3D Folding Axioms

3D Folding Axioms Roger C. Alperin∗

Abstract: We describe the axioms for 3D folding. This generalizes the HuzitaJustin axioms for foldings in the plane. In three dimensions the axioms involve a combination of foldings of point, line or plane to another point, line or plane for which there are finitely many possible solution planes when the given point, line, plane data is in general position.

1

Introduction

The Huzita-Justin axioms provide a wonderful set of prescriptions for making origami constructions which extend the familiar Euclidean ruler and compass construction methods in the plane [Alperin and Lang 09]. The most complicated of these axioms allows one to fold a crease line which serves as a reflection to place point P1 on line L1 and simultaneously point P2 on line L2 . There can be at most three possibilities for the location of the crease line for the given points and lines. These three lines are in fact the common tangents to two parabolas with foci at Pi and directrices Li for i = 1, 2 [Alperin 00]. Here we consider a generalization of these axioms to three and higher dimensions; in general we fold together affine subspaces (by using reflections in hyperplanes). In 3D origami we fold together a given set of points, lines, planes using the axioms; the axioms are described in terms of basic folds. Axioms are minimal sets of basic folds which yield a finite number of possible solutions (and not always zero). A basic fold denoted X | Y has its elements X,Y among the points (P), lines (L) and planes (π); the solutions to this basic fold is the set of planes for which an orthogonal reflection, ρ, in any one of those planes makes ρ(X) and Y incident. Our aim here is to enumerate the axioms; we leave the discussion of equations and other field theoretic aspects for a future project. We do illustrate with a few calculations the solutions to some of the axioms; besides linear algebra one may need to use the resultant or Groebner bases. Also we shall not investigate here the constructions one can make from new lines and points determined by the intersection of planes obtained from the axioms. The most complicated axiom in 3D folding is the generalization of the tangent lines to two parabolas to the folding of the seven possible tangent planes to three paraboloids formed by folding three given points (foci) to three given (directrix) ∗ Research

supported by NSF-EFRI

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Figure 1: Three Paraboloids With Seven Common Tangent Planes

planes. Using the notation as above, the axiom is denoted P1 | π1 , P2 | π2 , P3 | π3 where Pi are points and πi are planes, i = 1, 2, 3. In Figure 1 we illustrate the solutions to this complicated situation. Three paraboloids are obtained as the envelope of planes from the folding Pi | πi , i = 1, 2, 3 (see Section 2.5). The seven common solutions are shown. The paraboloids are tangent to the back solution plane and there are two ‘triangular’ sets of three planes which are also tangent to all three paraboloids. The given points are the foci of the paraboloids; the given planes are directrix planes of the paraboloids; both of these data items are not shown; they are described in Section 4.8. In Figure 2 we show the associated point-plane data and the solution (white plane) for Section 4.6 so that perpendicular reflection of the given points (red, green, blue) land on the given planes (red, green, blue). The three paraboloids are not shown.

2

Basic Folds in 3D

In this section we describe the basic folds which in combination will give the axioms.

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2.1 First Three Axioms Here are the simplest basic folds. Solutions are described. These three basic folds are in fact axioms. • 1) P1 | P2 ; perpendicular bisector plane. • 2) L1 | L2 ; if L1 , L2 lie in a plane π we get angle bisector planes perpendicular to π; otherwise there are no solutions. • 3) π1 | π2 ; angle bisector plane and its perpendicular plane through the intersection of π1 and π2 or the mid-plane if these are parallel. 2.2 Terminology and Finiteness In general when we describe a basic fold, elements with different indices are distinct. The other basic folds are listed below, labelled a)-f). Axioms will be numbered. Here is some terminology: • • • •

BP , the bundle of all planes passing through the point P is 2-dimensional; FL is the fan of all planes containing the line L is 1-dimensional; OL is the orthofan of all planes perpendicular to L is 1-dimensional; Oπ the orthobundle of all planes orthogonal (perpendicular) to π is 2-dimensional.

