Origami Polyhedra Designdl.booktolearn.com/ebooks2/handicraft/... · Montroll, John. Origami polyhedra design / John Montroll. p. cm. ISBN 978-1-56881-458-2 (alk. paper) 1. Origami

Origami Polyhedra Design

Origami Polyhedra Design

John Montroll

A K Peters, Ltd.Natick, Massachusetts

Editorial, Sales, and Customer Service Offi ce

A K Peters, Ltd.5 Commonwealth Road, Suite 2CNatick, MA 01760www.akpeters.com

Copyright © 2009 by A K Peters, Ltd.

All rights reserved. No part of the material protected by this copyright notice may be reproduced or utilized in any form, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the copyright owner.

Library of Congress Cataloging-in-Publication Data

Montroll, John. Origami polyhedra design / John Montroll. p. cm. ISBN 978-1-56881-458-2 (alk. paper) 1. Origami. 2. Polyhedra in art. 3. Polyhedra--Models. I. Title. TT870.M567 2009 736’.98--dc22

2008053787

Cover photograph: Models folded by John Montroll and John Szinger. Photograph by Gabor Demjen.

Printed in the United States of America

13 12 11 10 09 10 9 8 7 6 5 4 3 2 1

To Himanshu

Contents Symbols xi

Basic Folds xii

Introduction xiii

Part I: Designing Origami Polyhedra 1 Polyhedra Overview 3

Design Factors and Techniques 7

Math and Design 19

Design Method Examples 39

Polygons 45 Equilateral Triangle ✩ 46 Pentagon ✩ 48 Hexagon ✩ 52 Heptagon ✩ 54 Octagon ✩ 56 Silver Rectangle ✩ 57 Bronze Rectangle ✩ 58 Golden Rectangle ✩ 59

Part II: Platonic and Related Polyhedra 61 Tetrahedron Design 63 Tetrahedron ✩ 65 Duo-Colored Tetrahedron ✩✩ 67 Striped Tetrahedron ✩✩ 69 Tetrahedron of Triangles ✩✩ 72 Stellated Tetrahedron ✩✩ 76 Dimpled Truncated Tetrahedron ✩✩✩ 79

Cube Design 83 Cube ✩✩ 87 Striped Cube ✩✩ 89 Triangles on Cube ✩✩ 92 Cube with Squares ✩✩ 95 Stellated Octahedron ✩✩ 98 Cubehemioctahedron ✩✩✩ 100 Dimpled Rhombicuboctahedron ✩✩✩ 104 Stacked Cubes ✩✩✩✩ 108

Simple ✩

Intermediate ✩✩

Complex ✩✩✩

Very Complex ✩✩✩✩

viii Contents

Octahedron Design 113 Octahedron ✩✩ 115 Striped Octahedron ✩✩ 118 Duo-Colored Octahedron ✩✩ 121 Stellated Cube ✩✩ 124 Octahemioctahedron ✩✩✩ 127 Dimpled Truncated Octahedron ✩✩✩ 131

More Platonic Solids Design 135 Icosahedron ✩✩✩ 137 Dodecahedron ✩✩✩✩ 140

Sunken Platonic Solids Design 147 Sunken Octahedron ✩✩ 150 Sunken Tetrahedron ✩✩✩ 153 Sunken Cube ✩✩✩ 157 Sunken Dodecahedron ✩✩✩✩ 161 Sunken Icosahedron ✩✩✩✩ 166

Part III: Dipyramids and Dimpled Dipyramids 171 Dipyramid Design 173 Triangular Dipyramid 90° ✩ 178 Triangular Dipyramid ✩✩ 180 Triangular Dipyramid in a Sphere ✩✩ 183 Tall Triangular Dipyramid ✩✩ 186 Tall Square Dipyramid ✩✩ 189 Silver Square Dipyramid ✩✩ 192 Squat Silver Square Diamond ✩✩ 194 Pentagonal Dipyramid ✩✩ 197 Pentagonal Dipyramid 45° ✩✩ 201 Pentagonal Dipyramid in a Sphere ✩✩ 203 Golden Pentagonal Dipyramid ✩✩ 206 Squat Golden Pentagonal Dipyramid ✩✩ 208 Hexagonal Dipyramid ✩✩ 210 Silver Hexagonal Dipyramid ✩✩ 213 Hexagonal Dipyramid 36° ✩✩ 214 Hexagonal Dipyramid in a Sphere ✩✩ 215 Squat Silver Hexagonal Dipyramid ✩✩ 218 Heptagonal Dipyramid ✩✩✩ 219 Heptagonal Dipyramid 30° ✩✩✩ 223 Heptagonal Dipyramid in a Sphere ✩✩✩ 225 Octagonal Dipyramid ✩✩✩ 229 Octagonal Dipyramid 26° ✩✩✩ 232 Octagonal Dipyramid in a Sphere ✩✩✩ 234 Nonagonal Dipyramid ✩✩✩ 237 Decagonal Dipyramid ✩✩✩ 240

Contents ix

Dimpled Dipyramid Design 243 Tall Dimpled Square Dipyramid ✩✩ 247

Dimpled Silver Sqare Dipyramid ✩✩✩ 250

Heptahedron ✩✩✩ 254

Dimpled Squat Square Dipyramid ✩✩✩ 258

Tall Dimpled Hexagonal Dipyramid ✩✩✩✩ 262

Dimpled Silver Hexagonal Dipyramid ✩✩✩✩ 267

Dimpled Hexagonal Dypyramid ✩✩✩✩ 270

Dimpled Hexagonal Dipyramid in a Sphere ✩✩✩✩ 275

Octagonal Flying Saucer ✩✩ 280

Dimpled Octagonal Dipyramid in a Sphere ✩✩✩ 283

Dimpled Octagonal Dipyramid ✩✩✩ 287

Dimpled Silver Octagonal Dipyramid ✩✩✩ 288

xi

Symbols

Valley fold, fold in front.

Mountain fold, fold behind.

Lines

Crease line.

X-ray or guide line.

Arrows

Fold in this direction.

Fold behind.

Unfold.

Fold and unfold.

Turn over.

Sink or three-dimensional folding.

Place your finger between these layers.

xii Basic Folds

Sink Fold

In a sink fold, some of the paper without edges is folded inside. To do this fold, much of the model mustbe unfolded.

Inside Reverse Fold

In an inside reverse fold, some paper is folded between layers. Here are two examples.

Squash Fold

In a squash fold, some paper is opened and then made flat. The shaded arrow shows where to place yourfinger.

Basic Folds

1 2

Squash-fold. A three-dimensional intermediate step.

3

1 21 2

Reverse-fold.

1 2

Sink.

Reverse-fold.

Introduction xiii

Introduction

P olyhedra are beautiful shapes that have fascinated people throughout the ages. In ancienttimes, they were thought to possess magic powers. The Greeks knew of the five Platonicsolids and associated them with the nature of the universe. Over time, more polyhedrawere discovered, and they became better understood. Physical models of these geometricshapes have been formed in many media, including wood, metal, and paper. In this collection, you will learn to make many beautiful polyhedra, each by folding a single squaresheet of paper. Also, you will see how origami skill, mathematics, and polyhedra can becombined in an elegant way.

I am dedicated to the pursuit of creating each origami model from a single uncut square. I developedmethods for folding animals with that ideal at a time when using multiple sheets and cutting was prevalent.My techniques, discoveries, and ideals influenced many creators, leading to a more developed art. Now Iwish to do the same with polyhedra.

There has been a movement, over the past 15 years or so, of creating geometric shapes with modularorigami—the use of multiple sheets that are folded into identical units (modules) and then interlocked. Forexample, each edge of a polyhedron may be formed by one unit, and several units will interlock to form aface or vertex. I wanted to explore how these same shapes could be created from a single sheet of paper.Designing polyhedra from a single uncut square has led me to many new techniques and discoveries inorigami that I wish to share with you.

My models are designed to be folded from standard origami paper (sometimes called kami). Thispaper comes in squares, typically 6 to 10 inches wide, that are white on one side and colored on the other.In the diagrams in this book, the colored side is represented by gray shading. Most of the models end upwith only the colored side showing (and so single-colored paper could be used), but several designs createcolor patterns by exposing the white side as well as the colored side. Origami can also be folded from avariety of other papers, including standard notebook or printer paper, or from even thicker paper, using thetechnique of wet-folding.

This may be a good place to point out some folding techniques employed in this collection thatmay be less familiar to the casual folder. Most of the instructions begin with sequences of folds that are notfolds in the final model but rather are used to find particular landmarks, such as one seventh of the edgeof the paper or 30° through the center of the paper. Such sequences entail making small creases, such asfolding the paper in half but only creasing on the left edge. In addition, I aimed to keep the faces of the finalmodel with as few creases as possible, which led to some steps that require creasing only along part of afold.

After the crease pattern defining the faces of the polyhedron is formed, the model becomes three-dimensional, typically at a step to either puff out or push in at some central point on the paper. The foldinggenerally ends by tucking or locking one or more tabs. Throughout the folding instructions, symmetry playsa large role, which simplifies the folding.

The models range from simple to very complex, most being intermediate or complex. Still, noneof the models are too difficult, and even the most complex has fewer than 60 steps. The models have beentest-folded by many folders.

In Part I, “Designing Origami Polyhedra,” there are references to several polyhedra whosedirections are not given in this volume. My hope is to write a second volume of origami polyhedra that willinclude them.

The illustrations conform to the internationally accepted Randlett-Yoshizawa conventions. (See “Symbols” (page xi)). Origami paper can be found in many hobby shops or purchased by mail fromOrigamiUSA, 15 West 77th Street, New York, NY 10024-5192.

Many people helped with this project. I thank Brian Webb and John Szinger for their continuedsupport throughout this project. I thank Daniel Spaulding for his contributions. I thank John Szinger for folding some of the polyhedra featured on the cover. In particular, I thank my editors, Jan Polish, Charlotte Henderson, and Ellen Ulyanova. I also thank the many folders who proofread the diagrams.

John Montroll

www.johnmontroll.com

Designing OrigamiPolyhedra

PartI

3

Polyhedra OverviewPolyhedra are three-dimensional shapes composed of polygons, two-dimensional shapes composed ofstraight lines. Here we present some of the names associated with polyhedra.

Concave Polyhedra

If there are any lines connecting two points on the surface that are outside the polyhedron (as in the firstpicture below), even partially, then the polyhedron is concave, or nonconvex.

Sunken Cube(Sunk)

Platonic Solids

A Platonic solid is a convex polyhedron composed of identical regular polygons with identical vertices. Thus,all of the edges are the same length, the same number of edges meet at each vertex, and all the faces arethe same shape. There are five Platonic solids: the tetrahedron, cube, octahedron, icosahedron, anddodecahedron.

Tetrahedron

Cubehemioctahedron(Dimpled)

Stellated Cube(Star)

Convex Polyhedra

Take any two points on the surface of a polyhedron and picture a line connecting them (as in the first picturebelow). A polyhedron is convex if all such lines are either on the surface of the polyhedron or are completelyinside the polyhedron.

Icosahedron

Cube Octahedron Icosahedron Dodecahedron

HexagonalDipyramid

PentagonalPyramid

4 Part I: Designing Origami Polyhedra

Archimedean Solids

The Archimedean polyhedra are convex solids that are made from two or more types of regular polygonsand that have identical vertices. (This is similar to the definition of Platonic solids except all the faces arenot the same shape.) Here are some of the 13 Archimedean solids.

(A truncated polyhedron is created by slicing off the corners, leaving regular polygons in place of theoriginal vertices.)

CuboctahedronTruncated Tetrahedron Snub Cube

Triangular Prism Cube Pentagonal Prism Hexagonal Prism

Antiprisms

The two bases of a prism line up evenly with each other. For an antiprism, the two bases are offset, so thatit is composed of two identical n-gon bases and 2n lateral triangles going around its equator.

Truncated Octahedron Truncated IcosahedronIcosidodecahedron

Tall TriangularAntiprism

Golden PentagonalAntiprism

Tall HexagonalAntiprism

Rhombicuboctahedron

Prisms

A prism is a polyhedron where the cross sections are parallel to its base. It is composed of two n-gons andn identical rectangles.

Tall SquareAntiprism

Polyhedra Overview 5

Pyramids

A pyramid has a polygon base and triangles whose vertices meet at the top, somewhere above the centerof the base.

Triangular Pyramid Hexagonal Pyramid

Duals

The dual of a polyhedron is a polyhedron where the faces and vertices are switched. For many somewhatround-shaped polyhedra, the dual can be obtained by placing a point in the center of each face, connectingthe points to form a new polyhedron, and scaling it so both polyhedron and dual are inscribed in the samesphere.

The cube and octahedron are dual pairs. The icosahedron and dodecahedron are dual pairs.

The uniform pentagonal prism andpentagonal dipyramid are dual pairs.

The square antiprism and squaretrapezohedron are dual pairs.

The tetrahedron is its own dual.

Egyptian Pyramid

In general, the process of constructing the dual is called polar reciprocation. For nonrounded shapes, thedual is found by drawing lines from the center of the polyhedron through the centers of each face. The closerthe face is to the center of the solid, the further the vertex will be in the dual.

For example, consider the uniform pentagonal prism. Two sides are pentagons and five are squares.The dual of this somewhat squat shape is the pentagonal dipyramid, a longer shape.


Catalan Solids

The Catalan polyhedra are the duals of the Archimedean solids. Each one is convex with identical faces,though none of them are regular polygons. The vertices are not identical. Here are some of the 13 Catalansolids.

Triakis Tetrahedron Triakis CubeTriakis Octahedron

Dipyramids

Dipyramids, or diamonds, are the duals of prisms. Dipyramids are composed of a pair of identical pyramidsjoined at the base. All the sides are identical isosceles triangles.

Trapezohedra

Trapezohedra, or antidiamonds, are the duals of antiprisms. The sides are identical kite-shaped quadrilateralswhere three of the angles are the same.

DecagonalDipyramid

Rhombic Dodecahedron

HeptagonalDipyramid

HexagonalDipyramid

Squat SilverSquare Dipyramid

Triangular Trapezohedron(Cube)

SquareTrapezohedron

PentagonalTrapezohedron

OctagonalTrapezohedron


Design Factors and TechniquesDesigning origami polyhedra is an interesting challenge. It combines origami skill with mathematics. Thereare many factors to take into consideration, and it is difficult to satisfy all of these criteria at once:

• The faces should have no lines or creases, or as few as possible.

• The paper should be used efficiently to yield the largest possible result.

• The model should lock cleanly and retain its shape.

• The folding structure should maximize symmetry.

• The folding sequence should be as simple as possible and accessible to the general origami public.

Here is an outline of the design process.First, work out several two-dimensional layouts of the shape. Taking symmetry into account, (from best

to worst) use square symmetry, even/odd, odd, even, radial, or none. In general, each face of the polyhedrashould have no lines. But for many solids, this could be too restricting, so think of layouts where somesides are split (see my example of one of the cube layouts where one side has an X pattern). This couldsimplify symmetries and the folding. Using a computer graphics application, it is easy to draw manylayouts. Draw small pictures of each of the faces, and arrange them in different ways, saving all thearrangements.

Second, configure the layout on a square to form the crease pattern. For some, the vertices couldmeet the edge of the paper. Occasionally an edge from the layout will meet the edge of the paper. With agraphics application, simply draw several possibilities. For a given layout, try different rotations and sizingsto form the crease pattern. From these, make the crease pattern as large as possible and print it. Just cutout the square and fold using the lines. If it works, continue to the next phase.

Third, find the landmarks necessary to generate the creases. Sometimes the landmarks are easy tofind. Often some math is used and numbers are calculated. For some of the numbers, it is easy tofind the landmarks. But for many, Robert Lang’s ReferenceFinder software is used to generate the foldingprocedure.

Now comes the rest of the folding procedure, including lock. Generally, models with sunken sideshold easier than convex polyhedra, allowing for efficiency.

Once you have worked one out, you are ready for another. Some models can be used as bases tocreate more polyhedra. For example, a cube can be folded into a cubehemioctahedron by sinking thevertices.

Here we explore in more depth the general design concepts outlined above. Later, in “Math andDesign” (page 19), we will dive into details of how the designs are realized, as outlined in the third step.

a

Place the layoutson square paper.

Find thelandmarks.

Completethe folding.

Create relatedmodels.

Work out several2D layouts.


The folding pattern for a polyhedron begins with an “outline”: a two-dimensional layout of all of its faces,similar to what you would use if you were going to create the polyhedron by cutting out the layout andtaping certain edges to complete the model. One could find such a layout online, place it on a square, andfold it. Most likely, however, it would lead to a rather clumsy model. By understanding the structure better,a more elegant design can be realized.

Let’s define a layout to be the orientation of the faces of a polyhedron in two dimensions. A creasepattern is the layout on a square to be folded. For elegant designs, it is best not to split the faces. Also, wewill need to leave some of the paper for the tabs (see “Tabs” (page 11) for more details).

Symmetry is very important to the layout. Through symmetry, the model is easier to design and fold. Hereare different forms of symmetry for the crease pattern of polyhedra. They will be explained in detail.

No Symmetry

When the crease pattern is rotated or flipped andcannot be the same as its original pattern, thenthere is no symmetry. This form is the least useful.One of the few models I have designed with nosymmetry is the Tetrahedron shown above. Cube

tab

The layout of a tetrahedronshows a band of four triangles.

The crease pattern shows theposition of the layout on a square.

Tetrahedron

Radial Symmetry

A pattern has radial symmetry if it is the samewhen rotated by some angle (divisible by 360°)other than 90° or 180°. It is useful when thepaper is a polygon other than a square. However,when folding from a square, this method isgenerally not ideal. It could be useful for foldinga square into a polygon and then into a simplepolyhedron such as a pyramid.

Icosahedron

Hexagonal Dipyramid

Layout

Symmetry


Even Symmetry

The reflection along a line through the middle ofthe pattern is the mirror image when that patternhas even symmetry. This method is generally notvery useful. The main use for even symmetry isfor polyhedra whose sides are different shapes.Some of the pyramids and prisms I have designeduse even symmetry.

Cube

Odd Symmetry

For odd symmetry, the pattern is the same whenrotated 180°.

This is the single most important form ofsymmetry used for polyhedra. As I designmodels, I generally only consider odd symmetry.There are some polyhedra, such as the pyramidabove, that cannot use odd symmetry.

Most animals follow even symmetry,whereas odd symmetry is preferred forpolyhedra.

When using odd symmetry, there is aline going through the center at some angle. Thisis generally one of the first creases to be folded.

The number of identical faces of thepolyhedra has to be an even number.

All futher symmetries to be mentionedcontain odd symmetry.

Icosahedron

Heptahedron

The bold line goes through the centerfor the Golden Pentagonal Dipyramid.

Golden Pentagonal Antiprism

Hexagonal PrismGolden Pentagonal Pyramid


Even/Odd Symmetry

As the name suggests, with this symmetry thecrease pattern is both even and odd. Wheneven/odd symmetry applies, it can be veryuseful. The number of identical faces of thepolyhedron has to be a multiple of four.

Generally, this type of pattern reallyhas just odd symmetry: when it comes toincluding the tabs, the symmetry becomes odd.

Hexagonal Dipyramid in a Sphere

The tabs make anotherwise even/oddHexagonal Dipyramidhave odd symmetry.tab

tab

Square Symmetry

For square symmetry, the pattern is the samewhen rotated 90°. This is a special case of havingboth radial and odd symmetry.

When possible, this is often the bestchoice for a folding pattern from a square pieceof paper. It has the potential for producing themost efficient layout. Because of repetition, only1/4 of the folding needs to be diagrammed.Typically, there is a vertex at the center of the paper.

The number of identical faces of thepolyhedra has to be a multiple of four.

Octahedron

Dimpled Snub Cube

Triakis Cube

3/4 Square Symmetry

This unusual pattern at first appears to have nosymmetry. But when you consider the 1/4 of thepaper that is hidden in the final model, thecrease pattern has square symmetry.

It is useful for the cube and relatedpolyhedra. The number of identical faces of thepolyhedra has to be a multiple of three. The layout for the Cube shows 3/4 square symmetry. The

middle drawing shows the underlying square symmetry.


