Mathematics and Origami: The Ancient Arts Unite

Transcription

1 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 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 even paper airplanes. Although origami is commonly referred to as the art of paper folding, the study of origami reveals that there are many mathematical characteristics of it as well. Origami can be utilized in the study of geometry, calculus, and even abstract algebra. For students, origami could be the tangible key to their mathematical comprehension. Origami originated in China where it was known as Zhe Zhi. It later became popular in Japan, and is now considered a Japanese art. Although any kind of paper can be used to construct origami models, the most common is known as Kami, Japanese for paper. Just slightly thicker than tissue paper, Kami is usually colored on one side and white on the other. In Japan, the most widely used paper is called Washi, which is made of wood pulp [16]. There are various styles of origami such as traditional, rigid, and modular, to name a few. Traditional origami keeps with the belief that models should be made using a single sheet of square paper which cannot be cut or secured (glued) in any way. Paper cranes and planes would fall into this category. Rigid origami explores the idea of folding a single sheet of paper in such a way that it collapses easily without bending the regions between its creases. In other words, it can be folded flat with a rigid motion. The solar panel arrays used for space satellites were designed using the rigid map fold invented by Koryo Miura, a Japanese astrophysicists [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 units which then lock together to form a larger model. One defining characteristic of modular origami states that the larger model must be made up of identical units. Although models made with different units can be constructed and are often referred to as modular, they do not truly meet the requirements of the modular definition [15]. The first known example of modular origami can be dated back to It was a cube called the magic treasure chest. However, the traditional Kusudama, paper flowers strung together into a sphere, is considered to be the precursor to modern day modular origami. Modular origami did not truly become popular until the 1960 s [15]. Since then, mathematicians have discovered its uses in explaining a vast number of mathematical models. By inventing new folds and models they are continually contributing valuable information to the field of mathematics. 2 Relating Origami to Mathematics Understandably, it would not be odd to compare origami with geometry. Origami can, after all, be used to construct various geometric shapes. It may come as a surprise, however, to learn that origami has its own set 1

2 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 considered to be the most powerful known to date. The seventh was later discovered by Jacques Justin which has since been confirmed as accurate [3], [4], [9], [11]. POSTULATE 1 Given two points P 1 and P 2, 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 possible to draw (exactly) one straight line, which can be accomplished easily with a straightedge. POSTULATE 2 Given two points P 1 and P 2, one can fold a crease placing P 1 onto P 2. Straightedge and compass construction states we are able to construct a bisector of a line segment. In general, this postulate does just that. By placing P 1 onto P 2 we are simply locating the midpoint of the line segment P 1 P 2 and then folding a crease at that point which then becomes the bisector. POSTULATE 3 Given two lines L 1 and L 2, one can fold a crease placing L 1 onto L 2. When L 1 is not parallel to L 2, this origami move is equivalent to the bisecting of an angle using straightedge and compass. The resulting crease will go through the intersection of lines L 1 and L 2, resulting in the bisection of their vertical angles. POSTULATE 4 Given a point P 1 and a line L 1, one can fold a crease which will be to L 1 and pass through P 1. Obviously, this emulates the Euclidean construction allowing us to drop a line from a given point to a given line. POSTULATE 5 Given two points P 1 and P 2, and a line L 1, one can fold a crease that places P 1 onto L 1 and passes through P 2 This fold can be difficult to accomplish. It may be necessary to fold P 1 onto L 1 at a point which is off of the sheet of paper in order for the crease to pass through P 2. QUESTION 1 What exactly is this postulate accomplishing? The answer may astound you. It is actually solving a quadratic equation. The crease which is constructed by this postulate is actually tangent to the parabola with focus P 1 and directrix L 1 [4]. Let us prove it. Proof 1 Let L 1 be the line forming the bottom edge of our paper. Let P 1 be a point toward the middle, fairly close to L 1, and P 2 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 creased line L 2. Figure 1: Perform Postulate 5.

