By Thomas Hull
IntroductionActivity 1 Folding Equilateral Triangles in a sq. job 2 Origami Trigonometry task three Dividing a size into equivalent Nths: Fujimoto Approximation job four Dividing a size into equivalent Nths precisely job five Origami Helix job 6 Folding a Parabola job 7 Can Origami Trisect an Angle?Activity eight fixing Cubic Equations job nine Lill's procedure job 10 Folding Strips into Knots task eleven Haga's ""Origamics"" task 12 Modular superstar Ring task thirteen Folding a Butterfly Bomb job 14 Molly's Hexahedron job 15 enterprise Card Modulars task sixteen 5 Inter. Read more...
summary: IntroductionActivity 1 Folding Equilateral Triangles in a sq. task 2 Origami Trigonometry task three Dividing a size into equivalent Nths: Fujimoto Approximation job four Dividing a size into equivalent Nths precisely job five Origami Helix job 6 Folding a Parabola task 7 Can Origami Trisect an Angle?Activity eight fixing Cubic Equations job nine Lill's technique task 10 Folding Strips into Knots job eleven Haga's ""Origamics"" job 12 Modular famous person Ring job thirteen Folding a Butterfly Bomb task 14 Molly's Hexahedron task 15 company Card Modulars job sixteen 5 Inter
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Additional resources for Project Origami : Activities for Exploring Mathematics, Second Edition
I1 i2 i3 . )2 . What will T0 ( x ) be? How about T1 ( x )? Proofs? Question 6: As we perform Fujimoto’s method on our initial guess x, we can think of it as performing T0 and T1 over and over again to x. When approximating 1/5ths, what happens to the binary decimal of x as we do this? Use this to prove the observation that you made in Question 2. 28 Activity 3 (3) A number theory question. . (courtesy of Tamara Veenstra) In Question 3 you were asked to use Fujimoto to approximate 1/7ths, and you should have noticed that in doing so you do not make pinch marks at every multiple of 1/7, unlike when approximating 1/5ths.
2 . 075 left over. This is bigger than 1/16, so the fourth digit is a 1. 0125 left. This is smaller than 1/32 and 1/64. But wait, 1/80 = (1/5)(1/16) and we got 1/80 after removing the 1/16 term from it. This means that if we factor out a 1/16 from our 1/80 remainder, we get 1/5 and we’re back to where we started! 0011)2 . In Question 2, we know that in Fujimoto’s method to make 1/5ths, we had to fold the right side twice and then the left side twice. So we folded the sequence Right, Right, Left, Left.
Folding the triangles actually adds some logical deductions to the problems, where the student needs to observe that when one object (either an angle or a length) is folded onto another object, congruent objects can be created (either congruent angles or lengths). This is an important aspect of paper folding that results in “proofs by origami” and is employed in other activities in this book, like the Trisecting an Angle activity. Furthermore, it is the author’s opinion that many students have a very hard time internalizing the methods and utility of trigonometry.