The tabs and locking methods are among the trickiest of the problems that arise when designing origamipolyhedra. Everything hinges around them. A large, seemingly efficient design might not have enough tabto lock the faces in place. An efficient use would have just the right amount of tab.

The tabs are dependent on the symmetry used. Odd symmetry often allows for two tabs to make aknot. This is used for several of the dipyramids.

Square or 3/4 square symmetry leads to the twist lock, a very efficient method. Three or four sidesmeet at a vertex. The tabs spiral inside to lock the model.

Generally, the tabs are found on the corners of the square. Their size is arbitrary. In many designs,I work out a layout to be as large as possible. If the model does not hold well, then I work out a creasepattern with the same layout but either smaller or at a different angle to allow more tab.

Consider the octahedron. To make the largest model, the eight triangles are arranged using squaresymmetry so that four vertices meet the corners of the square. However, the crease pattern provides no tab.By rotating the layout a little, each corner has some excess to form a tab. To finish the folding, the four tabsmeet and close with a twist lock. This efficient lock allows for a large octahedron.

Octahedron This crease pattern with squaresymmetry would make the largestoctahedron, but there is no tab.

By rotating the layout, the fourcorners become tabs. This leadsto the twist lock.

This is one of the last stepsin the folding. The model isstill 2D. The tabs are shownon both sides. The modelnow opens and becomes 3D.

Tuck and interlock thetabs. The dots will meet.This type of lock formsa knot.

PentagonalDipyramid

tab

tab

The crease patternshows odd symmetrywith the two tabs.

The folding method for the Pentagonal Dipyramid shows a typical lock from a layout using odd symmetry.

A

tab tabtab

tab

A

Tabs

1/10 tab

tab tab

tab


Let’s look at two examples, the Cube and the Icosahedron, to better explore what it means to optimizea crease pattern.

Optimizing the Cube

Let’s optimize to find the crease pattern that generates the largest cube. The six faces suggest using even,odd, or 3/4 square symmetry. For each pattern, the length of the side will be given.

The sides have length .25 in these crease patterns.

Optimizing the Crease PatternOptimizing refers to finding the largest model with a good tab while keeping the folding as simple as possible.Symmetry plays an important part.

Given two crease patterns for the same polyhedron, the efficiency in terms of size can be found.Here are three ways of measuring that efficiency:

1. Measure the area of the faces of the model and divide by the area of the square paper. The larger the number, the larger the model.

2. Calculate the length of a side of one of the faces in a 1 × 1 square. Compare the lengths from two crease patterns. The larger length yields a larger model.

3. Represent both crease patterns using faces of the same size. The pattern embedded in a smaller square yields a larger model.

Compare the following two crease patterns for the octahedron. One uses odd symmetry, the other squaresymmetry. For a 1 × 1 square, the length s of a side of the faces is given. Also, the triangular faces are allthe same size.

Square symmetry. This creasepattern yields a larger octahedronOctahedron Odd symmetry.

Cross layout, even symmetry. 3/4 square symmetry.Odd symmetry.Cube

s =1 2 3( ) ≈ .2887

s

1/10

s = 123 30 ≈ .3697

Optimizing the Crease Pattern 13

s = 1/(4cos(22.5°) ≈ .2706

α = 22.5°

By rotating the layout with 3/4 square symmetry by 22.5°, the sides increase from .25 to .2706. This modelis diagrammed in “Cube Design” (page 83). It closes with a three-way twist lock.

α

In the rotated version, three vertices meet the edges of the square. The layout can be rotated a bit more sothat six vertices meet the edges. This would optimize the 3/4 symmetric layout with respect to size. The cubeit generates would be larger, but there would not be enough tab to hold it together. Thus, this crease patternis not used for the cube. However, it is used to optimize other related polyhedra. This includes the TriakisTetrahedron and Cubehemioctahedron.

s = 1/(4cos(arctan(.5))) ≈ .2795

Triakis Tetrahedron Cubehemioctahedron

Now consider placing a band of four square faces along the diagonal with even or odd symmetry. Thisincreases the sides of each square. Unfortunately, there is not enough tab.

Though we are used to placing sides next to each other in a layout, they do not have to be. By rotating twosquare faces so they only meet at a vertex, we can make another layout with the same efficiency. This creasepattern yields a cube that can lock. Thus, this is the most efficient cube. It will be diagrammed in “PrismDesign ” in a future volume on origami polyhedra.

Even symmetry. Odd symmetry.

Even/odd symmetry optimizes the cube.

s = 2 5 ≈ .283

s = 2 5 ≈ .283


Optimizing the Icosahedron

With its 20 sides and somewhat ball-like shape, this model is a challenge to design. Before taking symmetryinto account, let us try a few ideas.

The layout has three-way radial symmetry plus one extra face on the bottom. In general, placing a facein the center does not optimize the layout. The few times where it does is when there are an odd numberof faces or the model does not have odd symmetry.

Consider the standard layout given in books.

The difficulty with this is it does not fill out the square paper very well and the locking would be difficult.Let’s try radial symmetry.

Icosahedron

Radial symmetry.

Standard layout.

The darker triangle representsthe face in the center.

It would be very convenient to make a layout with one face in the center and the rest revolving around it.Here is a possible layout.


It would again be difficult to place this layout in a square. Also, the locking would involve five vertices tomeet, which does not hold together well.

Let’s try the band method, as with the cube. The equator has a band of ten triangles that will beused in the layout. Place the band along the diagonal in the crease pattern. There will be five triangles oneither side. This has odd symmetry and fills the square well. It also has plenty of space for tabs, allowingthe model to lock well. Because of the 15° lines starting from the origin, it is easy to set up the crease patternand fold. (We will discuss folding angles in more detail in “Math and Design” (page19).)

The equator is shaded. The 15° lines are boldedin the crease pattern.

Layout with a bandof ten triangles.

The layout could be made larger in the square paper, but the landmarks—key elements upon which the restof the folds are based, discussed in “Math and Design” (page 19)—would be difficult to find. Also, findinglandmarks could lead to extra creases that show on the faces, reducing the crisp look of the final model.Also, with a larger layout, there would be less tab. Thus, this is my optimal crease pattern for the Icosahedron.

15°

tab

tab

The dipyramids have a unified structure. I came up with a formula relating the angles of the triangular facesto the proportions of the finished model. This made it easy to design a large collection of dipyramids.

diameter

height

α

Let H = height divided by diameter, α = angle at top of each triangle, n = number of sides of the

and

α = 2 arcsin[sin(180°/n)cos(arctan(H))]

Groups of PolyhedraWe have seen that the band method optimizes the crease pattern for the cube and the icosahedron. Doesthe band method also work for other polyhedra? Can we sort polyhedra into groups so that each group hasa similar crease pattern? If so, then once we determine an efficient pattern for one polyhedron in that group,it may lead us easily to efficient patterns for the others. By understanding the structure of the two-dimensionallayout with respect to symmetry, tabs, and optimization, it is possible to develop methods of making relatedpolyhedra.

Dipyramids

H = tan arccossin(α/2)

sin(180°/n)( ( ((

polygonal base.


This led to two general crease patterns depending on whether the polygonal base has an even or odd numberof sides.

The Hexagonal Dipyramidhas even/odd symmetry.

The Heptagonal Dipyramidhas odd symmetry.

Golden Pentagonal Antiprismwith crease pattern.

Hexagonal Antiprismwith crease pattern.

Square Antiprismwith crease pattern.

The method for developing antiprism crease patterns goes as follows:Leave some space on the left and right sides of the square for the tabs. Fold vertical lines evenly spaced

and make a band of triangles, which make up the equator of the antiprism, going through the center of thesquare. Add the two remaining polygon faces, one on either side of the band. All of this is done using oddsymmetry.

Band of triangles.Vertical lines with some spaceon the left and right for tabs. Two polygons are added.

For antiprisms, I came up with a general structure. Using this structure, I easily designed several of them.

Antiprisms


The Tetrahedron has a band of four triangles in its crease pattern. The same band can be used to createtetrahedra with color patterns.

Tetrahedron withcrease pattern.

Duo-Colored Tetrahedronwith crease pattern.

Striped Tetrahedronwith crease pattern.

Tetrahedra with Color Patterns

Stellated OctahedronDimpled Truncated

Octahedron

Cubehemioctahedron Dimpled HexagonalDipyramid

Related to the Cube

This group of polyhedra has the same surface as the cube. Thus, by folding a cube and sinking vertices ormaking other changes, a variety of polyhedra can be created.

Cube

DimpledRhombicuboctahedron


Folding from a Crease PatternSuppose you have a crease pattern and you want to fold the polyhedron from it. It will take some experimentingto find the best way to do so. Here are a few ideas.

Start with the white side of the paper. The crease pattern shows the orientation of the faces. Valley-fold along the edges of the faces. There are places where faces meet at a vertex with some angle betweenthem. Bisect those angles with mountain folds.

Basically, the part with the faces will show, and the rest will be hidden inside the polyhedron.If there are lines that go through the exact center of the square, fold those as part of the first folds.

They will be valley or mountain folds depending on whether they are on the edges of the faces.As the model is being folded, try to avoid creasing over the faces. See if there is some excess paper

from opposite corners that can be used as tabs. Sometimes, I draw the edges of the faces so that when Ifold the model I will know what to show and what to hide.

Here is the crease pattern for a Pentagonal Dipyramid.

Begin by valley-folding alongthe crease going through thecenter. Then valley-fold alongthe two vertical lines.

Continue valley-folding along theedges. The lines can extend beyondthe vertices as long as they do notpass through the faces.

Bisect the angles between thefaces with mountain folds.

1 2 3

4

tab

tab

Collapse along the creases. Thereis some paper on oppositecorners that can form tabs.

Math and Design 19

Math and DesignOnce we have determined how to best arrange a layout on our square paper, we have to figure out how toperform the folds needed to generate the crease pattern. We will need to fold particular angles or find aparticular length along a side of the square. To do this, we employ mathematics. A square sheet of paperlends itself well to math, especially when designing or understanding origami polyhedra. In general, algebra,geometry, and trigonometry are used.

Typical uses of math in designing polyhedra are the following:

• to find the lengths of lines and the values of angles,

• to find the landmarks of a crease pattern,

• to divide the paper into nths (such as thirds).

We will first outline some basics of mathematics on the square.

(0, 0) (0, 1)

(1, 1)

(.5, .5)

(.75, .25)

(0, .25)

(1, 1)

Consider a square to be 1 unit by 1 unit in length. The length ofthe diagonal is found by the Pythagorean Theorem, using the righttriangle formed by two adjacent edges and the diagonal as thehypotenuse:

Any point on the square is defined by (x, y) where both x and yare values from 0 to 1. In the third image on the right are some examples.

Any crease defines a line. Lines can be expressed by

where m = slope, or the change in y divided by the change in x,and b = y-intercept, or where the line meets the left side of thepaper.

Equivalently, a line can be expressed as

where (x1,y1) is a point on the line.The line between points (0, .25) and (1, 1) is

When folding practically anything, each crease can be defined byan equation of a line. Fortunately, the process of folding is obliviousto the math.

1

1

Basics

1

1

2

y = mx + b,

y − y1 = m(x − x1),

a2 + b2 = c2

12 + 12 = c2

c = 2 ≈ 1.4142136

y = .75x + .25, since

m =y2 − y1

x2 − x1

=1− .25

1− 0= .75


When designing origami models, there is often a landmark that is the key to folding the model. Given acrease pattern, the landmark can be found mathematically. The landmark can be represented as a point(x, y) on a 1 × 1 square. Quite often, the landmark occurs on the edge of the square.

For many models, the landmarks are divisions of the paper into nths, such as thirds. Other landmarksdeal with angles, especially going through the center of the paper.

Examples:

In the crease pattern for the GoldenPentagonal Dipyramid, there is a line throughthe center at 18° from the horizontal line.The landmark a is calculated to be .33754.

1/7

2/7 1/7

The crease pattern for Triangleson Cube shows divisions of 1/7.

Other landmarks require more calculations.

Example:

The Sunken Rhombic Dodecahedron is composed of 48 triangles. Each face is an isoscelestriangle with an apex angle of 45°. Once the landmark a is found, the model can be folded.

18°

a

a

Landmarks

45°

Algebra is often used to find landmarks.To fold a Triakis Tetrahedron or Cubehemioctahedron, the following layout is used. To make the

most efficient use on a square, it is rotated so the outermost vertices meet the edges of the square sheet.

The problem is to find landmark a. Once known, the model can be folded. After some algebra, itis determined that a = .25. That landmark is easy to fold. Next is the math used and the first few diagrams.

Layout Layout rotated for bestplacement on square.

a

Calculation of Landmark Using Algebra

CubehemioctahedronTriakis Tetrahedron

Math and Design 21

To see how this landmark is used, let’s look at the first six steps of the Triakis Tetrahedron. Note the .25 marksin Step 1. In Step 6, we can make out the outline of the underlying layout.

T2

2s

s

c

T1

b

as

Triangles T1 and T2 are similar.

a

b

c b

ss

ssT1

T2

Find length a given a 1 × 1 square.Let s = length of the side of the small squares.

Triangles T1 and T2 are similar, so

Since b = a 2, then b = s 5 .For a 1 × 1 square,

a + c + b = 1

2s

5+ 5s +

s

5= 1

8

5s = 1

s =5

8. So,

a =2s

5=

2

5

5

8=

1

4a = .25.

c2 = s2 + (2s)2 = 5s2

c = 5s

a

b=

2s

sa = 2b

b =a

2and

a

s=

2s

cac = 2s2

a =2s2

c=

2s2

5s=

2s5

.

1 2 3

4 5 6

Make small marks byfolding and unfoldingin quarters.

Fold and unfold. Fold and unfold.

Unfold. Fold and unfold.

Triakis Tetrahedron


sin(a)

A=

sin(b)

B=

sin(c)

C

or

A

sin(a)=

B

sin(b)=

C

sin(c)

Any triangle.

a b

c A

C

B

Law of Sines

Triangular Dipyramid

2

3

2

Crease pattern and oneof the triangular faces.

P

Calculation of Landmark Using Trigonometry

Trigonometry is also used to calculate landmarks. Here are some basic trigonometric formulas that we useoften in such calculations.

P

α/22

1.5

β

α

β β

2

3

2

Example:

This Triangular Dipyramid is composed of sixtriangles. Each triangle has sides proportional to2, 2, and 3. In the crease pattern there are threelandmarks, shown by the dots. Knowing landmarkP is sufficient for finding the rest and folding themodel.

Given the crease pattern and the proportionsof each triangle to be 2, 2, and 3, find the coordinates of landmark P in a 1 × 1 square.

1. Find the angles of the triangular faces.

sin(α/2) = 1.5/2 = .75

α/2= arcsin(.75)

α ≈ 97.18°

α + 2β = 180°

β = 41.41°

Right triangle.

AdjacentOppositeHypotenusex

A

OH

To find length of a line: To find angle:

A2 + O2 = H2

sin(x) = O/H

cos(x) = A/H

tan(x) = O/A

arcsin(O/H) = x

arccos(A/H)= x

arctan(O/A) = x

Math and Design 23

We get more information by looking attriangle T.

• Find angle λ.

• Find angle φ.

• Use Law of Sines to find c.

Now we use the diagonal equation to findlength a.

Then,

3. Find the coordinates of P.

Since P is along the diagonal, x = y and c is thehypotenuse of an isosceles triangle:

c2 = x2 + y2 = x2 + x2 = 2x2

x = c/ 2

so, P = (c / 2, c/ 2) ≈ (.3171387, .3171387)

2. Find lengths a and c. We get some information from the diagonal:

Pc

a

c

Pc

a

c

45°

λ

φ

β

β

Ta

1 2 3

4 5 6

Fold and unfoldat the edges.

Bring the left edge to thelower dot. Crease at theintersection.

Unfold androtate 180°.

Repeat steps 2−4.Turn over and rotate.

Fold and unfold but donot crease in the center.

A folding sequence locates the landmark (.3171387, .3171387). Here are the first few steps of the TriangularDipyramid.

Fold and unfold.Rotate.

a + 2c = 2 a + 2c = 2

a + 2(.8671567)a = 2

a = 2/(1 + (2)(.8671567))

= .5172097

c = .8671567a

= .4485018

2β + λ = 180°

λ = α ≈ 97.18°

φ = 180° − 45° − λφ ≈ 37.82°

c/sin(φ) = a/sin(45°)

c = asin(φ)/sin(45°)

≈.8671567a


Robert Lang’s ReferenceFinder

Once the landmark is found, then a folding method is used to locate it. As an origami designer, I want mymodels to be realized by folding alone without a ruler or protractor, and even with no awareness to thefolder that there might have been pages of calculations. Sometimes, the landmarks are easy enough to findwith a few folds. But quite often, it would be difficult to think of an efficient folding procedure. So I useRobert Lang’s ReferenceFinder computer program, http://www.langorigami.com/referencefinder.htm.

ReferenceFinder gives an efficient folding sequence to locate any point on a square with a smallerror. The folding sequence typically takes only three to five steps.

When designing the Golden Pentagonal Antiprism, the following layout was used.

b a

ba

36°

b a

ba

36°

36°

Golden PentagonalAntiprism

1 2 3

Fold in half at the bottom. Unfold.

4

Fold and unfold.a ≈ .1367287

Landmark b, though difficult to calculate, is easy to find and did not require ReferenceFinder.

Fold and unfoldto the diagonal.

b = .41421356

Fold and unfoldalong the diagonal.

1 2

Landmarks a and b are all that is needed to set up the folds.Length a defines the line going through the center of the paper at 36° from a vertical line. It was

easy enough to calculate.

a = .5(1 − tan(36°)) ≈ .1367287

ReferenceFinder found a simple folding sequence to locate landmark a.Length b defines an isosceles (36°, 36°, 108°) triangle and pentagon below and above the center

band. It was difficult to calculate. Yet it was determined that

b = 2 − 1 ≈ .41421356

which is an easy landmark to find.According to ReferenceFinder, landmark a can be found this way:

Dividing a Square into nths 25

First, some specific cases will be shown. Then some general methods will be given.

Divisions of 1/3

Divisions of 1/5

1 2 3 4 5

Fold and unfoldin half on the left.

Crease atthe bottom.

Unfold. Fold and unfoldon the right.

1/3

The 1/3 mark.

1 2 3

Fold and unfoldcreasing along partof the diagonal.

Crease onthe left.

Unfold.

4 5

Fold to the creaseand unfold.

1/5

The 1/5 mark.

Method 1: This is an approximation with a division of 0.198912. This is quick and simple to fold andgenerally has good enough precision (99.5%).

Method 2: This is exact but slighly more difficult to fold.

Fold and unfoldon the right.

Fold and unfoldin half on the left. Unfold. The 1/5

and 2/5 locationsare found.

1/5

The 1/5 mark.

1 2 3 4 5

1/52/5

Dividing a Square into nths


Divisions of 1/6

1 2 3 4 5

Fold and unfoldin half on the left.

Crease at the top. Unfold. Fold and unfoldon the right.

The 1/6 mark.

1/6

Divisions of 1/7

Divisions of 1/8

51 2 3

Fold and unfoldby the top.

4

Fold and unfoldto find thequarter mark.

Fold and unfoldalong the diagonal.

The intersection atthe dot divides thesquare into 3/7 and4/7. Fold and unfold.

1

2 1/7

The 1/7 mark.

1 2

Fold on the left.Fold and unfoldat the top.

3 4

3/8

1/8 1/8

The 1/8 mark.Unfold. The 3/8mark is also found.

This is an example of the diagonal method for dividing the square into nths as shown on the next page.

The standard method for finding the 1/8 mark is to divide the paper in half three times. That would takethree folds. This clever method requires only two folds.

The first three steps are the same as those used to find the 1/3 mark.


Divisions of 1/9

4

1 2 3

Fold and unfoldon the left.

Unfold. The 2/9 and 4/9marks are found.

Bring the lower right corner to thetop edge and the bottom edge tothe left center. Crease on the left.

Fold and unfold.

5

The 1/9 mark.

This is an example of the edge method for dividing the square into n2 parts as shown on page 30.

1

23 3/8

Fold and unfold inhalf three times tofind the 3/8 mark.

1 2 3 4

Fold and unfold. Fold and unfold. The dot is at(8/11, 3/11).