3 From the point P 1 construct a line which is to the folded portion of L 1 using Postulate 4. Let X be the point where this line intersects L 2. Figure 2: Completion of Postulate 4. By opening our paper, we observe that the line segment XP 1 and the line segment from X to L 1, call it XA, are equal. You may want to fold and unfold L 2 to convince yourself of this. Figure 3: The segments from X to P 1 and L 1 are equidistant. Therefore, X is the point on L 2 which is equidistance to both P 1 and L 1. By definition, this point is on the parabola with focus P 1 and directrix L 1. By Postulate 3, L 2 is also the bisector of AXP 1. Therefore, any point on L 2 is equidistant to A and P 1 Choose a point, Y, on L 2 between P 2 and X. Using Postulate 4, construct the line to L 1 passing through Y, call it Y B. Note, that Y BA is right, thus Y B < Y A = Y P 1. Since all points of the parabola must be equidistant to both the directrix, L 1, and the focus, P 1, we know the parabola lies above L 2 at this point. Figure 4: The point Y lies below the parabola with focus P 1 and directrix L 1.

4 Similarly, we can show this is true for any point on L 2 from X to L 1. Thus, L 2 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 your paper to represent P 2. For each of these points, complete Postulate 5 to reveal the outline of the parabola. Figure 5: Tangent lines to the parabola with focus P 1 and directrix L 1. Construction with strictly straightedge and compass limits our ability to trisect an angle or double the volume 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 origami and we will discuss them in Section 3. The following postulate is the key to unlocking this ancient problem. POSTULATE 6 Given two points P 1 and P 2 and two lines L 1 and L 2, one can fold a crease that places P 1 onto L 1 and P 2 onto L 2. 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 P 1 and directrix L 1, the other with focus P 2 and directrix L 2 [4]. Let s see how the slope of this tangent line helps us solve the cubic equation x 3 + ax 2 + bx + c = 0 [3]. Proof 2 Let P 1 be the point (a, 1) and P 2 be (c, b). Let L 1 be defined by y = 1 and L 2 by x = c. Perform Postulate 6, placing P 1 and P 2 on L 1 and L 2 respectively, call this crease L 3. Since L 3 is not parallel to the axis, let it be defined by y = tx + u By proof 1, L 3 is tangent to the parabola with focus P 1 and directrix L 1.

5 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 (x 1, y 1 ) be a point of L 3. Thus, L 3 is tangent to 1 at 4y 1 = (x 1 a) 2 [A] Taking the derivative gives us y 1 = 1 2 (x 1 a), so the slope, t, of the line is t = 1 2 (x 1 a) [B] Using the point slope formula we write the equation of the line as y y 1 = 1 2 (x 1 a)(x x 1 ) which we will rewrite as y = 1 2 [(x 1 a)x (x 1 a)x 1 ] + y 1 Notice that this is equivalent to y = tx 1 2 (x 1 a)x 1 + y 1 and results in u = 1 2 (x 1 a)x 1 + y 1 By substituting in for y 1 from [A], and t from [B], we see u = tx (x 1 a) 2. Notice this can be reduced further to u = tx 1 + t 2. We insert x 1 from our slope formula, [B], above to get u = t(2t + a) + t 2, and finally u = t 2 at Similarly, the equation for the parabola, 2, with focus (c, b) and directrix x = c can be written as 4cx = (y b) 2 which intersects the parabola at By implicit differentiation we obtain y 2 = 4cx 2 = (y 2 b) 2 2c y 2 b. Therefore, the slope of our line is t = 2c y 2 b The point slope formula gives us the equation of the line as y y 2 = rewritten to show y = 2c y 2 b x 2cx 2 y 2 b + y 2 2c y 2 b (x x 2). This can be

6 Thus, we can say After substitutions of x 2 and t, this reduces to u = y 2 2cx 2 y 2 b u = 2c t + b c t = b + c t Therefore, u = t 2 at AND u = b + c t, therefore t2 at = b + c t. When c 0, t 3 + at 2 + bt + c = 0 However, when c = 0, this means that P 2 is on L 2 so either t = 0 or u = b. In this case, we see t 2 + 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 L 1 and L 2, one can fold a crease placing P onto L 1 which is to L 2 [3]. 3 Useful Origami Techniques There are several origami folds which prove to be very useful both for construction and for the solving of various mathematical problems. One of the most common techniques is the folding of a length into n ths. This is necessary for many origami models and should be considered a vital operation in origami construction. Dividing paper in 1 2, 1 4, 1 8, is not considered difficult. It may not be obvious, however, how one would n divide a paper into 1 3, 1 5, 1 7, n+1. Let s start with learning the case for 1 3 and then generalize it for 1 2n 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.

