Handshakes and Polygon Diagonals

Date: 09/12/2001 at 20:42:42
From: Brendan
Subject: If a polygon has # sides, how many diagonals will it have?

Dear Dr. Math,

I can't figure this one out!

If a polygon has 42 sides, how many diagonals does it have?

Date: 09/13/2001 at 05:40:04
From: Doctor Pete
Subject: Re: If a polygon has # sides, how many diagonals will it 
have?

Hi,

Let's look at a hexagon, with 6 sides. Pick a vertex. How many 
diagonals come from that vertex? (Three.) How many vertices in a 
hexagon? (Six.)  

So there should be 3*6 = 18 diagonals, but in fact if you count them 
there are only half as many. Why? Well, since a single diagonal joins 
exactly two vertices, for each vertex we counted the same diagonal 
twice.

So look at a 42-gon.

     How many diagonals come from a single vertex?
     How many vertices are there in a 42-gon?
     How many diagonals are in a 42-gon?

Let's look at something similar to this. If you have 100 people at a 
party, and everyone shakes hands with everyone else, how many 
handshakes take place? Well, pick a person. He shakes hands with 99 
others. In fact, each person shakes hands with 99 others, so there 
should be 100*99 = 9900 handshakes; but again, it takes two people to 
shake hands, so this is double the correct amount, which is 4950 
handshakes.

By now the pattern should be clear. If there are n people at a party, 
the number of handshakes is n(n-1)/2.

Now, there is a different way to count these handshakes. The first 
person shakes hands with 99 others. The second person shakes hands 
with 98 others, not including the first person, whom we have already 
counted. The third person shakes hands with 97 others, not including 
the first two. And so on, until the 99th person, who shakes hands with 
the 100th person. So there are

     99 + 98 + 97 + ... + 2 + 1

handshakes. But we have seen that this is equal to 100*99/2, and this 
suggests a formula. In general, if there are n people at a party, then 
counting the number of handshakes in two different ways gives the 
formula

     1 +  2 + ... + (n-2) + (n-1) = n(n-1)/2.

Finally, we can prove this summation in yet another way. Write

     S =     1 +     2 + ... + (n-2) + (n-1)
     S = (n-1) + (n-2) + ... +     2 +     1
    ----------------------------------------
    2S =     n +     n + ... +     n +     n,

and since there are n-1 "n"'s, then we have 2S = n(n-1), or

     S = n(n-1)/2.

However, the handshake analogy isn't exact.  At a party, each person gets to
shake hands with everyone but himself.  But in a polygon, each vertex
doesn't get to form a diagonal with himself... or with the two vertices
nearest him.  (He's already connected to them by edges.)  So for a 
polygon, we have to subtract the number of edges from the total for
the handshake problem: 

     S = n(n-1)/2 - n

       = [n(n-1) - 2n]/2

       = [n(n - 1 - 2)]/2

       = n(n-3)/2

- Doctor Pete, The Math Forum
  http://mathforum.org/dr.math/