Diagonal Method for Dividing the Square into nths

In this method, two lines are folded, and their intersection divides the paper into a desired fraction. Oneof the lines is along the diagonal of the square.

To use this method, we need to know the powers of 2. These are the numbers 21 = 2, 22 = 4, 23

= 8, 24 = 16, 25 = 32, and so on. (Note that we can always divide a square into a fraction of a power of 2by successively folding in half: see, for instance, Step 1 in the example.)

Example: Divide into 11ths.

We begin with two numbers that add up to 11. One of the numbers is the greatest power of 2 lessthan 11, which is 8. The other is 3 = 11 − 8.

Make a fraction of the two numbers. This would be 3/8. Find height 3/8 on the right edge, and folda crease from the lower left corner to the 3/8 mark. Fold a downward diagonal crease. These two creases intersect at (8/11, 3/11).

4/9

2/9

1/9


b/a

Fold and unfoldfrom the lower leftcorner to the rightedge at b/a.

b/a

(c, d)

c

d

Find the location b/aon the right edge.

2 3 41

Fold and unfoldalong the diagonal. The intersection is at

(a/(a + b), b/(a + b)).

Diagonal method for dividing the square into nths.

Given n, find a and b so n = a + b where a = 2m is the largest power of 2 < n.

Example: n = 11 then a = 8, b = 3.n = 25 then a = 16, b = 9.

Let us check that intersection point. Our two creases are the lines y = (3/8)x and y = −x + 1. They intersectat

and y = (3/8)(8/11) = 3/11.

This method does not work well if b/a is smallbecause of the small angle, which is physicallydifficult to fold.


n = a + b

27 = 16 + 11

b/a = 11/16

Fold and unfold inhalf four times tofind the 11/16 mark.

1 2 3 4

Fold and unfold. Fold and unfold. The dot is at(16/27, 11/27).

1

2

311/164

−x + 1 = (3/8)x

1= (11/8)x

= 8/11


Edge Method for Dividing the Square into nths

To use this method, first locate a key point on the left edge. Make a fold using that point, and the divisionis found on the top edge.


We begin with two numbers whose sum is 7 and one of the numbers is the greatest power of 2 lessthan 7. That power of 2 is 4. So one number is 4, and the other is 3.

Make a fraction of the two numbers. This would be 3/4. Find height 3/4 on the left edge. Bring thebottom right corner to the top edge and the bottom edge to the 3/4 mark on the left. The top edgeis divided into 7ths by the bottom right corner.

1 2 3

Unfold.

Bring the bottom right corner tothe top edge and the bottomedge to the 3/4 mark on the left.

4


3/4

Fold and unfold.

1/7

5

The 1/7 mark ison the top edge.

b/a

Find the intersection of the two lines:Given a 1 × 1 square, the equation of thedownward diagonal is

y = −x + 1

The equation of the line from the origin to(1, b/a) is

y =b

ax

Solve for x and y:

Derivation of the diagonal method for dividing the square into nths.

y = −x + 1,

y =ba

x, so

b

ax = −x + 1

b

ax + x = 1

a + ba

x = 1

x =a

a + b,

y =b

ax =

b

a

a

a + b=

b

a + b.

( (

( (( (


Edge Method for Dividing the Square into n2 Parts

In this method, locate a key point on the left edge. Make a fold using the key point, and divisions are foundon the left and right edges.


We begin with the square root of 25, which is 5. Find two numbers whose sum is 5 and one of thenumbers is the greatest power of 2 less than 5. That power of 2 is 4. So one number is 4, and the other is 1.

Make a fraction of the two numbers. This would be 1/4. Find height 1/4 on the left edge. Bring thebottom right corner to the top edge and the bottom edge to the 1/4 mark on the left. The left and right edges are divided into 25ths by the crease.

1 2 3

Unfold.Bring the bottom rightcorner to the top edgeand the bottom edge tothe 1/4 mark on the left.

4


2/25

8/25

Opposite sides are dividedinto 2/25 and 8/25.

1/4

4

3

Fold and unfold.Bring the lower right corner tothe top edge and the bottomedge to the left landmark.

5

Unfold.

b

2

a

Edge method for dividing the square into nths.

Given n, find aT and a B so n = aT + aB where aB = 2m is the largest power of 2 < n.

Example: n = 7 then aB = 4, aT = 3.n = 11 then aB = 8, aT = 3.

Find the locationof a = aT /aB onthe left edge.

1

b =1− a

1+ a


Derivation of the edge method.

b

c

b

1. Given b, find c.

1 2

Given landmark b, fold thelower right corner to b.

b

c

1− c

1− c

b c

1/2 3/8

1/4 9/32

3/4 7/32

1 2 1/4

3 2 1/8

Examples of b and c valuesfor some easy landmarks.

Edge method for dividing the square into n2 parts.

Given n2, find aT and aB so n = aT + aB where aB = 2m is the largest power of 2 < n.

Example: n2 = 25 then aB = 4, aT = 1.n2 = 81 then aB = 8, aT = 1.

3

2

Bring the lower right corner tothe top edge and the bottomedge to the left crease.

4

Unfold.

c

d

1

a

Find the location of a = aT/aBon the left edge.

c =2a

1+ a( )2

d =2a2

1+ a( )2

The hypotenuse of the triangle = 1− c, so

b2 + c2 = (1− c)2

b2 + c2 = 1− 2c + c2

b2 = 1− 2c

c = (1− b2)/2


b

cα

β1− b

1− a

α

β

a

b

c42 3

Fold and unfold.Bring the lower rightcorner to the top edge andthe bottom edge to a.

1

Unfold.

2. Given a, find b and c.

a

Label segmentsand angles.

a

b

c

1− b

1− a α

β

a

b

c

1− b

1− a α

βα

β

The dotted lines on the right represent the bottom two corners when the first fold is performed. The angle in the upper left triangle is 180° − 90° − β = α. The two triangles whose sides meet the upper edge are similar.

Because the triangles are similar, so

a =1− b

1+ b.

And solving for b,

(1+ b)a = 1− b

a + ab + b = 1

b(a + 1) = 1− a

b =1− a

1+ a.

Since c = (1− b2)/2 and

b =

1− a

1+ athen, after some calculation,

c =2a

(1+ a)2.

42 3

Fold and unfold.The dot locates a.

1

Unfold.

3. Given b, find d.

b

c

da

Given landmark b, fold thelower right corner to b.

b

1− a

1− b=

b

c.

Since c = (1− b2) / 2,

1− a

1− b=

2b

1− b2

1− a =2b(1− b)

1− b2

=2b(1− b)

(1− b)(1+ b)

=2b

1+ b,


a

b

c

d

e

1− c

α

βα

β

βα

d

Solve for d:

d + e = a

Because of similar triangles

d

e=

c

1− c

d

a − d=

c

1− c

d (1− c) = c(a − d )

d − dc = ca − cd.

Thus

d = ca

=1− b2

2

1− b

1+ b

=(1− b)(1+ b)(1− b)

2(1+ b)

= (1− b)2 2.

4. Given a, find b, c, and d.

d

b

c42 3

Fold and unfold.

Bring the lower rightcorner to the top edge andthe bottom edge to a.

1

Unfold.

a

Landmarks b, c, and dare determined by a.

a b c d

1/2 1/3 4/9 2/9

1/4 3/5 8/25 2/25

3/4 1/7 24/49 18/49

1/8 7/9 16/81 2/81

3/8 5/11 48/121 18/121

5/8 3/13 80/169 50/169

Final formula and examples of some values.

So,

d =

1−1− a

1+ a

2

2

=

1+ a − 1+ a

1+ a

2

2

=

2a

1+ a

2

2

=2a2

(1+ a)2 .

Given b,

c = (1− b2) 2,

d = (1− b)2 2.

Given a,

b =1− a

,1+ a

c =2a

(1+ a)2 .

Given a

b =1− a

1+ a

c =2a

1+ a( )2

d =2a2

1+ a( )2

( (( (

( (

( (

( (


Angles

For a 1 × 1 square,

a = tan(α)

45° and 22.5°

Fold along the diagonal.

30°

1 2 3

Unfold.Fold and unfold. Fold the dot to thecenter crease.

30°

4

30¡

.42265

.57735

Fold and unfoldto the diagonal.

22.5°

1 2

22.5°.4142135

1

.5

30° Note the dotted triangle from step 3.The hypotenuse is 1, and the oppositeside is .5. Thus the angle is 30°.

a

α

tan(22.5°) ≈ .4142135 ≈ 2 − 1

tan(30°) ≈ .57735 ≈ 3/3

When we design polyhedra, angles are needed. Here are folding sequences to form mainly integer angles.Some are key angles that generate others.

45°

3

Angles 35

15°

1 2 3

Fold and unfold. Fold to the crease. Unfold.

4

15°

Note the dotted triangle from step 3.The hypotenuse is 1, and the adjacentside is .5. Thus the angle is 60°, makingthe angle in step 4 to be 30°/2 = 15°.15°

.26795.5 1

60°

32°

5/8

32°

Since arctan(5/8) ≈ 32.00538°, thefolding method for divisions of 1/8 canbe used as a simple way to obtain 32°.

1 2

Fold on the right.Fold and unfoldat the top.

3 4

58°

8°16°32°

29°

Given 32°, several other integer anglescan be found. Note that 32° divided inhalf five times is 1°. So, in theory, all theinteger angles can be found from it!

Unfold.

32°

32°32°

26°

13°

1 2

Here is a list of some angles that are easyto find given 32°:

1°, 2°, 4°, 8°,13°, 16°, 22°, 26°, 29°.

3/8

5/8

tan(15°) ≈ .26795 ≈ 2 − 3

Folding method to make 32°:

Fold and unfold.

Here is a way to find more integer angles from 32°.


14°

1/414° 14° 7°

76°

38°

19°

More integer angles can be found from 14°,including 28° since 2 × 14° = 28°. Theseangles can be found given 14°:

7°, 19°, 28°, 38°, 76°.

Since arctan(1/4) ≈ 14.0362°,a simple way to obtain 14° isto mark 1/4 on the right edge.

24°

Since arctan(4/9) ≈ 23.96°, the foldingmethod for divisions of 1/9 can be usedas a simple way to obtain 24°.

4/924°

24°66°

33°

12°

Here is a list of some angles that are easyto find given 24°:

3°, 6°, 12°, 21°, 33°, 39°, 42°, 48°.

1 2 3


Unfold.Bring the upper right cornerto the bottom edge and thetop edge to the left center.Crease on the right.

4/924°

4

Fold and unfold at 24°.


Angles 37

36°

tan(36°) ≈ .72654



1 2 3 4

5

More integer angles can befound given 36°. Here areof some of them:9°, 18°, 27°, 54°.


Bring the top edge to the right dotand the top left corner to thebottom edge. Crease on the right.

Unfold.

36°

Fold and unfold.

36° 18° 9°

54°

27°

More integer angles can be foundgiven 25°. Here are of some of them:

5°, 10°, 20°, 35°, 40°, 50°.

36°

.72654

6

25°

tan(25°) ≈ .4663



1 2 3 4

5


Bring the top edge to the left dotand the top right corner to thebottom edge. Crease on the right.

Unfold.

Fold and unfold.

6

25° 25°

.4663 40° 50°25°

20°20°

10°

70°35°

36° 25°

The folding methods forfinding 25° and 36° arealmost the same.


Angles through the Center

In general, given angle α, calculate a.Locate the landmarks on opposite edges.

In the dotted triangle

tan(α) = (.5 − a)/.5,so

a = .5(1 − tan(α))

a

a

αα

a

.5 − a

.5

α

The crease patterns of many polyhedra have a line going through the center of the square paper at someangle.

Suppose you can fold an angle from the corner of the square and now want to fold that angle through thecenter. Here is a method.

α

Fold and unfold along the knownangle but only crease on the right.


Rotate 180°.

α

Repeat steps 1−2. Fold and unfold.

αα

The line goes throughthe center at angle α.

1 2 3

4 5 6

30° through the Center

1 2 3

Unfold.Fold and unfold. Fold the dot to the verticalline and crease on the right.

4

Fold and unfold onthe right. Rotate 180°.


30°

The lines goes throughthe center at 30°.

5 6 7

30°


Design Method Examples

Radial

Even/odd

Even/odd, butterfly layout

Even/odd

Even/oddOdd

Here are a few.

The lengths of the sides areproportional to 1, 2, 2.

Hexagonal Dipyramidin a Sphere

Here are some examples to illustrate the process behind designing origami polyhedra. We follow the stepsoutlined at the beginning of “Design Factors and Techniques” and implement tools discussed there and in“Math and Design.” In the end, I hope that I have given some insight into design concepts, as there is stillso much to be discovered and enjoyed.

Design Method for the HexagonalDipyramid in a Sphere

This solid is composed of twelve isosceles triangles. Some math is require to find the angles and lengthsof the sides of each triangular face. The formulas are given in “Dipyramid Design” (page 173).

Layouts

41.41°

69.3° 69.3°1

2 2


Here are just a few possibilities.

After experimenting with these and other crease patterns, I discovered that this one is the best. It has goodsymmetry and locks together well.

Given the angle of 41.41°, length a is calculated to be .3110. Robert Lang’s ReferenceFinder software isused to find a short folding procedure to locate point a.

a

α

α ≈ 41.41°

In each of these crease patterns, the size of the triangular faces are the same. Thus, the smaller the squareenclosing the crease pattern, the larger the finished model.

Crease Patterns

Landmarks


Design Method for the SunkenTetrahedron Using Math

Designing the sunken tetrahedron requiresa bit of math. First, we find the proportionsof the sides of each triangular face. Wethen develop a layout with landmarks.

The sunken tetrahedron iscomposed of 12 isosceles triangles. Theproportions of the sides of the triangle areto be found.

1. Begin with a tetrahedron where the edges are all 1 unit in length. Find the length of line D, on the base from a

vertex to the center.

1

1

1

1

a

b

c

D

a b

c

1/2

30°

D

2. Find height H.

1

1

a

b

c

H

1/ 3

1

1/ 3

H

a

3. Find length F, which represents one of the sides of the triangles of the sunken tetrahedron. Since triangles T1 and T2 are similar: F

1/2

1

F F4. The proportions are found, as shown

This shows the proportions of the sides. To make work for the values easier, the numbers are scaled so that the proportions are 3, 3, and 2 2.

Base of tetrahedron.

To solve for D:

1

a

b

c

F

1/2

1/2

1/ 3

2/3

1

a

b

c

F

F

1

2/3

1

a1/ 3

2/3

3 3

2 2

1

32 2

32 2

cos(30°) = 1 2D

D = 12cos(30°)

= 13

1=H 2 +13

2

H = 23

F1 2

= 12 3

F = 32 2

( (

T1

T2

on the right.


Now that the proportions of the triangles are known, a layout can be made. There are several possibilities.Here are a few.

Even symmetry. Odd symmetry.

3/4 square symmetry.3/4 square symmetry.

Layouts

Crease Patterns

After experimenting with these crease patterns,this one was chosen. It yields the largest modelthat has the fewest steps and holds together well.

If it does not maximize the size of themodel, then it is very close to it and appears tomaximize with respect to its simplicity. It uses 3/4square symmetry.


1. Find α.

2. Find a. (This crease is used in the folding.)

α

θ

α

θ

b a

.5

3. Find c.

αα

cd

.5φ

Given the crease pattern, the landmarks are to be found.

Triangular face.

Landmarks

α

3 3

2 2

α1

3

2

θ = (90° − 2α) 2= 45° − α

tan(θ) = b .5 = 2bb = .5tan(θ)

a = .5 − b= .5(1 − tan(θ))= .4142135= 2 − 1= tan(22.5°)

φ = 90° − 2αtan(φ) = d .5 = 2d

d = .5tan(φ)

c = .5 − d= .5(1 − tan(φ))= .3232233= (4 − 2) 8

tan(α) = 1 2

α = arctan(1 2)

= 35.26439°


4. Find e.

The landmarks are

a = .4142135 = tan(22.5°)

c = .3232233 = (4 − 2)/8

e = .1464466 = (2 − 2)/4

g = .2679491 = tan(15°)

Of these, a and g are easy to find and sufficient forcompleting the folding pattern.

Find landmark a by folding the angle 22.5°,since a =.4142135 = tan(22.5°).

Find landmark g by folding the angle 15°,since g = .2679491 = tan(15°).

5. Find g, first find k.

Proportion

αα

e

f

g

h

j j

k

.5

j

a = .4142135 = tan(22.5°)

a22.5°

g = .2679491 = tan(15°)

1 2

g15°

Fold and unfold.

tan(α) = f .5

f = .5tan(α)

e = .5 − f

= .5(1 − tan(α))

= .1464466

= (2 − 2) 4

k /h = .5/ ( h + 2j)

h = 3 j

j = h 3

h + 2j = h + (2 3)h

= h(1+ 2 3)

k =h

2h(1+ 2 3)

=1

2(1+ 2 3)

g = .5 − k

= .5 1−1

1+ 2 3

= .2679491

= tan(15°).

and

so

( (

Polygons 45

Rectangle

Silver RectangleBronze RectangleGolden Rectangle

Dimensions

1 x 2 ( ≈ 1.414)1 x 3 ( ≈ 1.732)1 x (1 + 5)/2 (= φ ≈ 1.618)

n

3456789191112

Polygon

TriangleSquarePentagonHexagonHeptagonOctagonNonagonDecagonHendecagonDodecagon

Angle at vertex

60°90°108°120°128.57°135°140°144°147.27°15°

α = 180°/5 = 36° The lines meet thevertices of a pentagon.

For a heptagon, α = 180°/7 ≈ 25.7°.

α

SilverRectangle

BronzeRectangle

1

1.618034

GoldenRectangle

α

PolygonsSince polygons are the faces of polyhedra, we present how to fold certain polygons before we move intofolding directions for polyhedra.

Many of the polyhedra in this collection have faces that are regular polygons, which are polygonswith congruent angles and sides. Each interior angle between two sides of a regular n-gon = (n − 2)180°/n.

Regular Polygons

One of the easiest ways to create a regular polygon in a square is by the book-fold symmetry method. Inthis method, start with a vertex centered on the top of the square. Create lines that all radiate from that top-center point and that divide the edge into n equal angles. Let the top-center point be the top vertex of aregular n-gon; all of the vertices will lie on the lines. All the lines will meet the vertices of the n-gon nomatter the size of the n-gon.

The angle between any two adjacent lines radiating from the top center of the square is α = 180°/n.

α

Rectangles are also used in polyhedra design. Folding directions will be given for rectangles based on threefamous ratios.

Rectangles

1

2

1

3


Equilateral Triangle

This shape has three sides with interior anglesof 60°. For sides of length 1, the height is 3 2.

(1/3)h

(2/3)h

The bisectors—lines that divide each angle in half—divide the height into thirds where they intersect.

Triangles placed in a layout for polyhedra come in different orientations. Two methods for folding triangleswill be given: book-fold and diagonal symmetry.

Book-fold Symmetry

The triangle can be oriented at different heights in the square. Here is the layout following the book-foldsymmetry method, with the top vertex meeting the top of the square.

Landmarks.

60°60°

30°

1

3 2≈ .8660254

1 − 3 2≈ .1339746

.5

1

h = 3 2 .866025460°

Tetrahedron of Triangles Octahemioctahedron Icosahedron

Equilateral triangles are used in several polyhedra. The tetrahedron of triangles, octahemioctahedron, andicosahedron are composed entirely of them.

Folding method:

Fold and unfold inhalf on two edges.

Bring the dot to the crease.Crease on the left.

Unfold.

1 2 3 4


Equilateral Triangle 47

Diagonal Symmetry

Here, the height of the triangle is along the diagonal fold instead of the book fold. This layout yields thelargest triangle in a square.

The length of the side of the triangle = 1/cos(15°) ≈ 1.0352762. For the book-fold method, the length is 1.

60°

15°

tan(15°) = 2 − 3 .2679491

1

Folding method:

Fold and unfold inhalf on two edges.

Bring the dots to the creases. Fold and unfold.

Fold behind. Fold inside.


1 2 3

4 5 6 7


5 6 7

Landmarks.


Pentagon

Pentagonal faces are on the dodecahedron, golden pentagonal prism, and golden pentagonal antiprism.

The angle at eachvertex = 108°.

αβ

h

β

hhb

108°

w

108°α α

1

Some calculations are required to determine the landmarks.

