HMMT 二月 2019 · 冲刺赛 · 第 8 题

HMMT February 2019 — Guts Round — Problem 8

[ 4 ] A regular hexagon P ROF IT has area 1. Every minute, greedy George places the largest possible equilateral triangle that does not overlap with other already-placed triangles in the hexagon, with ties broken arbitrarily. How many triangles would George need to cover at least 90% of the hexagon’s area? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2019, February 16, 2019 — GUTS ROUND Organization Team Team ID#

解析

[ 4 ] A regular hexagon P ROF IT has area 1. Every minute, greedy George places the largest possible equilateral triangle that does not overlap with other already-placed triangles in the hexagon, with ties broken arbitrarily. How many triangles would George need to cover at least 90% of the hexagon’s area? Proposed by: Yuan Yao Answer: 46 It’s not difficult to see that the first triangle must connect three non-adjacent vertices (e.g. P OI ), 1 1 which covers area , and leaves three 30-30-120 triangles of area each. Then, the next three triangles 2 6 1 1 cover of the respective small triangle they are in, and leave six 30-30-120 triangles of area each. 3 18 This process continues, doubling the number of 30-30-120 triangles each round and the area of each triangle is divided by 3 each round. After 1 + 3 + 6 + 12 + 24 = 46 triangles, the remaining area is 4 3 · 2 48 8 1 = = < 0 . 1, and the last triangle removed triangle has area , so this is the minimum 4 6 · 3 486 81 486 number necessary.