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How to check does polygon with given sides' length exist?

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3

$begingroup$

I have polygon with $n$ angles. Then I have got $n$ values, which mean this polygon's sides' length. I have to check does this polygon exist (means - could be drawn with given sides' length). Is there any overall formula to check that? (like e.g. $a+bge c$, $a+cge b$, $c+bge a$ for triangle)

geometry polygons

share|cite|improve this question

edited Jun 13 '14 at 21:48

Matt E

105k8220390

asked Jun 13 '14 at 10:19

TN888TN888

220312

$endgroup$

add a comment | 

3

$begingroup$

I have polygon with $n$ angles. Then I have got $n$ values, which mean this polygon's sides' length. I have to check does this polygon exist (means - could be drawn with given sides' length). Is there any overall formula to check that? (like e.g. $a+bge c$, $a+cge b$, $c+bge a$ for triangle)

geometry polygons

share|cite|improve this question

edited Jun 13 '14 at 21:48

Matt E

105k8220390

asked Jun 13 '14 at 10:19

TN888TN888

220312

$endgroup$

add a comment | 

$begingroup$

I have polygon with $n$ angles. Then I have got $n$ values, which mean this polygon's sides' length. I have to check does this polygon exist (means - could be drawn with given sides' length). Is there any overall formula to check that? (like e.g. $a+bge c$, $a+cge b$, $c+bge a$ for triangle)

geometry polygons

share|cite|improve this question

edited Jun 13 '14 at 21:48

Matt E

105k8220390

asked Jun 13 '14 at 10:19

TN888TN888

220312

$endgroup$

share|cite|improve this question

edited Jun 13 '14 at 21:48

Matt E

105k8220390

asked Jun 13 '14 at 10:19

TN888TN888

220312

share|cite|improve this question

edited Jun 13 '14 at 21:48

Matt E

105k8220390

asked Jun 13 '14 at 10:19

TN888TN888

220312

edited Jun 13 '14 at 21:48

Matt E

105k8220390

edited Jun 13 '14 at 21:48

Matt E

105k8220390

asked Jun 13 '14 at 10:19

TN888TN888

220312

asked Jun 13 '14 at 10:19

TN888TN888

220312

2 Answers
2

active

oldest

votes

5

$begingroup$

The only rigid polygons are triangles. If $n>3$, then there are many polygons with the same sequence of sides. This can be proved by cutting the polygon along a diagonal and using induction. The only restriction is the triangle inequality: each side is less than the sum of the other sides. Indeed, given a sequence of sides satisfying the triangle inequality you can find a triangle whose sides are two consecutives sides and the third is the remaining sides straightened flat. If you insist on angles less than $pi$, just perturb this a bit.

share|cite|improve this answer

edited Jun 13 '14 at 11:57

answered Jun 13 '14 at 11:52

lhflhf

166k11172402

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For the existence of a diagonal, see chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke.
  $endgroup$
  – lhf
  Jun 13 '14 at 11:56
  
  

  
  
  
  

add a comment | 

1

$begingroup$

No, there's no overall formula. There are some weak conditions (the sum of any $n-1$ sides' lengths must be greater than the length of the remaining side, for instance), but this is merely necessary for the existence of any polygon with those side-lengths, not one that has your desired angles. Is it sufficient? I'm not certain offhand.

Re-reading, perhaps when you said that you "have "n" angles" you meant "I'm looking for an $n$-angle polygon".

In that case, the inequalities I cited above are necessary, but are they sufficient? I suspect that they are, although they'd only guarantee a polygon with those side-lengths...not a non-self-intersecting polygon. For that latter condition, you'd have to do some additional work.

share|cite|improve this answer

edited Jun 13 '14 at 11:22

answered Jun 13 '14 at 11:10

John HughesJohn Hughes

64.8k24292

$endgroup$

add a comment | 

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5

$begingroup$

The only rigid polygons are triangles. If $n>3$, then there are many polygons with the same sequence of sides. This can be proved by cutting the polygon along a diagonal and using induction. The only restriction is the triangle inequality: each side is less than the sum of the other sides. Indeed, given a sequence of sides satisfying the triangle inequality you can find a triangle whose sides are two consecutives sides and the third is the remaining sides straightened flat. If you insist on angles less than $pi$, just perturb this a bit.

share|cite|improve this answer

edited Jun 13 '14 at 11:57

answered Jun 13 '14 at 11:52

lhflhf

166k11172402

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For the existence of a diagonal, see chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke.
  $endgroup$
  – lhf
  Jun 13 '14 at 11:56
  
  