Pentagon Dodecahedron Golden Pentagonal Prism Golden PentagonalAntiprism

w

sin(108°)=

1

sin(36°)w = sin(108°)/sin(36°)

= 1.618034

= (1+ 5)/2

= φ , the golden mean.

1. Find width w. Let the length of each side = 1.

Find α.2α + 108° = 180°

α = 36° By the Law of Sines

2. Find height h. First find angle β.

α + β = 108°/2 Since α = 36° then β = 18°.

tan(18°) = .5/hh = 1/(2tan(18°))

≈ 1.5388418

3. Find the ratio of hb to h, where hb isthe height of the bottom section.

β = 108° − 90° = 18°

hb = cos(18°) ≈ .9510565

hb/h = .95105651/.5388418

= .618034

= φ − 1

Pentagon 49

b = 1.

Then h = 1.618034 = φand

ht = .618034 = φ − 1

where ht is the height of the top section.

1

φ − 1 ≈ .6180344. Scale the pentagon so h

The pentagon can be folded with book-fold or diagonal symmetry.

Book-fold Symmetry

The pentagon can be oriented at different heights in the square. In this layout, the pentagon meets the topof the square.

Landmarks:a = .5tan(36°) ≈ .3632712b = (1 − a)tan(18°) ≈ .2068857

108°36°

18°

.5

a ≈ .3632712

b ≈ .2068857

1

Folding method:

Fold and unfold in halfon the corners and edge.

1 2

Fold and unfoldon the bottom.

3

Fold and unfoldon the bottom.

4

Repeat steps 2−3on the right.

5


6

Bring the bottom edge to the dot onthe left and the bottom right cornerto the top edge. Crease on the left.

φ ≈ 1.618034


7

Unfold.

8

Fold and unfold onthe left and right.

9

12

13 14 15

Fold and unfold.

Unfold. Pentagon

Diagonal Symmetry

This layout yields the largest pentagon in a square.

.284079

.4424634

.1133907

Length of side of the pentagon ≈ .6257378.For the book-fold method, the length is.6180339 = .52 + .36327122 .

Landmarks.

10 11

Crease on the left.

Pentagon 51

Folding method:

Fold and unfoldby the corners.

1 2 3

Fold and unfoldon the edges.

Bring the dot on the left tothe diagonal crease and thetop corner to the loweredge. Crease on the left.

Repeat steps3−4 on the right.

4

5 6 7

Fold and unfold tothe diagonal. Creaseon the right.

Unfold.

8

9 10 11 12

13 14 15 16

Pentagon

Unfold.

Bring the bottom cornerto the diagonal crease andthe right edge to the upperdot. Crease on the left.

Fold without creasing.

Unfold.

Repeat steps 7−8 inthe other direction.


Hexagon

The hexagon can be folded using book-fold and diagonal symmetry.

Book-fold Symmetry

For this layout, the hexagonis placed in the center.

.25.2886751 ≈ 3/6

.06698721 ≈ (2 − 3)/4

.5

1

Fold and unfold inhalf on the edges.

1

Folding method:

2

Fold and unfold.

3 4

Fold and unfoldin the center.

Fold the cornersto the creases.

5 6 7

Hexagon

12

3 ≈ 1.732

120°

The angle at each vertex = 120°.Let length of each side = 1.Then width = 2, height = 3.

Hexagonal AntiprismHexagonal Prism

Several polyhedra have hexagonal faces. These include the hexagonal prism and hexagonal antiprism.

Hexagon

Hexagon 53

Fold and unfold inhalf on the edges.

1

Folding method:

2

Crease on the left.

3 4

5 6 7

Hexagon

Diagonal Symmetry

This layout yields the largest hexagon in a square.

Length of side of hexagon ≈ .517638.For the book-fold method, the length is .5.

( 3 − 1)/2 ≈ .3660254

.1698729

Unfold. Repeat steps 2−3three more times.


Fold and unfold.

Fold and unfoldon the edges.

8

9 10 11 12

Landmarks.


Heptagon

128.57°

The angle at each vertex ≈ 128.57°.

w

d

β3β

5β

2β2ββ

1. Find width w. Let length of each side of the heptagon = 1. First find length d. By the Law of Sines:

d/sin(5β ) =1/sin(β )

d = sin(5β ) /sin(β )

≈ 1.80194 Now find w, again using the Law of Sines:

w/sin(3β ) = d/sin(2β )

w = (d )sin(3β ) /sin(2β )

≈ 2.24698

β

α

β = 180°/7 ≈ 25.71°α = 5β ≈ 128.57°

2. Find height h.tan(β/2) = 1/(2h)

≈ 2.190643

β/2h

.5

The heptagon meetsthe top of the square.

a ≈ .240787

b ≈ .3730198

c ≈ .2717565

Folding directions will be given for the Heptagon using book-fold symmetry.

Book-fold Symmetry

There are polyhedra that have heptagonal sides. This includes a heptagonal pyramid and antiprism. In thiscollection, there are some heptagonal dipyramids, but their faces are triangles.

Some calculations are required to determine the landmarks.

Landmarks:a = (.5)tan(180°/7) ≈ .240787

b = 1 − (.5)tan(360°/7) ≈ .3730198c = .5 − tan(180°/14) ≈ .2717565

h = 1/(2tan(β/2))

Heptagon 55

Heptagon

Fold and unfold inhalf on three edges.

1

Folding method:

2

Fold and unfold.

3 4

5 6 7

Bring the left edge to thecrease on the left and the topedge to the bottom rightcorner. Crease on the left.

Unfold.


Fold and unfoldcreasing lightly.

8

9 10 11 12

13 14 15 16

Crease on the left.

Unfold.

Unfold.


Octagon

Folding method:

1 2 3

Crease on the left.Fold and unfoldby the corners.

4 5 6

7 8

Unfold and rotate 90°. Repeat steps 2−4three more times.

Refold.

Octagon

135°

The angle at each vertex = 135°. Let the length of each side = 1. The heightof the octagon = 2.414236 = 1 + 2.

2.414236 ≈ 1+ 2

.7071067 ≈ 1/ 2

Some polyhedra have octagonal sides, including an octagonal pyramid and antiprism.

Silver Rectangle 57

Silver RectangleThe silver rectangle has sides proportional to 1 × 2. This is the same as 2/2 × 1, or .7071 × 1. Whenfolded in half, it keeps the same ratio. The European paper formats A4, A5, and A6 come in this proportion.

Folding method:

1 2 3

Unfold.Fold and unfold.

4 5

Silver Rectangle

The silver rectangle is associated with 2 and 45° lines. The Tall Triangular Antiprism has faces that are45° isosceles triangles.

Tall Triangular Antiprism with layout and some silver rectangles marked with dotted lines.

1

2/2 ≈ .707 A silver rectangle is dividedinto two silver rectangles.


Bronze RectangleThe bronze rectangle has sides proportional to 1 × 3. This is the same as 3/3 × 1, or .57735 × 1. It isassociated with equilateral triangles.

Folding method:

The bronze rectangle is associated with 3 and 30° or 60° lines. The folds for the Tetrahedron use the bronzerectangle.

1

3/3 ≈ .57735 The diagonals outlinetwo equilateral triangles.

1 2 3


4 5

Bronze Rectangle

Tetrahedron with layout and a bronze rectangle marked with dotted lines.

Golden Rectangle 59

Golden RectangleThe golden rectangle has sides proportional to 1 × 1.618034. This is the same as .618034 × 1.

The name comes from the golden mean (phi = φ), φ ≈ 1.618034 ≈ ( 5 + 1) / 2. It is the solution to

x − 1 = 1/x

φ − 1 ≈ .618034

Folding method:

The golden rectangle is associated with φ and pentagons. The folds for the Egyptian Pyramid use the goldenrectangle.

1

.618034

Fold and unfold.

5

Golden Rectangle

Egyptian Pyramid, one side, and layout with a golden rectangle marked with dotted lines.

The golden rectangledivides into a square anda smaller golden rectangle.

1 2 3

4


2

φ

This number is associated with nature and beauty.We saw one example with the regular pentagon:if the length of each edge is 1, then the width is φ.There are many examples of things in nature thatare divided by the golden mean, including thestructure of the human body. If an arm is 1.618units in length, then the lower arm is of length 1and the upper arm is of length .618. The sameapplies to the distance from elbow to wrist andthe length of a hand.


Let us look into the calculations behind the folding directions for these three rectangles.

Silver Rectangle

Golden Rectangle

BronzeRectangle

22.5°

a

1

Find a.tan(22.5°) = a/1

a = tan(22.5°)

≈ .4142135

= 2 −1

22.5°

a

1

b b

Find b.2b + a = 1

b = (1 − a)/2

≈ .2928932= (2 − 2)/2

22.5°

a

1

b b

cFind c.c = a + b

≈ .7071067

= 1/ 2

SilverRectangle

Bronze Rectangle

Find angle α.

By setting up these folds, a (in the third drawing) haslength 1. By symmetry, b also has length 1. An equilateraltriangle is formed. Since α bisects 60°, α = 30°.

1

60°a

b

cFind c.tan(30°) = c/ 1

c = tan(30°)≈ .57735

= 1/ 3

Find angle α.tan(α) = 1/2

α = arctan(1/2)

≈ 26.565°

Find angle β.α + 2β = 90

β ≈ 31.717°

Find c.tan(β) = c/ 1

c = tan(β)

≈ .618034

= ( 5 − 1) 2

= φ − 1

GoldenRectangle

1

.5

α

ββ

1

.5

c

ββ

α

Derivation of the Rectangles

α

Platonic andRelated Polyhedra

PartII

63

Tetrahedron Design

Crease Patterns

Let s be the length of a side of a triangle in a 1 × 1 square. A larger s value yields a larger tetrahedronfrom the same size paper.

The tetrahedron is composed of four equilateral triangles.

Even symmetrys = .5 Even symmetry

s ≈ .5176This makes the largest tetrahedron, but it does not have enough tab to hold.

Odd symmetrys = .5

Odd symmetrys = .4

Though this pattern would make asmaller tetrahedron, it could be used for effects such as a two-color versionand is used for the Striped Tetrahedron.

Even symmetrys = .5

Not symmetrics = .5

There is a tab on the right side.Without the tab, this would haveodd symmetry. Used for theTetrahedron (diagrammed).

Layout of the Tetrahedron

There are two main layouts. One is a triangle partitioned into four parts; the other is a band of four triangles.

tab

64 Part II: Platonic and Related Polyhedra

Fold the edges in so that a band of four triangles can be formedwhere each triangle shows both sides (colors) of the paper.

Striped Tetrahedron Fold the edges in so that the band of four triangles has a stripegoing through it.

Dimpled Truncated Tetrahedron Tetrahedron in an Octahedron

(Sink the four corners of a tetrahedron inand then back out but not all the way.)

Stellated Tetrahedron

Tetrahedron BaseSeveral polyhedra can be formed using the tetrahedron as a base. These models have the same surface asa tetrahedron and thus share the same underlying structure.

Color Patterns

It is interesting to fold a tetrahedron with color patterns. Fold the square to show both sides of the paperand then choose a layout to create the pattern. Fold the tetrahedron from the layout.

Duo-Colored Tetrahedron

Tetrahedron 65

Tetrahedron

1 2 3

4 5 6

Unfold.Fold and unfold onthe top and bottom.

Fold and unfold. Unfold.

The tetrahedron, a Platonic solid composed of four triangles, is one of the simplest polyhedra. The creasepattern shows that it is constructed with a band of four triangles.


7 8 9

10 11 12

13

15

Fold along the crease.

Fold along the crease.Unfold.

tab

Refold and tuck the tab inside. Thencrease along the edges of thetetrahedron to give it crisp edges.

Tetrahedron

14

Unfold.

Duo-Colored Tetrahedron 67


Each face of this tetrahedron shows both sides of the paper. Two opposite edges are folded a certain amountto show both sides of the paper. A band of four triangles with both colors gives the layout. The paper isdivided into thirds.


21 3

Unfold.Crease at the bottom. Fold and unfold.Rotate 180°.

Repeat steps 2−4. Fold and unfold. Bring the dots to thebold lines. Crease atthe bottom.

Unfold.

4

5 6 7 8


9 10 11

12 13 14

15 16 17

18 19

20

Unfold and rotate 180°.

Repeat steps 10−11.

Rotate 90°.Unfold.


Fold and unfold to divideinto thirds. Rotate 90°.

12

tab

Fold along hidden creases to formthe tetrahedron. Tuck the tab inside.

2

1

1/3 1/3 1/3

Striped Tetrahedron 69

Striped Tetrahedron

To fold this model, a striped cylinder is formed. With a few folds, the cylinder turns into the striped tetrahedron.The paper is divided into fifths.


21 3

Unfold.Crease at the bottom.

Fold and unfold.Rotate 180°.

4


5 6


7 8 9

10 11 12

13 14 15

16 17

Fold and unfoldat the bottom.


Repeat steps 7−12. Turnover and rotate 90°.

Fold and unfoldalong the crease.

Bring the lower right corner to thetop edge and the bottom edge tothe left center. Crease on the right.

Fold and unfold at the edges.

Fold and unfold allthe layers together.

Unfold.

Striped Tetrahedron 71

18 19

20 21 22

2324

25

Do not crease. Tuck inside but do not crease.The dots will meet. Curl themodel to make a cylinder.

Turn the cylinderinto a tetrahedron.

Striped Tetrahedron

Fold and unfold all the layerstogether along the creases.

Fold and unfold allthe layers together.


Tetrahedron of Triangles

1 2 3

Fold and unfold tofind the quarter mark.

Fold and unfoldby the diagonal.

4

Rotate.

1

2

3/7

4/7

5


6

Unfold all butthe thinnest flap.

Each face of this tetrahedron has a white inner triangle. The layout shows four triangles in a triangulararrangment, each with inner white triangles. The paper is divided into sevenths.

Tetrahedron of Triangles 73

7 8

Fold and unfold on theupper and lower parts.


18

9

10 11

Unfold. Fold along the creases.

12

Squash-fold.

13 14 15

16 17

Fold along a partiallyhidden crease.


19 20

Turn over and rotate 180°.

21

22 23

Unfold.


Fold to the center.

24

25 26 27

Squash-fold. Thedots will meet.

Fold and unfold. Unfold to step 25.

Tetrahedron of Triangles 75

28 29

30

Squash-fold at the top.

Lift up at the upperdot. Rotate so that Ais at the bottom.

The model is 3D. The view is of theinside of one completed corner of thetetrahedron. Tuck the tab inside. Thedots will meet.

31

32 33

Tuck inside the pocket.

Tetrahedron of Triangles

A



This model resembles four tetrahedra, each on the faces of a central tetrahedron. The crease pattern shows3/4 square symmetry.

2 3

Fold and unfold atthe top and bottom.

Crease at the topand bottom.

Unfold. Fold and unfold atthe top and bottom.

4 5 6

1

Fold and unfold.

Stellated Tetrahedron 77

Unfold.

10 11 12

13 14 15

16 17 18

Fold and unfold. Unfold and rotate 90°.

Repeat steps 10−15 twomore times. Rotate.

Bisect the angle andpush in at the dot.

Fold and unfold.

7 8


9


19 20

21 22

23 24

Bisect the angle andpush in at the dot. Rotate to view the outside so

the center dot is at the top.

Repeat two more times. Bring the dots together.Repeat all around. Rotateto view the bottom.

Each of the three tabs creates a pocket.Interlock the tabs into the pockets.


Dimpled Truncated Tetrahedron 79


Dimpled Truncated Tetrahedron

1 2 3

4 5 6

Unfold.


The dimpled truncated tetrahedron is formed from a tetrahedron. In the crease pattern, the darker regionsshow the sunken parts.


7 8 9

10 11 12

13 14


15

16 17 18

Fold and unfold. Fold and unfold. Fold and unfold.

Fold and unfold.Fold and unfold. Fold and unfold.

1 2

3

Dimpled Truncated Tetrahedron 81

19 20 21

22 23 24

25 26 27

28 29

Unfold.

Fold and unfold. Lift up but do not flatten.

Push in at the lower dot.The other dots will meet.

Lift up on the left and right. Fold thelayers behind on the right to form asunken triangle in the center (so thatthe inner pair of dots meet). Flatten theinside layers against the outside ones(so that the outer pair of dots meet).

Rotate to viewthe right side.

Locate the crease from step 22 andrepeat steps 24−26 two more times.

Dimpled Truncated TetrahedronForm a sunken triangle withthree interlocking reverse folds.

Fold along partiallyhidden creases.

83

Cube Design

Thus there are eleven nets of the cube.

There is only one arrangement containing only two squares in a row.

The cube is composed of six square faces. The different layouts of the six faces yield many design possibilitiesfor the cube and related polyhedra.

There are four arrangements of three squares in a row.

One arrangement of twocolumns. This is the mostefficient for a strip, such asfrom a dollar bill.

This layout shows 3/4square symmetry.

Familiar cross pattern,even symmetry.

Odd symmetry. No symmetry.Even symmetry.

Layout of the Cube

There are six possible combinations of a band of four with one on each side.


Even/odd symmetry. Sidesdo not need to meet.

Square symmetry. One face willhave lines going through it.

Waterbomb (a traditional origami model).

More Arrangements

s = .25

Waterbomb. Cross layout. 3/4 square symmetry.Odd symmetry.

Diagonal orientation increases efficiency. The first two do not have enoughtab to lock, but the third one does, making it my most efficient cube (to bediagrammed in a future volume).

Most efficient, though notab. Still, can be useful formodels using a cube base.

.25 < s < .283

Gives someroom for tab.

s = 1/(4cos(22.5˚) ≈ .2706

Layout for my secondmost efficient cube,diagrammed here.

a

a = 22.5º

a

Fours corners form thelock on one of the sideswith an X pattern. It isuseful as a cube base.Conveniently, a = 1/3.

s = 1/(4cos(arctan(.5))) ≈ .2795

Most efficient layout for arrangementusing 3/4 square symmetry (cleanfaces). Though not enough tab, usefulas a cube base. This crease pattern isused for the Cubehemioctahedron.

a

a = .25

Crease Patterns

Let s be the length of a side of a small square in a 1 × 1 square. A larger s value yields a larger cube fromthe same size paper.

s = 2 5 ≈ .283 s = 2 4 ≈ .3536

s = 1 13 ≈ .27735

Cube Design 85

Cross pattern.Two opposite faces are white. 5 × 5 grid.

Triangles on Cube

Striped Cube By simply folding two edges as shown, the arrangement of two columnsyields a cube with stripes on each face.

3/14

Fold the edges in so that a band goes through the center of the layout.

Color Patterns

It is interesting to fold a cube with color patterns. Fold the square to show both sides of the paper and thenchoose a layout to create the pattern. Fold the cube from the layout.


Stella Octangula

This is a union oftwo tetrahedra.

Triakis Tetrahedron

Stellated Octahedron Dimpled TruncatedOctahedron

Dimpled GreatRhombicuboctahedron

Cubehemioctahedron

Dimpled Rhombicuboctahedron Dimpled Hexagonal Dipyramid

a

a = (3/4) 45° = 33.75°

Hide the white paper to formthe Triakis Tetrahedron.

a

a = 30°

Hide the white paper toform the Stella Octangula.

These models are also related to the cube.

Cube Base

Several polyhedra can be formed using the cube as a base. These models are isomorphic to the cube andshare the same surface.

Cube 87

Cube

1 2 3 4

5 6



7 8

This cube was designed using trapezohedral (or antidiamond) symmetry. The crease pattern shows 3/4 squaresymmetry.

Fold along a partiallyhidden crease.


9

Unfold.

10

Fold and unfoldalong the creases.

11

12

Fold and unfold. Rotate 90°.

13 14

Puff out at the dot in the centeras the model becomes 3D.

15 16

Flatten the inside paper shown withthe x-ray lines. Repeat step 15 twomore times. Rotate the loose cornersto the top. One side has more layers.(We are looking slightly underneaththe model.)

Fold and unfold.Repeat on theother two flaps.

17

18

Interlock the three tabs toclose and lock the model.

19

Puff out at the dot in the centerand flatten inside. (We are lookingat an outside corner of the model.)

Cube

Note the orientation of thevertical crease by the bottom.Fold and unfold.

Repeat step 11three more times.

