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Polygon diagonals

How many of them are there?

Problem 1 (Polygon diagonals) Suppose the function \(f: \mathbb{N} \to \mathbb{N}\) denotes the number of distinct diagonals that can be drawn in a regular \(n\)-gon. For example, \[ f(3) = 0, \quad f(4) = 2, \quad f(5) = 5. \] Determine the value of \(f(9) - f(7)\).

Answer 1 Order the vertices in a counterclockwise manner and take the vertex \(v_1\) of an \(n\)-gon. A diagonal from this vertex connects \(v_1\) to some other vertex \(v_j\) such that \(j \neq 1, 2, n\). Hence, there are only \(m = n-3\) diagonals emanating from the first vertex. Similarly, from the second vertex \(v_2\), one can draw diagonals to all other vertices except \(v_1, v_2, v_3\).

However, once we reach \(v_3\), the diagonal connecting \(v_3\) to \(v_1\) has already been counted as a diagonal emanating from \(v_1\) so in addition to not being able to draw diagonals to \(v_2, v_3\), and \(v_4\), we also cannot draw a diagonal to \(v_1\). This pattern continues for all subsequent vertices; that is, for \(v_4\), we cannot draw diagonals to not three, not four, but \(5\) vertices.

In other words, after two vertices have been selected, we can draw diagonals to only \(m-1, m-2, \ldots, 1\) vertices until we exhaust all the vertices.

Hence, the total number of diagonals for an \(n\)-gon is given by \[ f(n) = 2(n-3) + \frac{(n-3-1)(n-3)}{2} = \frac{1}{2}n(n-3). \]

Therefore, we have \(f(9) - f(7) = \frac{1}{2}(9)(6) - \frac{1}{2}(7)(4) = 27 -14 = \boxed{13}\).

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