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Origami Geometry Projects for Math Fairs

Discover the intriguing intersection of origami and geometry in Robert Geretschläger's collection. This project compiles six advanced problems derived from a single origami fold, providing proofs for geometric identities, perimeters of triangles, and properties of circles. Participants will engage with profound challenges including tangent lines, perimeter equality, and inradius equivalences, all while enhancing their understanding of mathematical concepts through hands-on exploration. Suitable for math fairs and enthusiasts alike, this project promotes critical thinking and problem-solving skills.

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Origami Geometry Projects for Math Fairs

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  1. Origami Geometry Projects for Math Fairs Robert Geretschläger Graz, Austria

  2. 6 Problems from 1 Fold • Prove that C‘D‘ is a tangent of the circle with center C. passing through B and D. • Prove that the perimeter of triangle GAC‘ is equal to half the perimeter of ABCD. • Prove the identity AG = C‘B + GD‘ • Prove that the sum of the perimeters of triangles C‘BE and GD‘F is equal to the perimeter of triangle GAC‘. • Prove that the perimeter of triangle GD‘F is equal to the length of line segment AC‘. • Prove that the inradius of GAC‘ is equal to the length of line segment GD‘. 1. More Mathematical Morsels; Ross Honsberger 2. VIII Nordic Mathematical Contest 1994 4. 37th Slovenian Mathematical Olympiad 1993 6. classic Sangaku problem

  3. 6 Problems from 1 Fold Problem 1 Prove that C‘D‘ is a tangent of the circle with center C. passing through B and D.

  4. 6 Problems from 1 Fold Problem 2 Prove that the perimeter of triangle GAC‘ is equal to half the perimeter of ABCD. AC‘ + C‘G + GA = AC‘ + C‘P + GP + GA = AC‘ + C‘B + GD + GA = AB + DA

  5. 6 Problems from 1 Fold Problem 3 Prove the identity AG = C‘B + GD‘ AC‘ + C‘G + GA = AB + C‘D‘ = AC‘ + C‘B + C‘G + GD‘  AG = C‘B + GD‘

  6. 6 Problems from 1 Fold Problem 4 Prove that the sum of the perimeters of triangles C‘BE and GD‘F is equal to the perimeter of triangle GAC‘. GAC‘ ~ C’BE ~ GD’F AG = C’B + GD’  AC’ = BE + D’F  C’G = EC’ + FG AG + AC’ + C’G = (C’B + BE + EC’) + (GD’ + D’F + FG)

  7. 6 Problems from 1 Fold Problem 5 Prove that the perimeter of triangle GD‘F is equal to the length of line segment AC‘. AC‘ = D‘P = D‘G + GP = D‘G + GD = D‘G + GF + FD = D‘G + GD + FD‘

  8. 6 Problems from 1 Fold Problem 6 Prove that the inradius of GAC‘ is equal to the length of line segment GD‘. • C‘I = C‘III = x, GII = GIII = y, AI = AII = r • 2C‘D‘ = AC‘ + AG + GC‘ • = (r + x) + (r + y) + (x + y) • = 2(x + y + r) • 2(x + y + GD‘) = 2(x + y + r) • GD‘ = r

  9. The Pentagon Project The Golden Ratio a : 1 = 1 : (a-1)  a² - a = 1  a² - a – 1 = 0  a = f = a : 1

  10. The Pentagon Project Angles in a regular pentagon 36° 36° 108° 36° 72° 72°

  11. The Pentagon Project The Golden Triangle d : 1 = 1 : (d-1) 1  d = = f 1 72° d-1

  12. The Pentagon Project Placing the Pentagon on the Paper

  13. The Pentagon Project Step 1

  14. The Pentagon Project Step 2

  15. The Pentagon Project Step 3

  16. The Pentagon Project Step 4

  17. The Pentagon Project Step 5

  18. The Pentagon Project Step 6

  19. The Pentagon Project Step 7

  20. The Pentagon Project Step 8

  21. The Pentagon Project Additional challenges for advanced pentagonists: +++ Can a regular pentagon with sides a longer than 1/f be placed in the interior of a unit square? +++ Determine a folding sequence for a larger regular pentagon. +++ Determine the largest possible value of a. Prove that your value is the largest possible.

  22. The Pentagon Project Folding a pentagram Part 1 Folding method by Shuzo Fujimoto; from „The New Origami“ by Steve and Migumi Biddle

  23. The Pentagon Project Folding a pentagram Part 2 Folding method by Shuzo Fujimoto; from „The New Origami“ by Steve and Migumi Biddle

  24. The Pentagon Project Challenge Question: Decide whether the pentagram folded by this method is regular or not and prove your assertion. Answer: The pentagram is not regular! Why? Have a closer look at step 6!

  25. The Pentagon Project 1

  26. Axioms of Construction Straight-edge and Compass:

  27. Axioms of Construction Paper folding: (O1)

  28. Axioms of Construction

  29. Axioms of Construction (O6) (O7)

  30. Axioms of Construction (O7*)

  31. Folding Roots y 5 x -5 5 -5 Linear equation ax = b Solution: x = slope of the crease is

  32. Folding Roots Quadratic Equation x²+px+q = 0 x² - 2usx + 2uvs – 2uw = 0 u²s² - 2uvs + 2uw = 0 Parabola: x² = 2uy Tangent: y = s(x - v) + w u = 2, v = -p, w = q Parabola: x² = 4y (Focus F(0,1), directrix y = -1) P0(-p,q)

  33. Folding Roots t: y = cx + d p1: yy1 = ax + ax1 p2: xx2 = by + by2

  34. Folding Roots

  35. Folding Roots _ _ AC : CB =

  36. Folding Roots t: y = cx + d p1: (y-n)(y1–n) = a(x-m) + a(x1–m) p2: xx2 = by + by2 x³ + px² + qx + r = 0 p = -2m, q = 2n, r = a, b = 1

  37. Folding Roots angle trisection: cos 3a = 4cos³a - 3cos a or:

  38. Thanks for listening! robert.geretschlaeger@brgkepler.at http://geretschlaeger.brgkepler.at

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