Let $R$ denote the rectangle, and let $S$ denote the square with minimum length that covers $R$ . Let $s$ denote the length of a side of $S$. We claim that $R$ is inscribed in $S$. We also claim that $R $can only be inscribed in the two ways shown below. Let $S_1$ and $S_2$ denote the squares shown on the left-hand and right-hand sides, respectively. For $S_2$ , the sides of $R$ are parallel to the diagonals of $S2$. We can see that $s = a$ if $S = S_1$. It is easy to see that $s=\frac{\sqrt{2}(a+b)}{2}$ if $S=S_2$. Taking the minimum value of $(a,\frac{\sqrt{2}(a+b)}{2})$, we conclude that $s= \left\{ \begin{array}{ll} a & a<(\sqrt{2}+1)b \\ \frac{\sqrt{2}(a+b)}{2} & a\geq (\sqrt{2}+1)b \\ \end{array} \right. $ Now we prove our claim that these are only two ways. Let $R = ABCD$ and $S = XYZW$. WLOG, we place $XY$ horizontally. By the minimality of $S$, we can assume that at least one vertex, say $A$ , of $R$ lies on one side of $S$ , say $WX$ (see the left diagram). If neither $B$ nor $D$ lies on the sides of $S$, we can then slide $R$ down vertically so that one of them, say $B$, lies on side $XY$ (see the middle figure). If neither $C$ or $D$ lies on the sides of $S$ , then we can apply an enlargement, centered at $X$ with scale less than $1$ , to $S$ such that the image of $S$ still covers $R$ . This violates the minimality of $S$. Therefore, at least one of $C$ and $D$ lies on the sides of $S$, that is, three consecutive vertices of $R$ lie on the sides of $S$ (see the right-hand side figure). WLOG assume that they are $A, B, C$. If any of these three vertices coincide with any of the vertices of $S$, then we clearly have $S = S_1$. Hence we may assume that $A , B, C$ are on sides $WX$, $XY$ and $YZ$, respectively. By symmetry, we may also assume that $AB =a>b=BC$. NoteAll rectangular-like shapes are indeed rectangles If $D$ does not lie on line segment $ZW$, then we can slide $R$ up so both $B$ and $D$ lie in the interior of $S$ (see the left-hand slide) Let $O$ be the center of $R$. We can then rotate $R$ around $O$ with a small angle so that all four vertices lie inside $S$ (see the middle figure shown below). It is easy to see that we can use a smaller square to cover $R$ (by applying an enlargement centered at $O$ with a scale less that $1$) , violating the minimality of $S$. This means our assumption is wrong, and D must lie on side $ZW$ (see the righthand side figure shown below) , which is the case when $S = S_2$. NoteAll rectangular-like shapes are indeed rectangles We finish our proof that if $S = S_2$, then the sides of $R$ are parallel to the diagonals of $S_2$. By symmetry, it suffices to show that $AX = XB$. It is not difficult to see that $\angle XAB = \angle YBC = \angle ZCD= \angle WDA$. So triangles $ABX$, $BCY$, $CDZ$ and $DAW$ are similar. Set $AX = ax$ and $XB = ay$. Then $BY = bx$ and $CY = by$. Also, $DW = bx$ and $WA = by$, which gives $by + ax = WA + AX = WX = XY = XB + BY =ay+bx$, implying that $(a-b)x= (a-b)y$, or $x=y$, as desired $\blacksquare$