Counting Diagonals

Date: 03/14/98 at 14:16:52
From: Molly Thompson
Subject: Polygons

How many diagonals can be drawn for a polygon with 15 sides? How many 
diagonals can be drawn for a polygon with n sides? 

I tried to find a pattern in how many diagonals there are for a 
polygon with 4 sides, 5 sides, etc., but I couldn't extract a pattern. 
Please show me a method for this question.

Date: 03/16/98 at 13:59:36
From: Doctor Wilkinson
Subject: Re: polygons

Looking for a pattern with polygons of 4 sides, 5 sides, etc. was an
excellent idea. Sometimes, though, it's pretty hard to see a pattern 
in a sequence of numbers. It may be better to look for the pattern in 
the whole process of counting the diagonals. That is, look for a 
systematic way of counting diagonals as you look at specific cases 
like 4 and 5.

So let's take another look at polygons with 4 and 5 sides.

For 4 sides, we have a quadrilateral ABCD.  The diagonals are: 

   AC
   BD

There's one diagonal containing A, one containing B, one containing C, 
and one containing D. Let's leave it at that for a minute, and look at 
case 5.

Here we have a pentagon ABCDE.  The diagonals are:

   AC
   AD
   BD
   BE
   CE

Here we have two diagonals containing A, two containing B, two 
containing C, two containing D, and two containing E.

So for 4 sides there is one diagonal for each vertex and for 5 there 
are two. Why? Well, there is a diagonal from each vertex to each other 
vertex except the two adjacent vertices. For 4 vertices, there are 
three other vertices for each vertex, and two of them are adjacent, 
which leaves 1; for 5 vertices, there are four other vertices for each 
vertex, and two of them are adjacent, which leaves 2.

Okay now, we're starting to get somewhere. For four vertices there is 
one diagonal containing each vertex, but we don't end up with four 
diagonals but just two. For five vertices there are two diagonals 
containing each vertex, but we don't end up with ten diagonals but 
just five. What's going on?

Aha!  If you count one diagonal for every vertex of the quadrilateral,
you're counting every diagonal twice, because every diagonal contains 
two vertices. And similarly for the pentagon. So you have to divide by 
two.

I hope this helps you with the general problem.

-Doctor Wilkinson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/