Sum of a polygon's interior angles

In summary, the conversation discusses the formula for finding the sum of interior angles in a polygon with n sides (180n - 360) and whether or not a set of n numbers that sum up to this value can necessarily be represented as the interior angles of a polygon with n sides. The participants also discuss the difficulty of proving this formula and the potential for counter examples.
  • #36
I'll return to my idea in post #23

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

Let [tex]m_1, m_2, ...m_k[/tex], be defined by

[tex]m_i = 180 - n_i [/tex] for 1<=i<=k ( some [tex]m_i[/tex] may be negative)

the sum of all [tex]m_i[/tex] is 360

rotations when [tex]m_i[/tex]>0 will be clockwise and when [tex]m_i[/tex]<0 will be anti-clockwise

Start with a line segment [tex]A_0[/tex] to [tex]A_1[/tex]

at [tex]A_1[/tex] rotate through [tex]m_1[/tex] and draw a line segment [tex]A_1[/tex] to [tex]A_2[/tex]

at [tex]A_2[/tex] rotate through [tex]m_2[/tex] and draw a line segment [tex]A_2[/tex] to [tex]A_3[/tex]

repeat until

at [tex]A_k[/tex] rotate through [tex]m_k[/tex] and draw a line segment [tex]A_k[/tex] to [tex]A_{k+1}[/tex]

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

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 [tex]A_{k+1}[/tex] and [tex]A_0[/tex] 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
 
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  • #37
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
 
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  • #38
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" :)
 
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  • #39
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.
 
  • #40
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?
 

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