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Consider a spherical triangle, with side lengths $0 \leq a,b,c \leq \pi$. In texts on spherical geometry we see proofs that $a \leq b + c$ but this inequality is unnecessarily weak, in that some sets of lengths of sides satisfy the triangle inequality (and its cyclic permutations) but no triangle can be constructed with the given side lengths; for example $a=\pi, b=\pi, c=\frac{\pi}{2}$.

Stronger is $$a \leq \begin{cases}b + c, &b+c\leq \pi\\2\pi-b-c, &b+c > \pi\end{cases},$$ or more concisely, $$\cos a \geq \cos(b+c),$$ which does have the property that a triangle can be constructed from edges of length $a,b,c$ if and only if the lengths satisfy the triangle inequality.

Does this inequality generalize in a clean way to general polygons of side lengths $0 \leq a, b, c, d, \ldots \leq \pi$?

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up vote 1 down vote accepted

"There exists a spherical $n$-gon with sidelengths $2\pi \nu_1,\dots, 2\pi \nu_n$ if and only for every subset $P\subset \{1,\dots,n\}$ of odd cardinality $|P|$ the inequality $$\sum_{i\in P}\nu_i -\sum_{i\notin P}\nu_i - \frac{|P|-1}{2}\le 0$$ holds."

Source: Spherical polygons and unitarization by R. Buckman and N. Schmidt. Apparently, this REU paper was not formally published. See also the UMass REU site.

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Interesting! Thanks. – user7530 Aug 2 '12 at 20:46

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