4. $A B C D$ is a rectangle with area $6 \mathrm{~cm}^{2}$.
The point $E$ lies on $A B, F$ lies on $B C, G$ lies on $C D$ and $H$ lies on $D A$. The point $I$ lies on $A C$ and is the point of intersection of $E G$ and $F H$, and $A E I H$ and $I F C G$ are both rectangles. One possible diagram is shown to the right.
Given that the combined area of $A E I H$ and $I F C G$ is $4 \mathrm{~cm}^{2}$, find all possible values for the area of rectangle $A E I H$ in cm ${ }^{2}$.
Solution
Say that the lengths in $\mathrm{cm}$ of $A B$ and $A D$ are $x$ and $y$ respectively.
We are given that $x y=6$.
Note that $A E I H$ is similar to $A B C D$, because they share a diagonal (meaning the ratios of their sides are the same). Then we know that, for some value of $k$ (with $0<k<1$ ), $A E=k x$ and $A H=k y$.
Hence $E B=(1-k) x$ and similarly $H D=(1-k) y$.
We know that the shaded area is 4 , and so
$$
\begin{array}{r}
(k y)(k y)+(1-k) x(1-k) y=4 \\
k^{2} x y+\left(1-2 k+k^{2}\right) x y=4
\end{array}
$$

Since $x y=6$, this becomes:
$$
\begin{aligned}
6 k^{2}+6\left(1-2 k+k^{2}\right) & =4 \\
12 k^{2}-12 k+2 & =0 \\
6 k^{2}-6 k+1 & =0 \\
6\left(k-\frac{1}{2}\right)^{2}-\frac{1}{2} & =0 \\
k-\frac{1}{2} & = \pm \sqrt{\frac{1}{12}} \\
k & =\frac{1}{2} \pm \frac{\sqrt{3}}{6}
\end{aligned}
$$
(Since $0<\sqrt{\frac{1}{12}}<\frac{1}{2}$, both of these values of $\mathrm{k}$ are between 0 and 1.)
The area of $A E I H$ is $k^{2} x y=6 k^{2}$.
If $k=\frac{1}{2}+\frac{\sqrt{3}}{6}$, the area $=6 k^{2}=6\left[\frac{1}{4}+\frac{\sqrt{3}}{6}+\frac{3}{36}\right]=2+\sqrt{\mathbf{3}}$.
If $k=\frac{1}{2}-\frac{\sqrt{3}}{6}$, the area $=6 k^{2}=6\left[\frac{1}{4}-\frac{\sqrt{3}}{6}+\frac{3}{36}\right]=\mathbf{2}-\sqrt{\mathbf{3}}$.
It is straightforward to show that each of these areas is possible (e.g. by drawing a $6 \times 1$ rectangle).
Note: the same conclusion can be reached by equating the sum of the unshaded regions to 2 (they are, in fact, each equal to 1 ).