The idea of this book was suggested to me by Kindergarten Gift No. VIII. - Paper-folding. The gift consists of two hundred variously colored squares of paper, a folder, and diagrams and instructions for folding. The paper is colored and glazed on one side. The paper may, however, be of self-color, alike on both sides. In fact, any paper of moderate thickness will answer the purpose, but colored paper shows the creases better, and is more attractive. The kindergarten gift is sold by any dealers in school supplies ; but colored paper of both sorts can be had from stationery dealers. Any sheet of paper can be cut into a square as explained in the opening articles of this book, but it is neat and convenient to have the squares ready cut.

Title | Geometric Exercises In Paper Folding |

Author | Tandalam Sundara Row |

Publisher | The Open Court Publishing Co. |

Year | 1917 |

Copyright | 1917, The Open Court Publishing Co. |

Amazon | T. Sundara Row's Geometric Exercises in Paper Folding (Large Print Edition) |

Edited And Revised By Wooster Woodruff Beman, Professor Of Mathematics In The University Of Michigan And David Eugene Smith, Professor Of Mathematics In Teachers' College Of Columbia University

With 87 Illustrations

- Editors' Preface
- Our attention was first attracted to Sundara Row's Geometrical Exercises in Paper Folding by a reference in Klein's Vorlesungen uber ansgewahlte Fragen der Elementargeometrie. An examination of the bo...

- Introduction
- The idea of this book was suggested to me by Kindergarten Gift No. VIII. - Paper-folding. The gift consists of two hundred variously colored squares of paper, a folder, and diagrams and instructions f...

- I. The Square
- 1. The upper side of a piece of paper lying flat upon a table is a plane surface, and so is the lower side which is in contact with the table. 2. The two surfaces are separated by the material of the...

- II. The Equilateral Triangle
- 22. Now take this square piece of paper (Fig. 9), and fold it double, laying two opposite edges one upon the other. We obtain a crease which passes through the mid-points of the remaining sides and is...

- III. Squares And Rectangles
- 44. Fold the given square as in Fig. 13. This affords the well-known proof of the Pythagorean theorem. FGH being a right-angled triangle, the square on FH equals the sum of the squares on FG and GH. ...

- IV. The Pentagon
- 71. To cut off a regular pentagon from the square ABCD. Divide BA in X in median section and take M the mid-point of AX. Fig. 27. Then AB.AX=XB2, and AM=MX. Take BN=AM or MX. Then MN=XB. Lay of...

- V. The Hexagon
- 79. To cut off a regular hexagon from a given square. Fig. 29. Fold through the mid-points of the opposite sides, and obtain the lines A OB and COD. On both sides of AO and OB describe equilater...

- VI. The Octagon
- 87. To cut off a regular octagon from a given square. Obtain the inscribed square by joining the midpoints A, B, C, D of the sides of the given square. Fig. 32. Bisect the angles which the sides ...

- VII. The Nonagon
- 96. Any angle can be trisected fairly accurately by paper folding, and in this way we may construct approximately the regular nonagon. Fig. 35. Obtain the three equal angles at the center of an e...

- VIII. The Decagon And The Dodecagon
- 99. Figs. 36, 37 show how a regular decagon, and a regular dodecagon, may be obtained from a pentagon and hexagon respectively. Fig. 36. The main part of the process is to obtain the angles at th...

- IX. The Pentedecagon
- 101. Fig. 39 shows how the pentedecagon is obtained from the pentagon. Let ABCDE be the pentagon and O its center. Fig. 39. Draw OA, OB, OC, OD, and OE. Produce DO to meet AB in K. Take OF=½ of...

- X. Series
- Arithmetic Series 102. Fig. 40 illustrates an arithmetic series. The horizontal lines to the left of the diagonal, including the upper and lower edges, form an arithmetic series. Fig. 40. The in...

- X. Series. Continued
- 118. If BY and XY be given, to find the third term AY, we have only to describe any right-angled triangle on XY as the hypotenuse and make angle APX = angle XPB. 119. Let AX=a, AB = b, and AY= c. Ih...

- XI. Polygons
- 131. Find O the center of a square by folding its diameters. Bisect the right angles at the center, then the half right angles, and so on. Then we obtain 2 equal angles around the center, and the mag...

- XI. Polygons. Continued
- Let AB be a side of the first polygon, 0 its center, OA the radius of the circumscribed circle, and OD the apothem. On OD produced take OC= OA or OB. Draw AC, BC. Fold OA' and OB' perpendicular to AC ...

- XII. General Principles
- 147. In the preceding pages we have adopted several processes, e. g., bisecting and trisecting finite lines, bisecting rectilineal angles and dividing them into other equal parts, drawing perpendicula...

- XII. General Principles. Part 2
- Fig. 54. 162. Parallelograms have a center of symmetry. A quadrilateral of the form of a kite, or a trapezium with two opposite sides equal and equally inclined to either of the remaining sides, h...

- XII. General Principles. Part 3
- Fig. 60. Now produce AC and take CE = EF=FG.. .. = AC or BD. Draw DE, DE, DG....cutting AB in P3, P4, P5,,...... Then from similar triangles, P3B:AP3 = BD:AE. ... P3B: AB = BD: AF = 1:3. Simila...

- XIII. The Conic Sections
- Section I. - The Circle 181. A piece of paper can be folded in numerous ways through a common point. Points on each of the lines so taken as to be equidistant from the common point will lie on the ci...

- XIII. The Conic Sections. Part 2
- 2. And XOB =XVB +VBO, = 2 XVB. 3. ... XVB = ½XOB. 4. Similarly AVX=½AOX(each=zero in Fig. 62), and ...AVB = ½AOB. The proof holds for all three figures, point A having moved to X (Fig. 62), and th...

- XIII. The Conic Sections. Part 3
- 222. The angle between the polars of two points is equal to the angle subtended by these points at the polar center. 223. The circle described with B as a center and BC as a radius cuts the circle ...

- XIII. The Conic Sections. Part 4
- Take OT' in FO produced =4. OF. Bisect TN in M. Take Q in OY such that MQ = MN=MT. Fold through Q so that QP may be at right angles to OK Let P be the point where QP meets the ordinate of N. Then...

- XIII. The Conic Sections. Part 5
- 249. The above definition corresponds to the equation y2 = b2 / a2 (2ax - x2) when the vertex is the origin. 250. AN.NA' is equal to the square on the ordinate QN of the auxiliary circle, and PN : ...

- XIV. Miscellaneous Curves
- 262. I propose in this, the last chapter, to give hints for tracing certain well-known curves. The Cissoid 263. This word means ivy-shaped curve. It is defined as follows: Let OQA (Fig. 76) be a ...

- XIV. Miscellaneous Curves. Continued
- Fig. 81. The Ovals Of Cassini 272. When a point moves in a plane so that the product of its distances from two fixed points in the plane is constant, it traces out one of Cassini's ovals. The fix...

- Catalogue of Publications of the Open Court Publishing Company
- Chicago, Monon Building, 324. Dearborn St. London, Kegan Paul, Trench, Trubner & Co. Sundara Row, T. Geometrical Exercises In Paper-Folding With Half-Tones from Photographs of Exercises. Pages, x +...

- The Religion of Science Library
- Bi-monthly reprints of standard works, philosophical classics, etc. Yearly, $1.50. Separate copies according to prices quoted. No. 1. The Religion of Science. By Paul Carus. 25c (is. 6d.). 2. Three ...