It is easy to see that BP meets FL , OL , Oπ in a single plane as long as P does not lie on L or π. Also Oπ ∩ FL and Oπ ∩ OL have one common plane as long as L is not perpendicular or parallel to π. It is not difficult to see that the intersection of a fan and orthofan, two fans or two orthofans is finite in general. This finiteness property is important for seeing that certain sets of basic folds are axioms. 2.3

Fixed Basic Folds

• a) L | L; any plane in FL or OL ; this is 1-dimensional. • b) P | P; any plane in BP ; this is 2-dimensional. • c) π | π; any plane in Oπ or the trivial solution π; this is 2-dimensional. One can think of these basic folds as giving projective linear subvarieties of the Grassmannian of 2-planes in 3-space with the specified dimension. 2.4 Loci Basic Folds The next set of basic folds have loci solutions in general, that is they are tangent to a surface. • d) P | L: the locus is a parabolic cylinder, since we are reflecting across the tangent to the parabola in the plane of P, L with focus P and directrix L. All tangent planes are perpendicular to the plane of P, L. This is 1-dimensional. • e) L | π: i) If L, π meet at P then we get a cone (with vertex at P) of bisection planes reflecting L to π. All tangent planes pass through P.

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ii) If L, π are parallel then the solution planes form the tangent planes to a parabolic cylinder (with parabola focus P on L and directrix L0 in π, both lying on a plane perpendicular to π). All tangent planes are perpendicular to the plane of P, L0 . This is 1-dimensional. • f) P | π: the solutions are the tangent planes to the 3D analogue of parabola, an elliptic paraboloid, a surface of degree 2, Section 2.5; this is 2-dimensional. One can think of these loci basic folds as giving projective non-linear algebraic subvarieties of the Grassmannian of 2-planes in 3-space with the specified dimension. 2.5 Paraboloid Equation for Basic Fold f Here we explain how to obtain the equation of the paraboloid from the folding data of basic fold f). Say the point P = (a, b, c) and π has equation px + qy + rz = s. For any point P0 = (x0 , y0 , z0 ) of π, then the reflection plane π0 which folds P to P0 passes through their midpoint and its normal is parallel to PP0 . We thus obtain the family of fold planes π0 having the equations G(x0 , y0 , z0 ) : (a − x0 )x + (b − y0 )y + (c − z0 )z =

a2 + b2 + c2 − x02 − y20 − z20 . 2

We can obtain the equation of the surface in a well-known way, [Pottman 18], with this enveloping family of tangent planes. First eliminate z0 from the tangent plane equation G, using the equation px0 + qy0 + rz0 = s, to obtain F(x0 , y0 ) = G(x0 , y0 , s−px0r −qy0 ); then solve ∇F(x0 , y0 ) = 0 for its critical point. Using this critical point we can then eliminate x0 , y0 from F to get the equation for the associated paraboloid. Let N = p2 + q2 + r2 then the equation for the paraboloid is: (r2 + q2 )x2 + (p2 + r2 )y2 + (p2 + q2 )z2 − 2pqxy − 2prxz − 2qryz

= (2aN − 2sp)x + (2bN − 2sq)y + (2cN − 2sr)z + s2 − (a2 + b2 + c2 )N

3

Enumerating Axioms

We consider axioms where the elements are typically in general position: no incidences of elements in a basic fold unless the elements are equal; lines in an axiom are not parallel or perpendicular to another line or plane; planes specified in an axiom are not parallel or perpendicular; no three points involved in an axiom lie on a line; no two points in an axiom lie on a line perpendicular to a specified plane or parallel to a specified line. The dimensions for the basic folds allows us to make the following simplification in the enumeration process. 1. If a basic fold in an axiom is 1-dimensional then we will need only one other basic fold to make an axiom; this follows since another basic fold in general position will reduce the solution set to 0-dimensional and hence finite.

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2. If two of the basic folds in an axiom are 2-dimensional then we will need one other basic folds to make an axiom. These two folds reduce the dimension of solutions to one dimension so any subset from another basic fold in general position will be 0-dimensional and hence finite. But from the previous comment a 1-dimensional fold only needs to be combined with one other basic fold. Hence the case of three basic folds in an axioms only occurs if they are each 2-dimensional. The number of axioms can now be estimated by counting those with 1 ≤ i ≤ 3 basic folds. For i = 1 then we get the three axioms of Section 2.1. The 1dimensional basic folds are a, d, e; the 2 dimensional folds are b, c, f . For i = 2 we use any 1-dimensional basic fold and any other fold to get 15 possible axioms (i.e., 3 (same) +3 (different)+3 · 3). When i = 3 we choose any 3 with repeats from b, c, f to get 10 possible axioms (i.e., 3 (same) +3 · 2 (two same) +1 (different)). This gives at most 28 axioms. In the next subsection the axioms are enumerated in classes. We number the axioms by the class and then list which basic folds are used. For example axiom 4bbc occurs in the next section; the basic folds included in this axiom are two of type b and one of type c. Each occurrence of basic fold b in the axiom involves distinct data. 3.1

