same thing

I think our answers are the same, by altitude I mean height of the triangle. So the key to my solution is to match the height of the coin with one side of the hole when falling through,which is the same configuration in your solution. If the height of an equilateral triangle is 1,its side length should be 1/cos(30)=2/sqrt(3),not sqrt(5)/2.

Sorry for not being totally clear about the rotation part. When I say rotate coin around the matching vertex 30 degrees,the coin is kept on the same plane as the hole. That way,you can see that the height of the coin will match with one side of the hole after the rotation. Then tilting the coin any angle with respect to the hole plane, will let the coin fall through. The tilting angle does not have to be 90 degrees.