[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 degrees of freedom of a configuration can vary in particular cases. Let's look at a simple example.[br][br]Let there be 4 points C, D, E and F that are one unit away from two points A and B. We will assume that point A remains fixed at the origin of coordinates, while point B can slide along the Z axis [to simplify, between (0,0,0) and (0,0,2)]:[br][list][*]C[sub]x[/sub][sup]2[/sup] + C[sub]y[/sub][sup]2[/sup] + C[sub]z[/sub][sup]2[/sup] = 1[/*][*]D[sub]x[/sub][sup]2[/sup] + D[sub]y[/sub][sup]2[/sup] + D[sub]z[/sub][sup]2[/sup] = 1[/*][*]E[sub]x[/sub][sup]2[/sup] + E[sub]y[/sub][sup]2[/sup] + E[sub]z[/sub][sup]2[/sup] = 1[/*][*]F[sub]x[/sub][sup]2[/sup] + F[sub]y[/sub][sup]2[/sup] + F[sub]z[/sub][sup]2[/sup] = 1[/*][*]C[sub]x[/sub][sup]2[/sup] + C[sub]y[/sub][sup]2[/sup] + (C[sub]z[/sub] - B[sub]z[/sub])[sup]2[/sup] = 1[/*][*]D[sub]x[/sub][sup]2[/sup] + D[sub]y[/sub][sup]2[/sup] + (D[sub]z[/sub] - B[sub]z[/sub])[sup]2[/sup] = 1[/*][*]E[sub]x[/sub][sup]2[/sup] + E[sub]y[/sub][sup]2[/sup] + (E[sub]z[/sub] - B[sub]z[/sub])[sup]2[/sup] = 1[/*][*]F[sub]x[/sub][sup]2[/sup] + F[sub]y[/sub][sup]2[/sup] + (F[sub]z[/sub] - B[sub]z[/sub])[sup]2[/sup] = 1[br][/*][/list]We construct the spheres with centers A and B and radius 1. Their intersection will be, in general, a circle [b]c[/b], with a variable radius depending on the position of B. In it we place the four points, C, D, E and F.[br][br]The configuration thus obtained has (remember that we assume A is fixed), in general, 5 degrees of freedom (necessary to determine the position of each point B, C, D, E and F). Note that the above system has 8 equations and 13 unknowns.[br][br]But if we position B at two particular points, this can change.[br][br]If we set B = (0,0,2), the degrees of freedom are reduced to 0, since the four points C, D, E and F are forced to occupy the position (0,0,1), the only (real) solution of the previous system of equations (because an equation of the type x[sup]2[/sup] + y[sup]2[/sup] = 0 only has the null solution, if we consider x and y as real numbers). The configuration becomes rigid (in the field of real numbers). [The same would happen at (0,0,-2).][br][br]On the other hand, if we fix B = (0,0,0) the position of B coincides with that of A, so points C, D, E and F are free to move on the sphere with center A and radius 1. In total, 8 degrees of freedom. Note that the last four equations of the previous system coincide with the first four, so they do not provide more information.[br][br]In order to best solve the continuity problem of definition of points C, D, E and F, in the construction, every time we make B coincide with A, points C, D, E and F, move the position they had at [b]c[/b] (which remains indeterminate as the intersection of two spheres) to the equator of the sphere centered at A. And vice versa, every time B leaves the position (0,0,0), each of those points C, D, E and F, up to that free instant on the sphere, projects its position on the circle [b]c[/b].

[color=#999999]Authors of the construction of GeoGebra: [url=https://www.geogebra.org/u/carlosueno]Carlos Ueno[/url] and [color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url][/color][/color]