Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National olympiad Prove it South Africa

Let $S$ be a square with sides of length $2$ and $R$ be a rhombus with sides of length $2$ and angles measuring $60^\circ$ and $120^\circ$ . These quadrilaterals are arranged to have the same centre and the diagonals of the rhombus are parallel to the sides of the square. Calculate the area of the region on which the figures overlap.

Solution

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.,
$4 - 4 \cdot \frac{1}{2} \cdot 1 \cdot \frac{1}{\sqrt{3}} = 4 \cdot \left(1 - \frac{1}{2\sqrt{3}}\right).$

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