Diagonals of regular polygon?

The number of diagonals is n(n-1)/2, where n is the number of sides (or vertices). That is because, for the first vertex, you can make a diagonal, you can make n-1 diagonals with the other n-1 vertices. Then with the second vertex, you can make a diagonal with the other n-2, and so on. So

n-1 + n-2 + ... + 2 + 1 = (n-1)((n-1)+1)/2 = n(n-1)/2 diagonals (The sum of the first n whole numbers is n(n+1)/2.).

Diagonals Of Regular Polygons

I didn't know there were rules for diagonals of regular polygons. Do you mean that the regular polygons that are constructible with the classical tools (straightedge and compass) must have n be of the form 2^k (k >= 0) times a product of distinct Fermat primes (odd primes of the form ((2^(2^m)) + 1), m >= 0)?

In a totally unrelated matter, you can derive the length of a diagional of a regular polygon:

Picture a regular n-gon inscribed in a circle of radius one.

If a diagonal connects vertices that are k apart the shorter way around (where k = 1 means adjacent, k = 2 means one vertex in between, etc.) then the angle vertex - center - vertex will be ( 2 pi k / n ).

Drop a perpendicular bisector from the center to the diagonal (a chord of the circle) to get a right triangle with hypotenuse one (the radius), angle at the center half the above (therefore equal to ( pi k / n )), distance center to diagonal is cos(pi k / n), half the length of the diagonal is sin(pi k / n).

To calculate the diagonal or what? Here`s a little help...