HMMT 十一月 2022 · GEN 赛 · 第 6 题

HMMT November 2022 — GEN Round — Problem 6

In a plane, equilateral triangle ABC , square BCDE , and regular dodecagon DEF GHIJKLM N O each have side length 1 and do not overlap. Find the area of the circumcircle of △ AF N .

解析

In a plane, equilateral triangle ABC , square BCDE , and regular dodecagon DEF GHIJKLM N O each have side length 1 and do not overlap. Find the area of the circumcircle of △ AF N . Proposed by: Kevin Zhao √ Answer: (2 + 3) π ◦ ◦ ◦ Solution: Note that ∠ ACD = ∠ ACB + ∠ BCD = 60 + 90 = 150 . In a dodecagon, each inte- 12 − 2 ◦ ◦ ◦ rior angle is 180 · = 150 meaning that ∠ F ED = ∠ DON = 150 . since EF = F D = 1 and 12 ∼ ∼ DO = ON = 1 (just like how AC = CD = 1), then we have that △ ACD △ DON △ F ED and = = because the triangles are isosceles, then AD = DF = F N so D is the circumcenter of △ AF N . Now, √ √ 2 2 applying the Law of Cosines gets that AD = 2 + 3 so AD π = (2 + 3) π .