# Geometric construction of canonical 3D gadgets in origami extrusions

In a series of our three previous papers, we presented several constructions of positive and negative 3D gadgets in origami extrusions which create with two simple outgoing pleats a top face parallel to the ambient paper and two side faces sharing a ridge, where a 3D gadget is said to be positive (resp. negative) if the top face of the resulting gadget seen from the front side lies above (resp. below) the ambient paper. For any possible set of angle parameters, we obtained an infinite number of positive 3D gadgets in our second paper, while we obtained a unique negative 3D gadget by our third construction in our third paper. In this paper we present a geometric (ruler and compass) construction of our third negative 3D gadgets, while the construction presented in our third paper was a numerical one using a rather complicated formula. Also, we prove that there exists a unique positive 3D gadget corresponding to each of our third negative ones. Thus we obtain a canonical pair of a positive and a negative 3D gadget. The proof is based on a geometric redefinition of the critical angles which we introduced in constructing our positive 3D gadgets. This redefinition also enables us to give a simplified proof of the existence theorem of our positive 3D gadgets in our second paper. As an application, we can construct a positive and a negative extrusion from a common crease pattern by using the canonical counterparts, as long as there arise no interferences.

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## 1. Introduction

This is the fourth in a series of our papers on constructions of D gadgets in origami extrusions, where an origami extrusion is a folding of a D object in the middle of a flat piece of paper, and D gadgets are ingredients for origami extrusions which create faces with solid angles. In our three previous papers [2], [3] and [4], we studied D gadgets which create with two simple outgoing pleats a top face parallel to the ambient paper and two side faces sharing a ridge, where a simple pleat consists of a mountain and a valley fold which are parallel to each other. If the top face of a D gadget seen from the front side lies above (resp. below) the ambient paper, then the D gadget is said to be positive (resp. negative).

In our second paper [3], we presented a construction of positive D gadgets with flat back sides extending the construction in [2], which improve the D gadgets developed by Natan with a supporting pyramid on the back side [5], where we named our positive D gadgets (positive) origon gadgets or simply (positive) origons. Meanwhile, in our third paper [4], we presented three constructions of negative D gadgets in origami extrusions in addition to two known constructions before ours, of which we also extended Cheng’s construction [1]. Let us also call negative D gadgets by our three constructions negative origon gadgets or simply negative origons.

The constructions of the crease patterns of positive and our third negative origon gadgets are characterized by the ‘dividing point’ of the circular arc which has a radius of the same length as the ridge and subtains the angle formed between the side faces in the development. In our second paper [3], we introduced the critical angles by which the possible range of the dividing point is determined. Then we proved that for any possible set of angle parameters, there exist infinitely many positive origons which are distinguished by the choice of the dividing point. Meanwhile, in our third paper we proved that for any possible set of angle parameters, our third construction provides a unique negative origon. Let us call a negative origon gadget uniquely obtained in our third construction a canonical negative origon (gadget). The unique dividing point for a canonical negative origon is specified by the rotation angle from either endpoint of the arc. However, the angle was not given geometrically but numerically by a rather complicated formula.

In this paper we present a geometric (ruler and compass) construction of canonical negative origon gadgets in Construction 3.1. This solves the problem which came up in the conclusion of [4]. Also, we prove in Theorem 4.1 that for any canonical negative origon gadget, there exists a positive one with the same set of angle parameters and the same choice of the dividing point as the negative one. This result yields a canonical positive origon among infinitely many compatible ones, and a canonical pair consisting of a positive and a negative origon which are both canonical and correspond to each other. The proof is done by showing that the dividing point lies in the range of constructibility of positive origons, which is determined by the critical angles. For this purpose, we give in Definition 2.6 a geometric redefinition of the critical angles, which directly determines the range of constructibility. Using this redefinition, which is simpler than the original geometric definition and its numerical rephrasing given in [3], we can give a simplified proof of the existence theorem of positive origons in Theorem 2.16.

As an application of canonical pairs, we can construct a positive and a negative extrusion made of origon gadgets from a common crease pattern by replacing each origon with its canonical counterpart.

###### Convention 1.1.

We will use the following conventions.

1. As in our previous papers, we will use subscript for ‘L’ and ‘R’ standing for ‘left’ and ‘right’ respectively, and for the other side of , that is,

2. We will use ‘development’ for the flat piece of paper obtained by developing a D gadget, which includes not only the net of the extruded object but also the creases hidden behind after the folding.