Striped Cube 89

Striped Cube

1 2 3

4 5 6

Make small marks by foldingand unfolding in quarters.

Fold and unfold. Fold and unfoldin the center.


As seen with the tetrahedra designs, color patterns make for effective models; we now do the same withcube designs. The layout used here is two bands of three squares. By folding two edges toward the center,a layout with stripes is formed, as shown above.


7 8 9

10 11 12

13 14 15

16 17 18

Fold and unfold so thepairs of dots meet.

Unfold. Fold and unfold. Fold the edges tothe crease marks.


Fold and unfold.

Unfold.

Fold and unfold.

Fold along the creases.

Striped Cube 91


Fold and unfold. Puff out at the dot.Fold and unfold alongthe creases. Rotate.

Tuck inside.

Striped Cube

19 20 21

22 23

24


Triangles on Cube

1 2 3

4 56


Fold and unfold. Fold and unfold creasingalong the diagonal.

Fold and unfold creasingalong the diagonal.

The layout shows three rows of two squares. The paper is divided into sevenths.


Triangles on Cube 93

10 11 12

13 14 15

16 17 18

7 8 9

Unfold.

Unfold.

Fold and unfold.

Fold and unfold. Puff out at the dots. Rotate so the dot isfront and center.

Unfold and rotate 180°. Repeat steps 11−13.Turn over and rotate.


19 20

21

22

Turn over and repeat. Puff out at the dot. Begin to tuckinside. Turn over and repeat.Rotate to view the opening.

Triangles on Cube

A cB

Region A goes under B and tucksinto the pocket under C. Repeatbelow simultaneously.

Cube with Squares 95

Cube with Squares

1 2 3

4 5 6

Fold and unfoldalong the diagonals.

Fold and unfold butnot in the center.

Unfold.

Bring the lower right corner tothe top edge and the bottomedge to the left center. Creaseon the left and right.

Each face of this cube has a diamond color pattern. The paper is divided into ninths.

The 2/9 and 4/9marks are found.

4/9

2/9


7 8 9

10 11 12

13 14 15

16 17 18

Squash-fold. Squash-fold. Squash-fold.

Pull out the hidden corner. Squash-fold.

Repeat steps 11−12 onthe three other corners.

Turn over and rotate.

Repeat steps 14−15three more times.

Fold and unfold. Rotate 90°. Repeat step 17three more times.

Cube with Squares 97

19 20 21

22 23 24

25 26 27

28 29

Unfold.

Fold and unfold. Repeat steps 25−26on the right.

This is a view from the top.Tuck the layers inside.

Cube with Squares

Push in at the dots. Rotate.


Stellated Octahedron

1 2 3

4 5 6



Crease on the left.


The stellated octahedron has the same surface as a cube. The layout shows square symmetry. The paper isdivided into sixths.

1

2

Stellated Octahedron 99

7 8 9

10 11 12

13 14 15

16 18


Fold and unfold. Fold and unfold. Rotate 90°. Repeat steps 10−11three more times.

Puff out atthe dot.

Fold and unfold.

Repeat steps 13−15three more times. Stellated Octahedron

Tuck inside. The dots willmeet. Repeat three moretimes going around.

17

Fold and unfold. Rotate.


Cubehemioctahedron

1 2 3

4 5 6

The cubehemioctahedron is basically a cube with sunken corners. This model uses 3/4 square symmetry.The darker paper in the crease pattern shows the sunken sides.


Fold and unfold.

Unfold.

Fold and unfold.

Cubehemioctahedron 101

Repeat steps 4−6 inthe opposite direction.

7 8

Fold and unfold.Fold along the creases.

9

10 11 12

13 14 15

Fold and unfold.Fold and unfold. Repeat steps 10−11on two other sides.

16 17 18


Unfold.

Repeat steps 13−14 onthe three other sides.

Fold along hidden creases.


Unfold. Repeat steps 17−19on the other sides.


Turn over and rotatethe dot to the bottom.

19 20 21

22

Push in at the center toform a sunken triangle.The pairs of dots meet.

23 24

25 26 27

28 29 30

Push in at the upper dotto form a sunken triangle.

Fold both layers together. Rotate so the corner with thedot shows at the front left.

Push in at the dot toform a sunken triangle.

Repeat steps 27−28 onthe corner with the dot.

Rotate the dotto the center.

Tuck inside.

Cubehemioctahedron 103


31 32

33 34

35


Repeat steps 31−32 twomore times going around.

Begin to form the bottom sunkentriangle by folding toward theinside center and tucking.

Repeat step 34 two moretimes to complete thebottom sunken triangle. Cubehemioctahedron

36


Dimpled Rhombicuboctahedron

31 2

4 5 6

Fold and unfold butnot in the center.

Unfold.

Crease at the ends. Unfold and rotate.

The dimpled rhombicuboctahedron has the same surface as the cube. The dark regions of the crease patternshow the sunken sides. This model uses square symmetry. The paper is divided into tenths.

Dimpled Rhombicuboctahedron 105

7 8 9

10 11 12

13 14 15

16 17 18

The one tenthmark is found.

Unfold.


Repeat steps 7−9. Fold and unfold. Rotate 90°. Repeat steps 7−10.

Fold and unfold four times. Fold and unfold,extending the creases.

Fold and unfold on the left anda little on the right. Rotate 90°.

Repeat step 16 threemore times. Rotate.

Fold and unfold onthe left and in themiddle. Rotate 90°.

Fold and unfold.

1/10


19 20

21 22

23 24

25 26 27

Repeat step18three more times.

Fold and unfoldall the layers.

Repeat steps 20−23 three moretimes. Rotate the top to the bottom.

Fold and unfold. Repeat steps 25−26three more times.

Push in at the uppercenter dot to form asunken triangle. Theother pairs of dots meet.

Push in at the lower dot.Rotate a little to the right.

Tuck inside.

Dimpled Rhombicuboctahedron 107

28 29

30

Mountain-fold to forma flat square with fourwhite triangles.

Unfold back to step 28.

Tuck each tab to forma four-way twist lock.

31

Dimpled Rhombicuboctahedron

32

Rotate the topto the bottom.


Stacked Cubes

5 6

Unfold in the center.

1 2 3

Fold and unfold but donot crease by the top.

4

Fold and unfold to findthe quarter mark.

Fold and unfoldby the diagonal.

Rotate.

1

2

3/7

4/7

This model resembles two connected cubes. The paper is divided into sevenths. The crease pattern shows thatthe model uses even symmetry. The darker regions refer to the sunken triangle at the bottom, on which themodel stands.

Stacked Cubes 109

7 8

Unfold and rotate.

9

10 11 12

13 14 15

16 17 18

Repeat behind. Fold and unfold.Repeat behind.

Fold in thirds.Repeat behind.

Unfold.

Fold and unfold.Fold and unfold.

Fold and unfold.

Mountain-fold along the crease.


19 20 21

22 23

Fold and unfold.Fold and unfold. Unfold.


24

25 26 27

28 2930

Unfold.

Repeat steps 26−28 on the left. Repeat steps 26−27on the left.

Valley-fold alongthe creases.

1

2

2

3

Stacked Cubes 111

31 32 33

34 35

Unfold, but keepthe model 3D.

Unfold.

1

1

2

2

Continue wrappingaround. Rotate.

36 37

View from the top.Repeat below.

View of top.

View of bottom.

Push in on the bottom. This willform a sunken triangle, on whichthe model will stand.


38 39

40

Tuck inside, over thefull length of the model.Lift up at the dot.

Open some layers to connectand lock while refolding theedge from step 31.

Stacked Cubes

View of top.

View of bottom.

113

Octahedron Design

Layout of the OctahedronHere is a selection of some of the possible layouts.

The octahedron is composed of eight equilateral triangles. It can be used as a base to create morepolyhedra.

Even/odd symmetry.

s = 1/3 ≈ .33333

Odd symmetry.

Not enough tab. Not enough tab.

h

b

Odd symmetry.

Even/odd symmetry.

Square symmetry.Odd symmetry.

No symmetry.

Band of six triangles.

Crease Patterns

Let s be the length of a side of a triangle in a 1 × 1 square. A larger s value yields a larger octahedron fromthe same size paper.

If b = 1 thenh = 3 2

s = 2 3 + 3( ) ≈ .299 s = 1 2 3( ) ≈ .2887 s = 2 3 3( ) ≈ .3849


Color Patterns

Duo-Colored Octahedron The triangles shown by the dotted lines will soon be white.

Striped Octahedron

Octahemioctahedron Dimpled Truncated Octahedron Stellated Cube

By folding two edges as shown, the arrangement of two columnsyields an octahedron with stripes on each face.

These three models have the same surface as the octahedron. The octahemioctahedron and dimpled truncatedoctahedron are formed by sinking the six vertices.

Odd symmetry, bandalong diagonal.

s ≈ .3239

Square symmetry yields themost efficient layout but notenough tab. Still, it can beuseful as an octahedral base.

To be diagrammed in“Antiprism Design” ina future volume.

My most efficient design.By rotating the diagonala small amount, the fourcorners can lock themodel. Diagrammed.

1/10

Octahedral BaseSeveral polyhedra can be formed using the octahedron as a base.

s = 3 5 ≈ .34641 s = 1 6 ≈ .408 s = 123 30 ≈ .3697

Octahedron 115


Octahedron

4 5 6

The design for this octahedron has square symmetry. The model closes with four thin tabs interlocking atthe top, a concept called a twist lock. The thin tabs allow for efficient use of paper.

1 2 3

Unfold and rotate 180°.Bring the lower right corner to thetop edge and the bottom edge tothe left center. Crease on the right.

Repeat steps 2−3and rotate 90°.

Fold and unfold. Turnover and repeat.

1/10



7


8


9


10 11

Fold and unfold onthe left. Rotate 90°.

12


Fold and unfold.

13

Fold and unfold.

14 15

16 17 18


Unfold and rotate 90°. Repeat steps 5−7.

Octahedron 117

Push in at the dot.Squash-fold to form a triangle.

19

The angles of the white triangle are notimportant. Repeat steps 20−21 threemore times. Rotate to view the outside.

Open to fold inside andunfold. Repeat behind.

Open and flatten. Followthe dot in the next step.

Unfold the thin flaps.Repeat behind.

Octahedron

Open the model and bring thedots together. Close the modelby interlocking the four tabs. Thetabs spiral inward. This methodis called a twist lock. (Fold alongthe creases in step 25.)

20 21

22 23 24

Flatten. Turn over and repeat.

Fold and unfold.Repeat behind.

25 26 27


28 29 30


Striped Octahedron

1 2 3

4 5 6

Fold and unfold. Fold at the bottom. Unfold and rotate 180°.

Repeat steps 2−3.

Each face of this octahedron is striped, and this model is similar to the Striped Cube. The layout shows twobands of four triangles each.

Striped Octahedron 119

7 8 9

10 11 12

13 14 15

16 17 18

Unfold.

Fold and unfold.

Unfold. Repeat steps 9−10three more times.


Unfold. Fold and unfold. Fold the edges tothe crease marks.


19 20 21

23 24 25

26 27 28

Fold and unfold.

Fold and unfold. Puff out at the dot. Fold and unfold alongthe creases. Rotate.

Repeat steps 24−25. Tuck inside. The dots will meet.

Striped Octahedron

Fold and unfold.

22

Duo-Colored Octahedron 121

Duo-Colored Octahedron

1 2 3

4 5 6


Fold along the hidden crease.


Four sides of this octahedron are of one color, and the other four are white. Step 17 shows the formationof two white triangles. By step 27 all four white triangles are formed.

Layout. Step 17.


7 8 9 10

11 12 13 14

15 16 17 18

Unfold.

19 20 21

Unfold.

Fold and unfold. Fold and unfold alonghidden creases.

Fold the top layer. Fold and unfold.

Fold and unfold. Fold and unfold. Rotate 90°.

Duo-Colored Octahedron 123

22 23

24 25

26 27 28

29 30

Fold into the middle layer. Reverse-fold.

Wrap around.

Repeat steps 23−24. Fold all the layerstogether and unfold.

Fold all the layerstogether and unfold.

Open to view the inside.

Note the pocket. If the pocket isat the bottom, turn over and openagain. Fold, then flatten.

tab

Puff out at the upper dot and at thesimilar dot behind. Tuck the tab intothe pocket shown in the previousstep. The lower dots will meet.

31 32 33

Tuck inside.

Duo-Colored Octahedron


Stellated Cube

This model resembles eight pyramids, each on one face of a central cube. The layout shows square symmetry.The surface of this shape is the same as that of an octahedron. A side of the octahedron is represented bythe shaded part in the middle drawing.

2 3


Crease at the topand bottom.

Unfold. Fold and unfold atthe top and bottom.

4 5 6

1

Fold and unfold.

Stellated Cube 125

10 11 12

13 14 15

1617 18


Fold and unfold.Fold and unfold. Fold and unfold.

Fold and unfold. Fold and unfold. Unfold and rotate 90°.

7 8


9


19 20 21


Bisect the angle andpush in at the dot.

Repeat step 20 three moretimes. Rotate to view theoutside so that the dot isat the bottom.

22 23 24

Repeat on all the corners. Puff out at the dot in the center. Unfold and rotate.

25

26

27

Repeat steps 23−24 three more times.

Refold and tuck under the papershaded dark gray. Repeat three times.

Stellated Cube

Octahemioctahedron 127

Octahemioctahedron

1 2 3

4 5 6

Make small marks by foldingand unfolding in half.



Rotate 90°. Repeat steps 4−5three more times.

The octahemioctahedron comes from an octahedron where the six vertices are sunken. The structure issimilar to that of the Octahedron, but the tab is 1/16 instead of 1/10. This makes for a larger model. Thedarker regions represent the sunken sides.

1/16


7 8 9

10 11 12

13 14 15

Bring the pair of dots togetherand unfold. Rotate 90°.

Fold and unfold. Fold and unfold onthe left. Rotate 90°.





Unfold.

Octahemioctahedron 129

16 17

18


Fold to the crease line.Only crease on the left.

Unfold.

19 20

21

22

23

Fold to the crease line andunfold. Only crease on thebottom. Rotate 90°.

Repeat steps 17−19 threemore times. Rotate.



24 25

26

27

28

29 30



Push in at the dot.

Repeat step 26 three moretimes to form a sunken square.

Push in at the upper dot toform another sunken square.The other pairs of dots willmeet. Flatten inside, especiallyunder triangle A.

A

Repeat step 28 onthe three other sides.

The last sunken square isformed by four connectedreverse folds.

Octahemioctahedron

31

Dimpled Truncated Octahedron 131

Dimpled Truncated Octahedron

1 2 3

4 5 6

Make small marks by foldingand unfolding in half.


This design comes from the octahedron where the six vertices are sunken—but not all the way as in theoctahemioctahedron. A tab of 1/8 is used. The darker regions in the crease pattern represent the sunkensides.



1/8


7 8 9

10 11 12

13 14 15

16 17 18

Unfold.


Fold and unfold.


Slide the paper on the left sothat you only crease one layer.

Unfold back to step 12and rotate 90°.


Fold and unfold all thelayers along the creases.

Fold and unfold alongthe creases. Rotate.

Dimpled Truncated Octahedron 133

19

20

21

22

23

24

25

26

27

Fold and unfold.



Fold and unfold.

Fold and unfold.

Fold and unfold.Unfold and rotate 90°.


Dimpled Truncated Octahedron

28 29

30


Unfold, turn over,and rotate.

Push in at the dot as themodel becomes 3D.

31 32

33 34

35

Repeat step 30 threemore times to form asunken square on top.

Push in at the upperdot to form a sunkensquare. Flatten inside.


The last sunken square isformed by four connectedreverse folds.

135

More Platonic Solids DesignThe five Platonic solids are convex, have faces that are identical regular polygons, and have identical vertices.We have seen three of them so far, the tetrahedron, cube, and octahedron. The remaining two, the icosahedronand dodecahedron, are shown here.

Icosahedron

Band of 10 triangles with5 above and 5 below.

The icosahedron is composed of 20 equilateral triangles. Here are a few possible layouts and crease patterns.Several layouts have a band of ten triangles.

This version is diagrammed.


Designing an origami dodecahedron is quite a challenge. The dodecahedron is composed of 12 pentagons.Here are a few layouts.

This crease pattern is diagrammed.

The Five Platonic Solids

Icosahedron DodecahedronTetrahedron OctahedronCube

Dodecahedron

Icosahedron 137

Icosahedron

1 2 3

4 5 6

Unfold. Unfold.


The icosahedron is composed of 20 equilateral triangles. The layout shows a band of ten triangles downthe diagonal with five triangles on each side. Odd symmetry is used.


7

Fold and unfold at the top andbottom. Turn over and rotate.

8 9

10 11 12

13 14 15

16 17

Fold and unfold.

18



Fold and unfold at the ends.



Unfold.

Repeat steps 12−15. Push in at the dot.Bisect the angle.

Icosahedron 139

19 20 21

22 23 24

25 26 27

3028

Icosahedron

29

Valley-fold along thecreases for this squash fold.

Push in at the dot.Repeat steps 18−20 on the otherside. Rotate to view the outside sothe dot is in front and at the top.

Puff out at the lower dot. Puff out at the dot. Mountain-fold along theband to reinforce thecreases. Fold and unfold.

Triangle A will cover triangleB. Region C will go under D.The dots will meet.

Tuck inside by wrappingaround the layers.

Unfold back to step 22.

Repeat steps 22−26.Refold steps 22−26 behind.

A BC

D


Dodecahedron

1 2 3

4 5 6

To Plato, this dodecahedron, the quintessence (the “fifth being”), represented the whole universe. Thedodecahedron has 12 pentagonal faces. The crease pattern shows odd symmetry with a line going throughthe center at 36° from the horizontal line.


Repeat steps 2−3. Fold and unfold. Fold and unfold.

36°


Crease on the right.

Dodecahedron 141

7 8 9

Fold and unfold,creasing at the ends.

10 11 12

Fold and unfold.

Fold and unfold.

13 14 15

Fold along ahidden crease.

Fold along a hidden crease.

16 17 18


Repeat steps 11−15. Note the two pentagons.Fold and unfold.

Fold and unfold.

Fold and unfold.


19 20 21

Repeat steps 17−19.Fold and unfold. Rotate 180°.

22 23 24

Fold and unfold.

Note the pentagons withstars. Fold and unfold.


25 26 27

Fold and unfold.Note the pentagons withstars. Fold and unfold.

28 29

Repeat steps 25−27. Fold a thin strip atthe top and bottom.



Dodecahedron 143

30

The folding will become 3D.Bring the three dots together.

Unfold. The dots will meet inside.Rotate A to the center.

31

32 33

Rotate so that the boldcorner is at the top leftand repeat steps 30−33.

34

Unfold to step 30.

35

Bring the three dots together. Themodel will become 3D again.

36

37

38

Rotate and repeat steps36−37 on the opposite edge.

Unfold to step 36.

39

A


Only fold the top layer.Puff out at the dot.

41 42


43 44

45 46

47 48

(Mountain-fold alongan inner layer.) Pullout the second layer.

Squash-foldon the inside.

Tuck inside.

Mountain-fold the second layerwhile unfolding the thin strip.

40

The inner three dots will meet, andthe outer pair of dots will meet.

Spread some paperinside while doing thisfold. Flatten inside.

Dodecahedron 145

49 50 51

This forms the tab to ...(Fold, unfold, and turn over.)

... lock the model ...(Reverse-fold.)

... into this pocket.(Rotate the dot to the center.)

Fold one layer alongexisting creases.

52 53 54

55


DodecahedronTuck into the pocket of theadjacent pentagon. (See step 56.)

Refold along the creases.

56 57

58 59

Note the pocket.

Fold the layers inside andflatten against the outside.

A

Hide the dark paperunder pentagon A.

Sunken Platonic Solids Design 147

A sunken version of a polyhedron is created by taking the center of each face and sinking it towardthe center of the polyhedron. This replaces each n-sided face with n triangular faces. Sunken polyhedraare concave. In general, sunken models hold and lock better than the convex ones.

The group of sunken Platonic solids is interesting. It takes some unusual folds to capture their shapes.

Sunken Platonic Solids Design

A convenient layout has squaresymmetry. This crease patternproduces the largest model buthas no tab.

