Calculating Diagonals of a Polygon: What are the Values of p and q?

[Natasha1]

Homework Statement

A polygon with n sides has a total of 1/p . n . (n-q) diagonals, where p and q are integers.

(i) Find the values of p and q.

Homework Equations

The Attempt at a Solution

Can someone just help me to start on this? I know that q = 3 and p = 2 but how?

 

[Samy_A]

Homework Statement

A polygon with n sides has a total of 1/p . n . (n-q) diagonals, where p and q are integers.

(i) Find the values of p and q.

Homework Equations

The Attempt at a Solution

Can someone just help me to start on this? I know that q = 3 and p = 2 but how?

Take a polygon with n=10 (for example). How many diagonals can you draw from one of the corners?

 

[Natasha1]

9

 

[Samy_A]

9

No. A diagonal links the corner to one of the other non-adjacent corners. How many of these are there (still with n=10)?

 

[Natasha1]

8

 

[Samy_A]

8

No. A corner has 2 adjacent corners, and there is no diagonal linking a corner with itself. So, still with n=10, how many diagonals are there starting in one corner?

 

[Natasha1]

If you take a triangle with 3 sides (obviously), there are no diagonals, right?

 

[Samy_A]

If you take a triangle with 3 sides (obviously), there are no diagonals, right?

Yes, that is correct. A corner of a triangle has two adjacent corners, and that's it. No possibility to draw a diagonal there.

 

[Natasha1]

So a 4 sided polygon as only 1 diagonal, am I correct?

 

[Mark44]

The Attempt at a Solution

Can someone just help me to start on this? I know that q = 3 and p = 2 but how?

In future posts, you need to show more of an effort than this.

 

[Samy_A]

So a 4 sided polygon as only 1 diagonal, am I correct?

No. A 4 sided polygon has 2 diagonals.

 

[Natasha1]

But apart from trial and error is there any other way I can do this?

4 sides polygon
1/2 . 4 . (4-3) = 2 diagonals
5 sides polygon
1/2 . 5 . (5-3) = 5 diagonals
6 sides polygon
1/2 . 6 . (6-3) = 9 diagonals

And so on...

 

[Natasha1]

In future posts, you need to show more of an effort than this.

Apologies

 

[Samy_A]

But apart from trial and error is there any other way I can do this?

Yes, there is. I was trying to get you there by reasoning.

So, still with n=10, how many diagonals are there starting in one corner?

 

[Natasha1]

4 sided polygon
1/2 . 4 . (4-3) = 2 diagonals
5 sided polygon
1/2 . 5 . (5-3) = 5 diagonals
6 sided polygon
1/2 . 6 . (6-3) = 9 diagonals
...
10 sided polygon
1/2 . 10 . (10-3) = 35 diagonals

 

[Natasha1]

Is there anyway I can do this problem without doing it like i have, guessing what p and q are...

How can one prove it?

 

[Samy_A]

Is there anyway I can do this problem without doing it like i have, guessing what p and q are...

How can one prove it?

By reasoning how many diagonals there are. It is possible, it is easy, but you have to try it.

Outline:
Take one corner, and compute the number of diagonals it lies on. We were almost there: there is a diagonal linking that corner to any other corner, except itself and its two adjacent corners. That makes n - 1 - 2 = n -3 diagonals from that one corner.
Now, this is the case for each one of the n corners. So now you can compute the total number of diagonals, but don't forget that each diagonal connects two corners.

Good night.

 

[Natasha1]

Got it thanks!