回复：Thank you! But

I'll give it a shot. Basically, to optimize the situation, you want to match the shortest symmetric dividing line on the coin(in the triangle case the height), with the longest line formed by any two points on the hole(in this case one side of the hole). You can see this more clearly if both the coin and hole are squares. In fact, we can use this principle to generate solutions for both coin and hole to be any regular n polygans.