  
  
  
  

add a comment | 

5

$begingroup$

The only rigid polygons are triangles. If $n>3$, then there are many polygons with the same sequence of sides. This can be proved by cutting the polygon along a diagonal and using induction. The only restriction is the triangle inequality: each side is less than the sum of the other sides. Indeed, given a sequence of sides satisfying the triangle inequality you can find a triangle whose sides are two consecutives sides and the third is the remaining sides straightened flat. If you insist on angles less than $pi$, just perturb this a bit.

share|cite|improve this answer

edited Jun 13 '14 at 11:57

answered Jun 13 '14 at 11:52

lhflhf

166k11172402

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For the existence of a diagonal, see chapter 1 of Discrete and Computational Geometry by Devadoss and O'Rourke.
  $endgroup$
  – lhf
  Jun 13 '14 at 11:56
  
  

  
  
  
  

add a comment | 

$begingroup$

The only rigid polygons are triangles. If $n>3$, then there are many polygons with the same sequence of sides. This can be proved by cutting the polygon along a diagonal and using induction. The only restriction is the triangle inequality: each side is less than the sum of the other sides. Indeed, given a sequence of sides satisfying the triangle inequality you can find a triangle whose sides are two consecutives sides and the third is the remaining sides straightened flat. If you insist on angles less than $pi$, just perturb this a bit.

share|cite|improve this answer

edited Jun 13 '14 at 11:57

answered Jun 13 '14 at 11:52

lhflhf

166k11172402

$endgroup$

share|cite|improve this answer

edited Jun 13 '14 at 11:57

answered Jun 13 '14 at 11:52

lhflhf

166k11172402

answered Jun 13 '14 at 11:52

lhflhf

166k11172402

answered Jun 13 '14 at 11:52

lhflhf

166k11172402

1

$begingroup$

No, there's no overall formula. There are some weak conditions (the sum of any $n-1$ sides' lengths must be greater than the length of the remaining side, for instance), but this is merely necessary for the existence of any polygon with those side-lengths, not one that has your desired angles. Is it sufficient? I'm not certain offhand.

Re-reading, perhaps when you said that you "have "n" angles" you meant "I'm looking for an $n$-angle polygon".

In that case, the inequalities I cited above are necessary, but are they sufficient? I suspect that they are, although they'd only guarantee a polygon with those side-lengths...not a non-self-intersecting polygon. For that latter condition, you'd have to do some additional work.

share|cite|improve this answer

edited Jun 13 '14 at 11:22

answered Jun 13 '14 at 11:10

John HughesJohn Hughes

64.8k24292

$endgroup$

add a comment | 

1

$begingroup$

No, there's no overall formula. There are some weak conditions (the sum of any $n-1$ sides' lengths must be greater than the length of the remaining side, for instance), but this is merely necessary for the existence of any polygon with those side-lengths, not one that has your desired angles. Is it sufficient? I'm not certain offhand.

Re-reading, perhaps when you said that you "have "n" angles" you meant "I'm looking for an $n$-angle polygon".

In that case, the inequalities I cited above are necessary, but are they sufficient? I suspect that they are, although they'd only guarantee a polygon with those side-lengths...not a non-self-intersecting polygon. For that latter condition, you'd have to do some additional work.

share|cite|improve this answer

edited Jun 13 '14 at 11:22

answered Jun 13 '14 at 11:10

John HughesJohn Hughes

64.8k24292

$endgroup$

add a comment | 

$begingroup$

No, there's no overall formula. There are some weak conditions (the sum of any $n-1$ sides' lengths must be greater than the length of the remaining side, for instance), but this is merely necessary for the existence of any polygon with those side-lengths, not one that has your desired angles. Is it sufficient? I'm not certain offhand.

Re-reading, perhaps when you said that you "have "n" angles" you meant "I'm looking for an $n$-angle polygon".

In that case, the inequalities I cited above are necessary, but are they sufficient? I suspect that they are, although they'd only guarantee a polygon with those side-lengths...not a non-self-intersecting polygon. For that latter condition, you'd have to do some additional work.

share|cite|improve this answer

edited Jun 13 '14 at 11:22

answered Jun 13 '14 at 11:10

John HughesJohn Hughes

64.8k24292

$endgroup$

share|cite|improve this answer

edited Jun 13 '14 at 11:22

answered Jun 13 '14 at 11:10

John HughesJohn Hughes

64.8k24292

answered Jun 13 '14 at 11:10

John HughesJohn Hughes

64.8k24292

answered Jun 13 '14 at 11:10

John HughesJohn Hughes

64.8k24292