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Trying to make a 3D-model of a "ball" made up of squares and hexagons:

Given a square and four hexagons connecting to the square sides, what is the angle required between the square plane and the hexagon planes to make two sides of each hexagon connect with the sides of the two neighbouring hexagons?

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    $\begingroup$ The dihedral angle of a cuboctahedron is $\theta= \cos^{-1} (\frac{-1}{\sqrt{3}}) \simeq 125.26$. en.wikipedia.org/wiki/Cuboctahedron $\endgroup$ Commented Apr 29, 2017 at 12:52
  • $\begingroup$ @Donald Splutterwit Sorry but it's not a cuboctahedron (en.wikipedia.org/wiki/Cuboctahedron) but a truncated tetrahedron. $\endgroup$
    – Jean Marie
    Commented Apr 29, 2017 at 12:54
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    $\begingroup$ @JeanMarie Lets try truncated Octahedron en.wikipedia.org/wiki/Truncated_octahedron ... & amusingly the dihedral is the same as I previously stated $\ddot \smile$ $\endgroup$ Commented Apr 29, 2017 at 13:12
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    $\begingroup$ I agree, it is a truncated Octahedron. $\endgroup$
    – Jean Marie
    Commented Apr 29, 2017 at 13:22
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    $\begingroup$ The angle between the Hexagonal faces is $\phi= \cos^{-1} (\frac{-1}{3}) \simeq 109.47$ $\endgroup$ Commented Apr 29, 2017 at 13:23

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As already stated in the comment, you're looking at part of a truncated octahedron

The angles between the faces are stated as well:

4-6: arccos(−1/√3) = 125°15′51″
6-6: arccos(−1/3) = 109°28′16″

If you want to check them yourself or get e.g. edge angles, note that the truncated octahedron is a permutohedron: It's coordinates are all permutations of $(1,2,3,4)$, or if projected to 3D all variations of $(0,\pm1,\pm2)$ (note this gives you edges of length $\sqrt2$).

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