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I know the IFF condition holds for a triangle, but does the IFF condition hold true for any $n$-sided polygon?

The IFF condition is clear in the forward direction in that it's intuitive why any $n$-sided polygon would require that the sum of any $n-1$ sides be greater than the remaining side. But if the IFF condition is true, it's not intuitive to me why the sum of any $n-1$ sides being greater than the remaining side is a sufficient condition to ensure an $n$-sided polygon.

roulette01
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3 Answers3

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For an intuitive explanation, imagine that the side are thing hollow metal rods, with a chain running through all of them and joining them. (The chain forms a closed loop.)

Hold the longest rod parallel to the ground. The other rods hang down an close the polygon.

saulspatz
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  • The way you phrase it you seem to presuppose such a loop exists, which is a circular argument. I would have described it as one loose chain of all the shorter sides (in any order) that you attach to the ends of the horizontal longest side. – Jaap Scherphuis Apr 01 '21 at 15:40
  • @JaapScherphuis I don't see that that's really any different. There could be a length of chain inside the longest rod that you attach to the ends of the the loose chain. I'm thinking of just a bunch of rods with a chain running through them. They aren't assumed to lie in the same plane to begin with. – saulspatz Apr 01 '21 at 15:43
  • What I mean is that you already start by saying that the chain forms a closed loop, which is what you actually want to establish. I would start with an open loop (open at one or both ends of the longest rod), and then close it into a loop, justifying that this is possible because the longest rod is shorter than the rest of the chain together. – Jaap Scherphuis Apr 01 '21 at 15:48
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If there are 4 or more sides, you can always reduce the number of sides by joining together two adjacent ones into one long edge (essentially setting the vertex angle to 180). If that would result in a side longer than all the other sides put together, simply change your mind and choose an adjacent pair of those other sides instead (which when joined together clearly cannot form a side that is longer than the sum of the others). Repeat till you have only 3 sides left.

So if you can see why it is sufficient for a triangle, it is clear that it works for the original sequence of side lengths, as they form a triangle with some extra vertices on the sides of the triangle.

Jaap Scherphuis
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This can be rigorously proved by induction (which may or may not be intuitive to you).

I assume the result is clear for $n=3$, i.e. the triangle (let me know if it's not). For $n \geq 4$, let $l_1 \leq \cdots \leq l_{n-1} \leq l_n$ denote the sequence of lengths. Now let $$ \epsilon = \frac{1}{2}\min_j\left\{\sum_{i=1}^n l_i - 2l_j, l_1 \right\}, $$ and note $\epsilon > 0$ as it is a minimum of positive numbers (where positivity comes from the assumption on the $l_i$'s). Consider the sequence $a_1 = l_1 + l_2 - \epsilon, a_2=l_3, \cdots, a_{n-1}=l_n$; note $a_1 = l_1 + l_2 - \epsilon > l_2 > 0$. Thus we have a collection of $n-1$ lengths. Moreover, every sum of $n - 2$ of these is greater than the remaining length. Indeed, if $a_1$ is in the sum, we have $$ \sum_{\substack{1 \leq i \leq n - 1 \\ i \neq j}}a_i = \sum_{i = 1}^{n-1} a_i - a_i = \sum_{i = 1}^n l_i - l_{j+1} - \epsilon > l_{j+1} = a_{j} $$ because $\epsilon < \sum_{i=1}^n l_i - 2l_{j+1}$. For the remaining sum, first note $l_1 + l_2 \leq \sum_{i=3}^{n} l_i$; indeed, because $n \geq 4$ we have $l_1 \leq l_3$ and $l_2 \leq l_4$. Now we may compute $$ \sum_{i=2}^{n-1}a_i = \sum_{i=3}^n l_i \geq l_1 + l_2 > l_1 + l_2 - \epsilon = a_1 $$ Hence by induction we see that $a_1, a_2, \cdots, a_{n-1}$ are the side lengths of some polygon. Finally, note $l_1, l_2, l_1 + l_2 - \epsilon$ are the side lengths of a triangle (verify at your leisure). Thus we may attach the triangle with these side lengths (identifying the sides of length $l_1 + l_2 - \epsilon$) to our newly-constructed polygon to obtain a polygon with side lengths $l_1, l_2, \cdots, l_n$ and complete the proof!

User12345
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  • I made two assumptions while writing this proof: (1) you don't care if your polygons are convex or self-intersecting, and (2) the triangle case is clear to you. I can rewrite the proof if you like your polygons convex and non-intersecting--which does require some extra work after realizing the segments in the plane. I can also provide a proof for the case $n=3$ if that is unclear. – User12345 Apr 01 '21 at 17:29