Math Is Fun Forum

nsl22

Member

Registered: 2007-08-16

Posts: 8

Polygon Diagonals permutation problem

Ok, I did a search on this and found this equation, but it doesn't explain anything.

The question is, How many sides does a Polygon with 20 diagonals have. 

I've made drawings to figure it out, which obviously won't work for large polygons.  So that's no good.  I've also found a formula that says D = n(n-3)/2, where D is the number of diagonals and n is the number of sides.  The polygon does indeed have 8 sides, but this is a permutation problem.  Does anyone know of an analytical way of figuring this out? 

Here's an easier question that is similar:
How many heptagons can be drawn by joining the vertices of a polygon with 10 sides? 

A: heptagon = 7 sides.  It's a Combination problem.  You have 10 possible sides taken 7 at a time.  So, 10*9*8*7*6*5*4/7! = 120

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Polygon Diagonals permutation problem

The answer to your first question is that it’s an octagon.

Consider a vertex in an n-gon (n ≥ 4). It can’t form a diagonal with itself or the two vertices adjacent to it, but it can form diagonals with all the other (n−3) vertices. Thus each of the n vertices can form (n−3) diagonals with other vertices – so to find the total number of diagonals, multiply (n−3) by n and (to avoid double counting) divide by 2 . Hence

So if the polygon has 20 diagonals, set

, and solve for n.

Last edited by JaneFairfax (2007-08-17 06:17:31)

nsl22

Member

Registered: 2007-08-16

Posts: 8

Re: Polygon Diagonals permutation problem

Oh, ok.  That's the logical explanation.  I guess I got caught up with thinking that it was a combination/permutation type of problem. 

That's interesting however.  In order to discount duplicate diagonals, you divide.  You divide by two because in this case, we're talking about lines that connect a vertex.  In combination problems, division is often used to discount duplicates. 

Thanks!