. Using Lemma 6.12 from the lecture notes, $cos{\biggl (}{\pi \over 9}{\biggr )}$ is a constructible real number. Now if we substitute $\theta ={\pi \over 9}$ into $cos3\theta =4cos^{3}\theta -3cos\theta$, we obtain $cos{\biggl (}{3\pi \over 9}{\biggl )}=4cos^{3}{\biggl (}{\pi \over 9}{\biggl )}-3cos{\biggl (}{\pi \over 9}{\biggl )}$ where $cos{\biggl (}{3\pi \over 9}{\biggl )}={1 \over 2}$ .

If we let $x=cos{\biggl (}{\pi \over 9}{\biggr )}$, then $4x^{3}-3x={1 \over 2}$ . After rearranging and simplifying, we obtain $m{(x)}=x^{3}-{3 \over 4}x-{1 \over 8}=0$ . As $m(x)$ is monic and irreducible over Q (by applying Root Test), $m(x)$ is the minimal polynomial of $cos{\biggl (}{\pi \over 9}{\biggr )}$ . However $[\mathbb {Q} {\biggl (}{\pi \over 9}{\biggr )}:\mathbb {Q} ]=3\neq 2^{k}$ which contradicts Theorem 6.9. This means $\pi \over 9$ cannot be constructed.

Now, we ask the question: can we trisect the angle using origami?

It is, in fact, possible to trisect any given angle $\theta$ using origami. We can prove this using a method developed by Hisashi Abe published in 1980.

Given a square piece of paper, mark an arbitrary angle $\theta$ on the bottom left hand corner $P$ of the paper. This is produced by the line $l_{1}$ and the bottom edge of the paper as shown in figure 11a.

Figure 11a

Now, suppose point $Q$ is any constructible point on the left border of the square piece of paper. Applying axiom 4, we can form a line which passes through point $Q$ and is perpendicular to the left border. Using axiom 2, we can fold $P$ onto $Q$ producing a line $l_{2}$ that is in the middle to both the parallel lines passing through $P$ and $Q$. This is shown in figure 11b.  

Figure 11b

Next, we can apply axiom 6 to the lines $l_{1}$ and $l_{2}$ and points $P$ and $Q$ forming a fold line $m$ that maps $Q$ onto $l_{1}$ and $P$ onto $l_{2}$ illustrated in figure 11c.

Figure 11c

Now, define the point $P'$as the reflection of point $P$ about line $m$. This can be constructed as the intersection of the line perpendicular to $m$ through $P$ and the line $l_{2}$. $Q$ can be constructed in a similar way. Thus, an angle with measure of $\theta \over 3$ is formed by segment $PP'$ and the edge at the bottom of the paper.

In order to prove that the angle with measure of $\theta \over 3$ is formed by the segment $PP'$ and the edge at the bottom of the paper, we look at figure 11d.

Figure 11d

Let us define $R$ to be the intersection of segments $PQ'$ and $P'Q$. Similarly, we define $S$ to be the intersection of the line $l_{2}$ and the left border of the piece of paper. Let $T$ be the intersection of line $m$ and segment $PP'$. Let $\alpha$ be the angle formed by triangle $RPP'$ (i.e. $RPP'=\alpha$), $\beta$ be the angle formed by triangle $PP'S$ (i.e. $PP'S=\beta$ ) and $\gamma$ be the angle in triangle $RP'S$ (i.e. $RP'S=\gamma$).  As the reflections of $P$ and $Q$ about line $m$ are $P'$and $Q'$ respectively, $R$ must lie on line $m$. We can see that $P'QP$ is an isosceles triangle as $PS=QS$ and $P'SP$ and $P'SQ$ are two perpendicular right angles. Hence, $\beta =\gamma$. Likewise, $RPP'$ is also an isosceles triangle since we have the following properties: the reflection of $P$ about line $m$ is defined by $P'$, $PT=P'T$ and angles $RTP$ and $RTP'$ are two perpendicular right angles. Thus, $\alpha +\beta =\gamma$.

Notice, the bottom border of the piece of paper and line $l_{2}$ are parallel and so segment $PP'$ and the bottom border of the paper forms an angle of measure β. So we have:

$\theta =\alpha +\beta =(\beta +\gamma )+\beta =3\beta$.

We can see that angle $\theta$ has measures between 0 and ${\pi \over 2}$. In particular, when  $\theta ={\pi \over 2}$ , $Q=Q'$ and so line $m$ folds point $P$ onto $l_{2}$ through point $Q$. Segment $PP'$ trisects $\theta$ since $QPP'$ is an equilateral triangle. If angle $\theta$ has a measure between ${\pi \over 2}$ and $\pi$, we can divide $\theta$ into a right angle $\beta$ and an acute angle $\alpha$ such that $\theta =\alpha +\beta$. This can be seen using axiom 4, i.e. by constructing a line $l$ which is perpendicular to a side of $\theta$ with centre O as illustrated in figure 12.

Figure 12

We have ${\theta \over 3}={\alpha \over 3}+{\beta \over 3}={\alpha +\beta \over 3}$. Thus, the angle between lines $m$ and $n$ forms the trisection of the obtuse angle. Similarly, we can apply this method of trisecting an angle to any arbitrary angle.

Exploration of the field of origami numbers

So far, we have considered the seven single-fold operations. However, we can also explore multiple fold operations, by permitting the alignments of lines and points along multiple folds, as mentioned by Alperin and Lang (2009). We define a two-fold alignment as the minimal set of alignments that defines two simultaneous fold lines on a finite region of the Euclidean plane with a finite number of solutions. Also, a two-fold alignment is referred to as separable if and only if its alignments can be split into two single-fold sets. There are limitations with such two-fold alignments as practically it can be difficult to ensure that both folds occur simultaneously.

Further research studies on origami can be explored.

One such research study is considering what would happen if tracing was permitted in origami? In 2017, Lucero had mentioned how it is believed that all marked points and lines on one fold are also defined both on the folds below and above, making the paper appear ‘transparent’. So, we could consider whether the set of origami-constructible numbers would be altered from the construction of a new point or line by tracing another point or line onto the folded paper?

Another potential research study involves considering what would happen if cutting is permitted in origami mathematics? And whether this would result in the set of origami-constructible numbers being altered? Careful consideration would need to be taken regarding the definition of cutting; also, after cutting the paper, the types of folds that are permitted. For example, suppose the paper is cut along a line segment and it is not entirely split into two separate pieces of paper. One must then consider methods of folding the paper about the cut.

Moreover, we can also explore other types of field extensions derived from origami constructions but this time, beginning with something other than a unit distance and two points. For instance, we could consider what would happen if instead we begin with three points with a group of angles and/or distances between them?

Thus, it is evident that further extensions in origami can be explored.

${\displaystyle }$