By rotating 22.5°, there is enoughtab while not compromising thesize of the final model as much.This version is diagrammed.

Rotating by 45° produces aconvenient crease pattern,but the model becomes smalland thick.

Each side is an isosceles trianglewith an apex angle of 90°.

Sunken Octahedron

45° 22.5°

1 1

2


Even symmetry.

Crease pattern with square symmetry.

More samples of 3/4 square symmetry.

The sunken tetrahedron is composed of 24 isosceles triangles. The dimensions of each side is shown above.

The dots in the sunken cubedefine the triangle to the right.

Using 3/4 square symmetry and filling the squarewell, this version is the most efficient for use ofpaper. This version is diagrammed.

The sunken cube is composed of 24 isosceles triangles. The length of one side of those triangles is equalto half the length of the diagonal through the center of the cube. For a 1 × 1 × 1 cube, the length of thediagonal on a face (dotted line) is 2 (by the Pythagorean Theorem). So, the length of the diagonal throughthe center of the cube is 3, and the length of the side on the triangular face is 3/2.

11

2

1

2 3

2 2

3 3

Sunken Tetrahedron

Sunken Cube

1

3 2 3 2

Sunken Platonic Solids Design 149

The crease pattern shows odd symmetry.

A fan turns into a tower tocreate the sunken icosahedron.

Crease pattern.

Sunken Dodecahedron

Sunken Icosahedron

Each side is an isosceles trianglewith an apex angle of 90°.

1 1

2

This sunken dodecahedron is composed of 60 equilateral triangles. Difficult as it is, using equilateral trianglesdoes simplify the design and folding. This model has the same surface as the icosahedron. In the thirddrawing, the gray sides represent one side of the icosahedron. There are two other related shapes with thesame surface: one is a dimpled truncated icosahedron, the other is a dimpled soccer ball.

This shape is so complex I could not find a useful two-dimensional layout. But I imagined how a fan (step22) can turn into this model. The layout begins by dividing into twelfths. Also, because of its complexity,the folded model has a rugged beauty.


Sunken Octahedron

1 2 3

Fold and unfold.

4 5 6

Turn over and rotate.

Unfold and rotate.

Repeat steps 2−5.

All the faces meet at the center, and this model can be viewed as three intersecting squares correspondingto the x, y, and z planes. A polyhedron with zero volume, like this one, can be called a nolid (”not a solid”).This would be an octahedral nolid.

Sunken Octahedron 151

7

8 9

10 11 12

13 14 15

16 17 18

Turn over and repeat.

Squash-fold.

Turn over and repeat.

Mountain-fold along thecrease, sliding the hiddencorner at the dot under area A.

Fold and unfold. Turn over andrepeat steps 13−17.

Fold behind.

Fold along thecrease and rotate.

Open and follow the dot.

A


20 21

22 23 24

25 26

Tuck inside witha reverse fold.

Fold and unfold. Turn over andrepeat steps 20−23.

Repeat steps 25−26 onthe three remaining sides.

Sunken Octahedron

Bring the dots together and lift up onthe left. The model will become 3D.

19

Squash-fold.

Fold behind.

Tuck inside.

27

28

Sunken Tetrahedron 153

2 2

3 3

Sunken Tetrahedron

1 2 3

4 5 6

Fold and unfold. Bring the edge to the centercreasing on the right. Foldand unfold. Rotate 180°.

Fold and unfold.

Fold and unfold on the right.

Fold and unfold.

The sunken tetrahedron is composed of twelve isosceles triangles all meeting in the center. The sides of eachtriangle are proportional to 2 2, 3, 3. This can also be called a tetrahedral nolid. The crease patternshows 3/4 square symmetry.

Mountain-foldalong the crease.


7 8 9

10 11 12

13 14 15

1617

Valley-fold along the existingcrease. Turn over and repeat.

Mountan-fold along the existingcrease. Open and follow the dotinto the next step.

Valley-fold along the existingcrease. Turn over and repeat.

Turn over and repeat. Unfold.

Bring the corner to theline. Crease on the right.

Unfold. Fold and unfold. Rotate 90°.

Repeat steps 13−15 threemore times. Rotate.

Sunken Tetrahedron 155

18 19

20 21 22

23 24 25

26 27

Fold and unfold. Valley-fold along a hiddencrease and mountain-fold to lineup with the edge. Rotate 90°.


Crease to the leftof the middle dot.


Repeat steps 22−24 in theother direction. Rotate 90°.


Unfold.


Two layers.

28

29 30

31 32 33

34 35

36

Puff out at the upper dot asthe model becomes 3D.

Push in at the dot.Note the two layers.

Push in at the dot. Fold all the layersbehind on the crease.

Repeat steps 29−32 two moretimes beginning on the rightand cycling around.

Fold and unfold on threesides. Rotate to view the bottom.

tab

tab

tab

1

23

Fold paper 1, then fold 2 which covers tab 1. Then fold 3 to cover tab 2 andtuck tab 3 under 1. Pushin at the center. Rotate this side to the bottom.

Sunken Tetrahedron

Sunken Cube 157

1

3 2 3 2

Sunken Cube

The sunken cube is composed of 24 isosceles triangles all meeting at the center. The sides of each triangleare proportional to 1, 3/2, 3/2. This beautiful shape can also be called a hexahedral nolid, that is, itcomes from the cube but has no volume. The crease pattern shows square symmetry.z

1 2 3

4 5 6

Fold and unfold. Bring the edge to the centercreasing on the right. Foldand unfold. Rotate 180°.

Fold and unfold.


Fold and unfold. Push in at the dot.


Valley-fold on thecrease. Squash-fold tobisect the angle but onlycrease in the center.

9

10 11 12

13 14 15

1617 18

7 8

Repeat steps 6−7 threemore times and unfold.

Fold and unfold. Turnover and rotate 90°.

Fold and unfold.

Valley-fold on thecrease. Squash-fold tobisect the angle but onlycrease in the center.

Unfold. Fold and unfoldthe top layer alonga hidden crease.

Unfold.

Fold and unfold along a hiddencrease. Turn over and rotate.

Fold along the hiddenedge. Rotate 90°.

Do not crease

Do notcrease

Crease in the center.

Sunken Cube 159

19 20 21

22 23 24

25 26 27

28 29

Crease on the left.

Fold and unfold along the edge. Unfold.

Crease at the bottom. Fold and unfoldboth layers.


Repeat steps 9−24three more times. Form a sunken square on top.

Rotate the top to the bottom.

Push in at the dot as themodel becomes 3D.

Push in at the dot. Slide the paper. There is asmall squash fold at the top.


31

32 33

34

30

Sunken Cube

Push in at the dot. Fold and unfold.


tabtabtab

Fold the four corners inside toform the last sunken square.Interlock to cover the tabs.

Fold and unfold.

35

Sunken Dodecahedron 161

Sunken Dodecahedron

This sunken dodecahedron, one of several stellated icosahedrons, is composed of 60 equilateral triangles.This version has the same surface as an icosahedron. The crease pattern shows odd symmetry.


21 3

Unfold.


4


5 6


7 8 9

10 11 12

13 14 15

16 17 18

Fold and unfold.


Fold and unfold.

Unfold.

Fold and unfold.Fold and unfold. Fold and unfold.Rotate 180°.

Fold and unfold.

Fold and unfold.

Fold and unfold. Repeat steps 14−17.


19 20 21

22 24

25 26 27

23

28 29

Fold and unfold. Fold and unfold. Fold and unfold twice.Rotate 180°.

Repeat steps 20−21. Fold and unfold. Fold and unfold.Rotate 180°.

Repeat steps 23−24. Rotate 90°.Fold and unfold.

Push in at the dot. Squash-fold.

1 2


30

33

Turn over and repeat steps 32−34.

34 35

Push in at the dot toform a sunken pentagon.

36

37 38

Repeat steps 28−29 on the other side.Turn over to view the outside.

Push in at the dot andtuck under the left side toform a sunken pentagon.

Push in at the dot to forma sunken pentagon. Turnover and repeat.

Turn over and repeat.Rotate to view the bottom.

Tuck the tabsinto the pockets.

32

31

Push in at the upper dots to formtwo sunken pentagons. Rotatethe lower dot to the center.

Fold a little bit up at the bottom.Puff out at the upper dot.

tab tab

Tuck the tab, shownwith the dot, intothe pocket.

Here are two representations of the same step.


39 40

41 42 43

44

Sunken Dodecahedron

Note the pocket. This lock will loosen overthe next few steps. Be sure to put it backby step 43. Rotate to view the outside sothe dot is on the top at the front. Push in at the dot to form

a sunken pentagon. Turnover and repeat.

Push in at the dot and tuckinside. Turn over and repeat.


Tuck inside the pocket.Turn over and repeat.


Sunken Icosahedron

1 2 3

4 5 6

Fold and unfold. Fold and unfoldalong the diagonal.

This complex shape is composed of 60 isosceles right triangles. It is formed by first making a fan. I gave ita four-star (very complex) rating because of the step sequence 20−33 as the fan is turned into a tower, but the folds before and after are not so difficult. The paper is divided into twelfths.z

Fold and unfold.

Fold and unfold. Thisdivides the paper in thirds.

Fold and unfold.

1

21

2

Sunken Icosahedron 167

7 8 9

10 11 12

13 14 15

16 17 18

Unfold.



Mountain-fold along the creases.

Fold and unfold.

1

2

3

4





19 20 21

22 23 24

25 26 27

Tuck into thelowest layer.

Rotate to viewthe left side.

Make a sunken trianglealong the creases.

Wrap a layer underneath.


28 29


Repeat steps 22−24. Thisis where the edges meet.

This is the most difficult partas the fan becomes a towerin step 34. Until then, themodel will not hold together.

Pop out between thesecond and third bands.

Three layers

Sunken Icosahedron 169

31 32

33 34

35 36 37

38 39 40


View of top.

Bring the dots togetherand then let go.

Repeat step 34 four moretimes going around.

Bring paper frominside to the right side.

Repeat steps 36−37four more timesgoing around.

Rotate and find the sidewith the edge showing.

30

Pull out. Tuck inside.

Finally, the fan hasturned into a tower.Hopefully, the model should be able to hold together.

Fold along a crease.

Rotate so the dotgoes to the front.


41 42 43

44 45 46

47

Fold all the loose edgesgoing all around.

Form a sunken trianglebringing the extrapaper to the front.

Form a sunken trianglewhile folding the extrapaper inside.

Repeat step 43on the right.

Form the last two triangles on theback while bringing the extra paperto the front inside the front triangle.

Bring the dark paper to the frontto lock the folds. Also pushdown to keep the folds in place.

Sunken Icosahedron

Dipyramids andDimpled Dipyramids

PartIII

Dipyramid Design 173

Height/Diameter Formula

A measure of a dipyramid is the proportion of theheight to its diameter. One way to calculate thatis to find the ratio d/c, where d is the distance fromthe center of the polyhedron to the top vertex (i.e.,half the height) and c is the distance from thecenter to a vertex on the equator (i.e., half thediameter):

Let H = d/c

Assume that the length of each side of the polygonbase is 1. Length b of the triangle can be found bylooking at the right triangle created by splitting theface in half:

sin(α/2) = (1/2)/b

b = 12sin(α/2)

Length c on the polygon base is also one side ofan isosceles triangle and can be found by the sameprocess:

sin((360°/n)/2) = (1/2)/c

c = 12sin(180°/n)

With these two measurements, angle γ is found:

cos(γ) = c/b

γ = arccos(c/b)

tan(γ) = d/c = H

So given polygon n and angle α, the proportionof the dipyramid’s height to its diameter is

H = tan arccossin(α/2)

sin(180°/n))

If you want to find α given H and n, then

α = 2arcsin[sin(180°/n)cos(arctan(H))]

Dipyramid DesignA pyramid has a polygon base and triangular sideswhose vertices meet at the top. Dipyramids, ordiamonds, are composed of a pair of identicalpyramids joined at the base. The sides are identicalisosceles triangles. For a polygon with n sides,there are 2n triangles in a dipyramid. A dipyramidis defined by its polygon n and angle α at the topof the triangle. The variety of dipyramids is rich.

α

1c

db

α

γ

diameter

height

( ( ((

174 Part III: Dipyramids and Dimpled Dipyramids

2. Duals of uniform prisms (prisms with two regularn-sided bases and n square sides).

α = arccos[1 – 2(sin(180°/n))4]

I thank Peter Messer for this formula, found inhis paper “Closed-Form Expressions for UniformPolyhedra and Their Duals ” (Discrete &Computational Geometry 27:3 (2002), 353−375).

Groups of Dipyramids

1. Dipyramids inscibed in a sphere.

All of the vertices would rest on the surface ofa sphere and so H = 1.

Since cos(arctan(1)) = cos 45° = 1/ 2

α = 2 arcsin[sin(180°/n)/ 2]

For a given polygon n, here are the angles α:

n

3456789

101112

α

75.52°60°49.12°41.41°35.73°31.40°28.00°25.24°22.98°21.09°

Dipyramids inscribed in a spherefrom triangular to octagonal.

Dipyramids whose duals are uniformprisms from triangular to octagonal.

The uniform pentagonal prism andpentagonal dipyramid are dual pairs.

n

3456789

101112

α

97.18°60°40.42°28.96°21.70°16.84°13.44°10.96°9.11°7.68°

H

.57735011.376381.732052.076522.414212.747483.077683.405693.73205


3. Dipyramids of the same polygon n. For example,for a square base (n = 4),

4. Dipyramids with the same H value (H ≠ 1).Varying the polygonal base will change thevalue of angle α, as in the first group.

5. Dipyramids with the same angle α while varyingthe polygonal base.

Silver SquareDipyramid

Tall SquareDiamond

Squat SquareDiamond

Layouts and Crease Patterns of Dipyramids

Triangular base.(n = 3)

Even/odd symmetry Even/odd symmetry Square symmetry

Square base.(n = 4)

Square symmetry Even/odd symmetry

Base with odd number of sides.(n = 5, 7, ...)

Odd symmetry

Octahedron

H

1/ 21

22

α

70.53°60°48.19°36.87°


Base with even number of sides.(n = 6, 8, ...)

Even/odd symmetry

Landmarks

For dipyramids of polygons with an even numberof sides (n = 4, 6, 8, ...), the layouts have even andodd symmetry. Two lines going through the centerof the layout cross at angle α. Two horizontal linesdistance b from the top and bottom edges determinethe height of the faces. This value b defines theangle of the faces, α, and sets the layout.

Given α

tan(α/2) = (1/2 − a)/(1/2)a = (1/2)(1 − tan(α/2))

The value b should be set as small as possible (toincrease the size of the model) but large enoughto allow for a tab; b can be chosen as an easy-to-find landmark.

For dipyramids of polygons with an odd numberof sides (n = 5, 7, ...), the layouts have oddsymmetry. One line going through the center atan angle is related to the angles of the triangularfaces. Two vertical lines distance b from the leftand right edges are close to the center. This valueb defines the angles of the faces and sets the layout.

Given α

β = (180° – α)/2 = 90° – α/2γ = 90° – β = α/2tan(γ) = tan(α/2) = 1 – 2aa = (1/2)(1 – tan(α/2))

The value of b is between a and 1/2; it is chosenso that c is small but large enough to allow for atab; b can be chosen as an easy-to-find landmark.

a a

b

α

α

α1/2 − b

a

b c

a

β

γ

b

α

β


n

3456789

101112

α

90°60°45°36°30°25.714°22.5°20°18°16.3636°

Using the layouts above, another interesting groupof dipyramids would be those with crease patternsthat include the horizontal line.

Find α give a polygon n,

(n/2)α + β = 180°β = (180° – α)/2 = 90° – α/2

so

(n/2)α + 90° – α/2 = 180°

α = 180°/(n – 1)

From triangular to decagonal dipyramids.

α β


Triangular Dipyramid 90°

The angles of each of the six triangles are 90°, 45°, and 45°.

1 2 3

4 5 6

Unfold.

Fold and unfold. Fold and unfold atthe top and bottom.

Fold and unfold.

Triangular Dipyramid 90° 179

16

7 8 9

10 11 12

13 14

Fold and unfold.

Place your finger inside the model to open it. Bring the lower dots together and puff

Tuck inside the pocket.Turn over and repeat.

15

Note the pocket.




Triangular Dipyramid 90°



out at the upper dot.



1 2 3

4 5 6

Fold and unfold. Rotate.


Bring the left edge to the lowerdot. Crease at the intersection.

Unfold and rotate 180°.Repeat steps 3−4.Turn over and rotate. Fold and unfold but do

not crease in the center.

This dipyramid is the dual of the triangular uniform prism. Each of the six triangles has sides with proportions2, 2, and 3. The angles are 41.41°, 41.41°, and 97.18°.

Triangular Dipyramid 181

7 8 9

10 11 12

13 14 15

16 17 18

Bring the edge to the dotcreasing in the center.

Unfold.

Fold and unfold. Rotate 180°. Repeat steps 7−10.

Fold and unfold the top layeralong a hidden crease.

Unfold. Fold and unfold. Rotate 180°.


Unfold. Do not crease in the center.


19 20 21

22 23 24

25 26 27

28 29 30

Unfold. Mountain-fold along the creases.

Fold along a hidden crease.Turn over and repeat.

Unfold.

A

B

Lift the top layer up atthe dot. Bring corner Ain front of corner B.

Fold and unfoldseveral layers.

Turn over and repeatsteps 25−26.

Fold the layers inside. Tuck inside.


Triangular Dipyramid in a Sphere 183

Triangular Dipyramid in a Sphere

1 2 3

4 5 6

Fold and unfold. Fold and unfold on the left. Fold and unfold onthe left. Rotate 180°.

Repeat steps 2−3. Fold and unfoldcreasing at the edges.

Fold and unfold.

This triangular dipyramid is inscibed in a sphere. The crease pattern shows 3/4 square symmetry. Theangles of each of the six triangles are 75.52°, 52.24°, and 52.24°.


7 8 9

Fold and unfold. Rotate 90°. Repeat steps 6−7three more times.

10 11 12


Fold and unfold alongthe hidden crease.

Unfold.

13 14 15



Push in at the dot in thecenter and bisect the angle.Valley-fold along the crease.

16 17 18

Rotate 90°. Repeat steps 15−17two more times.

Triangular Dipyramid in a Sphere 185

19 20 21

Fold and unfold onthree of the four sides.

Fold and unfold all the layers.

22 23 24

Unfold. Tuck under the darker layer.

25 26 27

Lift up the two flaps.

tab

tab

tab

Bring the three dots togetherand close the model with athree-way twist lock.

Triangular Dipyramidin a Sphere

Valley-fold along the crease.


Tall Triangular Dipyramid

1 2 3

4 5 6

This is one of the simplest dipyramids in this section, but it uses several of the same folding techniques asother ones. The angles of each of the six triangles are 45°, 37.5°, and 37.5°.

Fold and unfoldcreasing at the ends.




Tall Triangular Dipyramid 187

7 8 9

10 11 12

13 14 15

16 17

Bring the lower left corner tothe crease. Fold on the left.


Crease underthe center line.

Fold and unfold along apartially hidden crease.


Repeat steps 10−12. Fold along the crease.

12


12


18

Unfold.


Lift up at the dot and the model willbecome 3D. Rotate to view the openingso that region A is at the top right.

Tall Triangular Dipyramid

Fold and unfold.

19 20 21

Tuck and interlock thetabs. The dots will meet.


Reverse folds.

22 23

2524

tab tab

A

A

tab

tab

Rotate.

Tall Square Dipyramid 189

1

2

Tall Square Dipyramid

1 2 3

Fold and unfold.

4 5 6

Valley-fold along the crease.Turn over and repeat.


The height of this square dipyramid is twice the diameter. According to the formulas for H and α derivedearlier, the small angle in each triangle is about 36.87° and a convenient landmark of 1/3 is used to achievethe dimensions.

Fold and unfold to findthe quarter marks.

Crease at the edges.

1/3


7 8 9

10 11 12

13 14 15

16 17 18

Extend the crease between thedots and mountain-fold alongit. Make four creases so thatthe pairs meet in the center.

Unfold.


Unfold. Repeat steps 11−14 on the leftand above on the left and right.


Push in at the dot.

Fold and unfold the toplayer along a hidden crease.


Unfold.