7 QUESTION 3 What has this series of folds done for us? The point where the two diagonals intersect, call it P, is exactly 2 3 from the left of our paper. Thus, the remainder is equal to 1 3. 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 3 rds. This gives us AB = 1, F D = x, DP = x, and DC = 1 x. Note ABC is similar to P DC. So, 3.2 Dividing a Length Into n ths AB P D = CB CD = 1 x = x = x = 2 2x = 3x = 2 x = 2/3 (Proof and graphics adapted from [7]) Proving this method in general is very similar to the previous proof. Since we are wanting to divide our paper into segments equal to 1 1 n the original length, we must first fold our paper by n 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 n ths. In this case n = 5.

8 As shown in proof 3 and by use of induction we can see, AB P D = CB CD = 1 x = 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 such as 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 can observe 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 This results in By definition of similarity we have Similarly, one can show = = = 90 1 = 3 2 = 4 A B A B C (Theorem, proof, and graphics adapted from [7])

9 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 the existing one in order to rid the people of a plague. Plato exclaimed that the gods only wanted to shame the Greeks for their neglect of mathematics and their contempt of geometry. Although the Greeks were unable to accomplish this task through straightedge and compass construction, other methods of doubling the cube were developed [12]. By utilizing the 6 th postulate, however, we are able to complete the task of doubling the cube with a few folds of a paper Peter Messer s Solution 1. Begin with a square sheet of paper. 2. Fold the paper into 3 rds from top to bottom. Open. 3. Let one of the creases be L 1 and let P 1 be the point where the other crease meets the paper s edge. 4. Let L 2 be the edge of the paper opposite P 1, while P 2 is the corner closest to P 1. See Figure 11 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 P 2 falls on L 2 is equal to X Y = 3 2 where X is the length of L 2 from P 2 to the corner, which includes the intersection with L 1 and Y is the remaining length of L 2 [7]. Figure 12: X Y = 3 2. Proof 6 Let Y = 1. This results in the sides of the square, s, being equal to X + 1.

10 Let A be the point where L 1 intersects X, B be the lower left corner, and C be the point where the 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 P 2 C = X + 1 d. Rewriting d via the Pythagorean Theorem we get, d = (X2 + 2X) (2X + 2) Also, notice that P 1 P 2 = 1 (X+1) 3s, which in terms of X is 3. We can also derive the value of AP 2 by taking X and subtracting 1 (2X 1) 3s, giving us a value of 3. Now, by Haga s Theorem, we know that P 2 AP 1 is similar to CBP 2. Therefore we can say, d X + 1 d = 2X 1 X + 1 = X2 + 2X X 2 + 2X + 2 = 2X 1 X + 1 = X 3 + 3X 2 + 2X = 2X 3 + 3X 2 + 2X 2 = X 3 = 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 be quite eloquent Koshiro Hatori s Rendition 1. Begin with a square sheet of paper and let the sides represent 8 units. 2. Divide the paper into 4 ths on the x-axis and 8 ths on the y-axis. Open. 3. Label your orgin as shown in Figure 14.

11 Figure 14: Where to label the orgin. 4. Let P 1 = (0, 1) and P 2 = ( 2, 0). 5. Let L 1 be the line y = 1 and L 2 be x = 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 listed above solved the equation x 3 2 = 0 where a and b were both equal to zero [3]. Therefore, the slope 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 You can see that you would derive the same result by simply setting the rise, Y, to approximately 5 8 run, X, approximately 1 2. However, the former set-up allows for easier reference to proof 2. and the 3.5 Trisecting the Angle Euclidean constructions enable us to easily bisect an angle, but the trisecting of an angle has proved to be an ancient problem. It simply cannot be done with straightedge and compass alone. By utilizing origami, however, we can trisect the angle Abe s Method 1. Begin with a square piece of paper. 2. Fold a crease, call it L 1, which passes through the bottom, left corner, P 2. This will form an angle, A, between the bottom edge of your paper and L 1. Open.

