The South African Mathematical Olympiad Third Round

Let $S$ be the square $ABCD$, with centre $O$. Let $R$ be the rhombus $EFGH$, with $EG$ the short diagonal, and $O$ the midpoint of $EG$. Since the diagonals of $R$ bisect the angles of $R$, we have that $\angle OFE=30^{\circ }$, so that $\sin 30^{\circ }=\frac{1}{2}$ forces $EG$ to have length $2$. We may therefore assume that $E$ is the midpoint of $AD$ and $G$ is the midpoint of $BC$. See Figure 1.

Figure 1

Consequently, $\angle AEJ=30^{\circ }$ so that $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$ implies that $AJ=\frac{1}{\sqrt{3}}$. The area of the region where $R$ and $S$ overlap is therefore, by symmetry, equal to the area of $S$ minus four times the area of triangle $AJE$, i.e., $$4-4\cdot \frac{1}{2}\cdot 1\cdot \frac{1}{\sqrt{3}}=4\cdot \left(1-\frac{1}{2\sqrt{3}}\right).$$