Another Tough Nut! Take any rectangle $ABCD$ such that $AB > BC$ and say the lengths of $AB$ and $CD$ are $S$ and $s$ respectively. The point $P$ is on $AB$ and $Q$ is on $CD$. For $APCQ$ to be a rhombus, the lengths $AP$ and $PC$ must be equal. Consider the point $P$ coinciding with $A$ (such that $AP=0$) and then $P$ moving along $AB$ so that the length $AP$ increases continuously from $0$ to $S$ while the length of $PC$ decreases continuously from $\sqrt{S^2 + s^2}$ to $s$. As $AP < PC$ initially (when P is at A) and $AP > PC$ finally (when $P$ is at $B$) there must be one point at which $AP = PC$. Similarly there is exactly one position of $Q$ such that $CQ = QA$ making $APCQ$ into a rhombus.

Now take $AP = PC = x$ than you can use Pythagoras' Theorem to find $x$ in terms of $S$ and $s$ so that you can find the ratio of the areas of the areas of the rhombus and the rectangle.