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Mathematics and Origami: The Ancient Arts Unite

Jaema L. Krier

Spring 2007

Abstract

Mathematics and origami are both considered to be ancient arts, but until the 1960s the two were

considered to be as different as night and day. Turns out they are not as different as everyone thought.

In fact, origami can be used to explain many mathematical concepts in fields such as geometry, calculus,

abstract algebra and others. The relationship between mathematics and origami has yet to be fully

explained. New folds and models are being developed by artists, architects, mathematicians, and other

origami enthusiasts. A set of Postulates, similar to those of Euclidean geometry, have been established

and some origami folds can even solve quadratic and cubic equations. An exploration of various techniques

and models can give a true understanding of how important origami is to the field of mathematics.

1 Introduction

When the word origami is mentioned, most people probably think of little paper cranes or maybe evenpaper airplanes. Although origami is commonly referred to as the art of paper folding, the study of origamireveals that there are many mathematical characteristics of it as well. Origami can be utilized in the studyof geometry, calculus, and even abstract algebra. For students, origami could be the tangible key to theirmathematical comprehension.

Origami originated in China where it was known as Zhe Zhi. It later became popular in Japan, and isnow considered a Japanese art. Although any kind of paper can be used to construct origami models, themost common is known as Kami, Japanese for paper. Just slightly thicker than tissue paper, Kami is usuallycolored on one side and white on the other. In Japan, the most widely used paper is called Washi, which ismade of wood pulp [16].

There are various styles of origami such as traditional, rigid, and modular, to name a few. Traditional

origamikeeps with the belief that models should be made using a single sheet of square paper which cannotbe cut or secured (glued) in any way. Paper cranes and planes would fall into this category. Rigid origamiexplores the idea of folding a single sheet of paper in such a way that it collapses easily without bendingthe regions between its creases. In other words, it can be folded flat with a rigid motion. The solar panelarrays used for space satellites were designed using the rigid map fold invented by Koryo Miura, a Japaneseastrophysicists [16].

Unfortunately, utilizing a single sheet of paper has its limitations, thus modular origami was invented.Modular origamimodels are made with more than one sheet of paper and are formed by constructing unitswhich then lock together to form a larger model. One defining characteristic of modular origami states thatthe larger model must be made up of identical units. Although models made with different units can beconstructed and are often referred to as modular, they do not truly meet the requirements of the modulardefinition [15].

The first known example of modular origami can be dated back to 1734. It was a cube called the magic

treasure chest. However, the traditionalKusudama, paper flowers strung together into a sphere, is consideredto be the precursor to modern day modular origami. Modular origami did not truly become popular until the1960s [15]. Since then, mathematicians have discovered its uses in explaining a vast number of mathematicalmodels. By inventing new folds and models they are continually contributing valuable information to thefield of mathematics.

2 Relating Origami to Mathematics

Understandably, it would not be odd to compare origami with geometry. Origami can, after all, be used toconstruct various geometric shapes. It may come as a surprise, however, to learn that origami has its own set

1

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of postulates much like geometry. The process of paper folding can be reduced to seven simple postulates.The first six were developed by a nuclear scientist by the name of Humiaki Huzita, and they are consideredto be the most powerful known to date. The seventh was later discovered by Jacques Justin which has sincebeen confirmed as accurate [3], [4], [9], [11].

POSTULATE 1 Given two pointsP1 andP2, one can fold a single crease which passes through them.

It is evident this parallels Euclids first postulate, through any two distinct points, it is possibleto draw (exactly) one straight line, which can be accomplished easily with a straightedge.

POSTULATE 2 Given two pointsP1 andP2, one can fold a crease placingP1 onto P2.

Straightedge and compass construction states we are able to construct a bisector of a linesegment. In general, this postulate does just that. By placing P1 ontoP2 we are simply locatingthe midpoint of the line segmentP1P2and then folding a crease at that point which then becomesthebisector.

POSTULATE 3 Given two linesL1 andL2, one can fold a crease placingL1 onto L2.

