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 Post subject: Nth term for diagonals of regular polygonsPosted: Sun Oct 18, 2009 7:22 pm

Joined: Thu Jan 10, 2008 9:53 am
Posts: 66
Location: kent
My ds has to find the number of diagonals in a regular polygon of 20 sides etc. He has drawn the diagonals for an equal lateral triangle, rectangle, pentagon hexagon heptagon . We have established that the diagonals are going up in the sequence of + 3,4,5, but how can we use this pattern to work out the nth term and then calculate it for a 50 or 100 sided shape?

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 Post subject: Re: Nth term for diagonals of regular polygonsPosted: Mon Oct 19, 2009 10:24 am

Joined: Fri Mar 20, 2009 7:34 pm
Posts: 37
kentish maid wrote:
My ds has to find the number of diagonals in a regular polygon of 20 sides etc. He has drawn the diagonals for an equal lateral triangle, rectangle, pentagon hexagon heptagon . We have established that the diagonals are going up in the sequence of + 3,4,5, but how can we use this pattern to work out the nth term and then calculate it for a 50 or 100 sided shape?

What do you mean by "diagonals" ? Triangles don't have diagonals; the term only applies to polygons with 4 or more sides. Do you mean mid-points ?

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 Post subject: Re: Nth term for diagonals of regular polygonsPosted: Mon Oct 19, 2009 11:08 am

Joined: Fri Jul 06, 2007 8:31 pm
Posts: 1209
kentish maid wrote:
My ds has to find the number of diagonals in a regular polygon of 20 sides etc. He has drawn the diagonals for an equal lateral triangle, rectangle, pentagon hexagon heptagon . We have established that the diagonals are going up in the sequence of + 3,4,5, but how can we use this pattern to work out the nth term and then calculate it for a 50 or 100 sided shape?

Well... you do seem have found a pattern.... now just express the answer in terms of the number of sides (n)...

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 Post subject: Posted: Mon Oct 19, 2009 11:45 am

Joined: Thu Jan 03, 2008 8:26 am
Posts: 1327
Location: Watford, Herts
A polygon of n sides also has n points (or vertices). You can't have a diagonal from a point to itself or to the two adjacent points, so from each of those n points you have diagonals to n-3 points. That gives us n(n-3) diagonals, but we've counted each diagonal twice (at its two end-points). Thus the number of unique diagonals is n(n-3)/2. Check with a few simple cases: triangles have no diagonals, quadrilaterals have 2, pentagons 5, etc.

(edited to correct the pentagon number)

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 Post subject: Posted: Mon Oct 19, 2009 12:17 pm
I think you'll find the answer is 0.5 nsquared -1.5n.

For a rectangle this would be (4x4) divided by 2 subtract 1.5 x 4
= 2
For a pentagon this would be (5x5) divided by 2 subtract 1.5 x 5
= 12.5 - 7.5 = 5
For a hexagon this would be (6x6) divided by 2 subtract 1.5 x 6
= 18 - 9 = 9

etc. etc.

So for 50, you would have (50x50) divided by 2 subtract 1.5 x 50
= 1250 - 75 = 1175 diagonals
So for 100, you would have (100x100) divided by 2 subtract 1.5 x 100
= 5,000 - 150
= 4850

For ones with changing difference, it is the nth term of a quadratic sequence rather than a linear one so you are expecting a squared in there.
You start off with rectangles (because you can't get diagonals in triangles as someone has already pointed out). Thus your n terms go 4, 5, 6, 7, 8. You have then worked out the diagonals for these ie. 2, 5, 9, 14. Their changing difference is 3, 4, 5, 6, etc... You then work out the difference between this changing difference which is 1, and you half that, and put it in front of your nsquared.

You then have to get from 4 to 8 to finally 2 which would be -6
from 5 to 12.5 to finally 5 -7.5
from 6 to 18 to finally 9 -9
from 7 to 24.5 to finally 14 -10.5

In each case this is take away 1.5 times the original n.

So you end up with 0.5nsquared - 1.5n.

I'm sorry if this is a bit obscure. I have always found it quite difficult to explain (even to myself!).

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 Post subject: Posted: Mon Oct 19, 2009 12:19 pm
Just read WP's explanation. So much better than mine although the answer is the same so I'd ignore mine.

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 Post subject: Posted: Mon Oct 19, 2009 4:09 pm

Joined: Thu Jan 10, 2008 9:53 am
Posts: 66
Location: kent
Thanks everyone. WP that's a fab explanation

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