... (ka + i) = f (i) + ka = (ni + k)a Besides the zero function, this is the general solution of the given functional equation To verify this, we plug in m = ka + i, n = la + j and obtain f (m + f ... rectangle can be placed within the second one with the angle α between AB and EF if and only if a cos α + b sin α ≤ c, a sin α + b cos α ≤ d (1) Hence ABCD can be placed within EF GH if and only ... If equality holds, there is exactly one way of placing This happens, for example, when (a, b) = (5, 20) and (c, d) = (16, 19) Second remark This problem is essentially very similar to (SL89-2)...