WhenL1 is not parallel to L2, this origami move is equivalent to the bisecting of an angle usingstraightedge and compass. The resulting crease will go through the intersection of linesL1 andL2, resulting in the bisection of their vertical angles.

POSTULATE 4 Given a point P1 and a line L1, one can fold a crease which will be to L1 and passthroughP1.Obviously, this emulates the Euclidean construction allowing us to drop a line from a givenpoint to a given line.

POSTULATE 5 Given two pointsP1 andP2, and a lineL1, one can fold a crease that placesP1 onto L1and passes throughP2

This fold can be difficult to accomplish. It may be necessary to fold P1 ontoL1 at a point whichis off of the sheet of paper in order for the crease to pass through P2.

QUESTION 1 What exactly is this postulate accomplishing?

The answer may astound you. It is actually solving a quadratic equation. The crease which isconstructed by this postulate is actually tangent to the parabola with focus P1 and directrix L1[4]. Let us prove it.

Proof 1 LetL1 be the line forming the bottom edge of our paper. LetP1 be a point toward the middle,

fairly close to L1, andP2 be a point on the left or right edge of our paper.

Perform Postulate 5, as shown below, leaving the paper folded in this position, call the creasedlineL2.

Figure 1: Perform Postulate 5.

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From the pointP1 construct a line which is to the folded portion ofL1 using Postulate 4.LetXbe the point where this line intersectsL2.

Figure 2: Completion of Postulate 4.

By opening our paper, we observe that the line segmentX P1 and the line segment fromX toL1,call itXA, are equal. You may want to fold and unfoldL2 to convince yourself of this.

Figure 3: The segments from X to P1 and L1 are equidistant.

Therefore,Xis the point onL2 which is equidistance to bothP1 andL1. By definition, this pointis on the parabola with focusP1 and directrixL1.

By Postulate 3, L2 is also the bisector ofAXP1. Therefore, any point onL2 is equidistantto A andP1

Choose a point, Y, on L2 between P2 and X. Using Postulate 4, construct the line to L1passing through Y, call it Y B. Note, thatY BA is right, thus Y B < Y A = Y P1. Since allpoints of the parabola must be equidistant to both the directrix, L1, and the focus, P1, we knowthe parabola lies aboveL2 at this point.

Figure 4: The point Ylies below the parabola with focus P1 and directrix L1.

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Similarly, we can show this is true for any point onL2 fromX to L1.

Thus, L2 is the tangent line to the parabola.

(Proof and graphics adapted from [4])

In order to see this more clearly, choose numerous points along the left and right edges of yourpaper to representP2. For each of these points, complete Postulate 5 to reveal the outline of theparabola.

Figure 5: Tangent lines to the parabola with focus P1 and directrix L1.

Construction with strictly straightedge and compass limits our ability to trisect an angle or double thevolume of a cube (constructing a line with length 3

2) [3], [4], [9], [12]. These feats cannot be accomplished

by simple Euclidean construction. Such problems can, however, be addressed with the application of origamiand we will discuss them in Section 3. The following postulate is the key to unlocking this ancient problem.

POSTULATE 6 Given two pointsP1 andP2 and two linesL1 andL2, one can fold a crease that placesP1 onto L1 andP2 onto L2.

QUESTION 2 How does this postulate help us in solving cubic equations?

In general, the crease created by this fold is the line which is tangent to two separate parabolas,one with focus P1 and directrix L1, the other with focus P2 and directrix L2 [4]. Lets see howthe slope of this tangent line helps us solve the cubic equation x3 + ax2 + bx + c= 0 [3].

Proof 2 LetP1 be the point(a, 1) andP2 be(c, b).

LetL1 be defined byy = 1 andL2 byx= c.Perform Postulate 6, placingP1 andP2 onL1 andL2 respectively, call this creaseL3. SinceL3is not parallel to the axis, let it be defined by

y= tx + u

By proof 1, L3 is tangent to the parabola with focusP1 and directrixL1.