3. We will determine the standard side from which a D gadget is seen so that the extruded faces lie above the ambient paper. Thus we will usually consider the development of a positive (resp. negative) gadget seen from the front (resp. back) side.

4. If we denote a circular arc with endpoints and by , then we will regard the arc as swept counterclockwise from to .

5. If we denote an angle by (resp. ), then we will regard the angle as the rotation angle around of the ray starting from through to the ray starting from through measured counterclockwise (resp. clockwise), which may take a negative value depending on the range of the angle under consideration.

To fix the conditions and the construction of typical points, lines and angles common to all of the later constructions of origon gadgets, we prepare the following.

###### Construction 1.2.

We consider a net on a flat piece of paper as in Figure 1.3, where . We require the following conditions on (or ), and .

1. , and , or equivalently, , and .

2. , or equivalently, .

To construct a negative gadget from the above net, we prescribe its simple outgoing pleats by introducing parameters for their changes from the direction of for , for which we further require the following conditions.

1. and , where we take clockwise (resp. counterclockwise) direction as positive for (resp. ).

2. and .

3. , or equivalently, .

In particular, if , then conditions (iii.a)–(iii.c) are simplified as

1. , or equivalently, .

Then we construct the creases of simple pleats and consisting of a mountain and a valley fold parallel to each other, which we prescribe as the outgoing pleats of positive and negative origons as follows, where we regard as taking both and .

1. Draw a ray starting from and going to the direction of followed by a clockwise rotation by . Also, draw a ray starting from and going to the direction of followed by a counterclockwise rotation by . Following Convention 1.1, , we can also define (resp. ) as a ray starting from (resp. ) such that (resp. ).

2. Draw a perpendicular to through for both , letting be the intersection point.

3. Draw a perpendicular bisector to segment for both , letting be the intersection point. Since is the excenter of , segment is a perpendicular bisector of segment and also a bisector of .

The resulting creases are shown as solid lines in Figure 1.4.

###### Remark 1.5.

Define by and , so that we have . In our previous papers [2], [3] and [4], we assumed in addition to in condition (iii.b). This was because angle is given by

 (1.1) tanγσ =1−cosγ+sinγtanδσ′sinγ+cosγtanδσ′+tanδσ,or tan(γσ−γ2) =tanδσ′−tanδσ2+(tanδσ+tanδσ′)/tan(γ/2),

and we avoided the appearance of . However, since we see from condition (iii.c) of Construction 1.2 that and do not take simultaneously, we can define continuously even at or by the limit of as or , so that we have

as expected. Hence in this paper we will not assume in condition (iii.b). Note that holds if and only if , in which case we have and .

We end this section with the organization of this paper. In Section 2, we give in Definition 2.6 a redefinition of critical angles originally introduced in [3], and then recall the construction of positive origon gadgets in Construction 2.8. Then after proving the equivalence of the original and the new definition of the critical angles in Theorem 2.13, we give in Theorem 2.16 a simplified proof of the existence theorem of positive origons using the new definition of critical angles. In Section 3, we give in Construction 3.1 a geometric (ruler and compass) construction of canonical negative origon gadgets. The equivalence of the original numerical and the new geometric definition of the canonical dividing point is proved in Theorem 3.7. Then in Section 4, we prove in Theorem 4.1 the existence of canonical positive origons. Finally, Section 5 gives our conclusion.

## 2. Redefinition of the critical angles

First we recall the original definition of the critical angles used in constructing positive origon gadgets, which is given in [3], Definition .

###### Definition 2.1.

Consider a development as in Figure 1.4, for which we assume conditions (i), (ii) and (iii.a)–(iii.c) of Construction 1.2. Then we define the critical angles for by the following construction, where all procedures are done for both .

1. Draw a ray starting from and going inside so that .

2. Let be the intersection point of ray and polygonal chain .

3. Then we define the critical angle by

 ζσ=∠BσAQσ.
4. Let be a point in minor arc such that . Let us call a critical dividing point.

5. The critical angles and the critical dividing points constructed here are shown in Figure 2.7, where the possible range of radius of minor arc with center is shaded, which is used in Construction 2.8, .

The critical angles are also defined numerically by the following result.

###### Proposition 2.3 ([3], Proposition 3.6).

The critical angle given in Definition 2.1 is numerically given by

 (2.1)

where we set

 bσ=tan(βσ−δσ),c=tan(γ/2),c′=tan(γ/2+δL+δR),% and dσ=tanδσ.

