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Greatest number of diagonals

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Greatest number of diagonals

by gmattesttaker2 » Wed Feb 19, 2014 6:31 pm
Hello,

Is there a formula that can be used for the kind of question given below? Is the following formula correct for finding the total number of diagonals:

The number of diagonals of a polygon of side x = x(x-3)/2


The greatest number of diagonals that can be drawn from one vertex of a regular 6-sided polygon is

(A) 2
(B) 3
(c) 4
(D) s
(E) 6

OA: B


Thanks,
Sri
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by Patrick_GMATFix » Wed Feb 19, 2014 10:27 pm
A diagonal goes from one vertex to any other vertex except itself and the two vertices adjacent to it. As a result, the max number of diagonals that can be drawn from any vertex in a polygon with x sides (x vertices) is x-3. This question is about a 6-sided polygon, so the answer is 6-3 = 3.

The number of diagonals can be found using combinations. A diagonal is a link between two vertices, so we can think of the number of diagonals as the number of ways to pick 2 vertices. However, since the links between adjacent vertices make up the sides (not the diagonals), we must subtract these from the total.

Let's test it with familiar figures

Eg: how many diagonals in a square?
Ways to pick 2 vertices from 4 = 4C2 = 4!/(2!2!) = 6 links.
Subtract the 4 links that are sides rather than diagonals and you end up with 6-4 = 2 diagonals.

Eg: how many diagonals in a triangle?
Ways to pick 2 vertices from 3 = 3C2 = 3!/(2!1!) = 3 links.
Subtract the 3 links that are sides rather than diagonals and you end up with 6-4 = no diagonals in a triangle

Eg: how many diagonals in a hexagon (6 sides) ?
Ways to pick 2 vertices = 6C2 = 6!(2!4!) = 15 links.
Subtract the 6 links that are sides rather than diagonals and you end up with 15-6 = 9 diagonals (see figure)

Thus we can derive a general formula: # of diagonals in an n-sided polygon is n!/((n-2)!*2!) - n

Image

The formula you provided, # of diagonals = x(x-3)/2, is the simplified version of what I derived above, so it is correct
Image

-Patrick
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by Brent@GMATPrepNow » Thu Feb 20, 2014 9:37 am
gmattesttaker2 wrote:Hello,

Is the following formula correct for finding the total number of diagonals:

The TOTAL number of diagonals of a polygon of side x = x(x-3)/2
Patrick's approach is great. Here's another approach.

Choose any vertex.
There are x vertices, so this step can be accomplished in x ways.

Choose a second vertex to create the diagonal.
If we want to create a diagonal, the second vertex CANNOT be the same as the first vertex we chose AND the second vertex cannot be one of the two ADJACENT vertices. So, among the n vertices, 3 of them are "off limits." So this step can be accomplished in x - 3 ways.

So, the number of ways to complete BOTH steps (and get a diagonal) = (x)(x - 3)

IMPORTANT: We have inadvertently counted each diagonal TWICE. For example, if we chose vertex A in the first step and vertex G in the second step, we get the EXACT SAME diagonal as choosing vertex G in the first step and vertex A in the second step.

Since we have counted each diagonal TWICE, we'll take (x)(x - 3) and divide it by 2 to get: (x)(x - 3)/2

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Thu Feb 20, 2014 9:51 am, edited 1 time in total.
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by Patrick_GMATFix » Thu Feb 20, 2014 9:48 am
Brent, that's really elegant. I'm jealous not to have thought of it.
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by Brent@GMATPrepNow » Thu Feb 20, 2014 9:52 am
Thanks Patrick :-)

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Brent
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by gmattesttaker2 » Thu Feb 20, 2014 11:17 am
Patrick_GMATFix wrote:A diagonal goes from one vertex to any other vertex except itself and the two vertices adjacent to it. As a result, the max number of diagonals that can be drawn from any vertex in a polygon with x sides (x vertices) is x-3. This question is about a 6-sided polygon, so the answer is 6-3 = 3.

The number of diagonals can be found using combinations. A diagonal is a link between two vertices, so we can think of the number of diagonals as the number of ways to pick 2 vertices. However, since the links between adjacent vertices make up the sides (not the diagonals), we must subtract these from the total.

Let's test it with familiar figures

Eg: how many diagonals in a square?
Ways to pick 2 vertices from 4 = 4C2 = 4!/(2!2!) = 6 links.
Subtract the 4 links that are sides rather than diagonals and you end up with 6-4 = 2 diagonals.

Eg: how many diagonals in a triangle?
Ways to pick 2 vertices from 3 = 3C2 = 3!/(2!1!) = 3 links.
Subtract the 3 links that are sides rather than diagonals and you end up with 6-4 = no diagonals in a triangle

Eg: how many diagonals in a hexagon (6 sides) ?
Ways to pick 2 vertices = 6C2 = 6!(2!4!) = 15 links.
Subtract the 6 links that are sides rather than diagonals and you end up with 15-6 = 9 diagonals (see figure)

Thus we can derive a general formula: # of diagonals in an n-sided polygon is n!/((n-2)!*2!) - n

Image

The formula you provided, # of diagonals = x(x-3)/2, is the simplified version of what I derived above, so it is correct
Image

-Patrick
Hello Patrick,

Thanks for the detailed explanation and for the excellent diagram. It is clear now. Thanks again.

Best Regards,
Sri
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