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Figure 6: Slope of the tangent line to1.

This parabola,1, is defined by4y= (xa)2, since its focus is(a, 1)and the directrix isy = 1.Let(x1, y1) be a point ofL3. Thus, L3 is tangent to 1 at

4y1= (x1 a)2 [A]

Taking the derivative gives usy1= 12(x1 a), so the slope, t, of the line is

t=1

2(x1 a) [B]

Using the point slope formula we write the equation of the line as y y1 = 12 (x1 a)(x x1)which we will rewrite as

y=1

2[(x1 a)x (x1 a)x1] + y1

Notice that this is equivalent to y= tx 12(x1 a)x1+ y1 and results in

u= 12

(x1 a)x1+ y1

By substituting in for y1 from [A], and t from [B], we see u =tx1+ 14 (x1 a)2. Notice thiscan be reduced further to u= tx1+ t2. We insertx1 from our slope formula, [B], above to getu= t(2t + a) + t2, and finally

u= t2 at

Similarly, the equation for the parabola, 2, with focus(c, b) and directrixx = c can be writtenas4cx= (y b)2 which intersects the parabola at

4cx2= (y2 b)2

By implicit differentiation we obtainy2= 2cy2b

. Therefore, the slope of our line is

t= 2c

y2 b

The point slope formula gives us the equation of the line asy y2 = 2cy2b (x x2). This can berewritten to show

y= 2c

y2 b x 2cx2y2 b+ y2

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Thus, we can say

u= y2 2cx2y2 b

After substitutions ofx2 andt, this reduces to

u= 2c

t + b c

t =b +

c

t

Therefore, u= t2 at ANDu= b + ct

, thereforet2 at= b + ct

. Whenc = 0,

t3 + at2 + bt + c= 0

However, whenc= 0, this means thatP2 is onL2 so eithert= 0 oru= b. In this case, we see

t2 + at + b= 0

(Proof based on [3])

Since the trisecting of angles and doubling of cubes reduces to the solving of a cubic equation,one can begin to realize how this postulate can be utilized for such problems.

POSTULATE 7 Given a pointPand two linesL1 andL2, one can fold a crease placingP onto L1 whichis to L2 [3].

3 Useful Origami Techniques

There are several origami folds which prove to be very useful both for construction and for the solving ofvarious mathematical problems. One of the most common techniques is the folding of a length intonths. Thisis necessary for many origami models and should be considered a vital operation in origami construction.Dividing paper in 1

2, 14

, 18

,... 12n

is not considered difficult. It may not be obvious, however, how one woulddivide a paper into 1

3, 15

, 17

,... 12n+1

. Lets start with learning the case for 13

and then generalize it for 12n+1

.

3.1 Dividing a Length Into Thirds

1. Start with a square piece of paper.

2. Fold paper in half. Open.

3. Fold the paper on the diagonal. Open.

4. Now fold one half of the paper on the diagonal such that it intersects the main diagonal.Open.

Your paper should resemble Figure 7.

Figure 7: After completion of Step 4.

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QUESTION 3 What has this series of folds done for us?

The point where the two diagonals intersect, call it P, is exactly 23

from the left of our paper.Thus, the remainder is equal to 13 . The series of folds has allowed us to identify this point [7].

Proof 3 LetP = (x, x) and the squares sides equal one unit.

Figure 8: Dividing a paper into 3rds.

This gives usAB= 1, F D= x, DP =x, andDC= 1 x.

NoteABC is similar toP DC. So,|AB||P D| =

|CB ||CD |

= 1x

=12

1 x= x= 2 2x

= 3x= 2 x= 2/3

(Proof and graphics adapted from [7])

3.2 Dividing a Length Into nths

Proving this method in general is very similar to the previous proof. Since we are wanting to divide ourpaper into segments equal to 1

nthe original length, we must first fold our paper by 1

n1 its length [7]. SeeFigure 9.

