Problem Solving

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

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



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


-Patrick

Brent, that's really elegant. I'm jealous not to have thought of it.

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



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


-Patrick

Hello Patrick,

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

Best Regards,
Sri