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During my research in combinatorial geometry, I have encountered the following elementary question which I am hoping to have some help on.

Let $\zeta = 27.22236\ldots$. Does there exist a set of $\gamma_i$'s such that $\sum\limits_{i=1}^{8} \gamma_{i} = \zeta$, where $\gamma_i \in \{2\pi - \arccos(1/3), 2\pi - 2\arccos(1/3), 2\pi - 3\arccos(1/3), 2\pi - 4\arccos(1/3)\}$?

There are other versions of the problem which I need to also solve for a case analysis involving spherical simplicial complexes on $\mathbb{S}^2$, with other values of $\zeta$ and a different number of $\gamma_i$'s, but if someone knows how to solve this particular question I think I will be able to generalize quite easily.

Let me know if you have any questions, and I can add additional motivation to the question if necessary, but it involves simplicial complexes, sphere packings, and other details I would prefer not to bother mention.


EDIT: A bit more information about the value for $\zeta$. Let $\Omega = 15\pi - 33\arccos(1/3)$, $\psi = 2\pi - 4\arccos(1/3)$, and $\beta = \arccos\left(\frac{\cos(\pi/3)-\cos(\pi/3)\cos(z)}{\sin(\pi/3)\sin(z)}\right)$ where $z = \arccos\left(\frac{1 + 3\cos(\psi)}{4}\right)$ Then, $\zeta = \Omega - 2\psi - 2\beta + 9\pi - 2\arccos(1/3) = 27.22236\ldots$

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    @QiaochuYuan: I'll edit my answer and include exactly what $\zeta$ is since you asked. :)2012-07-08

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All of your sums will be of the form, $16\pi-k\arccos(1/3)$ for some integer $k$ between 8 and 32, so compute $(16\pi-\zeta)/\arccos(1/3)$ and see if it looks like an integer.