Proof 4 LetP = (x, x) and the squares sides equal one unit.

Figure 9: Dividing a paper into nths. In this casen = 5.

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As shown in proof 3 and by use of induction we can see,

|AB||P D| =

|CB ||CD |

= 1x

=1

n1

1 x= x= (n 1)(1 x)

= x= n nx 1 + x= nx= n 1 x=

n 1n

(Proof and graphics adapted from [7])

This technique will allow us to expand our construction abilities and help to solve complicated tasks suchas the doubling of a cube.

3.3 Hagas Theorem

Hagas Theorem was named for Kazuo Haga, a retired professor of biology from the University of Tsukuba,Japan [7]. His theorem allows for the proof of more complicated origami constructions.

THEOREM 1 By choosing a point, P, on top of a square and setting the bottom corner onto P, we canobserve three similar triangles, A, B, andC.

Figure 10: Three similar triangles formed by placing a corner onto a point, P.

Proof 5 Observe from Figure 10 that2 + 3 = 90

1 + 2 = 90

3 + 4 = 90

This results in1 = 3

2 = 4

By definition of similarity we haveA B

Similarly, one can showA B C

(Theorem, proof, and graphics adapted from [7])

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3.4 Doubling the Cube

As stated previously, the doubling of a cube cannot be accomplished with straightedge and compass alone.The problem began in Greece when Eratosthenes wrote of the gods demanding an alter twice the size of theexisting one in order to rid the people of a plague. Plato exclaimed that the gods only wanted to shamethe Greeks for their neglect of mathematics and their contempt of geometry. Although the Greeks wereunable to accomplish this task through straightedge and compass construction, other methods of doublingthe cube were developed [12].

By utilizing the 6th postulate, however, we are able to complete the task of doubling the cube with a few

folds of a paper.

3.4.1 Peter Messers Solution

1. Begin with a square sheet of paper.

2. Fold the paper into 3rds from top to bottom. Open.

3. Let one of the creases be L1 and letP1 be the point where the other crease meets the papersedge.

4. Let L2 be the edge of the paper opposite P1, whileP2 is the corner closest to P1. See Figure11 for proper labeling.

5. Perform Postulate 6 without re-opening the paper.

Figure 11: Steps 1 - 4.

QUESTION 4 What has this fold done?

The point where P2 falls on L2 is equal to XY

= 3

2 where X is the length ofL2 from P2 to thecorner, which includes the intersection with L1 and Yis the remaining length ofL2 [7].

Figure 12: XY

= 3

2.

Proof 6 LetY = 1. This results in the sides of the square, s, being equal to X+ 1.

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LetA be the point whereL1 intersectsX, B be the lower left corner, andC be the point wherethe crease meets the bottom edge.

Figure 13: Solving for X, using Hagas Theorem.

LetBC=d. Since the bottom edge equalsX+ 1, this results inP2C=X+ 1 d. Rewritingdvia the Pythagorean Theorem we get,

d=(X2 + 2X)

(2X+ 2)

Also, notice thatP1P2= 13

s, which in terms of X is (X+1)3

.

We can also derive the value ofAP2 by takingXand subtracting 13s, giving us a value of

(2X1)3

.

Now, by Hagas Theorem, we know thatP2AP1 is similar toCB P2. Therefore we can say,d

X+ 1 d = 2X 1

X+ 1

=

X2 + 2X

X2

+ 2X+ 2

= 2X 1

X+ 1

= X3 + 3X2 + 2X= 2X3 + 3X2 + 2X 2

= X3 = 2

= X= 3

2

(Theorem, proof, and graphics adapted from [7])

This is not the only method for doubling a cube. Koshiro Hatori has another method that proves to bequite eloquent.

3.4.2 Koshiro Hatoris Rendition

1. Begin with a square sheet of paper and let the sides represent 8 units.

2. Divide the paper into 4ths on the x-axis and 8ths on the y-axis. Open.

3. Label your orgin as shown in Figure 14.

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Figure 14: Where to label the orgin.

