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### Counting Diagonals

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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
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/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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