HMMT 十一月 2025 · THM 赛 · 第 6 题
HMMT November 2025 — THM Round — Problem 6
题目详情
- Regular hexagon SAT U RN (with vertices in counterclockwise order) has side length 2. Point O is the ◦ reflection of T over S . Hexagon SAT U RN is rotated 45 counterclockwise around O . Compute the area its interior traces out during this rotation.
解析
- Regular hexagon SAT U RN (with vertices in counterclockwise order) has side length 2. Point O is the ◦ reflection of T over S . Hexagon SAT U RN is rotated 45 counterclockwise around O . Compute the area its interior traces out during this rotation. Proposed by: Derek Liu √ Answer: 5 π + 6 3 Solution: O ′ N ′ R ′ S S N ′ U ′ A A ′ T R T U ′ ′ ′ ′ ′ ′ Let S A T U R N be the rotated hexagon. Note that S is the point on the hexagon closest to O , and ′ ′ U is the farthest. Thus, the region traced between SU and S U (shaded in red above) is 1 / 8 of an √ 2 2 annulus with inner radius OS and outer radius OU . Note that OS = (2 3) = 12 and √ 2 2 2 2 2 OU = OT + T U = (4 3) + 2 = 52 , so this region has area (52 − 12) π/ 8 = 5 π . © 2025 HMMT The remaining traced area consists of half of the hexagon on each side; specifically, it consists of √ ′ ′ ′ ′ trapezoids SAT U and U R N S (the regions in green). The hexagon has area 6 3, so the answer is √ 5 π + 6 3 .