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 [Broken]
 
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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?
 
<h2>1. What is the formula for finding the sum of a polygon's interior angles?</h2><p>The formula for finding the sum of a polygon's interior angles is (n-2) * 180 degrees, where n is the number of sides in the polygon. This formula is also known as the angle sum property of polygons.</p><h2>2. How do you determine the number of sides in a polygon?</h2><p>To determine the number of sides in a polygon, count the number of line segments that make up the shape. Each line segment represents one side of the polygon.</p><h2>3. Why is the sum of a polygon's interior angles important?</h2><p>The sum of a polygon's interior angles is important because it helps us understand the properties and characteristics of polygons. It can also be used to solve problems and equations involving polygons.</p><h2>4. Can the sum of a polygon's interior angles be negative?</h2><p>No, the sum of a polygon's interior angles cannot be negative. Interior angles are always measured in degrees, which cannot be negative. If the calculated sum is negative, it means there was an error in the calculations.</p><h2>5. How can the sum of a polygon's interior angles be used to classify a polygon?</h2><p>The sum of a polygon's interior angles can be used to classify a polygon as convex or concave. If the sum is less than 180 degrees, the polygon is concave and if the sum is greater than 180 degrees, the polygon is convex.</p>

1. What is the formula for finding the sum of a polygon's interior angles?

The formula for finding the sum of a polygon's interior angles is (n-2) * 180 degrees, where n is the number of sides in the polygon. This formula is also known as the angle sum property of polygons.

2. How do you determine the number of sides in a polygon?

To determine the number of sides in a polygon, count the number of line segments that make up the shape. Each line segment represents one side of the polygon.

3. Why is the sum of a polygon's interior angles important?

The sum of a polygon's interior angles is important because it helps us understand the properties and characteristics of polygons. It can also be used to solve problems and equations involving polygons.

4. Can the sum of a polygon's interior angles be negative?

No, the sum of a polygon's interior angles cannot be negative. Interior angles are always measured in degrees, which cannot be negative. If the calculated sum is negative, it means there was an error in the calculations.

5. How can the sum of a polygon's interior angles be used to classify a polygon?

The sum of a polygon's interior angles can be used to classify a polygon as convex or concave. If the sum is less than 180 degrees, the polygon is concave and if the sum is greater than 180 degrees, the polygon is convex.

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