Count me if you can

$\text{Let us number the vertices of polygon as shown in figure below and denote the sides and diagonals of polygon by the vertices it contains.}\newline\text{For example, the diagonal formed by joining the vertices }1\text{ and }3\text{ will be denoted by }(1,3)\text{ and so on. }$ $\text{Let us try to make groups of parallel diagonals. }$ $\text{Let us start with diagonal }(1,3).\text{ Since the polygon is regular, by symmetry it is easy to deduce that }(n,4)\text{ and } ( n - 1 , 5 )$ $\text{ will be parallel to }$ $(1,3).$ $\text{We see that on decreasing one vertex and increasing the other by }1\text{ we get a new parallel diagonal. }(1\text{ is decreased by }1 \text{ to get }n)\newline\text{Given any diagonal }(p,q)\,,\,$ $(p\pm k,q\mp k)\text{ is parallel to }(p,q)\text{ for all }k\text{ such that }(p\pm k,q\mp k)\text{ is valid diagonal.}$ $\text{Making a group of diagonals }(1,3)\,,\, (n,4) \text{ and }(n-1,5)\text{ we see that sum of the vertex in each diagonal is either }4\text{ or }n+4 \text{. This is the defining}$ $\text{property of diagonals of this group.}$

$\text{Similarly making the group of }(1,4)\,,\, (n,5)\text{ and }(n-1,6)\text{ we see that sum of the vertex in each diagonal is either }5\text{ or }n+5\text{ and so on there will }$ $\text{be a group whose vertex sum is }6\text{ or }n+6.$ $\text{In general there will be a separate group for separate vertex sum. }\text{Denoting }G_m \text{ as the group of diagonals with vertex sum }m \text{ or }n+m\,,\,$ $\text{where }m\text{ ranges from }4\text{ to }n+3\,,\,m = n+4 \text{ and onwards will come in }G_4 \text{ and onward. So there will be total of }n\text{ groups from }G_4\text{ to }G_{n+3}.$ $\text{Also denoting by }N(m)\,,\,\text{the number of}\text{ diagonals in group }G_m.$

$\textbf{Case 1 : }n\text{ is odd}\newline \underline{\textbf{Group }G_m\,,\, 4\leq m\leq n}$

$\newline \text{Sum of the vertex will be }m\text{ or }n+m \newline \text{This group contains the diagonals : }\newline\hspace{150pt} (1,m-1)\hspace{150pt}(n,m)\newline \hspace{150pt}(2,m-2)\hspace{150pt}(n-1,m+1)\newline\hspace{170pt}\vdots \hspace{185pt}\vdots\newline\hspace{150pt}(l,m-l)\hspace{150pt}(n-k,m+k)\newline \text{Diagonals }(l,m-l)\text{ and }(n-k,m+k)\text{ are such that the difference between the vertex in these diagonals is minimum. This minimum difference}$ $\text{ for the diagonal is }2\text{ or }3\text{ depending upon parity of }m.$

$\newline \hspace{150pt} l = \large\frac{m-2}{2}$ $\hspace{20pt} \text{when }m\text{ is even}\hspace{20pt}\text{ and }\hspace{20pt}k = \large\frac{n-m-3}{2}$ $\hspace{20pt} \text{when }m\text{ is even}\newline \hspace{150pt} l = \large\frac{m-3}{2}$ $\hspace{20pt} \text{when }m\text{ is odd}\hspace{24pt}\text{ and }\hspace{20pt}k = \large\frac{n-m-2}{2}$ $\hspace{20pt} \text{when }m\text{ is odd}\newline \text{Counting all the diagonals we get }\newline \hspace{150pt}N(m) = l+k+1 = \large\frac{n-3}{2}$ $\,\,\,(\text{independent of }m)$

