Axioms are based on Huzita-Hatori Origami Axioms. All circle folds are based on one circle - the boundary of the paper.

Axiom 1

Given two points p1 and p2, there is a unique fold that passes through both of them.

Folding a line through two points

Axiom 2

Given two points p1 and p2, there is a unique fold that places p1 onto p2.

Folding a line putting one point on another

Axiom 3

Given two lines l1 and l2, there is a fold that places l1 onto l2.

Folding a line putting one line on another

This is equivalent to finding a bisector of the angle between l1 and l2.

Axiom 2 - 3 - C

Given two points p1 and p2 on the circle, there is a unique fold that places p1 onto p2 and places the half-circles l1 and l2 defined by the fold onto one another.

This is equivalent to finding a bisector of the angle between radii to p1 and p2.

Axiom 4

Given a point p1 and a line l1, there is a unique fold perpendicular to l1 that passes through point p1.

Folding through a point perpendicular to a line

This is equivalent to finding a perpendicular to l1 that passes through p1.

Axiom 4-C

Given a point p1, there is a unique fold that passes through point p1 and places the half-circles l1 and l2 defined by the fold onto one another.

This is equivalent to finding a diameter through p1.

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. If the distance between p1 and p2 is greater than the distance between p2 and l1, there are two such folds, if the distances are equal, one such fold. If the distance between p1 and p2 is smaller than the distance between p2 and l1, the fold is impossible.

This is equivalent to finding the intersection(s) of the circle centered in p2 with the radius equal to the distance between p1 and p2 with l1.

Axiom 5-C

Given two points p1 and p2, there is a fold that places p1 onto the circle and passes through p2. If the distance between p1 and p2 is greater than the distance between p2 and the circle, there are two such folds, if the distances are equal, one such fold. If the distance between p1 and p2 is smaller than the distance between p2 and the circle, the fold is impossible.

The two-solution situation is shown above.

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.

Huzita axiom 6.png

Axiom 1-6-C

Given two points p1 and p2, of which at least one is not on the circle, there are two and only two folds that place both points on the circle.

Need a drawing (Euclid's construction) here.

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.

Huzita-Hatori axiom 7.png

Axiom 7-C

Given one point p and a line l1, there is a fold that places p onto the circle and is perpendicular to l1. There is one such fold if p lies on the diameter perpendicular to l1, and two such folds otherwise.

This is equivalent to finding midpoints of segments between p and the intersections of the circle with the line parallel to l1 and passing through p.

## Circle Origami Axioms

Discussion participants

Truncated Icosahedron, by Bradford Hansen-Smith, wholemovement.com

Axioms are based on Huzita-Hatori Origami Axioms. All circle folds are based on one circle - the boundary of the paper.

## Axiom 1

Given two pointsp1 andp2, there is a unique fold that passes through both of them.## Axiom 2

Given two pointsp1 andp2, there is a unique fold that placesp1 ontop2.## Axiom 3

Given two linesl1 andl2, there is a fold that placesl1 ontol2.This is equivalent to finding a bisector of the angle between

l1 andl2.## Axiom 2 - 3 - C

Given two points

p1 andp2 on the circle, there is a unique fold that placesp1 ontop2 and places the half-circlesl1 andl2 defined by the fold onto one another.This is equivalent to finding a bisector of the angle between radii to

p1andp2.## Axiom 4

Given a pointp1 and a linel1, there is a unique fold perpendicular tol1 that passes through pointp1.This is equivalent to finding a perpendicular to

l1 that passes throughp1.## Axiom 4-C

Given a pointp1, there is a unique fold that passes through pointp1 and places the half-circlesl1 andl2 defined by the fold onto one another.This is equivalent to finding a diameter through

p1.## Axiom 5

Given two pointsp1 andp2 and a linel1, there is a fold that placesp1 ontol1 and passes throughp2. If the distance betweenp1andp2is greater than the distance betweenp2andl1, there are two such folds, if the distances are equal, one such fold. If the distance betweenp1andp2is smaller than the distance betweenp2andl1, the fold is impossible.This is equivalent to finding the intersection(s) of the circle centered in

p2with the radius equal to the distance betweenp1andp2withl1.## Axiom 5-C

Given two points

p1 andp2, there is a fold that placesp1 onto the circle and passes throughp2. If the distance betweenp1andp2is greater than the distance betweenp2and the circle, there are two such folds, if the distances are equal, one such fold. If the distance betweenp1andp2is smaller than the distance betweenp2and the circle, the fold is impossible.The two-solution situation is shown above.

## Axiom 6

Given two pointsp1 andp2 and two linesl1 andl2, there is a fold that placesp1 ontol1 andp2 ontol2.## Axiom 1-6-C

Given two pointsp1 andp2, of which at least one is not on the circle, there are two and only two folds that place both points on the circle.Need a drawing (Euclid's construction) here.## Axiom 7

Given one pointpand two linesl1 andl2, there is a fold that placespontol1 and is perpendicular tol2.## Axiom 7-C

Given one pointpand a linel1, there is a fold that placesponto the circle and is perpendicular tol1. There is one such fold ifplies on the diameter perpendicular tol1, and two such folds otherwise.This is equivalent to finding midpoints of segments between

pand the intersections of the circle with the line parallel tol1and passing throughp.