\section*{Problem 13 - V01013}

How many diagonals does a 4775-gon have?
}

For the number \(d\) of diagonals of an \(n\)-gon, the following formula applies:

\[
d(n)=\frac{n}{2} \cdot(n-3)
\]

Justification: From each of the \(n\) vertices, a diagonal can be drawn to \(n-3\) other vertices. With \(n \cdot(n-3)\), each diagonal is counted twice. Consequently, there are \(\frac{1}{2} n \cdot(n-3)\) diagonals.

If \(n=4775\) is substituted, then for the 4775-gon, there are exactly 11,393,150 diagonals.