HMMT-Nov 2022 gen, Problem 6

In a plane, equilateral triangle $A B C$, square $B C D E$, and regular dodecagon $D E F G H I J K L M N O$ each have side length 1 and do not overlap. Find the area of the circumcircle of $\triangle A F N$.

Solution: Solution: Note that $\angle A C D=\angle A C B+\angle B C D=60^{\circ}+90^{\circ}=150^{\circ}$. In a dodecagon, each interior angle is $180^{\circ} \cdot \frac{12-2}{12}=150^{\circ}$ meaning that $\angle F E D=\angle D O N=150^{\circ}$. since $E F=F D=1$ and $D O=O N=1$ (just like how $A C=C D=1$ ), then we have that $\triangle A C D \cong \triangle D O N \cong \triangle F E D$ and because the triangles are isosceles, then $A D=D F=F N$ so $D$ is the circumcenter of $\triangle A F N$. Now, applying the Law of Cosines gets that $A D^{2}=2+\sqrt{3}$ so $A D^{2} \pi=(2+\sqrt{3}) \pi$. $\fbox{(2+\sqrt{3}) \pi}$.

Answer: (2+\sqrt{3}) \pi