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Polygon diagonals formula

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by Naruto » Sun Jun 14, 2009 4:42 am
Well guys I searched if this has been posted before, and didnt find it. Btw whats the best way to search if a question has been posted before or not. I simply put the whole question in the search, is there a smarter way I am missing?

Anway, here's the question

A Polygon has 20 diagonals, how many sides does it have.

A 12
B 11
C 10
D 9
E 8


Now by taking general cases I was trying to solve and I came up with this equation where No. of diagonals of a polygon with n sides = (n-3)n/2
Its applicable here, but i want to know if its general, and if there is any other shorter way to solve the problem
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by tohellandback » Sun Jun 14, 2009 5:28 am
well this is how I solved:
0 diagonals-3 sides
2 diagonals-4 sides
5 diagonals-5sides
9 diagonals-6sides

you see 0+2=2,2+3=5,5+4=9. so next will be 9+5=14, 14+6=20
20 diagonals will have 8 sides
The powers of two are bloody impolite!!
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by Naruto » Sun Jun 14, 2009 5:32 am
tohellandback wrote:well this is how I solved:
0 diagonals-3 sides
2 diagonals-4 sides
5 diagonals-5sides
9 diagonals-6sides

you see 0+2=2,2+3=5,5+4=9. so next will be 9+5=14, 14+6=20
20 diagonals will have 8 sides
I didnt understand please elaborate what series are u continuing?
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by tohellandback » Sun Jun 14, 2009 5:40 am
0-2 (add 0 to 2)
2-5(added 3)
5-9 (added 4)
so next you will add 5
9-14(added 5)
14-20(added six)
The powers of two are bloody impolite!!
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Re: Polygon diagonals formula

by dtweah » Sun Jun 14, 2009 5:40 am
Naruto wrote:Well guys I searched if this has been posted before, and didnt find it. Btw whats the best way to search if a question has been posted before or not. I simply put the whole question in the search, is there a smarter way I am missing?

Anway, here's the question






Now by taking general cases I was trying to solve and I came up with this equation where No. of diagonals of a polygon with n sides = (n-3)n/2
Its applicable here, but i want to know if its general, and if there is any other shorter way to solve the problem
A Polygon has 20 diagonals, how many sides does it have.

A 12
B 11
C 10
D 9
E 8

What is a shorter way? What can be shorter and surer than factoring
n^2 -3n -40=0
(n-8)(n+5)=0
n=8.
With formula this takes less than 15 seconds. There is no shorter way order than already knowing the answer!!
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Re: Polygon diagonals formula

by Naruto » Sun Jun 14, 2009 6:00 am
dtweah wrote:
Naruto wrote:Well guys I searched if this has been posted before, and didnt find it. Btw whats the best way to search if a question has been posted before or not. I simply put the whole question in the search, is there a smarter way I am missing?

Anway, here's the question






Now by taking general cases I was trying to solve and I came up with this equation where No. of diagonals of a polygon with n sides = (n-3)n/2
Its applicable here, but i want to know if its general, and if there is any other shorter way to solve the problem
A Polygon has 20 diagonals, how many sides does it have.

A 12
B 11
C 10
D 9
E 8

What is a shorter way? What can be shorter and surer than factoring
n^2 -3n -40=0
(n-8)(n+5)=0
n=8.
With formula this takes less than 15 seconds. There is no shorter way order than already knowing the answer!!
You misunderstood me, I meant is this formula genuine because i derived it based on eg. for triangel, quadrilateral, pentagon and hexagon. I wanted to know if there is another approach, because deriving that formula took me 4 mins.
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by Vemuri » Sun Jun 14, 2009 10:53 pm
I solved the question the same way as tohellandback did. But, your question is valid. Even I am curious to know if there is a relationship between sides & diagonals of a polygon. It will greatly help in answering the questions quickly.
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by scoobydooby » Sun Jun 14, 2009 11:25 pm
there indeed is a general formula to find out the number of sides given the number of diagonals. (dug out my high school math book)

the number of diagonals of a polygon of n sides is: nC2-n
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by rah_pandey » Sun Jun 14, 2009 11:27 pm
A formal proof of n^2-3n-40=0

for a n sided polygon lets choose one point say A
now there are n-1 points except A itself from which lines can be drawn to A but not all will be diagonals. 2 such lines that can be drawn are the sides passing from A and joining adjoining points

thus effectively n-3 points can be joined to form a diagonal from A thus n-3 diagonals are possible from any given point

for n points total no of such lines =n*(n-3)

but of these n*(n-3) there is overlap(by 2 times)

thus total no of diagonal =n*(n-3)/2

using for 20 diagonals

we get n(n-3)/2=20
=> n=8 or n=-5
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by Vemuri » Mon Jun 15, 2009 3:01 am
scoobydooby wrote:there indeed is a general formula to find out the number of sides given the number of diagonals. (dug out my high school math book)

the number of diagonals of a polygon of n sides is: nC2-n
That's great Scoobydooby !!! Thanks a ton for providing the formula. Here goes my "thanks" to you :-)
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by Ian Stewart » Mon Jun 15, 2009 7:36 am
rah_pandey wrote:A formal proof of n^2-3n-40=0
There are a few good ways to prove the formula, and rah_pandey provided one above. An alternative proof - first starting with a simpler question:

* If you draw n dots on a page, how many different lines can be drawn connecting pairs of these dots? This is the same question as: If n people attend a party and all shake hands, how many handshakes take place? Since we need to choose 2 dots to make a line, and order doesn't matter, the answer is nC2.

* Now, looking at the n corners of an n-sided shape, we can draw nC2 different lines connecting pairs of these corners. All of these lines are diagonals *except* the n edges, so we have nC2 - n diagonals. That might look different from the formula in the post above, but it turns out to be equal if you expand.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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