4. LetP1= (0, 1) and P2= (2, 0).5. LetL1 be the line y= 1 and L2 be x = 2.6. Perform Postulate 6. Open. See Figure 15.

Figure 15: Steps 1 - 6 of Hatoris method.

QUESTION 5 How does this method solve the dilemma of doubling a cube?

Recallproof 2. There we showed that Postulate 6 helps us solve cubic equations. The steps listedabove solved the equation x3 2 = 0 where a and b were both equal to zero [3]. Therefore, theslope of the crease is our solution. Notice that the rise, Y, appears to be equal to 5 and the run,X, is approximately 4. Therefore,

Y

X 5

4 1.25 3

2

You can see that you would derive the same result by simply setting the rise, Y, to approximately 58

and therun, X, approximately 1

2. However, the former set-up allows for easier reference to proof 2.

3.5 Trisecting the Angle

Euclidean constructions enable us to easily bisect an angle, but the trisectingof an angle has proved to bean ancient problem. It simply cannot be done with straightedge and compass alone. By utilizing origami,however, we can trisect the angle.

3.5.1 Abes Method

1. Begin with a square piece of paper.

2. Fold a crease, call it L1, which passes through the bottom, left corner, P2. This will form anangle,A, between the bottom edge of your paper and L1. Open.

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3. Fold the paper into 4ths from top to bottom. Open.

4. LetL2 be the crease 14

from the bottom edge of the paper.

5. Label P1 as the point where the left edge and the 12

way mark intersect.

Figure 16: Set-up for trisecting the angle, A.

6. Perform Postulate 6. Do not re-open.

7. Extend L2 by folding a new crease, L3. See Figure 17. Open.

8. ContinueL3 so that it goes through the corner, P2.

Figure 17: Extend L2 to form L3.

QUESTION 6 IsL3 the line which trisectsA?

This answer is yes! The angle formed by L3 and the bottom edge of our paper is equal to 23

that ofA [4]. We can use Postulate 3 to fold the bottom edge of paper onto L3, thus forminganother crease,L4, which gives us all three angles. Lets prove that these three angles are in factequal to one another, thus each equal to 1

3A.

Proof 7 Beginning with Figure 17, letA be the point whereP1 lies onL1, B be the point whereL2

intersectsL3, andCbe the point whereP2 lies onL2.

Using Postulate 1, fold the crease which goes throughA, B, andC. Notice that this is the lineformed by the edge of your paper in Figure 17, therefore giving usP2BA = P2BC= 90

. Thiscan be seen in Figure 18.

Perform Postulate 4, constructing a line which is to the bottom edge of your paper, at sayD,that goes throughC.

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Figure 18: Angle trisection proof.

Now observe,AB= BC=C D

This is true, since each of them are 14

the width of the paper.

SinceP2BA = P2BCandP2B = P2B we have thatP2BA = P2BCby SAS. Therefore,

1 = 2

We can use the same logic to showP2BC= P2DC. This results in

2 = 3

= 1 = 2 = 3

(Theorem, proof, and graphics adapted from [4])

3.6 Constructing a 60 Angle

It seems intuitive to say, that in the process of constructing various models, the need will arise to constructan angle of a given degree. The formation of a 60 angle is one which will prove handy in the next section.Lets learn Francis Ows 60 Unit which was published in the December 1986 issue of British OrigamiMagazine (No. 121 p 32) [5].

1. Fold your paper in 4ths. Open.

2. Fold inward on the 14 and 34 creases so that the edges of your paper touch the

12 line.

Figure 19: Step 2.

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3. Using only the top layer of the right hand side, fold the paper in half again, call this creaseL1. You will only need to do this for the top quarter of your paper. Unfold this step to returnto the previous step.

4. Using Postulate 5, we place the upper left corner,P1, onto L1 so that it passes through P2,whereP2 is the point at the top of the

12

line. Call this L2.

Figure 20: Steps 3 and 4.

