Nicoleta Bostan Math IA - Origami Geometric Proofs

Origami Geometric Proofs Nicoleta Bostan Candidate Number: 000974-0013 Introduction: "Within the mathematical theory of origami geometric constructions, the seven Huzita-Justin axioms define what is possible to construct by making sequential single creases formed by aligning combinations of points and lines. It has been mathematically proven that there are only the seven axioms, and that those folds permit the construction of solutions to arbitrary equations up to degree 4: quadratics, cubics, and quartics—but no higher."(Lang, Robert J.) This is the quote by Robert J. Lang that led to my investigation into how origami solve equations up to degree 4 by solving Ancient Greek problems that were considered impossible for 2000 years, even though quartics were not achievable in this exploration due to the limitation of knowledge in the area. However, 4th degree equations in origami are claimed to be solved by the compass and straight edge method, described in the 7th axiom developed by the mathematician Justin. Origami, or paper folding has emerged as a tradition of ceremonial gifts made out of paper due to paper being a luxury item around the 6th century. These ceremonial tokens and also art form have proven to follow a set of axioms that were developed in the 20th century. Although valuable in terms of culture specific to Asian countries, the practical usage of paper has lead to theorems that in the end demonstrated a vast array of other applications in real life that are practical today. Origami are at the basis of car air bags, architectural structures, space satellite panels and others, which are entirely due to the application of theorems developed from observations of simple geometry in a plane. What makes origami special is that such incredible shapes are possible from one sheet of paper with no cuttings at all, using just folding techniques. On top of the usefulness of the application of origami, the art aspect is still preserved and it is both interesting through a mathematics lens and aesthetically. In Ancient Greece there were several geometric problems that could not be solved at the time because they were using only compass and straight edge methods for the creation of architectural models, but origami has proved to be more powerful than the ancient method as the problems of doubling the cube and trisecting an angle can be created through the art of paper folding. Although it is only possible to fold straight lines in origami in a plane, as curves are mathematically arbitrary in folding, there are a multitude of constructions that are yet to be discovered that use the power of origami, as also demonstrated by the quote at the beginning of this paper. The focus of this paper will be on investigating what kind of shapes are possible to construct using origami as demonstrated by geometric proofs. Contrary to Euclidean geometry or straight edge and compass method where it has been proven that trisecting the angle and doubling the volume of a cube is impossible, we can make 1

geometric constructions with origami, “using the side of the paper as the straight edge and folding up to an angle to simulate a compass”. (Gaur) Geometric proofs usually use straight edge and compass method, but as mentioned before, origami acts as both a compass when the side describing the straight edge is folded up. However, the method used by Euclid of Alexandria who is often called the father of geometry could not always solve all geometric problems. By using a straight-edge and compass, it was possible to perform a large number of geometric operations. Nonetheless, Euclidean geometry has its own limitations, as mentioned before, because the two famous problems of antiquity could not be solved using Euclidian axioms. On the other hand, problems such as doubling the volume of the cube, and trisection of an angle which cannot be solved with Euclidean geometry have been always able to be proved by origami, even before the discovery of the method. Origami as a mathematics branch allows to simultaneously align two separate points onto two different lines, as given by the first Huzita axiom. This alignment is the one that is responsible for the cubic equation and the according solving of the antiquity problems. Origami is regarded more like a hobby or an art form and I have always been drawn to this, in particular how one piece of paper is capable of creating incredible models of anything from the real world using angles and folds. This personal connection to origami has pushed me into thinking whether there is more to this and as I have discovered many inventions rely on origami structures such as the first space solar panel satellite that used the Miura fold in order to transport a huge structure by economy of space, or airbags that fold into small compartments to blow efficiently into safe cushions in case of accidents. Essentially, mathematics is meant to help in the understanding of rules and patterns, such as numbers, lines, or angles. This paper was a challenge as geometric proofs are not part of the curriculum. Nonetheless, I had the opportunity to come back to the knowledge I previously had from Romanian school, where we were taught geometric proofs in the last years of middle school, and apply that in this exploration. First of all, it is important to consider the basics of origami mathematics, which are the 7 Huzita-Justin Axioms that are as follows. These are essential in each construction as they are used throughout every fold in the procedures and proofs below.

Huzita’s Origami Axioms 1. Given two points P1 and P2 we can fold a line connecting them.

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2. Given two points P1 and P2 we can fold P1 onto P2.

3. Given two lines L1 and L2 we can fold line L1 onto L2.

4. Given a point P1 and a line L1 we can make a fold perpendicular to L1 passing through the point P1.

5. Given two points P1 and P2 and a line L1 we can make a fold that places P1 onto L1 and passes through the point P2.

6. Given two points P1 and P2 and two lines L1 and L2 we can make a fold that places P1 onto line L1 and places P2 onto line L2.

