One of the standard arguments for the formula for the sum of the interior angles of a polygon involves the exterior angles of the polygon. The argument goes smoothly enough when the polygon is convex. What about a concave polygon?
In what follows, I present the basic argument quickly and then describe how and why the argument becomes problematic when the polygon is concave. Then I resolve the problems by adapting the argument slightly so that we can be sure it applies to all polygons.
Consider the pentagon below:
The marked angles are called the exterior angles of the pentagon. Each exterior angle is paired with a corresponding interior angle, and each of these pairs sums to 180° (they are supplementary).
A pentagon has 5 interior angles, so it has 5 interior-exterior angle pairs. These pairs total 5*180=900°.
But the exterior angles sum to 360°. So we need to subtract that from the 900° total, leaving 540° for the interior angles of the pentagon.
There is nothing special about this being a pentagon. If we consider a polygon with n sides, then we have:
This formula corresponds to n pairs of supplementary interior and exterior angles, minus 360° for the total of the exterior angles.
A concave polygon, informally, is one that has a dent. More formally, a concave polygon has at least one interior angle greater than 180°.
The pentagon below is concave.
There is one exterior angle that is not marked. Can you find the exterior angle of this concave pentagon? Do you see why it's a problem?
The exterior angle appears to lie inside of the pentagon. So it doesn't seem to be exterior.
Furthermore, the exterior angle appears to have a measure of approximately 45°. And the interior angle has a measure greater than 180°. So the two angles do not seem to add to 180°. They don't appear to be supplementary.
If this pair of angles is not supplementary, then we don't have 5 pairs of 180°. And if we don't have 5 pairs of 180°, then the formula 5*180-360 doesn't hold. And if it doesn't hold for pentagons, then it doesn't hold for other figures and our formula is more limited than we thought.
So...does our formula apply only to convex polygons? Or can we fix things up so that it applies to concave polygons also?
To investigate that, let's take a walk.
Below is a satellite image of the courtyard of my workplace-Normandale Community College.
On top of the courtyard, we will superimpose a concave decagon (just as a decade has 10 years, a decagon has 10 sides).
In the video below, you join me on a walk around the courtyard. As you walk, pay attention to two things:
The walk begins at vertex A and ends at vertex J.
Source: http://maps.google.com
The walk starts at vertex A.
If you pay very careful attention to the direction you are facing in the video, you can verify that at vertex H, you turn through the direction you were facing when you started at vertex A.
But then you keep turning!
You turn at vertices I and J, so it all adds up to more than 360°, right?
Yes, but we can look at it a different way. Notice what happens at vertex J. You turn the other way.
So let's think about that as a negative angle measure. This fixes our two problems:
Therefore our formula holds even for concave polygons. We still have n pairs of supplementary angles and the sum of the measures of the exterior angles is still 360°. Therefore,
