Start with a rectangle $ASCT$ that's longer than it is tall. Now rotate it counterclockwise around its center $O$ by an angle $0<(2\theta) <\frac{\pi}{2}$ until vertices $T$ and $S$ overlap where vertices $A$ and $C$ were originally.
Rectangles $ASCT$ and $A'S'C'T'$ intersect at points $A$, $B$, $C$, and $D$. Note that $ABCD$ is a rhombus with a side length of $r\sec\theta$, where $r=|OA|$ and $\theta$ is the angle of the vector $\overrightarrow{OA}$ assuming $O$ is at the origin (and hence $A=re^{i\theta}$ in polar coordinates).
We now have a rhombus with vertices $ABCD$. With the help of similar triangles and properties of rhombuses:
$Area(ABCD)=2r^2\tan\theta$
Observe that if $\theta=\frac{\pi}{4}$ then $ASCT$ is actually a square (not just a rectangle) and thus it rotates perfectly back onto itself at all 4 vertices, resulting in $ABCD$ being a square (not just a rhombus). And so: $Area(ABCD)=2r^2\tan(\frac{\pi}{4})=2r^2$, which is the area of a square with diagonal of length $2r$.
OK now back to our rhombus $ABCD$: Draw the 4 altitudes as shown below: $\overline{BG}$, $\overline{BH}$, $\overline{DE}$, and $\overline{DF}$.
Note $\overline{BH}$ and $\overline{DE}$ intersect at point $X$ and $\overline{BG}$ and $\overline{DF}$ intersect at point $Y$. Observe $BXDY$ is a rhombus too.
Question: What is the area of Rhombus $BXDY$ in terms of $r$ and $\theta$?
It'd be interesting to identify the relationship between the new Rhombus $BXDY$ and the original Rhombus $ABCD$. It feels like this could be a geometric proof of some kind of infinite sum, but I need help determining the relationship between $ABCD$ and $BXDY$ first.
My intuition: Observe the diagonals of rhombus $ABCD$ are
$|AC|=2r$ and
$|BD|=2r\tan\theta$ which just equals $|AC|*\tan\theta$.
Now observe the diagonals of rhombus $BYDX$ are
$|BD|=2r\tan\theta$ and
$|XY|=???$.
My guess is that $|XY|=|BD|*\tan\theta$. Do you agree this process is "telescoping" the rhombuses by simply scaling the diagonals down by $\tan\theta$ each time?
If so, then:
$Area(BYDX)= \dfrac{|BD|*|XY|}{2} = \dfrac{(2r\tan\theta)*(2r\tan^2\theta)}{2} = 2r^2tan^3\theta$
From an iterative perspective, this would imply the area of the $k^{th}$ telescoping rhombus (where $k=1$ relates to rhombus $ABCD$ and $k=2$ relates to rhombus $BYDX$) is calculated as:
$Area(Rhombus_k)=2r^2\tan^{(2k-1)}(\theta)$
We have $\angle YBO=90^\circ-\angle OBA=\angle OAB=\theta$, so $YO=BO\tan\theta=r\tan^2\theta$ indeed.