Officially known as Hatori’s but actually Justin’s Seventh Axiom: 3

7. Given a point P1 and two lines L1 and L2, we can make a fold perpendicular to L2 that places P1 onto line L1. (The axioms diagrams and explanations of each diagram were taken from a study paper: Origami and Geometric Constructions cited as: - - -. "Origami and Geometric Constructions.") Even though these 7 points are called axioms, this is an arbitrary term, as they can be described best as fundamental operations in origami folding that act as points and lines. Folds can use basic operations such as fold a point to another point (axiom 2), fold a line to another line (axiom 3, also known as angle bisection) or make a crease through one or more points (axiom 4), all in different combinations, but in the 1970s, mathematicians and origami folders began to systematically enumerate the possible combinations of folds and what constructions were possible with the combinations. The first systematic study was by Humiaki Huzita, who described came up with six ways of folding and aligning points, described in the Huzita Axioms. What is exceptional about the last axiom, the Justin axiom, is that it is at the basis of both trisection of an acute angle and of doubling the cube, as it will be demonstrated in the following proofs. Before showing pictures and diagrams, it is important to note that the folding demonstrated in the method part of both trisection and doubling the cube are handmade and taken picture of by myself, Nicoleta Bostan. Trisecting an Angle Method and Proof Divisions of angles into fractions other than powers of 2 are considerably harder and as proved consistently by the use of straight edge and compass, impossible. Dividing an angle into two is called bisecting an angle, while into three it is called trisecting, meaning division into three equal parts. The trisection of an angle was impossible for millennia. In 1837, Pierre Wantzel, a French mathematician and expert in arithmetic has come to the conclusion that it was impossible to perform a trisection of an angle. Furthermore, Wantzel showed that the problem of trisecting an angle is equivalent to solving a cubic equation. Moreover, he showed that only very few cubic equations can be solved using the straightedgeand-compass method. By following the steps outlined below, it can be proven that an acute angle can be trisected though the origami method, contrasting to Euclidean. It is assumed that the angle is acute, although it has been proven that it can be extended to obtuse angles, but this will not be investigated in this paper. 4

1: Draw any angle PBC so that point B is in the corner of the square paper.

2: Make a horizontal fold anywhere across the square defining line EF|| BC using axiom #4 meaning that AB perp on BC

5: With the corner still folded, fold both layers to continue the crease that ends at point G all the way to J. (as shown by the dotted line). The crease passes through point B, thus dividing PBC in the ration 2:1. Unfold

3: Fold line4: BCBy upusing the 6th axiom, fold bottom left to line EF and corner (side AB of the unfold, creating square) so that point E line GH, which creates EG touches = GB, line BP and point that by axiom 3, B touches line GH. the two parallel creases are spaced evenly apart.

7, final step: The two 6: Fold along the crease that runs to point J,creases BJ and BK divide theB.original angle PBC extending it to point intoBC three equal parts. Fold the bottom edge The configuration is up to line BJ and unfold. symmetric along the line Define line BK.

Proof:

Figure 1: Trisecting an Angle Sketch (Newton, Liz.) The figure above, the angle between the blue line and the side BC of the square is the angle meant to be trisected, or as in the model, the angle PBC. Therefore, it must be shown that: 5

α =β=γ As observed in step 4 of the method section above, the crease that is made, represented by the red line in the above diagram is a line of symmetry. Now look at the triangle EBb (b being the corner B when folded onto line GH). As already described in the steps of the method, it is known that segment EG is equal to GB according to the 3rd axiom given that BC and EF have the same midpoint due to the folding of the line BC on EF. Accordingly, it is known that Gb is perpendicular to EB according to the 6th axiom, also being the height of the triangle EBb. Therefore, it can be inferred that the segment Gb of the triangle EBb divides the side of the square EB, as EG=GB. Therefore, triangle EBb follows the definition of a isosceles triangle, as the height bisects the segment. As mentioned the red line from the diagram is a symmetry line and therefore triangles and segments of the same length are reflected in the red crease. Therefore, the mirror triangle of EBb is ebB, which is in corcodance isosceles as well. As the triangles are similar, the height of the triangle is hence Bg, bisecting the angle B in two equal halves, as it is an isosceles triangle’s height, thus showing that α =β . By mirror symmetry through the red line

β

is equal to the angle δ

of the triangle GbB, as these triangles

would be similar, GbB ~ gBb. And since the line GH is parallel to the bottom edge BC, we have that γ =δ , as the angles are alternate interior. This proves that

β=γ .

