# Sum of a polygon's interior angles

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...

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mjsd
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do you know how to prove this formula (180n-360) for the n-polygon? knowing that prove will help you construct an arguement for either prove or disprove your conjecture.

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

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.

JasonRox
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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.

mjsd
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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.

I understood your question. strategy: start with n=3 (the smallest) and see what conclusions you get first.....note 180n-360 is in fact 180(n-2)

Gib Z
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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 60-e, and the other 6. You can't construct e since it is transcendental.

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.

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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 60-e, and the other 6. You can't construct e since it is transcendental.

I'm not sure what you mean, but the terms in the set are known and are invariable.

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 60-e, and the other 6. You can't construct e since it is transcendental.

I don't believe Werg22 meant constructible in the sense of being constructible with ruler and compass, but merely whether or not there would exist an n-sided polygon that had those n angles for its internal angles.

My thoughts on the problem are that if you can find n numbers that add to 180(n-2) then you can construct a polygon that has at least (n-2) 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 n-gon.

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 (n-1) angles which then guarentees you the final angle to be from the initial sequence.

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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.

Yeah, this is what I described in post 9.

Edit 2: Maybe you should disregard everything I just said, after a bit more thought it seems you may be able to choose arbitrarily (n-1) angles which then guarentees you the final angle to be from the initial sequence.

It's not this simple. For other angles, what guarantee do we have that the final line won't intersect the first line in such a way that it would have to cut through the other lines?

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It's not this simple. For other angles, what guarantee do we have that the final line won't intersect the first line in such a way that it would have to cut through the other lines?

Yes, you have a good point which you might try to construct a counter example from, however, there may be some restriction you could make on the set of angle measures that would make this always possible.

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

JasonRox
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Yes, you have a good point which you might try to construct a counter example from, however, there may be some restriction you could make on the set of angle measures that would make this always possible.

Yes, but if you think about this a little bit you will see that the restrictions heavily rely on the previous angles. Your last angle chosen will hardly be a choice!

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

JasonRox
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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.

Let's assume the lengths are adjustable as we construct.

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 < 180-a. For the third angle c, c=180-a-b.

So, technically the last angle you have no choice as I mentionned earlier.

Yes, but if you think about this a little bit you will see that the restrictions heavily rely on the previous angles. Your last angle chosen will hardly be a choice!

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

The last angle will obviously not be chosen. The difficulty here is in if there is a way to adjust the length of the constructed polygon so that the intersection of the last and first line will form a closed n sides polygon with the other lines. For example, there are constructions where the two lines intersect in such a way that the last line has to cut through other lines. And the set is not that random, it has the property that the sum of its terms is 180n - 360...

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Let's assume the lengths are adjustable as we construct.

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 < 180-a. For the third angle c, c=180-a-b.

So, technically the last angle you have no choice as I mentionned earlier.

This was already clear... Like I said, the difficulty is to prove this for any n.

JasonRox
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This was already clear... Like I said, the difficulty is to prove this for any n.

If the length's are fully adjustable then I say yes it is possible. I would say do a couple examples and create an algorithm on how to construct it where the algorithm provides rules to how long they are.

So, you have to do dirty work here.

mjsd
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if it works for n=3, i think it will work for all n, but for n >3 the way you permute your arbitrary positive numbers inside the polygon may not be totally arbitrary.

if it works for n=3, i think it will work for all n, but for n >3 the way you permute your arbitrary positive numbers inside the polygon may not be totally arbitrary.

ok, right about now, someone should be realizing that "inducting on n" will be a correct strategy to solve this problem.

mjsd
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ok, right about now, someone should be realizing that "inducting on n" will be a correct strategy to solve this problem.

i think most of us here know what is going on right from the start.... the point about understanding the proof for 180(n-2) is that in doing so , one would realise how to best cut the n-polygon up. n=3 is the base case.

Another approach

Instead of the interior angle $$n_1, n_2, .....n_k$$

Consider the exterior angles $$m_1, m_2, .....m_k$$,instead where

$$m_i = 180 - n_i$$ for 1<=i<=k

and sum of all $$m_i$$ is 360

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

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

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

repeat until

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

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

There cannot be any spiralling because the partial sums of the angles $$m_1$$ + .....+ $$m_p$$ for all p<k must always be <360.

I think by adjusting line segment lengths you can arrange for $$A_k$$ to $$A_{k+1}$$ and $$A_0$$ to $$A_1$$ to lie along the same straight line and a polygon formed.