Three Fixed Basic Folds

This first class has axioms with three fixed basic folds; recall basic folds b, c are 2-dimensional. The axiom 4ccc does not occur since there are no planes perpendicular to three planes in general position. • 4bbb) P1 | P1 , P2 | P2 , P3 | P3 ; the plane passing through P1 , P2 , P3 . • 4bbc) P1 | P1 , P2 | P2 , π1 | π1 ; the plane passing through P1 , P2 and perpendicular to π1 ; by our general position assumptions the line L = P1 P2 is not perpendicular or parallel to π1 . • 4bcc) P1 | P1 , π1 | π1 , π2 | π2 ; planes in OL and BP1 where L is the line of intersection of π1 and π2 . 3.2

Two Fixed Basic Folds

Next comes the axioms involving two fixed elements. • 5aa) L1 | L1 , L2 | L2 ; plane in FL1 ∩ FL2 or OL1 ∩ FL2 or OL2 ∩ FL1 or OL1 ∩ OL2 . • 5ab) L1 | L1 , P1 | P1 ; plane in BP1 and either FL1 or OL1 . • 5ac) L1 | L1 , π1 | π1 ; plane in Oπ1 and either FL1 or OL1 . If L1 is parallel to π1 there is a unique solution; while if L1 is perpendicular to π1 , then π1 is the solution. If L1 meets π1 at P1 , then the normal line L through P1 together with L1 span the solution plane in FL1 . • 5bbf) P1 | P1 , P2 | P2 , P3 | π3 ; planes in FL where L = P1 P2 ; planes also tangent to a paraboloid.

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• 5bcf) P1 | P1 , π1 | π1 , P3 | π3 ; plane in Oπ1 and BP1 ; planes also tangent to a paraboloid. • 5ccf) π1 | π1 , π2 | π2 , P3 | π3 ; the planes perpendicular to π1 , π2 have normal line L given by the cross-product of normals of the given planes, hence in OL ; planes also tangent to paraboloid. 3.3 One Fixed Basic Fold Next are axioms with one fixed basic fold. For basic fold e) we can assume the locus is tangent planes to a cone by our general position condition. • 6ad) L2 | L2 , P2 | L1 ; planes in FL2 or OL2 and tangent to parabolic cylinder. • 6ae) L2 | L2 , L1 | π1 ; planes in FL2 or OL2 and tangent to cone or parabolic cylinder. • 6af) L1 | L1 , P1 | π1 ; planes in FL1 or OL1 and tangent to paraboloid. • 6bd) P1 | P1 , P2 | L1 ; at most two planes passing through P1 so that the reflection of P2 lands on L1 . • 6be) P1 | P1 , L1 | π1 ; at most two planes passing through P1 which are tangent to the associated cone. • 6bff) P3 | P3 , P1 | π1 , P2 | π2 ; planes in the (double) tangent cone (through P3 ) on the two paraboloids. • 6cd) π1 | π1 , P1 | L1 ; planes perpendicular to π1 and tangent to parabolic cylinder so perpendicular to plane of parabola. • 6ce) π1 | π1 , L2 | π2 ; planes perpendicular to π1 and tangent to a cone (through the cone point). • 6cff) π3 | π3 , P1 | π1 , P2 | π2 ; planes perpendicular to π3 and tangent to two paraboloids. 3.4 No Fixed Basic Folds The last class contains axioms where no basic folds have a fixed element i.e. involving basic folds d), e), f ) only. The planes in case d) are tangent to the associated parabola; the planes in case e) are tangent to a bisection cone or a parabolic cylinder; by our generic assumption we can assume it is a cone. These are 1 dimensional so only one more basic fold is needed to make an axiom. This analysis yields the following axioms. • • • • • •

