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If we have given set of $n$ segments (or lengths of segments), how can we easily check if this set is forming polygon with $n$ sides.

Example

Let our set be $A$, $A = \{3, 6, 6\}, n = 3$

The answer here should be true, because we can arrange those segments to make triangle.

But the problem for me is when there are more segments.

So I tried to come with some statment, I started from triangle, it says: To form a triangle there shouldn't be side that is bigger than the sum of the other two sides, and I came to statement for more than 3 sides.

Is this correct: For any n-sided polygon, there shouldn't be side that is bigger than the sum of any two other sides.

Thanks in advance.

someone123123
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  • You've gotten the opposite condition for triangles. Three lengths form a triangle if and only if each length is less than the sum of the other two (a condition best known as the "triangle inequality" in the weakened form of less than or equal to the sum of the other two sides). – hardmath Sep 28 '17 at 17:20
  • I just changed the text, I think it is okay now. – someone123123 Sep 28 '17 at 17:23
  • In its current form this Question asks the same thing. However it was edited after two Answers were posted, both of which attempt to solve a more ambitious problem (probably due to ambiguity in the original), so that neither of them addresses the problem raised here. – hardmath Sep 28 '17 at 17:25
  • Thanks, keep in mind the boundary case where one length exactly equals the sum of the other two. Most people would say this degenerate case is not really a triangle, but it might serve your intended application to have the segments lie flat together. – hardmath Sep 28 '17 at 17:26

1 Answers1

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You are on the right track, but the correct statement is that $n$ lengths admit a (convex) polygon if and only if no one length is greater than or equal to the sum of the other lengths.

To sketch a proof of necessity, consider any side $E$ of a polygon. The other sides form a path from one endpoint to the other, and since the shortest distance between the endpoints is given uniquely by the straight line segment (edge length), the sum of remaining lengths exceeds that of edge $E$.

We can also say a little about a proof of sufficiency, e.g. the construction of a polygon given side lengths, no one of which equals or exceeds the sum of the remaining lengths. It is possible to proceed by induction from the base case $n=3$ of triangles. The previous Question on maximizing the area of such constructed polygons shows a more elegant result.

This is a straightforward generalization of the case for triangles, and again it suffices to check that the longest segment's length is less than the sum of the remaining lengths.

hardmath
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    Thanks for the answer, can you just help me little more, why we can't form polygon if one side is bigger than the sum of others. – someone123123 Sep 28 '17 at 17:28
  • @someone123123: If one side were longer than the sum of all others, we could not reach between the two endpoints of the longer side even using all of the other sides. Indeed if the longest side is equal to the sum of all others, we could only double back from one endpoint to the other along the very track of that longest side, so the "polygon" would just be a doubled up line segment (not really a valid polygon). – hardmath Mar 30 '24 at 19:53