Count me if you can

Calculate number of pairs of parallel diagonals in a 20 sided regular polygon . If this value is x then x = ? If your answer is 640 then answer as 480, if your answer is 600 answer as 0 and if your answer is 400 then answer as 10

Additional Challenge : Derive a formula to find pairs of parallel diagonals in n sided regular polygon and post it in explanation.


The answer is 480.

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2 solutions

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

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

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

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

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

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

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

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

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

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

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

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

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

P ( n ) = m = 4 n + 3 N ( m ) C 2 = n n 3 2 C 2 = n ( n 3 2 ) ( n 3 2 1 ) 2 = n ( n 3 ) ( n 5 ) 8 \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}

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

Sum of the vertex will be m or n + m This group contains the diagonals : ( 1 , m 1 ) ( n , m ) ( 2 , m 2 ) ( n 1 , m + 1 ) ( l , m l ) ( n k , m + k ) Again, difference between the vertex in diagonals ( l , m l ) and ( n k , m + k ) is minimum i.e. either 2 or 3 which gives us \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}

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

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

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

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

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

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

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

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

N ( m ) = { n 4 2 m is odd n 2 2 m is even \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}

Then number of pair of parallel diagonals, P ( n ) = m = 4 n + 3 N ( m ) C 2 = n 2 n 4 2 C 2 + n 2 n 2 2 C 2 = n 2 ( n 4 2 ) ( n 4 2 1 ) 2 + n 2 ( n 2 2 ) ( n 2 2 1 ) 2 \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}

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

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

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

P ( n ) = { n ( n 3 ) ( n 5 ) 8 n is odd n ( n 4 ) 2 8 n is even \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}

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

This is probably a good method. I did it logically....

I thought to count number of vertices left in making diagonals i.e in order of even points left and odd points left. Then I counted that how many pairs of diagonals can be drawn parallel in each of the cases... then multiplying it with number of sides and dividing by 2 (cause we would have counted every pair twice.)

Very Soon, I will post my solution. This is just an idea how to begin with problem. It's more fun to enjoy solving it.

@Shikhar Srivastava Your solution is quite intuitive.

Nikola Alfredi - 1 year, 4 months ago

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Thank you. This question is very nice. I would love to see your solution.

Shikhar Srivastava - 1 year, 4 months ago

@Shikhar Srivastava ,

Try this out PART 2

Nikola Alfredi - 1 year, 4 months ago

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Thanks, I will try this question. It seems similar to above question.

Shikhar Srivastava - 1 year, 4 months ago

@Shikhar Srivastava

Please verify the solution in my reply to @Winod DHAMNEKAR comment. :) Click Here !

And I have posted my solution to this question..

Nikola Alfredi - 1 year, 4 months ago

@Shikhar Srivastava

I would like if you check out my solution to this discussion. No one has replied in it for long... And I just got to know how you found the solution to @Winod DHAMNEKAR question, thank you for helping :)

Nikola Alfredi - 1 year, 4 months ago
Nikola Alfredi
Feb 5, 2020

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