7dd) P1 | L1 , P2 | L2 ; planes tangent to two parabolic cylinders. 7de) P1 | L1 , L2 | π2 ; planes tangent to parabolic cylinders and cone. 7df) P1 | L1 , P2 | π2 ; planes tangent to parabolic cylinder and paraboloid. 7ee) L1 | π1 , L2 | π2 ; planes tangent to two cones. 7ef) L1 | π1 , P2 | π2 ; planes tangent to cone and paraboloid. 7fff) P1 | π1 , P2 | π2 , P3 | π3 ; consider the dual equations which are also of degree two. These three degree two equations will have eight common solutions in projective space, Chapter X of [Pedoe 88]. The plane at infinity is tangent to any

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paraboloid so this can be reduced to at most seven planes which are the common tangent planes to three paraboloids.

4

Examples

We illustrate some of the algebra required to solve a small selection of the axioms. 4.1 Axiom 6be Let P = (2, 3, −1), π the xy-plane, L the line through the origin making an angle of 60◦ . We fold L onto π so that the reflection plane passes through P. A solution plane π0 : x0 x + y0 y + z0 z = 0 makes an angle of 30◦ or 120◦ with the xy-plane, passes through the origin and P. z2

In the first case 34 = (cos 30◦ )2 = x2 +y02 +z2 and 2x0 + 3y0 − z0 = 0. The solution 0√ 0 0 √ planes are (−6 ± 30)x + y + (−9 ± 2 30)z = 0. The details for the second case of 120◦ are left as an exercise for the interested reader. 4.2 Axioms 5bbf, 5bcf, 5ccf We use the point and plane of fold f) specifying the paraboloid with a = 0, b = 0, c = 1, p = 1, q = 0, r = 1, s = 0; this gives the equation G : x2 + 2y2 + z2 − 2xz − 4z + 2 = 0, The tangent planes to the paraboloid G at the general point (x0 , y0 , z0 ) have normals given by the gradient ∇G(x0 , y0 , z0 ) = (2x0 − 2z0 , 4y0 , 2z0 − 2x0 − 4) and the tangent plane is (2x0 − 2z0 )(x − x0 ) + 4y0 (y − y0 ) + (2z0 − 2x0 − 4)(z − z0 ) = 0. Specifying two given points for 5bbf) gives two sets of constants to be substituted for x, y, z of this tangent plane equation. Thus we get two quadratic curve equations in x0 , y0 , z0 and hence at most 4 solutions. Specifying two planes for 5ccf) gives two sets of linear conditions on the normal ∇G(x0 , y0 , z0 ) and by general position we can then solve for x0 , y0 , z0 . Specifying a point and plane for 5bcf) gives the problem of locating the intersections of a plane with a quadratic curve not in the plane (by general position). There are at most 2 solutions. 4.3 Axiom 6bff The axiom is P3 | P3 , P1 | π1 , P2 | π2 In this example we use the data a = 1, b = 0, c = 0, p = 1, q = 0, r = 0, s = 0, giving paraboloid F = y2 + z2 − 2x + 1 = 0, and a = 0, b = 0, c = 1, p = 1, q = 0, r = 1, s = 0, giving paraboloid G = x2 + 2y2 + z2 − 2xz − 4z + 2 = 0, corresponding to the basic folds P1 | π1 , P2 | π2 . In general the solutions planes tangent to two paraboloids can be expressed by a planar quartic, [Pedoe 88]. In this case the curve is 2x3 − x2 y2 − 5x2 + 8x + 4xy2 − 4y2 − 4 − y4 = 0 (giving points of tangency on the first paraboloid after solving F for z).

