Without loss of generality take the edges of the octahedron as 2 units. Consider the square-based pyramid forming half the octahedron with the 'special' vertex, let's call this

V

, as the vertex of this pyramid. Then the plane of symmetry cutting through

V

and the two coloured faces, cuts the pyramid in an isosceles triangle with sides

\sqrt{3}, \sqrt{3}

and 2.

Now imagine the 'ring' of octahedra glued together! The plane of symmetry for the ring contains the plane of symmetry of the pyramid just described.

The angle at $V$ in this plane of symmetry (to 3 decimal places) is

$2 \sin^{- 1} 1 / \sqrt{3} = 70 . 529^{o} .$

So there is 'room' for $360 / 70.529 = 5.1$ octahedra in the ring. Clearly this means the fifth octahedron can be fitted in but no more so the answer is 5.