$\underline{\textbf{Group }G_{n+1}}$

$\newline\text{Sum of the vertex will be }n+1\newline\text{This group contains the diagonals : }\newline\hspace{150pt}(2,n-1)\newline\hspace{150pt}(3,n-2)\newline\hspace{170pt}\vdots\newline\hspace{150pt}(l,n-l+1)\newline\text{Again, diagonal }(l,n-l+1)\text{ is such that the difference between the vertex of this diagonal is }2\text{ which gives }\newline\hspace{150pt}l = \large\frac{n-1}{2}$ $\newline\text{Counting all the diagonals we get }\newline \hspace{150pt}N(n+1) = l-1 = \large\frac{n-3}{2}$

$\underline{\textbf{Group }G_{n+2}}$

$\newline\text{Sum of the vertex will be }n+2\newline\text{This group contains the diagonals : }\newline\hspace{150pt}(2,n)\newline\hspace{150pt}(3,n-1)\newline\hspace{170pt}\vdots\newline\hspace{150pt}(l,n-l+2)\newline\text{Again, diagonal }(l,n-l+2)\text{ is such that the difference between the vertex of this diagonal is }3\text{ which gives }\newline\hspace{150pt}l = \large\frac{n-1}{2}$ $\newline\text{Counting all the diagonals we get }\newline \hspace{150pt}N(n+2) = l-1 = \large\frac{n-3}{2}$

$\underline{\textbf{Group }G_{n+3}}$

$\newline\text{Sum of the vertex will be }n+3\newline\text{This group contains the diagonals : }\newline\hspace{150pt}(3,n)\newline\hspace{150pt}(4,n-1)\newline\hspace{170pt}\vdots\newline\hspace{150pt}(l,n-l+3)\newline\text{Again, diagonal }(l,n-l+3)\text{ is such that the difference between the vertex of this diagonal is }2\text{ which gives }\newline\hspace{150pt}l = \large\frac{n+1}{2}$ $\newline\text{Counting all the diagonals we get }\newline \hspace{150pt}N(n+3) = l-2 = \large\frac{n-3}{2}$

$\text{Total number of diagonals in all the }n\text{ groups } = \sum_{m=4}^{n}N(m) + N(n+1) + N(n+2) + N(n+3) = (n-3)\large\frac{(n-3)}{2}$ $+ 3\big(\large\frac{n-3}{2}\big) = \large\frac{n(n-3)}{2}$ $= \text{total number of diagonals in }n\text{-sided polygon. This confirms that no diagonal has been missed and no diagonal has been counted twice.}$

$\text{Let }P(n) \text{ denote the number of pair of parallel diagonals in }n\text{-sided regular polygon. Then }$

$\newline P(n) = \large\sum_{m=4}^{n+3} \hspace{1pt}^{N(m)}C_2 = n ^{\frac{n-3}{2}}C_{2} = n\frac{(\frac{n-3}{2})(\frac{n-3}{2} - 1)}{2} = \frac{n(n-3)(n-5)}{8}$

$\textbf{Case 2 : }n\text{ is even}\newline\underline{\textbf{Group }G_m\,,\,4\leq m\leq n}$

$\text{Sum of the vertex will be }m\text{ or }n+m\newline \text{This group contains the diagonals :}\newline\hspace{150pt}(1,m-1)\hspace{150pt}(n,m)\newline\hspace{150pt}(2,m-2)\hspace{150pt}(n-1,m+1)\newline\hspace{170pt}\vdots \hspace{185pt} \vdots \newline\hspace{150pt}(l,m-l)\hspace{150pt}(n-k,m+k)\newline\text{Again, difference between the vertex in diagonals }(l,m-l)\text{ and }(n-k,m+k)\text{ is minimum i.e. either }2\text{ or }3\text{ which gives us}$

$\hspace{150pt}l = \large\frac{m-2}{2}$ $\hspace{20pt}\text{when }m\text{ is even}\hspace{20pt}\text{ and }\hspace{20pt}k = \large\frac{n-m-2}{2}$ $\hspace{20pt}\text{when }m\text{ is even}\newline\hspace{150pt}l = \large\frac{m-3}{2}$ $\hspace{20pt}\text{when }m\text{ is odd }\hspace{22pt}\text{ and }\hspace{20pt}k = \large\frac{n-m-3}{2}$ $\hspace{20pt}\text{when }m\text{ is odd}\newline\text{Counting all the diagonals we get}\newline \hspace{150pt}N(m) = l+k+1 = \begin{cases} \large\frac{n-4}{2}\,\,& m\text{ is odd}\\ \frac{n-2}{2}\,\, & m\text{ is even}\end{cases}$