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For the point P3 = (1, 2, −1) after solving for the possible values of x1 we obtain an octic polynomial f ; this polynomial factors into two quartics f = (8x4 − 72x3 + 96x2 − 144x + 81)(4x4 − 40x3 + 100x2 − 144x + 81) and has 4 real roots. Thus there will be 4 planes that pass through P3 tangent to both paraboloids . 4.4 Axiom 6cff The axiom is π3 | π3 , P1 | π1 , P2 | π2 . In this example we use the same data for P1 , P2 , π1 , π2 as in the previous example. Let π3 be the plane x + 2y − z = 0. The tangent planes along the curve on the first paraboloid with x, y coordinates satisfying 2x3 − x2 y2 − 5x2 + 8x + 4xy2 − 4y2 − 4 − y4 = 0 have normal vector parallel to (−1, y0 , z0 ) at the point (x0 , y0 , z0 ) giving the tangent plane −(x − x0 ) + y0 (y − yy ) + z0 (z − z0 ) = 0. We want this tangent plane to be normal to the given plane π3 so −1 + 2y0 − z0 = 0. Together with the curve equation we reduce then to solving the equation h = (40x3 − 73x2 + 160x − 96)(40x3 − 105x2 + 160x − 96). The polynomial h has two real roots; thus there are two real solutions to this instance of the axiom. 4.5 Axiom 7ef We use the data for the paraboloid as in a previous example a = 0, b = 0, c = 1, p = 1, q = 0, r = 1, s = 0, giving paraboloid G : x2 + 2y2 + z2 − 2xz − 4z + 2 = 0. For basic fold e) we use the normal line L1 to π1 : z = 0 at the point (1, 2, 0) The tangent planes to the paraboloid G at the general point (x0 , y0 , z0 ) have normals given by the gradient v0 = ∇G = (2x0 − 2z0 , 4y0 , 2z0 − 2x0 − 4). To solve the axiom condition we want this plane to make 45◦ or 135◦ angle with the z2 0 −4) axis so 21 = (2z0 −2x . This simplifies to give the condition f : y20 − x0 + z0 = 1. |v0 |2 In addition the solution plane must pass through (1,2,0) (vertex of cone) so g : (2x0 − 2z0 )(1 − x0 ) + 4y0 (2 − y0 ) + (2z0 − 2x0 − 4)(−z0 ) = 0. Eliminate z0 from G(x0 , y0 , z0 ) and g(x0 , y0 , z0 ) using f ; then taking their resultant we get the restriction y40 − 2y20 − 8y0 + 1 = 0 having only two real solutions; this yields two solutions for this instance of the axiom. 4.6 Easy Example 7fff As an explicit example we determine the plane tangent to three paraboloids with parameters given by S1 : a = 1, b = 0, c = 0, p = 1, q = 0, r = 0, s = 0, F1 = y2 + z2 − 2x + 1 = 0; S2 : a = 0, b = 1, c = 0, p = 1, q = 1, r = 0, s = 0, F2 = x2 + y2 + 2z2 − 2xy − 4y + 2 = 0; S3 : a = 0, b = 0, c = 1, p = 1, q = 0, r = 1, s = 0, F3 = x2 + 2y2 + z2 − 2xz − 4z + 2 = 0.

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We solve equations making the tangent planes to these three paraboloids the same by using a variable point (xi , yi , zi ) on Fi , i = 1 − 3. These equations obtained in this way for F1 , F2 are (1 − y1 )x2 − (1 − y1 )y2 + 2 = 0, z1 x2 − z1 y2 + 2z2 = 0, z1 (2x2 y2 − y22 − 2z22 − x22 + 2y2 ) = 2z2 (x1 − y21 − z21 ). Eliminating x2 , y2 , z2 leads to the relation 1 − x1 y1 − y21 = 0. Doing similarly with F1 , F3 we obtain the relation 2x13 − x12 y21 − 5x12 + 8x1 + 4x1 y21 − 4y21 − 4 − y41 = 0. Both of these relations in x1 , y1 must be satisfied; using the resultant to eliminate y1 we find 4x13 − 6x12 + 12x1 − 9 = 0; using a real root x1 we solve for y1 , z1 to obtain the common tangent plane. (There is also quartic polynomial which does not have any real roots.) The folding data is shown in Figure 2.