5. Fold the upper right corner onto L2 so that the crease passes through P2.

Figure 21: Step 5.

The resulting angle equals 60.

Figure 22: Francis Ows 60 Unit.

(Theorem and graphics adapted from [5])

Proof 8 Recall that we folded the paper into 4ths.

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Figure 23: Ows 60 Unit proof.

Therefore,

AB= AC= 1

4 =DE= EF

However, B andClie midway between the latter two segments, by construction. Thus,

BM= 1

8 =M C

= BC= 14

This results in an equilateralABC.

BAC= 60

4 Various Modular Models & Their Characteristics

Origami can be used to construct various mathematical models from 2-space, 3-space, and even fractionalspace. In 2-space, obviously origami can construct various polygons such as squares, rectangles, triangles,pentagons, hexagons, decagons, and dodecagons to name a few. In 3-space, it is p ossible to constructregular polyhedra models such as tetrahedrons, cubes, octahedrons, and dodecahedrons, as well as semi-regular polyhedra like the truncated tetrahedrons, truncated octahedrons, lesser rhombicosidodecahedrons,and others. It is even possible to construct models which represent fractional dimensions, obviously, referringto fractals. Here we will explore a few of the more popular models.

4.1 Buckyballs

Buckyballs are named after the American artist and architect, Richard Buckminster Fuller. Born in 1895,Fuller was considered a visionary who excelled in architectural design and inventions. He designed thegeodesic dome structure, which is known for its self supporting nature. It is the only man-made structurewhich gets proportionally stronger as it grows in size. Buckminsters contribution revolutionized the field ofengineering. The first geodesic structure was constructed in 1949 [14].

DEFINITION 1 A Buckyball is a polyhedron which has the following two properties:

i.) Every face consists of either a pentagon or a hexagon.

ii.) Every vertex is of degree 3.

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It can be proven, that EVERY buckyball contains 12 pentagons no matter how large or small ofa model we construct. However, we can construct a model with numerous hexagons.

If those 12 pentagons are evenly distributed across the surface, it is then called a SphericalBuckyball [7].

These models are constructed utilizing a unit designed by Thomas Hull of Merrimack College. He callsthe unit a Pentagon-Hexagon-Zig-Zag, or PHiZZ unit. The smallest structure which can be constructed withthis unit is a dodecahedron made with 30 PHiZZ units. However, by utilizing 90 PHiZZ units we see the

formation of a truncated icosahedron, also known as a Buckminster fullerene or a C60 molecule [7].

Figure 24: Example of a Buckyball constructed with 30 PHiZZ units.

4.2 Sonobe Units

The Sonobe Unit was created by Mitsubobu Sonobe and Kunihiko Kasahara. The unit itself consists of twotabs and two pockets. This unit is interesting in that it allows for us to cap any polyhedron with triangularfaces. They are capped with a pyramid constructed from three interlocking Sonobe units [6].

Each unit contributes to the construction of two pyramids. Therefore, for every three units used, twocomplete pyramids can be formed. So, a model made with 12 units, constructs 8 pyramids which cap aoctahedron.

QUESTION 7 What does a model constructed with 30 units cap?

Since every three units construct two pyramids, we first divide 30 by 3 to get 10 and thenmultiply this by 2 to get 20. Therefore, the model constructed by 30 Sonobe units is a cappedicosahedron.

Figure 25: Capped icosahedron (left) and a capped octahedron (right).

QUESTION 8 What does a model made with 6 units look like?

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Using the same method as above, we take 6 and divide it by 3 to get 2. Multiplying by 2 gives us4, which is a tetrahedron. Notice, however, that a capped tetrahedron results in the formationof a Cube .

Figure 26: A capped tetrahedron, aka a cube.

