# Visualizing a Cubic Linkage through the Use of CAS and DGS

^{1}

^{2}

^{3}

^{4}

^{*}

^{†}

## Abstract

**:**

## 1. Introduction

`Maple`. Indeed, two of the authors already approached the study of this elusive structure in a previous work [10], stating some open questions which we partially solve here.

## 2. A Mathematical Model for Linkages

#### 2.1. Selection of the Model

**Definition**

**1.**

- 1.
- (Existence) For which abstract linkages does there exist a realization in ${\mathbb{R}}^{n}$?
- 2.
- (Rigidity) Which abstract linkages admit just one realization in ${\mathbb{R}}^{n}$ up to congruence?
- 3.
- (Configuration space) How is the space of all possible realizations of an abstract linkage? What is its dimension, topology…?

Since (squared) distance constraints satisfied by the joint coordinates are given by quadratic polynomials, it is possible to try to analyze the set of all realizations of a given pair $(G,l)$ symbolically using computational algebra. However, a Gröbner basis for an ideal generated by quadratic polynomials in k variables can require generators of degree $O({2}^{{2}^{k}})$ so computations with the squared distance constraints may quickly become intractable. Moreover, care must be taken in applying results from algebraic geometry in this setting as we are interested in realizations over the real numbers, and many results in algebraic geometry require an algebraically closed ground field.

**Definition**

**2.**

**Remark**

**1.**

**Definition**

**3.**

**Example**

**1.**

`Maple`:

- >
- with(PolynomialIdeals):
- >
- IL1:=<(Bx − Ax)^2 + (By − Ay)^2 − 1, (Cx − Bx)^2 + (Cy − By)^2 − 1,(Ax − Cx)^2 + (Ay − Cy)^2 − 1>
- >
- HilbertDimension(IL1)

- >
- with(PolynomialIdeals):
- >
- ILQ1:=<(Bx − Ax)^2 + (By − Ay)^2 − 1, (Cx − Bx)^2 + (Cy − By)^2 − 1,(Ax − Cx)^2 + (Ay − Cy)^2 − 1, Ax, A_y, B_x − 1, B_y>
- >
- HilbertDimension(ILQ1)

`Maple`to find all possible solutions for the corresponding system of equations:

- >
- solve(Generators(ILQ1), {Cx, Cy})

#### 2.2. Some Issues concerning the Model

#### 2.2.1. An Algebraic Definition of Rigidity

**Definition**

**4.**

**Proposition**

**1.**

**Proof.**

#### 2.2.2. Complex Versus Real Realizations

**Example**

**2.**

`Maple`we obtain a different output:

- >
- with(PolynomialIdeals):
- >
- IL2:=<(Bx − Ax)^2 + (By − Ay)^2 + (Bz − Az)^2 − 1,(Cx − Bx)^2 + (Cy − By)^2 + (Cz − Bz)^2 − 1,(Cx − Ax)^2 + (Cy − Ay)^2 + (Cz − Az)^2 − 4,Ax, Ay, Az, Cx − 2, Cy, Cz>
- >
- HilbertDimension(IL2)

`Maple`is considering the complex model instead of the real one, and it is not hard to see that in this wider setting, where complex values of the variables are allowed, all possible realizations are of the form $A=(0,0,0)$, $C=(2,0,0)$ and $B=\left(\right)open="("\; close=")">1,t,\pm it$, which constitutes a 1-dimensional variety in ${\mathbb{C}}^{3}$.

#### 2.2.3. Faithfulness of the Model

**Problem**

**1.**

#### 2.2.4. Complexity of the Model

`Maple`our model for this free 3-linkage as follows and try to compute the dimension of the ideal ${V}^{3}\left(C\right)$.

