HMMT February, 2019

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:
A regular hexagon $P R O F I T$ 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?

Solution

Solution:
It's not difficult to see that the first triangle must connect three non-adjacent vertices (e.g. $P O I$ ), which covers area $\frac{1}{2}$ , and leaves three $30$ - $30$ - $120$ triangles of area $\frac{1}{6}$ each. Then, the next three triangles cover $\frac{1}{3}$ of the respective small triangle they are in, and leave six $30$ - $30$ - $120$ triangles of area $\frac{1}{18}$ each.

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 $\frac{3 \cdot 2^{4}}{6 \cdot 3^{4}}=\frac{48}{486}=\frac{8}{81}<0.1$ , and the last triangle removed triangle has area $\frac{1}{486}$ , so this is the minimum number necessary.

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