Therefore, it is proved that α =β=γ , as the initial angle of all three angles is trisected using origami and folds with the Huzita-Justin Axioms as a basis. Upon further research I have discovered that Wantzel has proved that the problem of trisecting an angle is impossible to solve using straight edge and compass method or the fields of the Galois theory, because it implies a solution of a certain cubic equation. However, when investigating for the proof of this using trigonometry that gets reduced to an algebraic problem that meant to show as the proof od the origami method that origamis are able to solve cubic equation of the form a x 3+ b x2 +cx + d=0 , I was unable to understand and grasp the concept as it is beyond what is taught in the curriculum. Doubling the Cube Method and Proof In the antiquity problem of doubling the cube, the objective was to build one side of a cube whose volume is twice as big than the given original cube, meaning that given a segment, somehow build another segment that is exactly

√3 2 as long.

The legend of doubling the cube tells that in 430BC Greece had been plagued and the oracle of the city advised the citizens to double the size of the altar, but upon doing so, they failed as the plague did not stop. This was a mathematical mistake because as the citizens doubled the sides of the altar, the volume increased eight times 6

(V = x3 becomes (2x)3 = 8x3). Therefore, according to the legend, they should have doubled the volume not the sides, but nonetheless, this was not possible to solve as they used the straight edge and angle method. In order to illustrate the method using origami, it is easy to show on a model, such as following. First, doubling the cube means that we are given a side of a cube of side length s 1 and volume V . The problem implies that a side of the cube s 2 which has the volume twice as big as the original cube, 2× V . This is illustrated in the method below using origami:

3: Make 1: 4: 2: Fold the a small crease low side fold B BC of connecting half way theup square point the right A over and and E, intersecting theaintersection C, side diagonal of the paper, ofXthe soAC at point X. point square that DE=EC, ofABCD. the lines as DAC and and C form BE aaccording side to axiom 4. to axiom 1. according

5: Fold the top side AD of the square to touch the mark X horizontally and unfold to form FG where DG =GI.

6: Fold corner C to lie on line AB while point I lies on line FG using the 6th axiom. Point C divides the edge AB into two segments. The ratio AC/CB multiplied by the side length of the initial square is the side for doubling the cube.

Proof:

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Figure 2: Sketch for Doubling the Cube (first three steps of method) (Newton, Liz.) Note: in the above diagram there is a mistake, as instead of y=1/2sx there should be written y=1/2x as shown in the following proof and reworking of the problem. In the origami method presented above first it is shown in steps 4 and 5, how one side of the square is divided in three equal parts according to the 3rd axiom. To ease the calculations, the origami is placed in a x and y plane, a coordinate system, thus corner B is at the point (0,0) and all other points as illustrated by figure 2. Now the line from A to C, the diagonal of the square has the equation

y=s−x (as a function in a coordinate

system) where s refers to the side of the square. The segment BE has the equation of

1 y= s as a function in a coordinate system. The two lines intersect at 2

point X marked in the models representation above. Therefore, the two functions are equalized as follows: 1 s−x= x 2 1 2 ∴ x + x=s 2 2 3 ∴ x=s 2 2 ∴ x= s 3 Therefore, by substituting x into the equation of the line

y=s−x , the y coordinate of the system can be

found: 2 1 y=s− s= s 3 3

The construction claims that the side length s2 of the new cube is

AC CB

multiplied by the side length s1 of the

given cube, given that s1 = s throughout this proof explanation. To solve the problem, we need to find s2 so that the ratio of the new side to the old side is the cube root of 2, meaning that s2 = AC/CB * s1 Since V2 = 2V1 8

s 2=√3 2 s 1 ∴ √3 2 s 1=

AC ∗s 1 CB

Therefore, the proof has to show that: AC 3 = √2 CB

Figure 3:Sketch for Doubling the Angle Mathematical Proof (Newton, Liz.) This is the model from the last step of the method of doubling a cube. In order to standardize the calculations, the length if CB can be attributed to 1, without affecting the proportions. The segment AC can be written as x, thus the side of the square would be AC+CB, in concordance with the previous statement, is equal to x+1, as illustrated in figure 3. The line segment BF corresponds to two thirds because as seen the intersection point gives that X is at 1/3 of the side therefore, as the lines are parallel according to axiom 3 the two parallel creases are spaced evenly apart, the line BF corresponds to 2/3 of a side of the square and therefore has length of Therefore as CF corresponds to BF-BC it can illustrated as

2( x+1) . 3

2( x+1) 2 x +2−3 2 x−1 −1= = 3 3 3

By writing d as the length of BJ, since the segment CJ is a part of the lower side of the square, its length would be

x+ 1−d .