- >
- with(PolynomialIdeals):
- >
- IC:=<(Ex − Ox)^2 + (Ey − Oy)^2 + (Ez − Oz)^2 − 1,(Fx − Ex)^2 + (Fy − Ey)^2 + (Fz − Ez)^2 − 1,(Ux − Fx)^2 + (Uy − Fy)^2 + (Uz − Fz)^2 − 1,(Ox − Ux)^2 + (Oy − Uy)^2 + (Oz − Uz)^2 − 1,(Dx − Ax)^2 + (Dy − Ay)^2 + (Dz − Az)^2 − 1,(Jx − Dx)^2 + (Jy − Dy)^2 + (Jz − Dz)^2 − 1,(Bx − Jx)^2 + (By − Jy)^2 + (Bz − Jz)^2 − 1,(Ax − Bx)^2 + (Ay − By)^2 + (Az − Bz)^2 − 1,(Ax − Ox)^2 + (Ay − Oy)^2 + (Az − Oz)^2 − 1,(Dx − Ex)^2 + (Dy − Ey)^2 + (Dz − Ez)^2 − 1,(Jx − Fx)^2 + (Jy − Fy)^2 + (Jz − Fz)^2 − 1,(Bx − Ux)^2 + (By − Uy)^2 + (Bz − Uz)^2 − 1>
- >
- HilbertDimension(IC)

`Maple`are not capable of determining the dimension of a ${V}^{3}\left(C\right)$. We will address this issue and related ones in Section 3.

#### 2.3. Introductory Examples

**Example**

**3.**

**Example**

**4.**

`Maple`to deal with these questions.

- >
- with(PolynomialIdeals):
- >
- IL4:=<(Ax − Bx)^2 + (Ay − By)^2 − 1,(Ax − Cx)^2 + (Ay − Cy)^2 − 1,Dx − Bx)^2 + (Dy − By)^2 − 1, (Dx − Cx)^2 + (Dy − Cy)^2 − 1,((Dx − Ax)^2 + (Dy − Ay)^2 − 2>
- >
- HilbertDimension(IL4)
- >
- IsRadical(IL4);PD:={PrimeDecomposition(IL4)};
- >
- for i from 1 to 4 do HilbertDimension(PD[i]) od;

`Maple`performs computations with the rationals as ground field) has four 3-dimensional ideals $PD\left(i\right)$ for $i=1,2,3,4$. Thus, we have ${V}^{2}\left({L}_{4}\right)={\cup}_{i=1}^{4}V\left(PD\left(i\right)\right)$. In fact, it can be checked that the realizations

`Maple`, we can obtain this elimination ideal as follows:

- >
- K:=<Ax − tx, Ay − ty, Bx − a − tx, By − c − ty, Cx − a − tx,Cy − c − ty, Dx − a − b − tx, Dy − c − d − ty,0, 0, 0, 0, 0,a^2 + c^2 − 1, b^2 + d^2 − 1, a*b + c*d, a*d − b*c − 1>;
- >
- H:=EliminationIdeal(K,{Ax,Ay,Bx,By,Cx,Cy,Dx,Dy});
- >
- PrimeDecomposition(H);

## 3. The Cubic Linkage

`Maple`CAS does not last long, for once again we get stuck and no output is produced. Therefore, we have to resort to a different approach in order to solve the dimension problem for this challenging linkage. Before proceeding to address this issue, we start by studying algebraically the available degrees of freedom for each vertex in the semifree linkage ${C}^{*}$.

#### 3.1. Degrees of Freedom for the Vertices

`Maple`about the elimination ideal of ${I}_{1}$ with respect the variables $\{{F}_{x},{F}_{y},{F}_{z}\}$ produces a quick answer:

- >
- with(PolynomialIdeals):
- >
- (Fx − Ex)^2 + (Fy − Ey)^2 + Fz^2 − 1, Ax^2 + Ay^2 + Az^2 − 1,Bx^2 + (By − 1)^2 + Bz^2 − 1, (Fx − Jx)^2 + (Fy − Jy)^2+ (Fz − Jz)^2 − 1, (Ex − Dx)^2 + (Ey − Dy)^2 + Dz^2 − 1,(Bx − Ax)^2 + (By − Ay)^2 + (Bz − Az)^2 − 1, (Bx − Jx)^2+ (By − Jy)^2 + (Bz − Jz)^2 − 1, (Dx − Ax)^2 + (Dy − Ay)^2+ (Dz − Az)^2 − 1, (Dx − Jx)^2 + (Dy − Jy)^2 + (Dz − Jz)^2 − 1,Ax, Ay, Az − 1, Bx, By − 1, Bz − 1, Jx − Fx, Jy − Fy,Jz − Fz − 1, Dx − Ex, Dy − Ey, Dz − 1>
- >
- IF:= EliminationIdeal(I_1,{Fx, Fy, Fz})
- >
- HilbertDimension(IF)