Tall Square Dipyramid 191

19 20

21 22

25

Fold and unfold alongthe crease. Rotate 180°.

Repeat steps 17−19.Then flatten.

Reverse folds.

Lift up at the dot to openthe model. Rotate so thatregion A is at the top right.


Fold and unfold.

23

Tall Square Dipyramid

24

tab tab

Atab

tab

A


Silver Square Dipyramid

1 2 3

Fold and unfold.

The ratio of the height of this square dipyramid to its width is 2 to 1. The small angle in each triangle isabout 48° to achieve the dimensions.

Fold and unfold creasinglightly at the bottom.


4 5 6


Repeat steps 2−3.

1

2

Silver Square Dipyramid 193

7 8 9

Unfold and rotate 90°. Fold and unfoldon the edges.

Continue with steps 11−20of the Tall Square Dipyramid.

10 11

Fold along the crease onthe right and rotate 180°.

Push in at the dot. Repeat step 10 andflatten. Then continue with step 24 throughthe end of the Tall Square Dipyramid.

12

Silver Square Dipyramid


Squat Silver Square Diamond

1 2 3

4 5 6

Fold and unfold. Bring the edge to the centerand crease on the left.

Fold and unfold in halftwo times. Rotate 90°.

Bring the edgesto the center.

The ratio of the height of this square dipyramid to its diameter is 1 to 2. The angles of each of the eighttriangles are 70.53°, 54.74°, and 54.74°. The crease pattern shows square symmetry. the the the the the t

Unfold.

1

2


1

2

Squat Silver Square Diamond 195

7 8 9

10 11 12

13 14 15

16 17 18


Valley-fold along the crease. Pushin at the dot and bisect the angle.

Flatten to form atriangle. Rotate 90°.

Repeat steps 15−16 three moretimes. Rotate to view the outside.

Flatten.

Repeat steps 6−9.


Fold and unfold alonga hidden crease.




19 20

21 22

23 24

25

Open to fold inside andunfold. Repeat behind.

Open and flatten. Followthe dot in the next step.

The model becomes 3D as the four dotsare brought together. Close the modelby interlocking the four tabs. These tabsspiral inward with a twist lock.

Turn over and repeat. Fold and unfold. Turnover and repeat.

Unfold the thin flaps.Turn over and repeat.

Squat Silver Square Dipyramid

Pentagonal Dipyramid 197

Pentagonal Dipyramid

4 5 6

1 2 3


Unfold.Bring the lower right corner to thetop edge and the bottom edge tothe left center. Crease on the left.

Repeat steps 2−4. Rotate 90°. Fold and unfold.

This ten-sided dipyramid is the dual of the uniform pentagonal prism. The small angle in each triangle isabout 41°.



7 8 9

10 11 12

13 14 15

16 17 18


Repeat steps 6−7.



Unfold.

Valley-fold along a hiddencrease and mountain-foldthe dot to the left edge.

Fold and unfold all thelayers along a partiallyhidden crease.

Unfold and rotate 180°. Repeat steps 13−16. Crease on the left.


Pentagonal Dipyramid 199

19 20 21

22 23 24

25

27 28

1

2

Fold and unfold the toplayer along hidden creases.

Fold and unfold both layers alongthe hidden crease between the dots.


Repeat steps 18−21. Fold and unfoldalong the creases.

Push in at the dotand bisect the angle.


Reverse folds.

On the right, fold along thecrease between the dots. On theleft, bring the edge to the dot.

26

Open while folding down. Thenflatten to form a triangle. Rotate 180°.


29

Fold the inside layers together by this method: Hold at the upper dot and pull the inside flap at the lower dot toward the center. The model locksat the upper dot. This fold is used in other polyhedra, especially in moredipyramids. This will be called the spine-lock fold. Turn over and repeat.

Inside view.

Lift up at the dot and the model willbecome 3D. Rotate to view the openingso that region A is at the top right.

30

Refold as it was in step 26.Turn over and repeat.

31

Fold and unfold.

32

33

34


Pentagonal Dipyramid

A

tab tab

tab

tab

A

Pentagonal Dipyramid 45° 201

Pentagonal Dipyramid 45°

4 5 6

1 2 3

Fold and unfold onthe top and bottom.

Repeat steps 2−4. Fold and unfoldcreasing lightly.

This ten-sided dipyramid is similar in folding to the Pentagonal Dipyramid. The crease pattern shows aneasier layout since the small angle in each triangle is 45°.

Fold and unfold on theleft creasing lightly.




7 8 9

10 11 12

13 14 15

16 17 18

Fold and unfold at the bottom. Fold and unfold at thebottom. Rotate 180°.

Repeat steps 7−8.



Fold and unfold.

Continue with step 15through the end of thePentagonal Dipyramid. Pentagonal Dipyramid 45°

Unfold.

Mountain-fold alongthe crease and bringthe dot to the left edge.


Pentagonal Dipyramid in a Sphere 203

Pentagonal Dipyramid in a Sphere

4 5 6

1 2 3

Fold and unfold. Rotate 180°. Repeat steps 2−4. Rotate 90°. Fold and unfold on the left.

This is the pentagonal dipyramid inscibed in a sphere. The ratio of the height to the diameter is 1 to 1. Theangles of each of the ten triangles are 49.12°, 65.44°, and 65.44°.


Fold and unfold on the left. Fold and unfold at the top.

1

1


7 8 9

10 11 12

Bring the lower rightcorner to the top edgeand crease on the right.

Unfold. Fold and unfold onthe right. Rotate 180°.

Repeat steps 6−9. Rotate 90°.Continue with steps 10−23of the Pentagonal Dipyramid.

13 14

15 16


Fold on the crease betweenthe dots for this squash fold.

Make a thin squash fold. Fold and unfold.Rotate 180°.

Pentagonal Dipyramid in a Sphere 205

17 18


Reverse folds.

Pentagonal Dipyramidin a Sphere

Fold the inside layers together for thisspine-lock fold. Turn over and repeat.

Lift up at the upper dot. The tabgoes under A as the dots meet.

19 20

Tuck inside.

2221

tabA


Golden Pentagonal Dipyramid

4

5 6

The ratio of the height to the diameter of this pentagonal dipyramid is 1.61803 to 1. The small angle ofeach triangle is 36°. The folding is similar to that of the Pentagonal Dipyramid.

1 2 3




Fold and unfold. Fold and unfoldat the top.

7 8

Fold and unfold atthe top. Rotate 180°.

Repeat steps 6−7.

1

2

1

2

1

2

Golden Pentagonal Dipyramid 207



Unfold. Unfold.


9 10 11

12 13 14

15 16 17 18


Golden PentagonalDipyramid


Repeat steps 13−18. Continue with step 18 through theend of the Pentagonal Dipyramid.

19 20 21



Squat Golden Pentagonal Dipyramid

All the sides of this pentagonal dipyramid are equilateral triangles. The ratio of the height to the diameteris .61803 to 1.

1 2 3

4 5 6

Fold and unfold.

Unfold, turn over, and rotate 90°.

Fold and unfold. Fold and unfold. Turnover and rotate 90°.

Squat Golden Pentagonal Dipyramid 209

7

Fold and unfold.

Squat GoldenPentagonal Dipyramid

8

Fold and unfold.

10 11

Fold and unfold.

12

Reverse folds.

13 14 15

16 17

9

Reverse-fold. Turnover and repeat.

Note the small pocket on theleft. Tuck the corner at the dotinto the similar pocket underA. Turn over and repeat.

The folding is done underregion A. Puff out at the upperdot. The two lower dots willmeet the ones at the edges.Turn over and repeat.

A

There are little pockets on theleft and right. Tuck inside them.Turn over and repeat.

A


Hexagonal Dipyramid

1 2 3

4 5 6

Fold and unfold. Fold and unfold at the bottom. Fold and unfold at the top.



This 12-sided dipyramid is the dual of the uniform hexagonal prism. The small angle in each triangle isabout 29°, and the ratio of its height to the diameter is 3 to 1.

1

3

Hexagonal Dipyramid 211

7 8 9

10 11 12

13 14 15

16 17 18


Unfold. Valley-fold along a hiddencrease and mountain-foldthe dot to the right edge.



Note the intersections at the dots.Fold and unfold along the creases.

Squash-fold and rotate 180°.Push in at the dot.

Fold and unfold tobisect the angle.



19 20 21

22 23 24

25 26


Reverse folds. Fold the inside layers togetherfor this spine-lock fold. Turnover and repeat.


Fold and unfold.

Hexagonal Dipyramid

A

Bring the edge to the dotand tuck under region A.Turn over and repeat.

A

tab tab

A

tab

tab

Lift up at the dot and the modelwill become 3D. Rotate to viewthe opening so that region A isat the top right.

Silver Hexagonal Dipyramid 213

Silver Hexagonal Dipyramid

The ratio of the height of this dipyramid to its diameter is 2 to 1. The small angle in each triangle is about34° to achieve the dimensions. The folding is similar to the Hexagonal Dipyramid.

1 2 3

Fold and unfold. Fold and unfoldcreasing lightly.

Fold and unfold on the left.

Fold and unfold at thebottom. Rotate 180°.

4 5 6

Repeat steps 2−4. Continue with step 4 through theend of the Hexagonal Dipyramid.

1

2


Hexagonal Dipyramid 36°

1 2 3

4 5 6

Fold and unfold. Fold on the right.

Continue with step 4through the end of theHexagonal Dipyramid.

The geometry of this dipyramid makes it easier to fold since the horizontal creases are part of the creasepattern. This is achieved by setting the small angle of each triangle equal to 36°. The folding is similar tothe Hexagonal Dipyramid.

Unfold.


Repeat steps 2−5.Fold and unfold onthe top. Rotate 180°.

7

Hexagonal Dipyramid in a Sphere 215

Hexagonal Dipyramid in a Sphere

5 6

This dipyramid is inscibed in a sphere. The angles of each of the twelve triangles are 41.41°, 69.3°, and69.3°, and the lengths of the sides are proportional to 1, 2, and 2.

1 2 3


Fold and unfold on the right.


4


Repeat steps 2−5.

1

1


7 8 9

10 11 12

13 14 15

16 17 18


Valley-fold along a hiddencrease and mountain-foldthe dot to the right edge.

Fold and unfold all the layersalong a partally hidden crease.


Crease in the center.Fold and unfold tobisect the angle.

Fold and unfold.

Fold and unfold.


Extend the creasealong the dots.

Hexagonal Dipyramid in a Sphere 217

Fold the inside layers togetherfor this spine-lock fold. Turnover and repeat.

19 20 21

22 23 24

25 26 27



Valley-fold along the crease forthis squash fold. Rotate 180°.


Push in at the dot.

Reverse folds.

28

Hexagonal Dipyramidin a Sphere

Fold the edge to the dot.Turn over and repeat.

1. Lift up at the dot.2. Fold under.3. Bring the ends3. together.Turn over and repeat.

Only the top half is drawn.Tuck the tab under A. Bgoes into the lower halfwhen the same folds arerepeated below.

29 30

1

2

3

tabA

B


Squat Silver Hexagonal Dipyramid

1 2 3

Fold and unfold. Fold and unfold creasinglightly on the left.


4 5

Repeat steps 2−3. Continue with step 7through the end of theHexagonal Dipyramid ina Sphere but skip step 27.

Squat Silver HexagonalDipyramid

6

The ratio of the height of this hexagonal dipyramid to its diameter is 1 to 2. The small angle in each triangleis about 48°, same as in the Silver Square Dipyramid. The folding is similar to the Hexagonal Dipyramidin a Sphere.

1

2

Heptagonal Dipyramid 219

Heptagonal Dipyramid

4 5 6

1 2 3

Fold and unfold in halfon the edges. Makelonger vertical creases.

This fourteen-sided dipyramid is the dual of the uniform heptagonal prism. The small angle in each triangleis about 22°.

Crease at the bottom. Unfold.

Fold and unfold at the bottom. Fold and unfold. Rotate 180°. Repeat steps 2−5.


7 8 9

10 11 12

13 14 15

16 17 18 19



Unfold.

Valley-fold along a hiddencrease and mountain-foldthe dot to the left edge.

Crease at the top. Unfold and rotate 180°. Repeat steps 7−8.


Fold and unfold. Fold and unfold all thelayers along a partiallyhidden crease.


Heptagonal Dipyramid 221

20 21 22

23 24 25

26 27 28

29 30


Repeat steps 16−20. Crease on the left.


Repeat steps 22−28. Fold and unfoldalong the creases.


31

1

2

Fold and unfold both layersalong a hidden crease.

Fold and unfold both layersalong the crease. Turn overaround the horizontal line.

Unfold and rotate 180°.Fold and unfold bothlayers along the crease.

Fold and unfold bothlayers along the crease.




Valley-fold along the creasebetween the dots for thissquash fold. Rotate 180°.


Reverse folds.

Lift up at the dot to open the model.Tuck and interlock the tabs.

Heptagonal Dipyramid

32 33 34

35 36 37

38 39 40

Bring the edge to the dot. All ofthe folds will be under region A.Turn over and repeat.

Fold and unfold.

tab tab

A

Heptagonal Dipyramid 30° 223

Heptagonal Dipyramid 30°

4 5 6

1 2 3

Fold and unfold creasing lightly.

This fourteen-sided dipyramid is similar to the Heptagonal Dipyramid. The small angle in each triangle is30°, which simplifies the geometry.


Fold and unfold.

Fold and unfold in halfat the top and bottom.

Fold and unfold at bottom. Fold and unfold atbottom. Rotate 180°.


Repeat steps 5−6.

7 8 9

10 11 12

13 14


Fold and unfoldalong the crease. Unfold. Fold and unfold.

Continue with step 16 through the endof the Heptagonal Dipyramid. Skip step18 since the crease is already there.

Heptagonal Dipyramid 30°

Heptagonal Dipyramid in a Sphere 225

Heptagonal Dipyramid in a Sphere

1 2 3

4 5 6

Unfold.Fold and unfold onthe top and bottom.


Fold and unfold at the top. Fold and unfold. Rotate 180°. Repeat steps 2−5.

This dipyramid is inscibed in a sphere. The angles of each of the fourteen triangles are 36°, 72°, and 72°.


7 8 9

10 11 12

13 14 15

16 17 18 19


Repeat steps 7−8.



Unfold.

Unfold.


Fold and unfold.

Fold and unfold onthe top. Rotate 180°.

Fold along the crease. Fold and unfold all thelayers along a partiallyhidden crease.

Heptagonal Dipyramid in a Sphere 227

20 21 22

23 24 25

26 27 28


29 30



31


Squash-fold.


Repeat steps 14−20. Crease on the left.

12


Fold and unfold both layersalong the crease. Turn overaround the horizontal line.




Fold the inside layerstogether for this spine-lockfold. Turn over and repeat.

Fold the inside layerstogether for this spine-lockfold. Turn over and repeat.

Heptagonal Dipyramidin a Sphere

32 33 34

35 36 37

38

Fold and unfold. Fold and unfold.Rotate 180°.


Reverse folds.


39

40

To finish the dipyramid:1. Lift up at the upper dot.2. Fold A inside.3. Bring the tab behind B3. so the dots meet.4. Repeat on the other side and4. tuck and interlock the tabs.

tab

AB

Octagonal Dipyramid 229

Octagonal Dipyramid

1 2 3

4 5 6

Fold and unfold. Fold and unfold at the bottom.

This 16-sided dipyramid is the dual of the uniform octagonal prism. The small angle in each triangle is about17°. The ratio of the height to the diameter is (1 + 2) to 1.


Repeat steps 2−3.



7 8 9

10 11 12

13 14 15

16 17 18


Unfold.






Fold and unfoldthe top layer.



Octagonal Dipyramid 231



19 20 21

22 23

24 25

26 28

Squash-fold and rotate 180°. Repeat steps 19−20.Then flatten.

Push in at the dot.

Reverse folds.

Fold and unfold.

Octagonal Dipyramid

Bring the edge to the dot. All ofthe folds will be under region A.Turn over and repeat.

27

A

tab tab

Lift up at the dot toopen the model. Tuckand interlock the tabs.


Octagonal Dipyramid 26°

1 2 3

4 5 6

Fold and unfold.

The folding of this 16-sided dipyramid is similar to the Octagonal Dipyramid. The small angle in each triangleis about 25.7°. That angle makes the geometry convenient since the horizontal lines are part of the creasepattern.

Repeat steps 2−3.


Fold and unfold on the leftand along the vertical crease.

Fold and unfold on thebottom. Rotate 180°.

Octagonal Dipyramid 26° 233

7 8 9

10 11

12


Fold and unfold at the bottom. Fold and unfold on the left.


Continue with step 10 through the end ofthe Octagonal Dipyramid. Skip step 14 sincethe crease is already there. In step 20, foldalong the lowest crease on the right.

13

Octagonal Dipyramid 26°


Octagonal Dipyramid in a Sphere

1 2 3

4 5 6

Fold and unfold.



Repeat steps 2−3.

This dipyramid is inscibed in a sphere. The angles of each of the 16 triangles are 31.4°, 74.3°, and 74.3°.

Fold and unfold alongthe vertical crease.

Octagonal Dipyramid in a Sphere 235

7 8 9

10

11 12

13 14 15

16 17 18


Unfold.




Fold and unfoldthe top layer.



Crease in the center.




19 20 21

22 23 24

25 26

28


Squash-fold and rotate 180°.


Push in at the dot.

Reverse folds.

27


1

2

3

1. Lift up at the dot.2. Fold under.2. (There are no guides.)3. Bring the ends together.Turn over and repeat.

29

tabA

B

Only the top half is drawn.Tuck the tab under A. B goesinto the lower half while thesame folds are repeated below.

Octagonal Dipyramidin a Sphere



Nonagonal Dipyramid 237

Nonagonal Dipyramid

1 2 3

4 5 6

Fold and unfoldcreasing lightly. Rotate.


The angle at the top of the triangle in this 18-sided dipyramid is 22.5°.

Bring the right edge to the upperdot and the lower dot to thediagonal. Crease on the bottom.

Unfold. Fold and unfold.Rotate 180°.

Repeat steps 3−5.


7 8 9

10 11 12

13 14

15 16 17



Fold and unfold at the bottom.

Repeat steps 7−9.



Unfold.

Unfold.

Nonagonal Dipyramid 239

18 19 20

21

23

Mountain-fold alongthe crease and bringit to the left edge.

Fold and unfold all the layersalong a partially hidden crease.

Now you are on your own. Thefolding is similar to that of theHeptagonal Dipyramid.

Nonagonal Dipyramid

Fold along the crease.Mountain-fold alongthe crease and bringit to the left edge.

22

Unfold and rotate 180°.Repeat steps 15−21.


Decagonal Dipyramid

1 2 3

4 5 6

Fold and unfold. Fold and unfold at the bottom. Fold and unfold theedge to the crease andcrease at the bottom.

Repeat steps 2−4.Fold and unfold. Rotate 180°.

The angle at the top of the triangles in this 20-sided dipyramid is 20°.

Decagonal Dipyramid 241

7 8 9

10 11 12

13 14 15

16 17

18


Unfold.

Unfold.


Fold and unfold four times,bisecting the angles.

Mountain-fold alongthe crease and bringit to the right edge.

Fold and unfoldall the layers.

Repeat steps 13−17 onthe left and above on theleft and right. Now youare on your own.

Decagonal Dipyramid



Mountain-fold alongthe crease and bringit to the right edge.

243

Dimpled Dipyramid DesignDimpled dipyramids are polyhedra where alternatesides of the dipyramids are sunken. They are eachcomposed of three different triangular faces. TriangleA represents the flat faces (isosceles triangles),which are the same as the faces of the nondimpledmodels. Triangles B are the (typically smaller)isosceles triangles by the equator. Triangles C arethe longer triangles that complete the sunken faces.Formulas from “Dipyramid Design” (page 173) will be used to relate the proportion of the height to the diameter to the angle α of triangle A.

Given polygon n and angle α, the proportion ofthe height of the dipyramid to its diameter, H, is

sin(α/2)sin(180°/n)

Given H and polygon n, α is found by

α = 2arcsin[sin(180°/n)cos(arctan(H))] A

BC

α

diameter

height

For a dimpled dipyramid with a given polygonbase n, and angle α in triangle A, the triangles Band C can vary. The models in this section aremade with square, hexagonal, and octagonal bases.(Note that, in general, n must be even.) Since theseshapes are complex and thus difficult to design,I worked out methods to define triangles B and C,given the polygon base, so as to simplify the folding.

