[color=#999999][color=#999999][color=#999999]This activity belongs to the [i]GeoGebra book[/i] [url=https://www.geogebra.org/m/h3gbmymu]Linkages[/url].[/color][/color][/color][br][br]The [url=https://www.geogebra.org/m/h3gbmymu#material/ycc5txth]previous construction[/url] 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.[br][br]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:[br][list][*]J = Sphere(U, [math]\sqrt{2}[/math]) ∩ Sphere(E, [math]\sqrt{2}[/math]) ∩ Sphere(A, [math]\sqrt{2}[/math])[br][/*][*]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)[br][/*][/list]

[color=#999999]Author of the construction of GeoGebra: [color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url][/color][/color]