4.3 Fractals

Fractals are complex in nature and can be difficult to model. By utilizing modular origami in the studyof fractals, mathematicians are able to create 2 and 3-Dimensional representations of various fractals. TheMenger Sponge, for example, can be easily represented by constructing cubes and interlocking them to formeach iteration of the model. One can see that it would be difficult (but not impossible), not to mentiontime consuming, to construct anything larger than a level three Menger Sponge. By using business cards, amoderate size level one model is constructed measuring approximately 6 1

4inches cubed. Such a model takes

192 business cards.

Figure 27: Level one Menger Sponge constructed with 192 business cards.

It is also possible to construct a Sierpinski tetrahedron. An example of a 2-Dimensional representationof a fractal model would be the Koch Snowflake or the Sierpinski triangle [10]. Modular origami allows usto represent numerous other fractal models which may be difficult or even impossible in other mediums.

4.4 Butterfly Bombs

Building a bomb is probably not your idea of fun; at least I hope not. Building Butterfly Bombs on theother hand can be fun and is not dangerous OR illegal! A butterfly bomb is a cuboctahedron (6 square facesand 8 triangular faces) constructed in such a way that the units do not lock. In other words, the model isunstable . Most modular origami models are constructed with the use of a locking mechanism; holding theunits in place. The Butterfly Bombs triangular faces are actually triangular cavities [7].

The unstable nature of the Butterfly Bomb makes it a fun puzzle to put together and even more fun totake apart. Construction of the model can prove to be difficult, especially if you have small or uncoordinated

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hands. Each piece of the model must be gently weaved into place. The last piece must be in place, beforethe model will stay together. Construction of the model is only half the fun. By throwing the model intothe air and hitting it with your palm, a beautiful explosion occurs, similar to that of your local fireworkshow.

Figure 28: Small and large versions of the Butterfly Bomb.

It is also possible to construct a bomb from the Butterfly Bombs Dual. The capped octahedron is

similar to that constructed by the Sonobe unit in Section 4.2. This model is constructed in the same weavingfashion as the Butterfly Bomb giving it the same unstable quality. Another interesting quality possessed bythis model is its ability to tessellate 3-dimensional space. The pyramid of the Dual fits perfectly into thetriangular cavity of the Butterfly Bomb [7].

Figure 29: A Butterfly Bomb (bottom) with a Butterfly Bomb Dual (top).

4.5 Five Intersecting Tetrahedrons (FIT)

One of the most fascinating models is that of the Five Intersecting Tetrahedrons, or FIT. The symmetry iswhat gives this model its complex and unique characteristics.

We begin with a dodecahedron. By connecting four equidistant corners we find the formation of aninscribed tetrahedron. Since the dodecahedron has twenty corners, it is then possible to inscribe a total offive tetrahedrons within it [5].

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Figure 30: Formation of the FIT model. (Graphics used with permission from [5])

In order to represent the intersection of the pyramids, we must construct the framework of each oneand then intertwine them symmetrically. This will require the use of Francis Ows 60 Unit discussed inSection 3.6, with a slight modification.

After completion of the 60 unit using a 1 x 3 inch paper (forming angles on the short end), openthe two flaps. The flap on the left, must be inverted to form a pocket. The right hand flapmust then be folded again, this time to the existing crease. You will then repeat all these stepson the opposite end of your paper, being sure to preserve the right-handedness of the unit [5].

Figure 31: A modified 60 unit. (Adapted from [5])

By connecting three units, we form a 180 vertex of one tetrahedron. Another three units will completethe first tetrahedron. The remaining pyramids must be constructed within the original to achieve the desiredresult.

The finished product is a mind boggling display of beauty.

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Figure 32: A completed FIT model constructed with 2 x 6 inch paper.

5 Conclusion

Mathematics and origami can both be considered beautiful forms of artwork in their own unique way. Morepowerful and mysterious, however, is the beauty hidden within their unique relationship. By exploring theserelationships further, mathematicians stand to learn more about the world we live in, while the artists willcreate new ways to represent lifes beauty. There is much to be learned from mathematical origami and muchmore to be discovered. They may be considered ancient arts, but it has taken modern day mathematicians(and artists of course) to unite them for eternity.

References

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