#### 3.2. Dimension of the Cube

`Maple`,

`CoCoA`or

`Singular`, but none of them was able to provide us with an straightforward computational result. Therefore, we had to resort to more involved arguments, which are partially supported by a CAS but at the same time require the intervention of some human guidance in the process. We will first show

**Theorem**

**1.**

**Proof.**

`Maple`, and it is the expected dimension $dim\left({I}_{A}\right)=6$:

- >
- with(PolynomialIdeals):
- >
- IA:=<Ex^2 + Ey^2 − 1, Fx^2 + (Fy − 1)^2 + Fz^2 − 1,(Fx − Ex)^2 + (Fy − Ey)^2 + Fz^2 − 1,Bx^2 + (By − 1)^2 + Bz^2 − 1, (Fx − Jx)^2 + (Fy − Jy)^2 + (Fz − Jz)^2 − 1,(Ex − Dx)^2 + (Ey − Dy)^2 + (Dz)^2 − 1,(Bx − Jx)^2 + (By − Jy)^2 + (Bz − Jz)^2 − 1,(Dx − Jx)^2 + (Dy − Jy)^2 + (Dz − Jz)^2 − 1 >;
- >
- HilbertDimension(IA)

**Theorem**

**2.**

**Proof.**

**Theorem**

**3.**

**Proof.**

**Corollary**

**1.**

**Proof.**

#### 3.3. Rigidification of the Cube

`Maple`about the dimension of the associated ideal of this linkage ${C}_{R}$ with the added conditions

- >
- with(PolynomialIdeals):
- >
- RC:=<Ex^2 + Ey^2 − 1,Fx^2 + (Fy − 1)^2 + Fz^2 − 1,(Fx − Ex)^2 + (Fy − Ey)^2 + Fz^2 − 1, Ax^2 + Ay^2 + Az^2 − 1,Bx^2 + (By − 1)^2 + Bz^2 − 1, (Fx − Jx)^2 + (Fy − Jy)^2 + (Fz − Jz)^2 − 1,(Ex − Dx)^2 + (Ey − Dy)^2 + (Dz)^2 − 1, (Bx − Ax)^2 + (By − Ay)^2+(Bz − Az)^2 − 1, (Bx − Jx)^2 + (By − Jy)^2 + (Bz − Jz)^2 − 1,(Dx − Ax)^2 + (Dy − Ay)^2 + (Dz − Az)^2 − 1, (Dx − Jx)^2 + (Dy − Jy)^2+(Dz − Jz)^2 − 1, Ex^2 + (Ey − 1)^2 − 2, Ax^2 + (Ay − 1)^2 + Az^2 − 2,Jx^2 + (Jy − 1)^2 + Jz^2 − 2, (Jx − Ex)^2 + (Jy − Ey)^2 + Jz^2 − 2,(Ax − Ex)^2 + (Ay − Ey)^2 + Az^2 − 2, (Ax − Jx)^2 + (Ay − Jy)^2+(Az − Jz)^2 − 2>
- >
- HilbertDimension(%)

- >
- PP := {PrimaryDecomposition(CR)}:nops(%)

- >
- PRP:=PrimeDecomposition(CR)
- >
- for i from 1 to 64 do IdealContainment(PP[i],PRP[i],PP[i]) od,

- we can also check that the primary ideals in the primary decomposition of $I\left({C}_{R}\right)$ are actually prime. Each of these 0-dimensional 64 prime ideals corresponds to a realization of the rigidified cube. They can be explicitly obtained in
`Maple`with

- >
- for i from 1 to 64 do H[i]:=Basis(PP[i],tdeg) od:for i from 1 to 64 do S[i]:=solve(H[i],{Ex,Ey,Ax,Ay,Az,Jx, Jy, Jz, Fx, Fy, Fz, Bx, By, Bz, Dx, Dy, Dz}) od;for i from 1 to 64 do subs(S[i], CR) od:

## 4. A Visual Modelization of the Cube in Geogebra

#### 4.1. Problems Arising in Dynamic Geometry Models

#### 4.1.1. The Hierarchy of Dg Constructions

#### 4.1.2. The Problem of Continuity

#### 4.2. Description of Our Visual Models

- 1.
- Our model can be positioned in any admissible realization, including degenerate ones, after fixing the position of the observer by determining the vertices O, U and E as described in Section 3.
- 2.
- The model is continuous, and joints and bars behave as expected in a real linkage model.
- 3.
- When selecting a vertex to drag it, it has the maximum geometrically possible degree of freedom, as shown in Table 1.
- 4.
- It is reasonably easy to obtain some distinguished configurations, including degenerate ones such as those described in [23].

