# Parameterized rigid cube

- Author:
- Rafael Losada Liste

- Topic:
- Congruence, Constructions, Cube, Geometry, Quadratic Equations

This activity belongs to the

*GeoGebra book*Linkages. Since the prism we have seen was not rigid, let's expand the constraints. Now we place bars of length square root of 2 on each of the faces of the cube, forming a tetrahedron, as shown in the following construction. With this, we get that the number of internal degrees of the cube is 0. Now, does the rigidity of the cube under these conditions imply that there is only ONE configuration (barring congruences)? That is, is the cube*globally*rigid? The answer is NO: the cube positions of the vertices can be parameterized like this:- O = (0, 0, 0)
- U = (0, 1, 0)
- E = (2e - 1, 0, 0)
- A = (0, 0, 1 - 2a)
- J = ((2e - 1) (3 - 4j), 3 - 4j, (1 - 2a) (3 - 4j)) / 3
- F = ((2e - 1) (3 - 2f) (3 - 2j), (3 - 2f) (3 - 2j), (1 - 2a) (6f + 4f j - 6j)) / 9
- B = ((2e - 1) (6b + 4b j - 6j), (3 - 2b) (3 - 2j), (1 - 2a) (3 - 2b) (3 - 2j)) / 9
- D = ((2e - 1) (3 - 2d) (3 - 2j), 6d + 4d j - 6j, (1 - 2a) (3 - 2d) (3 - 2j)) / 9

**e**,**a**,**j**,**f**,**b**and**d**can be 0 or 1 depending on whether each respective box is activated or not. If we fix E, A and J, that is, the basic tetrahedron UEAJ, which acts as the "skeleton" of the articulated cube, the vertices F, B and D admit exactly 2 positions (isomers). Two to the power of three is 8. Now, if we liberate E and A, each change in the state of**e**and**a**will reflect the cube, respectively, in the YZ and XY planes. So 8 times 2 times 2 is 32. Finally, if we change the state of parameter**j**, the basic tetrahedron UEAJ is reflected on its shaded face UEA, doubling all 32 positions to a total of 64.Note that the .
For example, from the initial position in the application (e=1, a=j=f=b=d=0), the action of changing the state of

**e**parameter changes the sign of the X component of each point, and the**a**parameter changes the sign of the Z component. But some points have that null coordinate, so they are not affected by this reflection. It is for this reason that some points take fewer possible positions than others:- E=(2e-1, 0, 0) depends on
**e**. Therefore it has 2 positions: (±1, 0, 0)

- A=(0, 0, 1-2a) depends on
**a**. Therefore it has 2 positions: (0, 0, ±1)

- J=((2e-1)(3-4j), 3-4j, (1-2a)(3-4j))/3 depends on
**e**,**a**and**j**. Therefore it has 8 positions: (±1, 1, ±1) and (±1, -1, ±1)/3

- F=((2e-1)(3-2f)(3-2j), (3-2f)(3-2j), (1-2a)(6f+4fj-6j))/9 depends on
**e**,**a**,**j**and**f**. But its Z coordinate vanishes when f=j=0, so it doesn't change when the state of**a**changes. Therefore it has 10 positions: (±1, 1, 0), (±1, 1, ±2)/3 and (±1.1, ±4)/9. Geometrically, this happens because one of the basic positions of F lies on the XY plane, so it is symmetry invariant with respect to that plane.

- B=((2e-1)(6b+4bj-6j), (3-2b)(3-2j), (1-2a)(3-2b)(3-2j))/9 depends on
**e**,**a**,**j**and**b**. But its X coordinate vanishes when b=j=0, so it doesn't change when the state of "e" changes. Therefore it has 10 positions: (0, 1, ±1), (±2, 1, ±1)/3 and (±4.1, ±1)/9. Geometrically, this happens because one of the basic positions of B lies on the YZ plane, so it is symmetry invariant with respect to that plane.

- D=((2e-1)(3-2d)(3-2j), 6d+4dj-6j, (1-2a)(3-2d)(3-2j))/9 depends on
**e**,**a**,**j**and**d**. Therefore D does reach the 16 positions: (±1, 0, ±1), (±1, ±2, ±1)/3 and (±1, 4, ±1)/9

**s=f+b+d+j**. This value s indicates the number of vertices of the typical unit cube with six square faces (that is, not articulated) that are reflected from one face of the tetrahedron, passing to the "inside" of the cube (or, if you prefer to put it that way, it indicates the number of "concavities"):- s=0 (unit cube of six square faces, 4*1 congruent positions)
- s=1 (a vertex is reflected in a face of the tetrahedron, passing inside the unit cube, 4*4 congruent positions)
- s=2 (two vertices pass into the unit cube, 4*6 congruent positions)
- s=3 (three vertices pass into the unit cube, 4*4 congruent positions)
- s=4 (four vertices pass into the unit cube, 4*1 congruent positions)

**e**and**a**) the combinatorial number**j**:- e=j=1, a=f=b=d=0

**f**:- e=f=1, a=j=b=d=0

Author of the construction of GeoGebra: Rafael Losada