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 1960’s the two wereconsidered 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 fullyexplained. New folds and models are being developed by artists, architects, mathematicians, and otherorigami enthusiasts. A set of Postulates, similar to those of Euclidean geometry, have been establishedand some origami folds can even solve quadratic and cubic equations. An exploration of various techniquesand 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. Traditionalorigami keeps 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 origami models 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 ‘magictreasure chest.’ However, the traditional Kusudama, 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 the1960’s [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

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 points P1 and P2, one can fold a single crease which passes through them.

It is evident this parallels Euclid’s 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 points P1 and P2, one can fold a crease placing P1 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 onto P2 we are simply locatingthe midpoint of the line segment P1P2 and then folding a crease at that point which then becomesthe ⊥ bisector.

POSTULATE 3 Given two lines L1 and L2, one can fold a crease placing L1 onto L2.

When L1 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 lines L1 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 passthrough P1.

Obviously, this emulates the Euclidean construction allowing us to drop a ⊥ line from a givenpoint to a given line.

POSTULATE 5 Given two points P1 and P2, and a line L1, one can fold a crease that places P1 onto L1

and passes through P2

This fold can be difficult to accomplish. It may be necessary to fold P1 onto L1 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 Let L1 be the line forming the bottom edge of our paper. Let P1 be a point toward the middle,

fairly close to L1, and P2 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 creasedline L2.

Figure 1: Perform Postulate 5.

From the point P1 construct a line which is ⊥ to the folded portion of L1 using Postulate 4.

Let X be the point where this line intersects L2.

Figure 2: Completion of Postulate 4.

By opening our paper, we observe that the line segment XP1 and the line segment from X to L1,call it XA, are equal. You may want to fold and unfold L2 to convince yourself of this.

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

Therefore, X is the point on L2 which is equidistance to both P1 and L1. By definition, this pointis on the parabola with focus P1 and directrix L1.

By Postulate 3, L2 is also the ⊥ bisector of ∠AXP1. Therefore, any point on L2 is equidistantto A and P1

Choose a point, Y , on L2 between P2 and X. Using Postulate 4, construct the ⊥ line to L1

passing through Y , call it Y B. Note, that 4Y 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 above L2 at this point.

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

Similarly, we can show this is true for any point on L2 from X 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 represent P2. 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 points P1 and P2 and two lines L1 and L2, one can fold a crease that placesP1 onto L1 and P2 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]. Let’s see howthe slope of this tangent line helps us solve the cubic equation x3 + ax2 + bx + c = 0 [3].

Proof 2 Let P1 be the point (a, 1) and P2 be (c, b).

Let L1 be defined by y = −1 and L2 by x = −c.

Perform Postulate 6, placing P1 and P2 on L1 and L2 respectively, call this crease L3. Since L3

is not parallel to the axis, let it be defined by

y = tx + u

By proof 1, L3 is tangent to the parabola with focus P1 and directrix L1.

Figure 6: Slope of the tangent line to ℘1.

This parabola, ℘1, is defined by 4y = (x−a)2, since it’s focus is (a, 1) and the directrix is y = −1.

Let (x1, y1) be a point of L3. Thus, L3 is tangent to ℘1 at

4y1 = (x1 − a)2 [A]

Taking the derivative gives us y′1 = 12 (x1 − a), so the slope, t, of the line is

t =12(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 asy =

12[(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 this

can be reduced further to u = −tx1 + t2. We insert x1 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 directrix x = −c can be writtenas 4cx = (y − b)2 which intersects the parabola at

4cx2 = (y2 − b)2

By implicit differentiation we obtain y′2 = 2cy2−b . Therefore, the slope of our line is

t =2c

y2 − b

The point slope formula gives us the equation of the line as y − y2 = 2cy2−b (x − x2). This can be

rewritten to showy =

2c

y2 − bx− 2cx2

y2 − b+ y2

Thus, we can say

u = y2 − 2cx2

y2 − b

After substitutions of x2 and t, this reduces to

u =2c

t+ b− c

t= b +

c

t

Therefore, u = −t2 − at AND u = b + ct , therefore −t2 − at = b + c

t . When c 6= 0,

t3 + at2 + bt + c = 0

However, when c = 0, this means that P2 is on L2 so either t = 0 or u = 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 point P and two lines L1 and L2, one can fold a crease placing P 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 into nths. Thisis necessary for many origami models and should be considered a vital operation in origami construction.Dividing paper in 1

2 , 14 , 1

8 ,... 12n is not considered difficult. It may not be obvious, however, how one would

divide a paper into 13 , 1

5 , 17 ,... 1

2n+1 . Let’s start with learning the case for 13 and then generalize it for 1

2n+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.

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 Let P = (x, x) and the square’s sides equal one unit.