2

1 −2 z

0

−1 0 y

−1 1 −2 2

1

0 x

−1

−2

Figure 2: Folding Given (Red, Green Blue) Points Across White Plane to Given (Red, Green Blue) Planes

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Figure 3: 3 Paraboloids Having 7 Common Tangent Planes

4.7

Seventh Order Example 7fff

Using S1 : a = 20, b = 1, c = 0, p = 0, q = 1, r = 1, s = 0, S2 : a = 0, b = 20, c = 1, p = 1, q = 0, r = 1, s = 0, S3 : a = 1, b = 0, c = 20, p = 1, q = 1, r = 0, s = 0, we obtain the three paraboloids forming a tripod as shown in Figure 3. The equations are P1 : x2 + y2 + 2z2 − 2xy − 4x − 80z + 802 = 0, P2 : 2x2 + y2 + z2 − 2yz − 4y − 80x + 802 = 0, P3 : x2 + 2y2 + z2 − 2xz − 4z − 80y + 802 = 0. By elimination of variables we obtain a 7th degree polynomial using Groebner bases, f = (8x − 7)(1523x6 − 4361904x5 + 4133293582x4 − 1297833733306x3 + 3732681749811x2 − 3246090685082x + 852552189699); there are 7 real roots. The roots of this are used to determine the equations of the tangent planes. In this case the paraboloids are divided into two parts, a base tripod with three ‘knobs’ containing the vertices and the ‘faces’ above the intersections. We can easily see the planes are either: tangent to the three paraboloid vertices; 3 planes tangent to 2 knobs and a face of the third; 3 planes tangent to two faces and one knob.

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4.8 Seven Planes Example 7fff The three paraboloids described by S1 : a = 10, b = 9, c = −20, p = 1, q = 1, r = 0, s = 0, S2 : a = −20, b = 10, c = 9, p = 0, q = 1, r = 1, s = 0, S3 : a = 9, b = −20, c = 10, p = 1, q = 0, r = 1, s = 0, do not intersect. There are 7 common tangent planes as shown in Figure 1. which have been determined from the roots of the polynomial f = (8x−53)(71669691x6 − 16324843092x5 +1346409319714x4 −51262913571518x3 +914740982982727x2 − 6479651759393546x + 9482118760046363). The root represents the x-coordinate for a point on the third paraboloid whose tangent plane there is also tangent to the first and second paraboloids.

5

Conclusions

There are 9 basic folds: the first three are in fact axioms; the next three have a fixed element, either a point, line or plane (fixed folds); the last three have a locus of solutions. We have enumerated the axioms in classes depending on the number of fixed basic folds. The counts for each class is • • • • •

A) 3 axioms which are basic folds; B0 ) 6 axioms which do not involve a fixed fold; B1 ) 9 axioms which involve exactly one fixed fold; B2 ) 6 axioms which involve exactly two fixed folds; B3 ) 3 axioms which involve exactly three fixed folds.

Thus the total number is 27. Counting them differently we see that there are • i) 3 axioms determined from 1 basic fold; • ii) 15 axioms determined from 2 basic folds; • iii) 9 axioms determined from 3 basic folds. For future work we will investigate the constructions which are possible using these axioms, starting from the set of rational points in space. We believe that one should be able to construct all algebraic points which are in any combination of extensions generated by the roots of a polynomial of degree at most 7. Often the axioms will work (that is there are finitely many solutions) for positions which are not in general position, but other times special positions leads to infinitely many solutions. One can also consider axioms for point, line, plane data which is not in general position. Are there any new interesting axioms? Investigating these cases requires some care. For example, suppose we consider the partial axiom P1 | P1 , P2 | P2 , π1 | π1 where the line L = P1 P2 is perpendicular π1 . Can this be extended to an axiom by adding in more basic folds? No, In this case the

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solution set is FL ; so any extension is not minimal since P1 | P1 , P2 | P2 also has the solution set FL .

References [Alperin and Lang 09] Roger C. Alperin and Robert J. Lang. “One-, two-, and multi-fold origami axioms.” In Origami4 , pp. 371–393. A K Peters, Natick, MA, 2009. [Alperin 00] Roger C. Alperin. “A Mathematical Theory of Origami Constructions and Numbers.” New York Journal of Mathematics 6 (2000), 119–133. [Pedoe 88] Dan Pedoe. Geometry, Second edition. Dover Books on Advanced Mathematics, Dover Publications, Inc., New York, 1988. [Pottman 18] H. Pottman. “Envelopes-Computational Theory and Applications.” https: //dmg.tuwien.ac.at/geom/ig/papers/pot109.pdf, 2018.

Roger C. Alperin San Jose State University, e-mail: [email protected]