Origami Using an Axiomatic System of Paper Folding

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Origami: Using an Axiomatic System of Paper Folding to Trisect the Angle

Origami: Using an Axiomatic System of Paper Folding to Trisect the Angle

Agenda • Euclidean Geometry • Euclidean Constructions • Origami and the Axiomatic System of

Agenda • Euclidean Geometry • Euclidean Constructions • Origami and the Axiomatic System of Humiaki Huzita • Trisecting the Angle • Proof of Trisection

Euclid’s Postulates 1. Between any two distinct points, a segment can be constructed. 2.

Euclid’s Postulates 1. Between any two distinct points, a segment can be constructed. 2. Segments can be extended indefinitely.

Euclid’s Postulates (cont. ) 3. Given two points and a distance, a circle can

Euclid’s Postulates (cont. ) 3. Given two points and a distance, a circle can be constructed with the point as the center and the distance as the radius. 4. All right angles are congruent.

Euclid’s Postulates (cont. ) 5. Given two lines in the plane, if a third

Euclid’s Postulates (cont. ) 5. Given two lines in the plane, if a third line l crosses the given lines such that the two interior angles on one side of l are less than two right angles, then the two lines if continued will meet on that side of l where the angles are less than two right angles. (Parallel Postulate)

Euclidean Constructions

Euclidean Constructions

Origami: Humiaki Huzita’s Axiomatic System 1. Given two constructed points P and Q, we

Origami: Humiaki Huzita’s Axiomatic System 1. Given two constructed points P and Q, we can construct (fold) a line through them. 2. Given two constructed points P and Q, we can fold P onto Q.

Origami: Humiaki Huzita’s Axiomatic System (cont. ) 3. Given two constructed lines l 1

Origami: Humiaki Huzita’s Axiomatic System (cont. ) 3. Given two constructed lines l 1 and l 2, we can fold l 1 onto l 2. 4. Given a constructed point P and a constructed line l, we can construct a perpendicular to l passing through P.

Origami: Humiaki Huzita’s Axiomatic System (cont. ) 5. Given two constructed points P and

Origami: Humiaki Huzita’s Axiomatic System (cont. ) 5. Given two constructed points P and Q and a constructed line l, then whenever possible, the line through Q, which reflects P onto l, can be constructed. 6. Given two constructed points P and Q and two constructed lines l 1 and l 2, then whenever possible, a line that reflects P onto l 1 and also reflects Q onto l 2 can be constructed.

Trisecting the Angle Step 1: Create (fold) a line m that passes through the

Trisecting the Angle Step 1: Create (fold) a line m that passes through the bottom right corner of your sheet of paper. Let be the given angle. Step 2: Create the lines l 1 and l 2 parallel to the bottom edge lb such that l 1 is equidistant to l 2 and lb. Step 3: Let P be the lower left vertex and let Q be the intersection of l 2 and the left edge. Create the fold that places Q onto m (at Q') and P onto l 1 (at P').

Trisecting the Angle (cont. ) Step 4: Leaving the paper folded, create the line

Trisecting the Angle (cont. ) Step 4: Leaving the paper folded, create the line l 3 by folding the paper along the folded-over portion of l 1. Step 5: Create the line that passes through P an P'. The angle trisection is now complete

Proof of Angle Trisection We need to show that the triangles ∆PQ'R, ∆PP'R and

Proof of Angle Trisection We need to show that the triangles ∆PQ'R, ∆PP'R and ∆PP'S are congruent. Recall that l 1 is the perpendicular bisector of the edge between P and Q. Then, → → Segment Q'P' is a reflection of segment QP and l 3 is the extension of the reflected line l 1. So l 3 is the perpendicular bisector of Q'P'. ∆PQ'R = ∆PP'R (SAS congruence).

Proof of Angle Trisection (cont. ) Let R` be the intersection of l 1

Proof of Angle Trisection (cont. ) Let R` be the intersection of l 1 and the left edge. From our construction we see that RP`P is the reflection of R`PP` across the fold created in Step 3. → → → RP'P = R'PP' and ∆P'PR' = ∆PP'S (SSS congruence). ∆PP'S = ∆PP'R (SAS congruence). ∆PP'S = ∆PP'R = ∆PQ'R

Other Origami Constructions • Doubling a Cube (construct cube roots) • The Margulis Napkin

Other Origami Constructions • Doubling a Cube (construct cube roots) • The Margulis Napkin Problem • Quintinsection of an Angle