12 3. Fold the paper into 4 ths from top to bottom. Open. 4. Let L 2 be the crease 1 4 from the bottom edge of the paper. 5. Label P 1 as the point where the left edge and the 1 2 way mark intersect. Figure 16: Set-up for trisecting the angle, A. 6. Perform Postulate 6. Do not re-open. 7. Extend L 2 by folding a new crease, L 3. See Figure 17. Open. 8. Continue L 3 so that it goes through the corner, P 2. QUESTION 6 Is L 3 the line which trisects A? Figure 17: Extend L 2 to form L 3. This answer is yes! The angle formed by L 3 and the bottom edge of our paper is equal to 2 3 that of A [4]. We can use Postulate 3 to fold the bottom edge of paper onto L 3, thus forming another crease, L 4, which gives us all three angles. Let s prove that these three angles are in fact equal to one another, thus each equal to 1 3 A. Proof 7 Beginning with Figure 17, let A be the point where P 1 lies on L 1, B be the point where L 2 intersects L 3, and C be the point where P 2 lies on L 2. Using Postulate 1, fold the crease which goes through A, B, and C. Notice that this is the line formed by the edge of your paper in Figure 17, therefore giving us P 2 BA = P 2 BC = 90. This can 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.

13 Figure 18: Angle trisection proof. Now observe, This is true, since each of them are 1 4 AB = BC = CD the width of the paper. Since P 2 BA = P 2 BC and P 2 B = P 2 B we have that P 2 BA = P 2 BC by SAS. Therefore, 1 = 2 We can use the same logic to show P 2 BC = P 2 DC. This results in 3.6 Constructing a 60 Angle 2 = 3 = 1 = 2 = 3 (Theorem, proof, and graphics adapted from [4]) It seems intuitive to say, that in the process of constructing various models, the need will arise to construct an 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 Origami Magazine (No. 121 p 32) [5]. 1. Fold your paper in 4 ths. Open. 2. Fold inward on the 1 4 and 3 4 creases so that the edges of your paper touch the 1 2 line. Figure 19: Step 2.

14 3. Using only the top layer of the right hand side, fold the paper in half again, call this crease L 1. You will only need to do this for the top quarter of your paper. Unfold this step to return to the previous step. 4. Using Postulate 5, we place the upper left corner, P 1, onto L 1 so that it passes through P 2, where P 2 is the point at the top of the 1 2 line. Call this L 2. Figure 20: Steps 3 and Fold the upper right corner onto L 2 so that the crease passes through P 2. Figure 21: Step 5. The resulting angle equals 60. Proof 8 Recall that we folded the paper into 4 ths. Figure 22: Francis Ow s 60 Unit. (Theorem and graphics adapted from [5])

15 Figure 23: Ow s 60 Unit proof. Therefore, AB = AC = 1 = DE = EF 4 However, B and C lie midway between the latter two segments, by construction. Thus, BM = 1 8 = MC This results in an equilateral ABC. = BC = 1 4 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 construct regular polyhedra models such as tetrahedrons, cubes, octahedrons, and dodecahedrons, as well as semiregular polyhedra like the truncated tetrahedrons, truncated octahedrons, lesser rhombicosidodecahedrons, and others. It is even possible to construct models which represent fractional dimensions, obviously, referring to 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 the geodesic dome structure, which is known for its self supporting nature. It is the only man-made structure which gets proportionally stronger as it grows in size. Buckminster s contribution revolutionized the field of engineering. 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.

16 It can be proven, that EVERY buckyball contains 12 pentagons no matter how large or small of a 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 Spherical Buckyball [7]. These models are constructed utilizing a unit designed by Thomas Hull of Merrimack College. He calls the unit a Pentagon-Hexagon-Zig-Zag, or PHiZZ unit. The smallest structure which can be constructed with this 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]. 4.2 Sonobé Units Figure 24: Example of a Buckyball constructed with 30 PHiZZ units. The Sonobé Unit was created by Mitsubobu Sonobé and Kunihiko Kasahara. The unit itself consists of two tabs and two pockets. This unit is interesting in that it allows for us to cap any polyhedron with triangular faces. They are capped with a pyramid constructed from three interlocking Sonobé units [6]. Each unit contributes to the construction of two pyramids. Therefore, for every three units used, two complete pyramids can be formed. So, a model made with 12 units, constructs 8 pyramids which cap a octahedron. 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 then multiply this by 2 to get 20. Therefore, the model constructed by 30 Sonobé units is a capped icosahedron. Figure 25: Capped icosahedron (left) and a capped octahedron (right). QUESTION 8 What does a model made with 6 units look like?

