How many coins do you need to remove so that the centers of all remaining coins form no equilatoral triangles?
Here is a simple puzzle that came from Martin Gardner's The Colossal Book of Short Puzzles and Problems
. Here we see ten circles arranged in the classic triangle bowling configuration. The question is "what is the minimum number of circles that can be removed such that there is no equilateral triangle formed by the centers of the remaining circles?"
I was originally having difficulty convincing myself of the optimality of my answer, but I am reasonably certain of it now. Still, it took a bit of pondering to justify. I'll let you all sit on it for a day or two, then give my reasoning through it.
Addendum: If you found that easy, try to answer the same question for the triangle with sides of length 5. And then length 6.