**Construction 1.**In [25], a cubic linkage is shown with our usual initial conditions for its positioning in ${\mathbb{R}}^{3}$: vertices O, U fixed and vertex E lying on plane $z=0$. In this construction point, A moves freely on the 2-dimensional unit sphere with center O, and J has 3 degrees of freedom. Since E is constrained to move along the horizontal unit circle with center O, it has one degree of freedom. Therefore, the distribution of degrees of freedom corresponds to the sequence O0-U0-E1-F0-A2-B0-J3-D0, and all the positions of the cube (except those that imply the coincidence of two or more vertices) are thus determined by the positions of E, A and J. Indeed, a vertex such as B is built as the intersection of the unit spheres with centers A, U and J. Since this intersection in general contains two points, there are two possibilities for choosing B, and this leads to two isomer realizations with respect to B. The same happens with vertices D and F. This explains the isomer options available in the construction. See Figure 14.

**Construction 2.**In [26], a more sophisticated model has been designed, starting from a distribution of degrees of freedom O0-U0-E1-F1-A2-B1-J0-D1 for general positions and allowing degenerate constructions when some vertices coincide. Next, we proceed to describe this model in more detail. See Figure 15.

`Point(<object>)`. On the other hand, vertices F, B and D determine only two possible positions for the vertex J, giving rise to two isomer realizations (with respect to J) that can be chosen by marking a checkbox.

**Black:**Fixed vertices.**Grey:**Vertices with 0 degrees of freedom.**Blue:**Vertices with 1 degree of freedom.**Green:**Vertices with 2 degrees of freedom.

#### 4.3. Related Constructions in Geogebra

**Planar linkages:**- This section describes the fundamentals of planar linkages construction, especially the rhombus (4-bar linkage), as it is the simplest closed and flexible planar configuration available. The intrinsic problems that appear with the use of DGS and are commented on in Section 4.1 are also introduced. Finally, examples of the use of scripts are shown to try to solve them, such as the one that enables the transmission of movement between vertices—what we call the dragging effect—in contrast with traditional geometric constructions in DGS.
**Other planar linkages:**- Although the main objective of this GeoGebra book is the study of the structure of the cubic linkage, which corresponds to a graph without one-degree vertices, we considered also chains with either free or fixed extremes (partly because of their interest in connection with applications such as simulations of robotic arms), improving their modeling thanks to the dragging effect mentioned above. In this section, we can see some simple planar examples.
**3D linkages:**- This section serves as an introduction to the study of the articulated cube. It shows, in simpler structures, some examples of the problems that will appear in the construction of the cube, such as visualizing certain linkages to decide on their rigidity (as in [28]) or the sudden changes in degrees of freedom mentioned in Example 3.
**Articulated cube:**- This is the main section of the GeoGebra book. In addition to constructions 1 and 2 in Section 4.2, we can observe the cube with 6 added bars as described in Section 3.3, with all its 64 possible realizations obtained by means of six parametrized variables. In the opposite direction, a cube is also shown in which no constraints are imposed on its vertices apart from those determined by the fixed length of the bars ([29]). Thanks to the dragging effect already mentioned, which goes beyond the algebraic treatment of linkages developed here, this construction behaves in a very realistic way, offering even an apparent physical behaviour connected to the inner workings of the GeoGebra software.

