Origami Geometry Projects for Math Fairs

Origami Geometry Projects for Math Fairs Robert Geretschläger Graz, Austria

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

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.

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

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 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)

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‘

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

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

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

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

The Pentagon Project Placing the Pentagon on the Paper

The Pentagon Project Step 1

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.

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

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

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!

The Pentagon Project 1

Axioms of Construction Straight-edge and Compass:

Axioms of Construction Paper folding: (O1)

Axioms of Construction

Axioms of Construction (O6) (O7)

Axioms of Construction (O7*)

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

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)

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

Folding Roots

Folding Roots _ _ AC : CB =

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

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

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

Origami Geometry Projects for Math Fairs