Sum of a polygon's interior angles

[jing]

Ok Doodle Bob, I see the flaw in my method, halving the angles does not work as you have to take exactly 180 from the n angles so that ther n-1 form a polygon. Mea Culpa

 

[StatusX]

hey, folks, y'all are making this problem *much* harder than it is.

First of all, you need to solidify what the conjecture is. I would point out as mjsd already has (albeit elliptically) that there are restrictions to the possible interior angle measures, e.g. there is no quadrilteral with angle measures (in degrees) 1, 1, 1, 357.

Also, this spiralling business is a hobgoblin: we're only looking at finite stages at any given time.

If you stay within the category of convex polygons, there is a perfectly rigorous -- not all that technical -- proof by induction that, given n numbers all within 0 and 180 and which add up to 180*(n-2), there is an n-sided convex polygon with those as its angle measures.

Who said the polygon needed to be convex? And you can have a quadrilateral with angles 1,1,1,357, it looks like a squished boomerang. Admittedly the proof doesn't guarantee the sides won't cross, but it seems like as long as all the angls are less than 360 it does turn out that way. I'll have to think about how you'd prove this.

 

[Werg22]

Who said the polygon needed to be convex? And you can have a quadrilateral with angles 1,1,1,357, it looks like a squished boomerang. Admittedly the proof doesn't guarantee the sides won't cross, but it seems like as long as all the angls are less than 360 it does turn out that way. I'll have to think about how you'd prove this.

SatusX, that's exactly what I'm trying to think about. Doodle, I have to admit that you lost me with your "arrange the angles in a certain way".

 

[mjsd]

NB: the problematic angles are 0 and 180.

 

[jing]

you need to arrange the n given numbers in a certain to be able to do this. .

Giving it more thought this arranging causes problems with the inductionstill restricting to the case when all angles <180

for n>3

from the set of n angles choose 3, $$a_{1}, a_{2}, a_{3}$$ such that

$$a_1 + a_2 + a_3 = 180 + d_1 + d_2$$

so that

$$a_1-d_1 + a_2 -d_2+ a_3 = 180$$

take the set of n-1 angles formed from the 'n set' with $$a_{1}, a_{2}, a_{3}$$ removed and $$d_{1},d_{2 }$$ added in.

Form the n-1 polygon. However since the induction method requires a choice of three angles in a certain arrangement how do you guarantee that $$d_{1}$$ and $$d_{2}$$ are adjacent so that the triangle with angles $$a_1-d_1, a_2 -d_2, a_3$$ can be adjoined to it to form the required n polygon?

 

[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?