$\underline{\textbf{Group }G_{n+1}}$

$\newline\text{Sum of the vertex will be }n+1\newline\text{This group contains the diagonals : }\newline\hspace{150pt}(2,n-1)\newline\hspace{150pt}(3,n-2)\newline\hspace{170pt}\vdots\newline\hspace{150pt}(l,n-l+1)\newline\text{Again, diagonal }(l,n-l+1)\text{ is such that the difference between the vertex of this diagonal is }3\text{ which gives }\newline\hspace{150pt}l = \large\frac{n-2}{2}$ $\newline\text{Counting all the diagonals we get }\newline \hspace{150pt}N(n+1) = l-1 = \large\frac{n-4}{2}$

$\underline{\textbf{Group }G_{n+2}}$

$\newline\text{Sum of the vertex will be }n+2\newline\text{This group contains the diagonals : }\newline\hspace{150pt}(2,n)\newline\hspace{150pt}(3,n-1)\newline\hspace{170pt}\vdots\newline\hspace{150pt}(l,n-l+2)\newline\text{Again, diagonal }(l,n-l+2)\text{ is such that the difference between the vertex of this diagonal is }2\text{ which gives }\newline\hspace{150pt}l = \large\frac{n}{2}$ $\newline\text{Counting all the diagonals we get }\newline \hspace{150pt}N(n+2) = l-1 = \large\frac{n-2}{2}$

$\underline{\textbf{Group }G_{n+3}}$

$\newline\text{Sum of the vertex will be }n+3\newline\text{This group contains the diagonals : }\newline\hspace{150pt}(3,n)\newline\hspace{150pt}(4,n-1)\newline\hspace{170pt}\vdots\newline\hspace{150pt}(l,n-l+3)\newline\text{Again, diagonal }(l,n-l+3)\text{ is such that the difference between the vertex of this diagonal is }3\text{ which gives }\newline\hspace{150pt}l = \large\frac{n}{2}$ $\newline\text{Counting all the diagonals we get }\newline \hspace{150pt}N(n+3) = l-2 = \large\frac{n-4}{2}$

$\text{In general, it can be said that for any }m \in \{ 4,5,6,\cdots n+2,n+3\}$

$\hspace{150pt}N(m) = \begin{cases} \large\frac{n-4}{2}\,\,& m\text{ is odd}\\ \frac{n-2}{2}\,\, & m\text{ is even}\end{cases}$

$\newline \text{Then number of pair of parallel diagonals, }P(n) = \large\sum_{m=4}^{n+3} \hspace{1pt}^{N(m)}C_2 = \large\frac{n}{2}\hspace{1pt}^{\frac{n-4}{2}}C_2 + \frac{n}{2}\hspace{1pt}^{\frac{n-2}{2}}C_2 = \large\frac{n}{2}\frac{\big(\frac{n-4}{2}\big)\big(\frac{n-4}{2} - 1\big)}{2} + \frac{n}{2}\frac{\big(\frac{n-2}{2}\big)\big(\frac{n-2}{2} - 1\big)}{2}$

$\newline\hspace{170pt}\Rightarrow P(n) = \large\frac{n(n-4)}{8}\bigg\{ \frac{n-6}{2} + \frac{n-2}{2}\bigg\}$

$\newline\hspace{170pt}\Rightarrow P(n) = \large\frac{n(n-4)^2}{8}$

$\text{Hence number of pair of parallel diagonals in }n\text{-sided regular polygon is given by }$

$\large \hspace{150pt} P(n) = \begin{cases} \large\frac{n(n-3)(n-5)}{8}\,\,& n\text{ is odd}\\\frac{n(n-4)^2}{8}\,\, & n\text{ is even}\end{cases}$

$\text{Putting }n = 20\text{ we get }P(n) = 640$

SOLUTION :