3. [4] Circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ are centered at $M, N$, and $O$, respectively. The points of tangency between $\omega_{2}$ and $\omega_{3}, \omega_{3}$ and $\omega_{1}$, and $\omega_{1}$ and $\omega_{2}$ are tangent at $A, B$, and $C$, respectively. Line $M O$ intersects $\omega_{3}$ and $\omega_{1}$ again at $P$ and $Q$ respectively, and line $A P$ intersects $\omega_{2}$ again at $R$. Given that $A B C$ is an equilateral triangle of side length 1 , compute the area of $P Q R$.
Answer: $\mathbf{2 \sqrt { 3 }}$. Note that $O N M$ is an equilateral triangle of side length 2 , so $m \angle B P A=m \angle B O A / 2=$ $\pi / 6$. Now $B P A$ is a $30-60-90$ triangle with short side length 1 , so $A P=\sqrt{3}$. Now $A$ and $B$ are the midpoints of segments $P R$ and $P Q$, so $[P Q R]=\frac{P R}{P A} \cdot \frac{P Q}{P B}[P B A]=2 \cdot 2[P B A]=2 \sqrt{3}$.