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Math 1C – Project ORIGAMICS Origami & Math??

Math 1C – Project ORIGAMICS Origami & Math??. Garfield, Lester, Michelle, Satomi. Origami = 折り紙 = paper-folding. 折り “ Ori ” = fold + 紙 “ kami/gami ” = paper Creating a representation of an object Using geometric folds and crease patterns

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Math 1C – Project ORIGAMICS Origami & Math??

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  1. Math 1C – ProjectORIGAMICSOrigami & Math?? Garfield, Lester, Michelle, Satomi

  2. Origami = 折り紙 = paper-folding • 折り “Ori” = fold + 紙“ kami/gami” = paper • Creating a representation of an object • Using geometric folds and crease patterns • Preferably without gluing or cutting the paper

  3. What Can Origami Do?? • Give deep understanding of math by allowing to visualize ex. problem solving, ratios, angles, combinatorics, conic sections, Euler's formula, tangent line approximations, congruence, negative curvature 

  4. What Can Origami Do?? • Rigid Origami, Deployable Structures ex. Miura Map Fold (Foldable solar sails), Airbag Packing, Foldable Space Telescope • Paper Product Designs • Solve Polynomial Equations • Prove Geometry Construction Problems

  5. What Can Origami Do??-Science, Engineering- • Miura Map Fold

  6. Huzita-Hatori Axioms

  7. Axiom 1 Given two points p1 and p2, there is a unique fold that passes through both of them. This is equivalent to using a straight-edge and compass to connect two points (solving a first degree equation).

  8. Axiom 2 Given two points p1 and p2, there is a unique fold that places p1 onto p2. This creates the perpendicular bisector to a line that contains the two points.

  9. Axiom 3 Given two lines l1 and l2, there is a fold that places l1 onto l2. This is equivalent to bisecting the angle between the two lines.

  10. Axiom 4 Given a point p1 and a line l1, there is a unique fold perpendicular to l1 that passes through point p1. This is equivalent to finding the perpendicular to a line that goes thru a given point.

  11. Axiom 5 Given two points p1 and p2 and a line l1, there is a fold that places p1 onto l1 and passes through p2. This is equivalent to solving a second degree equation.

  12. Axiom 6 Given two points p1 and p2 and two lines l1 and l2, there is a fold that places p1 onto l1 and p2 onto l2. This is equivalent to solving a third degree equation.

  13. Axiom 7 Given one point p and two lines l1 and l2, there is a fold that places p onto l1 and is perpendicular to l2.

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