Problem 438

Suppose that a circle with centre $O$ has radius 2 and that equilateral $\triangle PQR$ has its vertices on the circle.

Join $OP$, $OQ$ and $OR$.

Join $O$ to $M$, the midpoint of $PQ$.

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Since the radius of the circle is 2, then $OP=OQ=OR=2$.

By symmetry, $\angle POQ = \angle QOR = \angle ROP$.

Since these three angles add to $360^\circ$, then $\angle POQ = \angle QOR = \angle ROP = 120^\circ$.

Since $\triangle POQ$ is isosceles with $OP = OQ$ and $M$ is the midpoint of $PQ$, then $OM$ is an altitude and an angle bisector.

Therefore, $\angle POM = \frac{1}{2} \angle POQ = 60^\circ$ which means that $\triangle POM$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle.

Since $OP = 2$ and is opposite the $90^\circ$ angle, then $OM = 1$ and $PM = \sqrt{3}$.

Since $PM = \sqrt{3}$, then $PQ = 2PM = 2\sqrt{3}$.

Therefore, the area of $\triangle POQ$ is $\frac{1}{2}\cdot PQ \cdot OM = \frac{1}{2} \cdot 2\sqrt{3} \cdot 1 = \sqrt{3}$.

Since $\triangle POQ$, $\triangle QOR$ and $\triangle ROP$ are congruent, then they each have the same area.

This means that the area of $\triangle PQR$ is three times the area of $\triangle POQ$, or $3\sqrt{3}$.