[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]Before dealing with the case of the hinged cube, in which each of the six faces behaves like a hinge, let's look at a simpler case: each of the six faces is the [url=https://www.geogebra.org/m/h3gbmymu#material/sj6npzsa]planar rhombus[/url] that we have already discussed.[br][br]Then let O=(0, 0, 0), U=(0, 1, 0) and E be a point in the XY plane. We establish the conditions given by the fixed distances (the 11 unitary bars, since the OU bar already is):[br][list][*]OE: E[sub]x[/sub][sup]2[/sup] + E[sub]y[/sub][sup]2[/sup] = 1[br][/*][*]OA: A[sub]x[/sub][sup]2[/sup] + A[sub]y[/sub][sup]2[/sup] + A[sub]z[/sub][sup]2[/sup] =1[br][/*][*]UF: F[sub]x[/sub][sup]2[/sup] + (F[sub]y[/sub] -1)[sup]2[/sup] + F[sub]z[/sub][sup]2[/sup] = 1[br][/*][*]UB: B[sub]x[/sub][sup]2[/sup] + (B[sub]y[/sub] -1)[sup]2[/sup] + B[sub]z[/sub][sup]2[/sup] = 1[br][/*][*]EF: (F[sub]x[/sub] - E[sub]x[/sub])[sup]2[/sup] + (F[sub]y[/sub] - E[sub]y[/sub])[sup]2[/sup] + F[sub]z[/sub][sup]2[/sup] = 1[br][/*][*]ED: (D[sub]x[/sub] - E[sub]x[/sub])[sup]2[/sup] + (D[sub]y[/sub] - E[sub]y[/sub])[sup]2[/sup] + D[sub]z[/sub][sup]2[/sup] = 1[br][/*][*]AB: (B[sub]x[/sub] - A[sub]x[/sub])[sup]2[/sup] + (B[sub]y[/sub] - A[sub]y[/sub])[sup]2[/sup] + (Bz - A[sub]z[/sub])[sup]2[/sup] = 1[br][/*][*]AD: (D[sub]x[/sub] - A[sub]x[/sub])[sup]2[/sup] + (D[sub]y[/sub] - A[sub]y[/sub])[sup]2[/sup] + (Dz - A[sub]z[/sub])[sup]2[/sup] = 1[/*][*]BJ: (J[sub]x[/sub] - B[sub]x[/sub])[sup]2[/sup] + (J[sub]y[/sub] - B[sub]y[/sub])[sup]2[/sup] + (Jz - B[sub]z[/sub])[sup]2[/sup] = 1[/*][*]DJ: (J[sub]x[/sub] - D[sub]x[/sub])[sup]2[/sup] + (J[sub]y[/sub] - D[sub]y[/sub])[sup]2[/sup] + (Jz - D[sub]z[/sub])[sup]2[/sup] = 1[/*][*]FJ: (J[sub]x[/sub] - F[sub]x[/sub])[sup]2[/sup] + (J[sub]y[/sub] - F[sub]y[/sub])[sup]2[/sup] + (Jz - F[sub]z[/sub])[sup]2[/sup] = 1[br][/*][/list]Also, we add the condition that each face is flat. To do this, on each face, the sum of each pair of vectors with the same origin vertex must coincide with the vector that determines the diagonal:[br][list][*]U - O + E - O = F - O (so it is determined F = E + U - O)[/*][*]U - O + A - O = B - O (so it is determined B = A + U - O)[/*][*]E - O + A - O = D - O (so it is determined D = A + E - O)[/*][*]F - E + D - E = J - E (so it is determined J = A + E - O + U - O)[/*][/list](The equations corresponding to the other two faces follow from these four.) These last four equations reduce the first 11 to just these two:[list][*]OE: E[sub]x[/sub][sup]2[/sup] + E[sub]y[/sub][sup]2[/sup] = 1[br][/*][*]OA: A[sub]x[/sub][sup]2[/sup] + A[sub]y[/sub][sup]2[/sup] + A[sub]z[/sub][sup]2[/sup] =1[/*][/list]Therefore, E has 1 degree of freedom (it is determined by one parameter) and A has two degrees of freedom (it takes 2 parameters to determine it). The cube then has 3 degrees of freedom.[br][br]If, in addition, we do not want degenerate cases to occur, we can add the black [url=https://www.geogebra.org/m/h3gbmymu#material/app2yr7y]auxiliary bars[/url] that appear in the construction.

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