Planar facets cube
This activity belongs to the GeoGebra book Linkages.
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 planar rhombus that we have already discussed.
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):
- OE: Ex2 + Ey2 = 1
- OA: Ax2 + Ay2 + Az2 =1
- UF: Fx2 + (Fy -1)2 + Fz2 = 1
- UB: Bx2 + (By -1)2 + Bz2 = 1
- EF: (Fx - Ex)2 + (Fy - Ey)2 + Fz2 = 1
- ED: (Dx - Ex)2 + (Dy - Ey)2 + Dz2 = 1
- AB: (Bx - Ax)2 + (By - Ay)2 + (Bz - Az)2 = 1
- AD: (Dx - Ax)2 + (Dy - Ay)2 + (Dz - Az)2 = 1
- BJ: (Jx - Bx)2 + (Jy - By)2 + (Jz - Bz)2 = 1
- DJ: (Jx - Dx)2 + (Jy - Dy)2 + (Jz - Dz)2 = 1
- FJ: (Jx - Fx)2 + (Jy - Fy)2 + (Jz - Fz)2 = 1
- 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)
- OE: Ex2 + Ey2 = 1
- OA: Ax2 + Ay2 + Az2 =1
Author of the construction of GeoGebra: Rafael Losada