Sum of a polygon's interior angles

[jing]

I'll return to my idea in post #23

Let $$n_1, n_2, ...n_k$$ be a set of k "angles" such that their sum is 180(k-2)

Let $$m_1, m_2, ...m_k$$, be defined by

$$m_i = 180 - n_i$$ for 1<=i<=k ( some $$m_i$$ may be negative)

the sum of all $$m_i$$ is 360

rotations when $$m_i$$>0 will be clockwise and when $$m_i$$<0 will be anti-clockwise

Start with a line segment $$A_0$$ to $$A_1$$

at $$A_1$$ rotate through $$m_1$$ and draw a line segment $$A_1$$ to $$A_2$$

at $$A_2$$ rotate through $$m_2$$ and draw a line segment $$A_2$$ to $$A_3$$

repeat until

at $$A_k$$ rotate through $$m_k$$ and draw a line segment $$A_k$$ to $$A_{k+1}$$

As the sum of all $$m_i$$'s is 360 $$A_k$$ to $$A_{k+1}$$ will be parallel to $$A_0$$ to $$A_1$$

Following Status X, pick two non-parallel sides whose length you adjust. This corresponds to shifting the final point by a linear combination of two linearly indpendent vectors and so arrange for $$A_{k+1}$$ and $$A_0$$ to be coincident.

Any problems with the above please let me know

However you may end up with a complex polygon as in

http://en.wikipedia.org/wiki/Polygon

I am fairly certain you can take a complex polygon and as above by repeatedly picking two non-parallel sides whose lengths to adjust you can transform the complex polygon into a simple one, either concave or convex

 

[jing]

Below is a link to an image showing what I mean by transforming a complex polygon to a simple one. I believe that sufficient repeations of this process on even a very overlapping complex polygon would result in a simple polygon.


http://img101.imageshack.us/img101/5027/comptosimpps4.png

 

[kmwest]

I know this is an old thread, but... the only justificiation that the sum of a n-gon's interior angles = 180(n-2) is because any n-gon can have (n-2) triangles inscribed within it. I have yet to see any proof (albeit with little searching) for this assertion. In some doodling I think I've started on a decent proof; I'm just wondering if anybody has a good link to a proof of this so I can see if I'm going in the right direction. Thanks -

Edit: This isn't a homework question; I'm not even in school anymore. Just for "fun" :)

 

[mXSCNT]

My graph theory class covered a proof of triangulation by ear clipping. Basically, you find an "ear" which is 3 consecutive vertices ABC, which form a triangle completely contained within the polygon. This ear then becomes a triangle in the triangulation, and the process is repeated on the remaining polygon (with the ear removed). This works because every polygon has at least one ear, a statement whose proof I forget.

 

[kmwest]

OK, here's what I got.

1) Take a simple polygon (can be convex or concave). Draw a line segment between any two nonadjacent vertices such that the line segment does not
touch a boundary. This decomposes the original n-gon into two adjacent polygons that share a common side.
2) Define the two adjacent polygons as an (x+1)gon and a (y+1)gon, where x+y=n, the number of sides in the original polygon. As such, we can define
the decomposition as (forgive me if the notation is improper, I don't know any better)
n-gon = (x+1)gon + (y+1)gon
3) Define these new adjacent polygons as p1-gon and p2-gon, where p1=x+1 and p2=y+1

thus x=p1-1, y=p2-1
x+y=n
p1 - 1 + p2 - 1 =n
thus p1 + p2 = n+2 when a n-gon = p1gon + p2gon

4) Work from n=4, and "create" polygons of increasing n using the last equation
3gon + 3gon =4gon thus 4gon= 2(3gon)
3gon + 4gon =5gon thus 5gon= 3gon+2(3gon)=3(3gon)
4gon + 4gon =6gon thus 6gon= 4(3gon)
4gon + 5gon =7gon thus 7gon=2(3gon)+3(3gon)=5(3gon)
5gon + 5gon =8gon thus 8gon=2*3(3gon)=6(3gon)

By induction,
Any n-gon is made of (n-2) triangles, and thus the sum of the interior angles is (n-2)*pi


Any holes?