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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. $a+b\ge c$, $a+c\ge b$, $c+b\ge a$ for triangle)

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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.

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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.

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This is more a long comment to ihf's answer than a separate answer. I want to stress here that the statement made by ihf is wrong. To see why, take the side lengths $a_1,\dots,a_n$ all equal, with $n \geq 5$. Clearly $a_1,\dots,a_n$ satisfy the triangle inequality, but you cannot choose two consecutive side lengths $a_i, a_{i+1}$ (where $a_{n+1}=a_1$) so that there exists a triangle having as side lengths $a_i, a_{i+1}$ and the sum of all the remaining side lengths.

However the following is generally true. Given a sequence of positive side lengths $a_1,\dots,a_n$ satisfying the triangle inequality, that is $2a_i \leq \sum_{j=1}^{n} a_j$ for $i=1,\dots,n$, let us set $a_m=a_k$ for every integer $m \in \mathbb{Z}$ such that $k \equiv m$ (mod n). Then there exist $m_0, m_1, m_2$, with $m_0 <m_1<m_2 \leq m_0 +n$ such that there is a triangle whose sides have lengths $a_{m_0}+\dots+a_{m_1 -1}$,$a_{m_1}+\dots+a_{m_2 -1}$,$a_{m_2}+\dots+a_{m_0+n}$.

We can easily show this by induction. For $n=3$ the thesis is trivial. So consider $n \geq 4$. We state that there are $a_i, a_{i+1}$ such that $2(a_i + a_{i+1}) \leq \sum_{j=1}^{n} a_j$. Indeed, if this were not the case, then by summing up all the inequalities $2(a_i + a_{i+1}) > \sum_{j=1}^{n} a_j$ for $i=1,\dots,n$, we would get $4 \sum_{j=1}^{n} a_j > n \sum_{j=1}^{n} a_j$, an absurd. Then, if $2(a_i + a_{i+1}) \leq \sum_{j=1}^{n} a_j$, by setting $b_1=a_i + a_{i+1}$, $b_2=a_{i+2},\dots, b_{n-1}=a_{i+n-1}$, we get a sequence of $n-1$ positive numbers satisfying the triangle inequality, and so the inductive hypothesis applies to give the desired triangle.

Obviously this does not give a complete proof that, given a sequence $a_1,\dots,a_n$ of positive numbers satisfying the triangle inequality, that is $2a_i \leq \sum_{j=1}^{n} a_j$ for $i=1,\dots,n$, there is a convex polygon having as consecutive side lengths $a_1,\dots,a_n$, since we have also to slightly deform the triangle found above in order to assure that each interior angle is less than $\pi$. This is evidently possible. Actually, Boni Bogoșel has proved muc more in his page Cyclic polygons largest area: given a sequence $a_1,\dots,a_n$ of positive numbers satisfying the triangle inequality, that is $2a_i \leq \sum_{j=1}^{n} a_j$ for $i=1,\dots,n$, there is a $\mathbf{cyclic}$ $\mathbf{polygon}$ having $a_1,\dots,a_n$ as consecutive side lengths.

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