HMMT November

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

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 $A D^{2} \pi = (2 + \sqrt{3}) \pi$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.