Semi-rigid cube
- Author:
- Rafael Losada Liste
- Topic:
- Congruence, Constructions, Cube, Geometry, Quadratic Equations
This activity belongs to the GeoGebra book Linkages.
The previous construction shows that once you fix the 6 bars of the tetrahedron, the cube is rigid (although not globally). We will now liberate three of the bars.
Now E (blue) moves freely in the circle, in the XY plane, with center O and radius 1 (1 degree of freedom). For its part, A (green) moves freely along the sphere with center O and radius 1 (2 degrees of freedom). Therefore, the entire cube has 3 degrees of freedom, since once the positions of E and A are determined, the rest of the vertices of the cube are immediately determined (unless vertices coincide) by the intersections (two isomers) of the corresponding spheres:
- J = Sphere(U, ) ∩ Sphere(E, ) ∩ Sphere(A, )
- B = Sphere(U, 1) ∩ Sphere(A, 1) ∩ Sphere(J, 1)
- D = Sphere(E, 1) ∩ Sphere(A, 1) ∩ Sphere(J, 1)
- F = Sphere(U, 1) ∩ Sphere(E, 1) ∩ Sphere(J, 1)
Author of the construction of GeoGebra: Rafael Losada