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Sum of a polygon's interior angles 
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#1
Mar2407, 09:50 PM

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For a polygon with n sides, the sum of the interior angles is 180n  360. If we find n positive numbers that sum up to 180n  360, does that necessarily mean these numbers can be represented as the inside angles of a polygon with n sides? I can't prove this right or wrong...



#2
Mar2407, 11:43 PM

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do you know how to prove this formula (180n360) for the npolygon? knowing that prove will help you construct an arguement for either prove or disprove your conjecture.



#3
Mar2507, 01:33 AM

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Hummm... actually I don't think it's this simple to prove.



#4
Mar2507, 01:39 AM

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Sum of a polygon's interior angles



#5
Mar2507, 01:52 AM

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Huh thanks but I don't see how this helps... I'm not learning anything new here. Maybe my question wasn't clear. If if we have set of n terms and the sum of these terms is given by 180n  360, does that necessarily mean that a polygon with these terms as its interior angles can be constructed? It's not about the fact that the sum of the interior angles of a polygon is 180n  360... it's about going the other way around.



#6
Mar2507, 02:25 AM

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Well, if you can't work forward, than working backwards will be even more difficult.
So that's why knowing the proof to that formula will help. Because then, you know the forward direction and can lay down some logic on how to go backwards if even possible. 


#7
Mar2507, 04:49 AM

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#8
Mar2507, 05:12 AM

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No it doesn't nessicarily mean that, with regards to what mjsd just quoted you on, werg. Say n=3, one angle could be 60 + e, another 60e, and the other 6. You can't construct e since it is transcendental.



#9
Mar2507, 09:48 AM

P: 1,520

Well it's pretty obvious with n = 3, you can choose a point arbitrarily, draw a line then draw another line with one of the angle in between. Then you draw another line starting at the end of the second so that the angle is the second angle. Then at the intersection of the third line and the first the angle in between will obviously be what we are looking for. This method however is not as obvious going up, and when one of the number is greater than 180, we have to deal with convex polygons which makes this kind of construction difficult to imagine. Not to mention that you get really unusual shapes.



#10
Mar2507, 10:04 AM

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#11
Mar2507, 01:48 PM

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My thoughts on the problem are that if you can find n numbers that add to 180(n2) then you can construct a polygon that has at least (n2) of these same angles, I haven't proved this, but it makes some bit of sense, however I am not sure as to whether the final two angles are guaranteed to be equal to the original two numbers in your sequence, because there are an infinite number of other numbers that will sum to the remaining angle measure that must be present in the ngon. Edit: For example with a triangle you can arbitrarily choose one angle, and then the other two angles depend on the lengths of the sides that form the original angle, I believe it will be a similar situation for larger n. Edit 2: Maybe you should disregard everything I just said, after a bit more thought it seems you may be able to choose arbitrarily (n1) angles which then guarentees you the final angle to be from the initial sequence. 


#12
Mar2507, 02:07 PM

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#13
Mar2507, 02:15 PM

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#14
Mar2507, 02:19 PM

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I was thinking about angles between 90 and 180... then again what guarantees that we won't be constructing a spiral?



#15
Mar2507, 03:47 PM

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So, the chance of having a random set of angles contruct a polygon is pretty much nil it seems. 


#16
Mar2507, 03:52 PM

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So, basically choosing any three angles that totals to 180 degrees can be constructed into a triangle. Each angle has a boundary to not allow the sum to go over 180. Example, for the first angle a we have the choice 0 < a < 180. For the second angle b, b < 0 < 180a. For the third angle c, c=180ab. So, technically the last angle you have no choice as I mentionned earlier. 


#17
Mar2507, 03:57 PM

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#18
Mar2507, 03:59 PM

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