In particular, if , then we have and , so that

 ζσ=tan−1(12/tan(γ/2)+1/tanβσ)<γ2.

To introduce a new geometric definition of the critical angles, we shall prepare the following.

###### Definition 2.4.

Let be the circle with center through and , and let (resp. ) be the minor (resp. major) arc of circle with endpoints and (see Convention 1.1, ). We define angle functions on minor arc with range and on major arc with range by

 ϕσ(D) ϕ′σ(D)

Then we have on minor arc and on major arc . We can regard (resp. ) as a distance from on minor arc (resp. major arc ). Also, we define angle functions with range and on circle by

 ψσ(D) ρσ(D)

Then we have and on minor arc , and . Note that can be expressed in terms of by

 (2.2) ρσ=sinψσr−cosψσ.

where is defined by and represented in terms of and by

 (2.3) r=1cosγσ−sinγσtanδσfor δσ≠π2.

Thus if , then we have by condition (iii.c) of Construction 1.2, so that

 r=1cosγ−sinγtanδσ′

is well-defined. The angles defined here are shown in Figure 2.5.

Now we shall give a new geometric definition of the critical angles.

###### Definition 2.6.

We consider a net as in Figure 1.3. As in Definition 2.4, let be a circle with center through and , and be a minor and a major arc of respectively, and be angle functions. Then we define the critical angles for as follows, where we regard as taking both and .

1. Let be the intersection point of circle and a perpendicular to through . Equivalently, we can define a point so that . Then let be the intersection point of segments and .

2. Let be the intersection point of major arc and an extension of segment . Alternatively, we can define a point so that by in Proposition 2.15.

3. If lies between and in major arc , then we define to be the intersection point of minor arc and segment . If lies between and in major arc , then we define . We call a critical dividing point. (We can also skip procedure and define as the intersection point of minor arc and segment if it exists, and otherwise, although the construction of in tells us in advance whether segment intersects minor arc or not. See also Proposition 2.15 for the condition that intersects .)

4. Then we define the critical angle by .

5. The critical angles and the critical dividing points constructed here are shown in Figure 2.7, where the possible range of radius of minor arc with center is shaded, which is used in Construction 2.8, .

Definition 2.6 has advantages over Definition 2.1 in that it gives directly the possible range of the dividing point used in Construction 2.8, , and the construction of in Definition 2.1 is simpler than that of in Definition 2.1. We will prove in Theorem 2.13 that the critical angles defined in Definitions 2.1 and 2.6 are identical.

Then the construction of positive origon gadgets is given as follows.

###### Construction 2.8.

Consider a development as in Figure 1.4, for which we assume conditions (i), (ii) and (iii.a)–(iii.c) of Construction 1.2. Then the crease pattern of a positive origon gadget with prescribed simple outgoing pleats and is constructed as follows, where we regard as taking both and .

1. Let and be the critical angles given in either Definition 2.1, Proposition 2.3 or Definition 2.6. Alternatively, let and be the points given in either Definition 2.1 or Definition 2.6. Choose a dividing point in minor arc with center so that either of the following conditions holds:

• satisfies

 ϕL∈[γ−2ζR,2ζL]∩(0,γ),or equivalently,ϕR∈[γ−2ζL,2ζR]∩(0,γ);or
• .

If , or equivalently, for either , then the resulting origon gadget is said to be critical.

2. Let be the intersection point of and the bisector of , and redefine to be a ray starting from and going in the same direction as .

3. Determine a point on segment such that .

4. If , then determine a point on segment such that .

5. The crease pattern is shown as the solid lines in Figure 2.9 for a general case, and Figure 2.10 for a critical case. Also, the assignment of mountain and valley folds is given in Table 2.11.

According to [3], Setion , the condition that a positive origon gadget is constructible on the side of is given by

 (2.4) ∠ABσEσ⩽βσ−δσ.

We can rewrite to state the constructibility theorem of positive origons as follows.

###### Theorem 2.12 ([3], Theorems 4.1 and 4.3).

Let be a dividing point chosen in procedure of Construction 2.8, and let , and be as in Definition 2.4. The constructibility condition of a positive origon gadget is equivalent to both of the following conditions:

 (2.5) ϕσ⩽2ζσ;and (2.6) βσ+γσ2+ψσ2+ρσ⩾π2,

where is the critical angle given either geometrically by Definition 2.1 or 2.6, or numerically by Proposition 2.3. In particular, the equality of is equivalent to that of .

###### Theorem 2.13.

The critical angles and given in Definition 2.6 are the same as those given in Definition 2.1.

