Expert: Paul Klarreich - 10/23/2006

Question
I am a working on inductive proofs and am stuck.  For all n>2, the sum of the angle measures of the interior angles of a conves polygon of a n sides is (n-2) * 180. I know it works for n= 3, but I can't go further.

Answer
Hi, Jeanne,

Actually, the polygons don't have to be convex, but you have to deal with 'reflex' angles -- angles that are between 180 and 360, so we'll stay with convex polygons.

Scheme for any induction proof:

A. State the theorem clearly.

B. State the instance of the theorem for n = 1, or 2, or whatever the 'starting' case is. Prove it.

C1. State the instance of the theorem for  n = k.  This statement will be ASSUMED true.

C2. State the instance of the theorem for  n = k+1. This statement is what you must prove, and YOU WILL USE C1 SOMEWHERE IN THE PROOF.

Here we go:

A. For all n>2, the sum of the angle measures of the interior angles of a convex polygon of n sides is (n-2) * 180.  [Got that from your example -- good.]

B. Case n = 3.  The sum of the angle measures of the interior angles of a convex polygon of 3 sides is (3-2) * 180.  Proof: This is a triangle, and it is well known, [see page 492 of your high school geometry text, or whatever.] that the sum of the angles is 180, which is 1*180 = (3-2)*180.

[This part of an induction proof is usually ridiculously simple.]

C1. Case n = k:  [I'm going to compress a little, here.]

The sum of the interior angles of a convex polygon of k sides is (k-2) * 180.  ASSUMED.

C2. Case n = k+1:

The sum of the interior angles of a convex polygon of k+1 sides is (k+1-2) * 180, or (k-1)*180.  TO BE PROVED.

Now how do we prove it?  We do some geometry, of course.  Take our (k+1)-gon and draw a diagram of it.  

--- I mean it; get your paper and draw!

It has vertices labeled   A1, A2, ..., Ak+1.  Since it is a CONVEX polygon, that means that any segment that connects two points on the boundary is totally contained in the figure.  [That's the mathematical definition of convex.]

OK, then -- join A1 and Ak.  That splits the (k+1)-gon into two pieces:

-- Really! You have to draw it.

A triangle; vertices are A1-Ak+1-Ak.  
A k-gon: vertices are A1-A2-...Ak. [and Ak connects back to A1]

The sum of the angles of the triangle is 180.
The sum of the angles of the k-gon is (k-2)*180. [USE OF THE ASSUMPTION HERE.]

Now the rest is easier when you are looking at a diagram,

-- So draw it!

which I can't make for you, so I will just outline it.

All the original angles of the (k+1)-gon are present in either the triangle or the k-gon, EXCEPT the angles at A1 and Ak:

Part of angle A1 is in the triangle and the other part is in the k-gon.
  The two parts sum to the original angle A1

Part of angle Ak is in the triangle and the other part is in the k-gon.
  The two parts sum to the original angle Ak

So the original angle A1 and original angle Ak are both in the new sum, which is: (tada!)

180 + (k-2)*180 = (1)*180 + (k-2)*180 =

(1+k-2)*180 = (k-1)*180.

END OF PROOF.