17 Using the same method as above, we take 6 and divide it by 3 to get 2. Multiplying by 2 gives us 4, which is a tetrahedron. Notice, however, that a capped tetrahedron results in the formation of a Cube. 4.3 Fractals Figure 26: A capped tetrahedron, aka a cube. Fractals are complex in nature and can be difficult to model. By utilizing modular origami in the study of fractals, mathematicians are able to create 2 and 3-Dimensional representations of various fractals. The Menger Sponge, for example, can be easily represented by constructing cubes and interlocking them to form each iteration of the model. One can see that it would be difficult (but not impossible), not to mention time consuming, to construct anything larger than a level three Menger Sponge. By using business cards, a moderate size level one model is constructed measuring approximately inches 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 representation of a fractal model would be the Koch Snowflake or the Sierpinski triangle [10]. Modular origami allows us to 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 the other hand can be fun and is not dangerous OR illegal! A butterfly bomb is a cuboctahedron (6 square faces and 8 triangular faces) constructed in such a way that the units do not lock. In other words, the model is unstable. Most modular origami models are constructed with the use of a locking mechanism; holding the units 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 to take apart. Construction of the model can prove to be difficult, especially if you have small or uncoordinated

18 hands. Each piece of the model must be gently weaved into place. The last piece must be in place, before the model will stay together. Construction of the model is only half the fun. By throwing the model into the air and hitting it with your palm, a beautiful explosion occurs, similar to that of your local firework show. 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 is similar to that constructed by the Sonobé unit in Section 4.2. This model is constructed in the same weaving fashion as the Butterfly Bomb giving it the same unstable quality. Another interesting quality possessed by this model is its ability to tessellate 3-dimensional space. The pyramid of the Dual fits perfectly into the triangular 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 is what gives this model its complex and unique characteristics. We begin with a dodecahedron. By connecting four equidistant corners we find the formation of an inscribed tetrahedron. Since the dodecahedron has twenty corners, it is then possible to inscribe a total of five tetrahedrons within it [5].

19 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 one and then intertwine them symmetrically. This will require the use of Francis Ow s 60 Unit discussed in Section 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), open the two flaps. The flap on the left, must be inverted to form a pocket. The right hand flap must then be folded again, this time to the existing crease. You will then repeat all these steps on 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 complete the first tetrahedron. The remaining pyramids must be constructed within the original to achieve the desired result. The finished product is a mind boggling display of beauty.

20 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. More powerful and mysterious, however, is the beauty hidden within their unique relationship. By exploring these relationships further, mathematicians stand to learn more about the world we live in, while the artists will create new ways to represent life s beauty. There is much to be learned from mathematical origami and much more 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 [1] R. Alperin, A mathematical theory of origami constructions and numbers, New York Journal of Mathematics 6 (2000), pp. [2] K. Burczyk, Polyhedra classification, Krystyna Burczyk s Origami Gallery - Regular Polyhedra, burczyk/index-en.html [3] K. Hatori, Origami construction, K s Origami, [4] T. Hull, A comparison between straight edge and compass constructions and origami, Origami and Geometric Constructions, thull/omfiles/geoconst.html [5], Five Intersecting Tetrahedra, Origami and Geometric Constructions, thll/fit.html [6], Handouts for using origami in undergraduate math classes; MAA minicourse, Joint Meetings, New Orleans, LA, [7], Project Origami; Activities for Exploring Mathematics, A K Peters, Wellesley, [8] T. Ishida, Kusudama, kanzasi/en/e-kusu.html [9] L. Kinsey and T. Moore, Symmetry, Shape, and Space; An Introduction to Mathematics Through Geometry, Key College Publishing, Emeryville, [10] M. Kosmulski, Fractals; IFS, Modular Origami, mikosmul/origami/fractals.html

21 [11] D. Lister, Humiaki Huzita, The History of Origami, huzita.htm [12] J. O Connor and E. Robertson, Doubling the cube, history/histtopics/doubling the cube.html [13] T. Row, Geometric Exercises in Paper Folding, Dover Publications, New York, [14] Wikipedia, Buckminster Fuller, Wikipedia, The Free Encyclopedia, Fuller&oldid= [15], Modular origami, Wikipedia, The Free Encyclopedia, origami&oldid= [16], Origami, Wikipedia, The Free Encyclopedia,