To prove the theorem, we will use the following two results.

###### Proposition 2.14.

Let be a point in major arc with center given in Definition 2.6, , so that . Suppose that there exists a point in minor arc with center satisfying

 (2.7) βσ+γσ2+ψσ(D)2+ρσ(D)=π2.

Then we have .

###### Proof.

Let and . Define and by

 (2.8) t=tanψσ2andτ=tan(π2−βσ−γσ2)=cos(βσ+γσ/2)sin(βσ+γσ/2).

Recall from [3], Theorem that

 (2.9) tan(ψσ2+ρσ)=r+1r−1⋅tanψσ2,so thatτ=r+1r−1⋅t.

Thus if satisfies , then we calculate using , and as

 tanρσ(D) =sinψσr−cosψσ=2t/(1+t2)r−(1−t2)/(1+t2) =2t(1+t2)r−(1−t2)=2(r−1)/(r+1)(r−1)+(r−1)2/(r+1) =2τr+1+(r−1)τ2=2τr(1+τ2)−(τ2−1) =2τ/(1+τ2)r−(τ2−1)/(1+τ2)=sin(2βσ+γσ)r−cos(2βσ+γσ)=tanρσ(D′σ),

which gives as desired. This completes the proof of Proposition 2.14. ∎

###### Proposition 2.15.

Let and be points in major arc with center given in procedures and of Definition 2.6 respectively. Then we have

 (2.10) ϕ′σ(B′σ)=2δσ<2βσ=ϕ′σ(D′σ).

Also, either one of the following cases holds for :

1. , which holds if and only if , or equivalently, ;

2. , which holds if and only if

 ϕ′σ(B′σ)<ϕ′σ(D′σ)<ϕ′σ(B′σ′)

or equivalently, lies strictly between and in major arc ; or

3. , which holds if and only if

 0=ϕ′σ′(Bσ′)<ϕ′σ′(D′σ)<ϕ′σ′(B′σ′),

or equivalently, lies strictly between and in major arc .

###### Proof.

Regarding , the second equality is obvious from the construction of in of Definition 2.6. Thus it suffices to prove the first equality because then the middle inequality follows from condition (iii.b) of Construction 1.2. First suppose , so that . Then we have

 ∠BσB′σA=∠B′σBσA=π/2−δσ,

so that

 ϕ′σ(B′σ)=π−∠BσB′σA−∠B′σBσA=2δσ.

Next suppose , so that . Then we have

 ∠AB′σBσ=∠ABσB′σ=π−∠ABσC=π−(2π−(π/2+δσ))=δσ−π/2,

so that

 ϕ′σ(B′σ)=2π−∠BσAB′σ=2π−(π−2(δσ−π/2))=2δσ.

Thus we proved . Also, noting that , we see that is equivalent to

 2δσ′ =ϕ′σ′(B′σ′)⪌ϕ′σ′(D′σ) =(2π−γ)−ϕ′σ(D′σ)=2π−γ−2βσ,

which gives . Combining and , we obtain the equivalent conditions for in cases (i)–(iii). This completes the proof. ∎

###### Proof of Theorem 2.13.

Here let be as in Definition 2.1, so that is given numerically by of Proposition 2.3. Also, let be as in Definition 2.6. Then we shall prove that .

First suppose . Then Proposition 2.3 gives , while Proposition 2.15 gives , so that . Thus we have as desired.

Next suppose . Then we see from Proposition 2.15 that lies strictly between and in major arc , so that . Meanwhile, Definition 2.1 or Proposition 2.3 gives that . Now let be a point in minor arc such that . Then it follows from Theorem 2.12 that satisfies

 βσ+γσ2+ψσ(D)2+ρσ(D)=π2,

so that we have by Proposition 2.14. Since also satisfies , we must have from the monotonicity of on minor arc . Hence we have as desired.

This completes the proof of Theorem 2.13. ∎

Now we recall the following existence theorem of positive origons.

###### Theorem 2.16 ([3], Theorem 4.6).

The critical angles and given in either Definition 2.1, Proposition 2.3 or Definition 2.6 always satisfy . Equivalently, the critical dividing points and given in Definition 2.1 or 2.6 always satisfy (and also ). Thus intervals and given in procedure of Construction 2.8 include infinitely many points.

Here we shall give a geometric proof of this theorem using Definition 2.6, which simplifies the two proofs given in [3].

###### Proof.

It suffices to prove that . Since we see from