As the form is a square, angle B’s measure is 90 ° , therefore Pythagora’s theorem can be used as follows: x+ 1−d ¿ d 2 +1=¿

2

∴ d 2+1=x2 +2 x+ 1−2 d ( x +1 ) +d 2 ∴ d=

2

x +2 x 2( x+1)

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Considering the two right-angled triangles CFI and CBJ and also the 180 degree angle formed at C by the vertical edge of the square, and also considering that angle ICJ is right angle because of the 4th axiom meaning that the sides of the square form 90 degree angles, we get: α =180−δ−90 ∴ α=90−δ As known, the angles in a triangle add up to 180 degrees, therefore we infer that: γ =180−90−δ ∴ γ =90−δ And since the previous calculation demonstrates that α =90−δ , the two angles are equal: ∴ γ =α In order to determine the similarity of the triangles CFI and CBJ, we can notice the two equal angles one of which is was demonstrated above to be equal γ =α : the right angle and the angle α , therefore the given triangles are similar. Concluding from the similarity properties, the ratios of corresponding sides of the two similar triangles are equal. As mentioned before in the investigation, CI of the triangle SFI is 1/3 of the side of the square, when explaining where the two lines meet at X, and therefore has length of

x +1 , therefore according to the 3

similar triangle formula: 2 x−1 d 3 = x +1−d x+1 3 ∴

d 2 x−1 = x+ 1−d x+1

x2 +2 x and substituting d that was expressed some steps above as d= : expand, don’t simplify 2( x +1) x2 +2 x 2 ( x +1 ) 2 x−1 = 2 x +1 x +2 x x +1− 2 ( x+1 ) 2

x +2 x 2 x+ 2 2 x−1 ∴ = 2 x +1 x ( 2 x +2 ) +2 x+ 2−x −2 x 2 x+ 2 ∴

x 2+ 2 x 2 x+ 2 2 x−1 × = 2 x +2 2 x2 +2 x+ 2 x +2−x 2−2 x x +1

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∴

x 2 +2 x 2 x−1 = 2 x + 2 x +2 x +1

by cross multiplying the two equations leads to: 3

2

2

3

2

2

∴ x + x + 2 x +2 x=2 x +4 x + 4 x−x −2 x −2 3

∴ x =2 This proving the result that as

AC x = =x , CB 1

3 x=√ 2 .

What gives origami the power to solve problems that Euclidean geometry cannot is the fact that the problems discussed above are only solvable using a cubic equation, which simple plane geometry cannot usually solve. To demonstrate why doubling the cube problem is a cubic equation we first look at the side length s 2 so that: s 32=2∗V 1 where V1 is the volume of the initial cube, and since s 31=V 1 , where s 1 is the side length of the original cube. Therefore, 3

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s 2=2 s 1 ∴ s2 =√3 2 s1 And this is precisely the number that was found in the construction above (method and proof of doubling a cube). Conclusion: Origami brings 2D geometry to another level. It is used to construct many wonderful objects, including polyhedra, tessellations, air bags and many more. On top of the fact that origami opens a vast array of application in real life in combination with its constructional properties, origami could be explored and studied even more in the mathematical field. Some mathematicians also attempt to use origami to prove other geometric theorems, including construction of higher degree polynomials or topology. Since origami is a fairly new topic, people have yet to find its limit and potential in constructions even though they already have a pretty large array of applications. “For example, Humiaki Huzita, a Japanese-Italian mathematician and origami artist, discovered six of the seven axioms of origami in 1992, but ten years later, Hatori summarized another axiom. It is possible that the seven axioms mentioned previously are not complete.” (Yin 17) New properties could always be discovered, in parallel with the constant problem of knowledge being ever expanding, even in mathematics. Computational origami has allowed for the creation of diverse applications of knowledge in origami. These discoveries in the constructional properties of origami took nearly 2000 years to be discovered and applied to realize how much is possible using square paper the side of which as a straight edge and the angle when folding the corner as a compass. The reason for the impossibility of solving the antiquity problems is that straight edge and compass method can only solve quadratic equations. Therefore, we cannot trisect an angle or double the volume of a cube, as the 11

ancient problems suggest since the solution of those require solving a cubic equation, unattainable using the traditional Euclidean method. On the other hand, the cubic equations from these problems can be solved using origami as demonstrated above. By the use of the last axiom, it is possible to trisect an angle and double a cube’s volume. As a result, we can claim that origami construction is more powerful although more complex in comparison to straight-edge compass constructions.

Works Cited

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Gaur, Shiv. "Origami and Mathematics (And My Experience with Class Nine)." Journal of the Krishnamurti School. N.p., n.d. Web. 19 Feb. 2016. . Lang, Robert J. "Angle Quintisection." Lang Origami. N.p., n.d. Web. 18 Feb. 2016. . - - -. "Origami and Geometric Constructions." 1996. File last modified on 2015. PDF file. Newton, Liz. "Doubling the Cube Using Origami — Proof." Plus Magazine. N.p., 1 Dec. 2009. Web. 18 Feb. 2016. . - - -. "Trisecting the Angle Using Origami — Proof." Plus Magazine. N.p., 1 Dec. 2009. Web. 18 Feb. 2016. . Yin, Sheri. "The Mathematics of Origami." 3 June 2009. PDF file.

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