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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.
Rhombus in Rectangle
Age 14 to 16
Challenge Level 
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.
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