The 50 Cent Coins Problem
Brett Berry
4910

I got the right answer much quicker than your short answer in my head. I looked at the interface between the two coins and formed a triangle on the outside of both coins. The two sides formed are equal length which at least makes it an isosceles triangle. The angle opposite the third side that isn’t drawn in is not a right angle so that eliminated the possibility that the third side was a different length from the other two sides. 60+60+60 was the only option left. QED

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