[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 following construction distributes the 6 degrees of freedom homogeneously among 6 of the vertices of the cube, so that all of them obtain one degree of freedom except for two, which are determined except for isomers.[br][br]This construction covers all general cases, but does not impose restrictions (that is why sometimes the structure “breaks”). Special cases, such as the coincidence of two or more vertices, are not considered either. Its main virtue is that it distributes freedom as homogeneously as possible between its vertices.[br][br]Let us observe that the restriction imposed on point A forces the circle AUE to have a radius less than or equal to unity, which makes possible the existence of O (or its isomer O').

The construction steps are detailed below.[br][list][*]The segment (-1, 0, 0)-(1, 0, 0) is created and point A is placed on it.[/*][/list][list][*]The circle with center (0, 0, 0) and radius OA is created, and points U and E are placed on it.[/*][/list][list][*]Point O and its isomer are determined by O[sub]z[/sub][sup]2[/sup] +A[sub]x[/sub][sup]2[/sup] = 1, that is, as (0, 0, [math]\pm\sqrt{1-abs\left(A\right)^2}[/math]).[br][/*][/list][list][*]Point B is a point on the intersection circle of the spheres of radius 1 and centers U and A.[/*][/list][list][*]Point D is a point on the intersection circle of the spheres of radius 1 and centers E and A.[/*][/list][list][*]Point F is a point on the intersection circle of the spheres of radius 1 and centers U and E.[/*][/list][list][*]Finally, the point J and its isomer are determined as the intersection of the spheres of radius 1 and centers B, D and F.[br][/*][/list]

[color=#999999]Author of the construction of GeoGebra: [url=https://www.geogebra.org/u/carlosueno]Carlos Ueno[/url][/color]