Here is a trio of dimpled dipyramids whereangle α is the same for all (α = 36.87°).

Tall Dimpled Square DipyramidH = 2

Dimpled Hexagonal DipyramidH = 1.5

Octagonal Flying SaucerH = 1/ 2

α

αα

H = tan arccos( ( ))


A

B

CC

45°

α

A

A

B

CC

Square Dimpled Dipyramids

For these models, triangle B is chosen to be a 45°right triangle. The crease patterns have oddsymmetry.

Since n is 4, given H, α is found by

α = 2arcsin[cos(arctan(H))/ 2]

1

2

Tall Dimpled Square Dipyramid, H = 2, α = 36.87°

Dimpled Silver Square Dipyramid, H = 2, α = 48.19°

Heptahedron, H = 1, α = 60°

Dimpled Squat Square Dipyramid, H = 1/ 2, a = 70.53°

1

1

1

2

1

2

Dimpled Dipyramid Design 245

α

A

A

Hexagonal Dimpled Dipyramids

To simplify the folding for these models, trianglesB and C are chosen by the following:

1. Draw triangle A twice as shown in the first picture.

2. Draw a line connecting the dots. Given angle α, angles γ and δ can be found.

3. Draw a line at angle δ from the base of triangle A. Thus triangles B and C are now determined.

The crease patterns have odd symmetry. Since nis 6, given H, α is found by

α = 2arcsin[cos(arctan(H))/2]

Tall Dimpled Hexagonal Dipyramid, H = 3, α = 28.96°

Dimpled Silver Hexagonal Dipyramid, H = 2, α = 33.56°

Dimpled Hexagonal Dipyramid, H = 1.5, α = 36.87°

Dimpled Hexagonal Dipyramid in a Sphere, H = 1, α = 41.41°

1

1

1.5

1

1

2

1

3

A

A

α γ

δ

A BC

B

C

δ

α


Octagonal Flying Saucer, H = 1/ 2, α = 36.87°

Dimpled Octagonal Dipyramid in a Sphere, H = 1, α = 31.4°

1

1

Dimpled Octagonal Dipyramid, H = 1.5, α = 28.96°

Dimpled Silver Octagonal Dipyramid, H = 2, α = 25.53°

A

α

A A

A A

A

α γγ

γγ

A

α γγ

γγ

B

C

C

C

CA A

Octagonal Dimpled Dipyramids

To simplify the folding for these models, trianglesB and C are chosen by the following:

1. Draw triangle A three times as shown in the first picture.

2. Add lines for angles γ such that α + 2γ = 90°.

3. Connect the three dots. Thus triangles B and C are now determined.

The crease patterns have square symmetry. Sincen is 8, given H, α is found by

α = 2arcsin[sin(22.5°)cos(arctan(H))]

1

2

1

2

1

1.5

Tall Dimpled Square Dipyramid 247

1

2

Tall Dimpled Square Dipyramid

1 2 3

4 5 6

Fold and unfold. Fold and unfold on the left andright by the edges. Rotate 90°.

Fold and unfold. Fold and unfold. Fold and unfold onthe left and right.

The height of this tall dimpled square dipyramid is twice the diameter. Four alternating sides are sunken.According to the formula I derived, the small angle in each nonsunken face is about 36.87° to achieve thedimensions.

Fold and unfold.

1/41/3


7 8 9

10 11 12

13 14 15

16 17 18


Fold and unfold alonga hidden crease.


Fold and unfold. Rotate 180°. Unfold.

Fold to the edge.

Fold and unfold.

Tall Dimpled Square Dipyramid 249

19 21

22 23

24 25

26 27

Fold and unfold alongthe creases. Rotate 90°.

Rotate 180° and repeatsteps 20−22. Rotate thebottom to the top.

Push in at the lower dot. The otherdots will meet. Turn over and repeat.


tab tab

Interlock the tabs to close the model.

Tall DimpledSquare Dipyramid

Tuck inside.Puff out at the upper dot and pushin at the center dot. The other twodots will meet. Flatten inside.

20


Dimpled Silver Square Dipyramid

4 5 6

1 2 3


Unfold and rotate 180°.Bring the lower left corner to thetop edge and the bottom edge tothe right center. Crease on the left.


The ratio of the height of this square dipyramid to its width is 2 to 1. Four alternating sides are sunken.The small angle in each nonsunken face is about 48º to achieve the dimensions.

1

2

Dimpled Silver Squre Dipyramid 251

7 8 9

10 11 12

13 14 15

16 17 18

Fold and unfold. Bring the lower right corner to thetop edge and the bottom edge tothe left crease. Crease on the left.

Unfold.

Fold and unfold the top layer. Fold and unfold bothlayers along the crease.

Unfold. Fold and unfold. Fold and unfold. Turnover and rotate 180°.

Repeat steps 6−15. Fold and unfold. Fold and unfold atthe top and bottom.


19 20 21

22 23 24

25 26 27

28 29 30

Fold and unfold. Fold and unfold. Push in at the upper dot.Mountain-fold along thecrease. Bring the lowerdot to the crease.

Puff out at the dot.

Fold and unfoldall the layersalong the crease.

Unfold.

Fold along the crease. Fold and unfold alonga hidden crease.

Unfold.

Fold and unfold. Mountain-foldalong the crease. Crease lightlyalong the valley-fold line.

Repeat steps 21−29.Rotate 90°.

1. Fold and unfold.2. Fold and unfold to bisect the angle.Turn over and rotate 180°.

12

Dimpled Silver Squre Dipyramid 253

tab tab

31

Dimpled SilverSquare Dipyramid

32 33

34 35 36

37 38 39

40 41

42


Fold and unfold tobisect the angles.

Puff out at the dot.

Push in at the upper dot. The twoother dots will meet. Flatten inside.

Rotate 180° and repeatsteps 33−37. Rotate thebottom to the top.

Push in at the lower dot.The other dots will meet.Turn over and repeat.


Interlock the tabs.

Tuck inside.

12




Heptahedron

For this polyhedron, four faces are indented toward the center. It is named a heptahedron for its sevensides: four outer sides (same as the octahedron) and three center sides representing the x, y, and z axes. Itcan also be called a tetrahemihexahedron. This interesting shape combines equilateral triangles withisosceles right triangles.

1 2 3

Fold and unfold.

4 5 6

Valley-fold along the crease.Turn over and repeat. Unfold.

Unfold the two edges.

1

1

Heptahedron 255

9

10 11

Fold and unfold.

12

13 14 15

16 17

Fold and unfold.

18

Fold and unfold,creasing at the ends.

7 8


1

2

3

4

Fold and unfold. Fold and unfold onthe right. Rotate 180°.


Fold and unfold.

Fold and unfold.

1

2

3

4

Fold and unfold. Fold and unfold six times.


19 20

Fold along hidden creases. Fold and unfold alongan existing crease.

21

22 23 24

25


26

27 28

Push in at the upper dot to form a sunkentriangle. The other dots will meet.

Turn over andrepeat steps 25−26.

Unfold. Fold and unfold. Fold along the creases.

Heptahedron 257

29

Note how some paper comesout at A . Flatten at the dotsso each pair meets.

32

Fold all the layers together.

33

Turn over andrepeat steps 28−30.

37

Tuck the left flap into theright ones and flatten. Turn over and

repeat step 35.

Heptahedron

Unfold.

34

35 36

Bring the dotstogether. Rotate.

Flatten the flaps together. Foldand unfold all the layers.

3130

A


Dimpled Squat Square Dipyramid

4 5 6

1 2 3



Fold and unfold.


Repeat steps 2−4. Fold and unfold. Rotate.

The ratio of the height of this dimpled squat square dipyramid to its diameter is 1 to 2. The crease patternshows odd symmetry.

1

2

Dimpled Squat Square Dipyramid 259

7 8 9

10 11 12

13 14 15

16 17 18

Fold at the bottom. Repeat behindat the top for the mountain fold.

Unfold.

Fold and unfold by the diagonal. Fold and unfold onthe left. Rotate 180°.


Fold and unfold. Rotate 180°. Fold and unfold. Fold and unfold.Rotate 180°.

Fold and unfold. Fold and unfold. Fold and unfold.


19 20 21

22 23 24

25 26 27

28 29 30

Fold and unfold. Crease at the bottom.

Fold and unfold. Rotate 180°. Repeat steps 18−23.

Fold and unfold.

Unfold.



Fold and unfold the edgeto the dot. Rotate 180°.

Repeat steps 25−28. Fold and unfold tobisect the angles.

Dimpled Squat Square Dipyramid 261

31

Dimpled SquatSquare Dipyramid

32

33 34

35 36

37

38

39

40

Push in at the centerdot to form a sunkentriangle. Puff out at thetwo outer dots. The twolower dots will meet. On the second layer,

fold the edge to the dot.

Rotate 180° andrepeat steps 31−33.

Puff out at the centerdot and push in at theother two dots.

Rotate the lower dotto the bottom and theupper dot to the top.

Tuck the tab inside.The dots will meet.Turn over and repeat.


Interlock the tabs.

tab


Tall Dimpled Hexagonal Dipyramid

1 2 3

Fold and unfold.

4 5 6




Repeat steps 3−4. Turnover and rotate 90°.


The ratio of the height of this dimpled dipyramid to its diameter is 3 to 1. The small angle in each nonsunkenface is about 29° to achieve the dimensions.

1

3

Tall Dimpled Hexagonal Dipyramid 263

7 8 9

10 11 12

13 14 15

16 17 18

Repeat steps 6−7.


Unfold.

Fold and unfold by the leftmostvertical line. Rotate 180°.

Valley-fold along the crease andmountain-fold the dot to the edge.Crease below the center line anda little above it.

Unfold.

Repeat step 10.

Unfold.

Fold along the crease. Fold and unfold the toplayer along a hidden crease.


Unfold.

19 20 21

22 23 24

25 26 27

28 29 30

Fold along the crease. Fold and unfold the toplayer along hidden creases.









12

Bisect the angle.

Tall Dimpled Hexagonal Dipyramid 265


31 32 33

34 35 36

37 38 39

40 41

42



Unfold.



Fold and unfold. Push in at the dotto bisect the angle. Rotate 180°.

Repeat step 37. Fold on the left. Puff outat the two upper dots andpush in at the lower dot.

1. Begin at the right dot and fold a little below the left dot.2. The dots will meet. Flatten inside.

Rotate 180° and repeat steps39−40. Then rotate the dot tothe front and center.

Push in at the upper dot.The other dots will meet.

1

2


43 44 45

46 47 48

49 50

51

52

53

Turn over and repeat steps42−43. Then rotate the bottomto the top and bring the dot tothe front and center.

Tuck and interlock the tabs.

1

2


Squash-fold on the left andbring the top edge to the crease.

Turn over and repeat steps45−49. Then rotate the dotto the front and center.


Tall DimpledHexagonal Dipyramid

1. Make a small fold.2. Fold along the crease.

Dimpled Silver Hexagonal Dipyramid 267

Dimpled Silver Hexagonal Dipyramid

1 2 3

Fold and unfold.

4 5 6

Unfold.Crease on the right.

Bring the lower right corner to thetop edge and crease on the left.

Unfold. Fold and unfold.Rotate 180°.

The ratio of the height of this dimpled dipyramid to its diameter is 2 to 1. The small angle in each nonsunkenface is about 34° to achieve the dimensions. The folding is similar to that of the Tall Dimpled HexagonalDipyramid.

1

2


7 8 9

10 11 12

13 14 15

16 17 18

Repeat steps 2−6. Fold and unfoldon the left.


Fold and unfold by thehorizontal line. Rotate 180°.

Repeat steps 8−10and rotate 90°.

Continue with steps 9−38of the Tall DimpledHexagonal Dipyramid.

Fold on the left. Puff outat the two upper dots andpush in at the lower dot.

1. Begin at the right dot and fold a little below the left dot.2. The dots will meet. Flatten inside.



Turn over and repeat steps16−17. Then rotate the bottomto the top and bring the dotto the front and center.

1

2

1. Make a small fold.2. Fold along the crease.

1

2

Dimpled Silver Hexagonal Dipyramid 269

19 20 21

22 23 24

25 26


Squash-fold on the left sothe edge meets the dot.




Dimpled SilverHexagonal Dipyramid

12


Dimpled Hexagonal Dipyramid

1 2 3

Fold and unfold.

This model comes from a cube where six of the eight corners are sunken, as shown in the second picture.However, it was difficult to design a convenient model by folding a cube first. The angles in the nonsunkentriangles are 36.87°, 71.565°, and 71.565°, and the ratio of the height to diameter is 1.5 = 1.225. Thefolding is similar to that of the Tall Dimpled Hexagonal Dipyramid.

4 5 6

Fold and unfold to findthe quarter marks.


Crease on the left. Unfold.

Fold and unfoldat the edges.

1/3

1

1.5

Dimpled Hexagonal Dipyramid 271

7 8 9

10 11 12

13 14 15

16 17 18


Repeat steps 5−7. Fold and unfold atthe center and ends.

Fold and unfold. Fold and unfold.Rotate 180°.


Fold and unfold. Fold and unfold on the left. Bring the corner to thecrease and fold on the left.

Unfold. Fold and unfold atthe intersection.

Fold and unfold at the topand bottom. Rotate 180°.


19 20 21

22 23 24

25 26 27

28 29 30

Repeat steps 13−18. Fold and unfold the top layeralong a hidden crease.



Fold and unfold Fold and unfold.

Fold and unfold. Fold and unfold. Push in at the dot by the center.

Fold and unfold the top layeralong a hidden crease.

Dimpled Hexagonal Dipyramid 273

31 32 33

3435 36

37 38 39

40 41 42

Unfold. Push in at the dot by the center.

Unfold and rotate 180°. Repeat steps 30−34. Fold on the left. Puff outat the two upper dots andpush in at the lower dot.

Reverse-fold. The dots will meet.Flatten inside.



Divide roughly into 1/3. Fold along the crease.


Dimpled Hexagonal Dipyramid

43 44 45

46 47 48

49 50 51

52

On the second layer,bring the edge to the dot.

Turn over and repeat steps 40−43. Thenrotate the bottom to the top and bringthe dot to the front and center.

Fold along the creases.Turn over and repeat.


Fold both layers togetheralong the crease.

Tuck the white tabinside the pocket.

The hidden tab is shown withthe dotted lines. Turn over andrepeat steps 46−48. Then rotatethe dot to the front and center.



Dimpled Hexagonal Dipyramid in a Sphere 275

Dimpled Hexagonal Dipyramid in a Sphere

1 2 3


4 5 6


Repeat steps 3−5.Fold and unfoldat the bottom.



This dimpled dipyramid is inscibed in a sphere. The angles of each of the nonsunken triangles are 41.41°,69.3°, and 69.3°. The folding is similar to that of the Tall Dimpled Hexagonal Dipyramid. the the the the

1

1


7 8 9

10 11 12

13 14 15

16 17 18



Fold and unfold by the leftmostvertical line. Rotate 180°.

Valley-fold along the crease andmountain-fold the dot to the edge.Crease below the center line anda little above it.

Unfold.Repeat step 13.

Unfold.


Unfold.Bring the top left corner to thebottom edge and the top edge tothe right dot. Crease on the left.


Unfold.

Unfold.

19 20 21

22 23 24

25 26 27

28 29 30


Fold along the crease. Fold and unfold the toplayer along a hidden crease.







1 2

Bisect the angle.



31 32 33

34 35 36

37 38 39

40 41 42






Unfold. Fold and unfoldalong the crease.

Repeat steps 30−39. Fold and unfold. Push in at the dotto bisect the angle. Rotate 180°.

Repeat step 41.



43 4445

46 47 48

49 50 51

52 53

Fold on the left. Puff outat the two upper dots andpush in at the lower dot.



Fold along a hidden crease.Rotate the dot to the front and

Mountain-fold along the creaseand tuck between the layers.

Repeat steps 43−47 on the back.Then rotate the bottom to the top.

Push in at the lower dot.The other dots will meet.

Fold a thin strip and hide itbehind region A. Rotate thedot to the front and center.

The dots will meet.

5455

56


Tuck and interlock the tabs. Dimpled HexagonalDipyramid in a Sphere

A

center.


Octagonal Flying Saucer

1 2 3

4 5 6

Fold and unfold. Fold and unfold four times.

This model has an octagonal base. The ratio of the height to the diameter is 1/ 2. The crease pattern showssquare symmetry.



1

2

Octagonal Flying Saucer 281

7 8 9

10 11 12

13 14 15 16

17 18 19


Fold and unfold. Rotate 90°. Repeat step 10three more times.

Fold and unfold.

1

2

1. Unfold.2. Turn over around the dotted line.





20 21 22

23 24 25

26 27 28

29 30



Puff out at the upper dot andpush in at the center dot toform a sunken triangle. Theother pairs of dots will meet.

Repeat step 24 three more times,with the dot here matching theupper dot in step 24.

Push in at the upperdot. The other twodots will meet.

The folding is done on theinside. Refold the hidden flapas in steps 20−21 and flattenagainst the back of triangle A.

Repeat steps 26−27three more times. Rotatethe top to the bottom.

Bring the fourdots together.

Fold along the creasesand tuck inside. Repeatthree more times.

Octagonal Flying Saucer

31

A

Dimpled Octagonal Dipyramid in a Sphere 283

Dimpled Octagonal Dipyramid in a Sphere

1 2 3

4 5 6

Fold and unfold.


Fold and unfold on the left. Fold and unfold at thebottom. Rotate 180°.

Repeat steps 2−3.

This dimpled dipyramid is inscibed in a sphere. The angles of each of the eight (nonsunken) triangles are31.4°, 74.3°, and 74.3°.

1

1


7 8 9

10 11 12

13 14 15

16 17 18


Repeat steps 5−7.



Fold and unfold seven times.


Fold and unfold.

Fold and unfold seven times. Fold and unfold. Fold and unfold.

Fold and unfold. Note the right anglein the next step.

Dimpled Octagonal Dipyramid in a Sphere 285

19 20 21

22 23 24

25 26 27

28 29 30



Open to makethe model 3D.






Puff out at the upper dot andpush in at the center dot toform a sunken triangle. Theother pairs of dots will meet.

Repeat step 32 three more times,with the dot here matching theupper dot in step 32.

Push in at the upper dot. Theother two dots will meet.

The folding is done on theinside. Refold the hidden flapas in steps 28−29 and flattenagainst the back of triangle A.

Repeat steps 36−37 threemore times. Rotate thetop to the bottom.

Bring the four dots together.

Fold along the creases and tuckinside. Repeat three more times.

Dimpled OctagonalDipyramid in a Sphere

31 32 33

34 35 36

37 38 39

The folding is done on theinside. Refold the hiddenflap as in steps 23−24.


40 41

A

Dimpled Octagonal Dipyramid 287

Dimpled Octagonal Dipyramid

1 2 3 4

5 6

Fold and unfold.

Continue with step 6through the end of theDimpled OctagonalDipyramid in a Sphere.

Fold at the top.

Repeat steps 2−5.

The ratio of the height of this dimpled dipyramid to its diameter is 1.5 to 1. The small angle in eachnonsunken triangle is about 29° to achieve the dimensions. The folding is similar to that of the DimpledOctagonal Dipyramid in a Sphere.

Unfold. Fold and unfoldon the left.


87

1

1.5


Dimpled Silver Octagonal Dipyramid

1 2 3

4 5 6

Fold and unfold.

Continue with step 6 through theend of the Dimpled OctagonalDipyramid in a Sphere.

Fold and unfold at the bottom. Fold and unfold at thebottom. Rotate 180°.

Repeat steps 2−3.

The ratio of the height of this dimpled dipyramid to its diameter is 2 to 1. The small angle in each nonsunkentriangle is about 25.5° to achieve the dimensions. The folding is similar to that of the Dimpled OctagonalDipyramid in a Sphere.

1

2

Documents

Origami Polyhedra Designdl.booktolearn.com/ebooks2/handicraft/... · Montroll, John. Origami polyhedra design / John Montroll. p. cm. ISBN 978-1-56881-458-2 (alk. paper) 1. Origami