Figure 8: Dividing a paper into 3rds.

This gives us AB = 1, FD = x, DP = x, and DC = 1− x.

Note 4ABC is similar to 4PDC. So,

|AB||PD| =

|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

n the original length, we must first fold our paper by 1n−1 its length [7]. See

Figure 9.

Proof 4 Let P = (x, x) and the square’s sides equal one unit.

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

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

|AB||PD| =

|CB||CD|

=⇒ 1x

=1

n−1

1− x

=⇒ x = (n− 1)(1− x)

=⇒ x = n− nx− 1 + x

=⇒ nx = n− 1

∴ x =n− 1

n

¥

(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 Haga’s Theorem

Haga’s 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, and C.

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

Proof 5 Observe from Figure 10 that∠2 + ∠3 = 90◦

∠1 + ∠2 = 90◦

∠3 + ∠4 = 90◦

This results in∠1 = ∠3

∠2 = ∠4

By definition of similarity we haveA ∼ B

Similarly, one can showA ∼ B ∼ C

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

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 fewfolds of a paper.

3.4.1 Peter Messer’s 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 let P1 be the point where the other crease meets the paper’sedge.

4. Let L2 be the edge of the paper opposite P1, while P2 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 of L2 from P2 to the

corner, which includes the intersection with L1 and Y is the remaining length of L2 [7].

Figure 12: XY = 3

√2.

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

Let A be the point where L1 intersects X, B be the lower left corner, and C be the point wherethe crease meets the bottom edge.

Figure 13: Solving for X, using Haga’s Theorem.

Let BC = d. Since the bottom edge equals X + 1, this results in P2C = X + 1− d. Rewriting dvia the Pythagorean Theorem we get,

d =(X2 + 2X)(2X + 2)

Also, notice that P1P2 = 13s, which in terms of X is (X+1)

3 .

We can also derive the value of AP2 by taking X and subtracting 13s, giving us a value of (2X−1)

3 .

Now, by Haga’s Theorem, we know that 4P2AP1 is similar to 4CBP2. Therefore we can say,

d

X + 1− d=

2X − 1X + 1

=⇒ X2 + 2X

X2 + 2X + 2=

2X − 1X + 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 Hatori’s 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.

Figure 14: Where to label the orgin.

4. Let P1 = (0, 1) and P2 = (−2, 0).

5. Let L1 be the line y = −1 and L2 be x = 2.

6. Perform Postulate 6. Open. See Figure 15.

Figure 15: Steps 1 - 6 of Hatori’s method.

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

Recall proof 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 the

run, X, approximately 12 . 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 trisecting of 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 Abe’s 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.

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

4. Let L2 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. Continue L3 so that it goes through the corner, P2.

Figure 17: Extend L2 to form L3.

QUESTION 6 Is L3 the line which trisects ∠A?

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

that of ∠A [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. Let’s prove that these three angles are in factequal to one another, thus each equal to 1

3A.

Proof 7 Beginning with Figure 17, let A be the point where P1 lies on L1, B be the point where L2

intersects L3, and C be the point where P2 lies on L2.

Using Postulate 1, fold the crease which goes through A, B, and C. Notice that this is the lineformed by the edge of your paper in Figure 17, therefore giving us ∠P2BA = ∠P2BC = 90◦. Thiscan be seen in Figure 18.

Perform Postulate 4, constructing a line which is ⊥ to the bottom edge of your paper, at say D,that goes through C.

Figure 18: Angle trisection proof.

Now observe,AB = BC = CD

This is true, since each of them are 14 the width of the paper.

Since ∠P2BA = ∠P2BC and P2B = P2B we have that 4P2BA ∼= 4P2BC by SAS. Therefore,

∠1 = ∠2

We can use the same logic to show 4P2BC ∼= 4P2DC. 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.Let’s learn Francis Ow’s “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 3

4 creases so that the edges of your paper touch the 12 line.

Figure 19: Step 2.

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,where P2 is the point at the top of the 1

2 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 Ow’s “60◦ Unit”.

(Theorem and graphics adapted from [5])

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

Figure 23: Ow’s “60◦ Unit” proof.

Therefore,

AB = AC =14

= DE = EF

However, B and C lie midway between the latter two segments, by construction. Thus,

BM =18

= MC

=⇒ BC =14

This results in an equilateral 4ABC.

∴ ∠BAC = 60◦

¥

4 Various Modular Models & Their Characteristics

Origami can be used to construct various mathematical models from 2-space, 3-space, and even “fractional”space. 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 possible 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. Buckminster’s 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.

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 theformation 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?

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

4 inches cubed. Such a model takes192 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 Bomb’s 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

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 Bomb’s Dual. The capped octahedron issimilar 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].

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 Ow’s “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.

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 life’s 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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