Might be a start here if someone wants to apply a little more rigour

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JasonRox
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I'm surprised werg22 hasn't tried random examples for 4 or 5 sides. He's the one who wants to solve the problem.

Another approach

Instead of the interior angle $$n_1, n_2, .....n_k$$

Consider the exterior angles $$m_1, m_2, .....m_k$$,instead where

$$m_i = 180 - n_i$$ for 1<=i<=k

and sum of all $$m_i$$ is 360

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

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

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

repeat until

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

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

There cannot be any spiralling because the partial sums of the angles $$m_1$$ + .....+ $$m_p$$ for all p<k must always be <360.

I think by adjusting line segment lengths you can arrange for $$A_k$$ to $$A_{k+1}$$ and $$A_0$$ to $$A_1$$ to lie along the same straight line and a polygon formed.

Might be a start here if someone wants to apply a little more rigour

That is... everything that we are unsure about! First your method is equivalent to the one dleet and me firstly described, the only difference being that you used the supplementary angles; the construction is still the same.

"There cannot be any spiralling because the partial sums of the angles $$m_1$$ + .....+ $$m_p$$ for all p<k must always be <360."

This is unjustified. We are not sure whether or not it is possible to have a construction where the last $$A_{k+1}$$ corresponds to $$A_{1}$$ let alone the uncertainty that constructions that answer this criterion do not have overlapping lines. And Jason, I'm looking for something analytical, so trying random stuff is not my objective.

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StatusX
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jings description can be made rigorous. You can start with all sides having unit length, and then just pick two non-parellel sides whose length you adjust. This corresponds to shifting the final point by a linear combination of two linearly indpendent vectors, ie, anywhere, and when it lies on top of the starting point you have your shape. There's a little subtelty making sure the lengths of each side is positive, but this shouldn't be an insurmountable problem.

"There cannot be any spiralling because the partial sums of the angles $$m_1$$ + .....+ $$m_p$$ for all p<k must always be <360."

This is unjustified.

I was thinking convex polygons and so thought that $$n_1, n_2, .....n_k$$ were all <180.
Reading your posts again I now realise you are thinking concave or convex polygon and so some $$n_i$$ >180.

and in this case $$m_i = 180 - n_i$$ is not true.

But when $$n_1, n_2, .....n_k$$ are all <180 then it looks like the result holds.

Will have to reconsider for some $$n_i$$ >180.

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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.

Induct on n starting with n=3.

The inductive part uses the proof that Jason was trying to shove in your face (justifiably): to show that the desired n-gon exists, imagine splitting it up into a triangle and an (n-1)-convex-gon. This is whre the angle restriction applies: you need to arrange the n given numbers in a certain to be able to do this. Once that part is out of the way, induction tells us that both the triangle and the (n-1)-gon exists and hence the n-gon exists.

Oh, you're worried about matching the sides of these two in order to create the n-gon? That's the beauty of Euclidean geometry: we can shrink or stretch any figure so that one of its sides matches any given length.

You're worried about spiralling? I'm not. All I care about is that the result is true for each finite n, which induction takes care of for me.

If you stay within the category of convex polygons, there is a perfectly rigorous - proof by induction ...... you need to arrange the n given numbers in a certain to be able to do this.

When all the angles are <180 you do not need them in a certain order.

For n>3

take the angles $$a_1, a_2, a_n$$

form the n-1 polygon with angles

$$a_{2}/2, a_{n}/2$$ and the angles $$a_3,...... a_{n-1}$$

form the triangle with angles $$a_1, a_{2}/2, a_{n}/2$$ and add to the n-1 polygon placing the half angles together to form the required n polygon.

EDIT sorry folks this does not work as halving $$a_{2}, a_{n}$$ will not produce n-1 angles that add to a multiple of 180 and so you cannot get a n-1 polygon

If some angles>180 then arrange three < 180 in positions $$a_1, a_2, a_n$$ so that the added triangle does not intersect the n-1 polygon and I think the induction still works

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When all the angles are <180 you do not need them in a certain order.

no, that is incorrect: if the polygon is convex (which was the condition that i am assuming), then any three consecutive angles of the polygon need to add up to more than (or equal to) 180 degrees.

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

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StatusX
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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.

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
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NB: the problematic angles are 0 and 180.

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 induction

still 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 guarentee 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?

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