16. $\triangle A B C$ is equilateral with side length $2 . O$ is a point inside $\triangle A B C$, and $P, Q, R$ are points on the plane such that $O A P, O B Q$ and $O C R$ are all isosceles triangles (with vertices named in clockwise order) with vertical angles $\angle O A P, \angle O B Q$ and $\angle O C R$ equal to $15^{\circ}$. Find the area of $\triangle P Q R$.
(2 marks)
$A B C$ is an equilateral triangle, with side length $2$. $O$ is a point inside $\triangle A B C$, and $P, Q, R$ are points on the plane such that $O A P, O B Q$ and $O C R$ are all isosceles triangles (with vertices named in clockwise order), and their vertical angles $\angle O A P, \angle O B Q$ and $\angle O C R$ are all equal to $15^{\circ}$. Find the area of $\triangle P Q R$.
16. $2 \sqrt{3}-3$
16. Note that $O A P, O B Q$ and $O C R$ are similar isosceles triangles with base angles $\left(180^{\circ}-30^{\circ}\right) \div 2=75^{\circ}$. Hence $\triangle A O B \sim \triangle P O Q$ since $\frac{A O}{P O}=\frac{O B}{O Q}$ and $\angle A O B=$ $\angle P O Q$ (which follows from $\angle A O P=\angle B O Q$ ). Hence we have $\frac{A B}{P Q}=\frac{O B}{O Q}$, or $P Q=\frac{2 O Q}{O B}=4 \cos 75^{\circ}$. In the same way we see that $Q R$ and $R P$ have the same length. Hence $\triangle P Q R$ is indeed equilateral with area
$$
\begin{aligned}
\frac{\sqrt{3}}{4}\left(4 \cos 75^{\circ}\right)^{2} & =4 \sqrt{3} \cos ^{2} 75^{\circ} \\
& =2 \sqrt{3}\left(1+\cos 150^{\circ}\right) \\
& =2 \sqrt{3}-3
\end{aligned}
$$

Remark. It would be easy to guess the answer if one assumes $O$ to be the centre of $\triangle A B C$.