Label each unit grid square Q_(K,L) = {(x,y) in R^2 | |x-K| <= 1/2 and |y-L| <= 1/2} in the plane by the pair of integers (K, L) that are the coordinates of its center. Think of the plane R^2 as R^2 x {0} in R^3, an infinite tabletop. At time t = 0, place a unit cube C on Q(0,0). Assume the cube C rests on Q(K,L) at time t = N. A legal move, taken once per unit time step, is to rotate C by an angle of 90º about one of its 4 edges that are touching the tabletop at time t = N, so that it ends up at time t = N+1 resting on one of Q(K+1,L), Q(K-1,L), Q(K,L+1), or Q(K,L-1), with the side down at t = N+1 being one of the 4 sides of C that were vertical at t = N. Puzzle: ------- a) What is the smallest time t at which C can end up resting on Q(0,0) with its initially down side now facing up? Now what if we had used instead b) an octahedron or c) an icosahedron, on the triangular grid, with the corresponding question? (I.e., on each move the solid is rotated by π - (dihedral), so that it ends up on an adjacent triangle, resting on one of its faces adjacent to the one it had just been resting on.) d) The same question for a dodecahedron! (It doesn't matter that there is no tiling of the plane by regular pentagons; the question makes sense anyway.) —Dan