Rhombus in Rectangle
Take any rectangle $ABCD$ such that $AB > BC$. The point $P$ is on $AB$ and $Q$ is on $CD$. Show that there is exactly one position of $P$ and $Q$ such that $APCQ$ is a rhombus.
Show that if the rectangle has the proportions of A4 paper ($AB=BC$ $\sqrt 2$) then the ratio of the areas of the rhombus and the rectangle is $3:4$.
Show also that, by choosing a suitable rectangle, the ratio of the area of the rhombus to the area of the rectangle can take any value strictly between $\frac{1}{2}$ and $1$.
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