#### 4.4. Conclusions and Future Work

- (i)
- “Yet, we have to report that some jumps occur between isomer positions, near singular placements. For instance, when $a={270}^{\circ}$, the parallelogram $AOBU$ collapses. In view of the large bibliography on the continuity problem for dynamic geometry, it seems a non-trivial task to model a cube avoiding, if possible at all, such behavior.”Even though we did not completely solve the continuity problem for linkages, which constitutes an intrinsic difficulty in DGS geometric constructions, through the use of GeoGebra scripts, we have been able to simulate continuity in some of them (see, for example, [29]), although in the process we have lost some of the cleanliness of more formal, purely geometric constructions.
- (ii)
- “We remark that the cube we have modeled has six internal degrees of freedom, one for each free parameter we have introduced. But its distribution has not been homogeneous. For instance, the final vertex has been constructed without any degrees of freedom, by imposing some constraints: being simultaneously in a sphere and in two planes perpendicular to some diagonals. This difficulty to make a model where all semi-free vertices behave homogeneously is apparently similar to the planar parallelogram case, but now we cannot conclude that it is impossible to make such a construction, since, after fixing O and U we still have six vertices and six degrees of freedom. It is probably a consequence of our approach and not an intrinsic characteristic.”With respect to the dimension of the cube and the possible degrees of freedom for its vertices, we have completely solved the associated algebraic problem in Section 3, and with respect to distributing homogeneusly the six available degrees of freedom among six vertices of the cube, we have found a construction (see [30]) that achieves this, although without keeping fixed the adjacent vertices O and U.
- (iii)
- “Could you fix (say, by pasting some rigid plates) one, two,…facets in the cube and still have some flexibility on the cube? How many internal degrees of freedom will remain?"Some examples of adding bars to limit the flexibility of the cube and the corresponding algebraic discussion have been shown in Section 3.3.
- (iv)
- “For a planar parallelogram, one can feel the one-degree of freedom by checking that once you fix one semi-free vertex, the whole parallelogram gets fixed. The same applies for the spatial parallelogram. You have to fix, one after another, the two semi-free vertices. For the cube, how can you feel its six degrees of freedom? Can you fix five semi-free vertices and still move the cube?"Construction [30] (see also Figure 16) shows an example of a semi-free linkage for the cube where the six degrees of freedom of its internal configurations have been evenly distributed among six of its vertices (U, A, E, B, D and F), leaving two of them (O and J) with 0 degrees of freedom. Notice that, in non-degenerate configurations, dragging any of the vertices B, D or F will leave fixed all the other five vertices with one degree of freedom. This answers (at least partially) in the affirmative the question above.

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

## Abbreviations

DGS | Dynamic geometry software |

CAS | Computer algebra system |

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Vertex | Variables | Restrictions | Degrees of Freedom |
---|---|---|---|

O | — | 0 | |

U | — | 0 | |

A | $\{{A}_{x},{A}_{y},{A}_{z}\}$ | $B=A+U$ | 2 |

$F=E+U$ | |||

$J=D+U$ | |||

B | $\{{B}_{x},{B}_{y},{B}_{z}\}$ | $B=A+U$ | 2 |

$F=E+U$ | |||

$J=D+U$ | |||

D | $\{{D}_{x},{D}_{y},{D}_{z}\}$ | $B=A+U$ | 3 |

$F=E+U$ | |||

$J=D+U$ | |||

E | $\{{E}_{x},{E}_{y}\}$ | $A=O+Z$ | 1 |

$B=U+Z$ | |||

$J=F+Z$ | |||

$D=E+Z$ | |||

F | $\{{F}_{x},{F}_{y},{F}_{z}\}$ | $A=O+Z$ | 2 |

$B=U+Z$ | |||

$J=F+Z$ | |||

$D=E+Z$ | |||

J | $\{{J}_{x},{J}_{y},{J}_{z}\}$ | $B=A+U$ | 3 |

$F=E+U$ | |||

$J=D+U$ |

$({\mathit{A}}_{\mathit{x}},{\mathit{A}}_{\mathit{y}},{\mathit{A}}_{\mathit{z}})$ | $({\mathit{B}}_{\mathit{x}},{\mathit{B}}_{\mathit{y}},{\mathit{B}}_{\mathit{z}})$ | $({\mathit{D}}_{\mathit{x}},{\mathit{D}}_{\mathit{y}},{\mathit{D}}_{\mathit{z}})$ | $({\mathit{E}}_{\mathit{x}},{\mathit{E}}_{\mathit{y}},{\mathit{E}}_{\mathit{z}})$ | $({\mathit{F}}_{\mathit{x}},{\mathit{F}}_{\mathit{y}},{\mathit{F}}_{\mathit{z}})$ | $({\mathit{J}}_{\mathit{x}},{\mathit{J}}_{\mathit{y}},{\mathit{J}}_{\mathit{z}})$ |
---|---|---|---|---|---|

$(0,0,1)$ | $(0,1,1)$ | $(1,0,1)$ | $(1,0,0)$ | $(1,1,0)$ | $(1,1,1)$ |

$(0,0,1)$ | $({\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,1)$ | $(1,0,0)$ | $(1,1,0)$ | $(1,1,1)$ |

