Question You are given a regular polygon with $2⋅n$ vertices (it's convex and has equal sides and equal angles) and all its sides have length $1$. Let's name it as $2n-gon$.
Your task is to find the square of the minimum size such that you can embed $2n-gon$ in the square. Embedding $2n-gon$ in the square means that you need to place $2n-gon$ in the square in such a way that each point which lies inside or on a border of $2n-gon$ should also lie inside or on a border of the square.
You can rotate $2n-gon$ and/or the square
Doubt My solution:The minimum side would be the diagonal of the polygon .. Every side will make an angle of $2\pi/2n$ at the center ....consider the triangle formed points of any side and center ....bisect it (perpendicular) ...now using basic trigonometry...we have to find the hypotenuse and double it to get the diagonal ...so the angle formed at the center would be $\pi/2*n$ hence answer is $1/\sin(\pi/(2n))$.
Nut the solution tells me the answer is $\cos(\pi/(4n))/sin(\pi/(2n))$.
Remember that to be the minimum, then "no smaller square can contain this 2n-gon". Your writeup has not shown this.
The given answer claims that such a smaller square exists, which is why your value isn't the minimum.