HMMT November, 2022

Combinatorics Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let $A B C D E F$ 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?

Solution

Solution:

Without loss of generality, let $O$ be blue. Then we can't have any two adjacent blues on the perimeter of $A B C D E F$ . However, because of the two larger equilateral triangles $A C E$ and $B D F$ , 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 breaking the rule, so we must have two. With this, they must be diametrically opposite. So, in total, there are $2 \times 3 = 6$ good colorings.

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