$(0,0,1)$ | $(0,1,1)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,0)$ | $(1,1,0)$ | $(1,1,1)$ |

$(0,0,1)$ | $(0,1,1)$ | $(1,0,1)$ | $(1,0,0)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{2}{3}})$ | $(1,1,1)$ |

$(0,0,1)$ | $({\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}})$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,0)$ | $(1,1,0)$ | $(1,1,1)$ |

$(0,0,1)$ | $(0,1,1)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{2}{3}})$ | $(1,1,1)$ |

$(0,0,1)$ | $({\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,1)$ | $(1,0,0)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{2}{3}})$ | $(1,1,1)$ |

$(0,0,1)$ | $({\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}})$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{2}{3}})$ | $(1,1,1)$ |

$(0,0,1)$ | $(-{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}})$ | $({\textstyle \frac{1}{3}},-{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}},-{\textstyle \frac{2}{3}})$ | $(-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}})$ |

$(0,0,1)$ | $({\textstyle \frac{4}{9}},{\textstyle \frac{1}{9}},{\textstyle \frac{1}{9}})$ | $({\textstyle \frac{1}{3}},-{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}},-{\textstyle \frac{2}{3}})$ | $(-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}})$ |

$(0,0,1)$ | $(-{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}})$ | $({\textstyle \frac{1}{9}},{\textstyle \frac{4}{9}},{\textstyle \frac{1}{9}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}},-{\textstyle \frac{2}{3}})$ | $(-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}})$ |

$(0,0,1)$ | $(-{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}})$ | $({\textstyle \frac{1}{3}},-{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{9}},{\textstyle \frac{1}{9}},{\textstyle \frac{4}{9}})$ | $(-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}})$ |

$(0,0,1)$ | $({\textstyle \frac{4}{9}},{\textstyle \frac{1}{9}},{\textstyle \frac{1}{9}})$ | $({\textstyle \frac{1}{3}},-{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{9}},{\textstyle \frac{1}{9}},{\textstyle \frac{4}{9}})$ | $(-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}})$ |

$(0,0,1)$ | $({\textstyle \frac{4}{9}},{\textstyle \frac{1}{9}},{\textstyle \frac{1}{9}})$ | $({\textstyle \frac{1}{9}},{\textstyle \frac{4}{9}},{\textstyle \frac{1}{9}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}},-{\textstyle \frac{2}{3}})$ | $(-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}})$ |

$(0,0,1)$ | $(-{\textstyle \frac{2}{3}},{\textstyle \frac{1}{3}},{\textstyle \frac{1}{3}})$ | $({\textstyle \frac{1}{9}},{\textstyle \frac{4}{9}},{\textstyle \frac{1}{9}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{9}},{\textstyle \frac{1}{9}},{\textstyle \frac{4}{9}})$ | $(-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}})$ |

$(0,0,1)$ | $({\textstyle \frac{4}{9}},{\textstyle \frac{1}{9}},{\textstyle \frac{1}{9}})$ | $({\textstyle \frac{1}{9}},{\textstyle \frac{4}{9}},{\textstyle \frac{1}{9}})$ | $(1,0,0)$ | $({\textstyle \frac{1}{9}},{\textstyle \frac{1}{9}},{\textstyle \frac{4}{9}})$ | $(-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}},-{\textstyle \frac{1}{3}})$ |

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## Share and Cite

**MDPI and ACS Style**

Recio, T.; Losada-Liste, R.; Tabera, L.F.; Ueno, C.
Visualizing a Cubic Linkage through the Use of CAS and DGS. *Mathematics* **2022**, *10*, 2550.
https://doi.org/10.3390/math10152550

**AMA Style**

Recio T, Losada-Liste R, Tabera LF, Ueno C.
Visualizing a Cubic Linkage through the Use of CAS and DGS. *Mathematics*. 2022; 10(15):2550.
https://doi.org/10.3390/math10152550

**Chicago/Turabian Style**

Recio, Tomás, Rafael Losada-Liste, Luis Felipe Tabera, and Carlos Ueno.
2022. "Visualizing a Cubic Linkage through the Use of CAS and DGS" *Mathematics* 10, no. 15: 2550.
https://doi.org/10.3390/math10152550