HMMT 十一月 2022 · 冲刺赛 · 第 6 题

HMMT November 2022 — Guts Round — Problem 6

[6] Let ABCDEF be a regular hexagon and let point O be the center of the hexagon. How many ways can you color these seven points either red or blue such that there doesn’t exist any equilateral triangle with vertices of all the same color? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2022, November 12, 2022 — GUTS ROUND Organization Team Team ID#

解析

[6] Let ABCDEF be a regular hexagon and let point O be the center of the hexagon. How many ways can you color these seven points either red or blue such that there doesn’t exist any equilateral triangle with vertices of all the same color? Proposed by: Evan Erickson Answer: 6 Solution: Without loss of generality, let O be blue. Then we can’t have any two adjacent blues on the perimeter of ABCDEF . However, because of the two larger equilateral triangles ACE and BDF , we need at least two blues to keep us from having an all red equilateral triangle. We can’t have three blues on the perimeter without break the rule, so we must have two. With this, they must be diametrically opposite. So, in total, there are 2 × 3 = 6 good colorings.