Think in 3D of a flower made from regular octahedra with the vertices of the octahedra meeting at a point and faces in contact. How many octahedra can be joined together in this way around one centre?

Here is a solution from Andrei of Tudor Vianu National College, Bucharest, Romania.

First, I observe that the octahedron is symmetrical about a plane that passes through the 4 coplanar vertices (see figure), so for the problem I may take into account only a pyramid.

I shall calculate angle $MVN$.

As all exterior sides have equal lengths, without loss of generality I may assume $AB = 1$. In this case $VN$ is the height of equilateral triangle $VBC$, so it has a length of $\sqrt 3 /2$. $MN$ is equal to $AB$, having a length of 1.

To calculate angle $MVN$, I shall calculate its sine.

\begin{matrix} \sin MVN & = 2 \sin \frac{∠ MVN}{2} \cos \frac{∠ MVN}{2} \\ = 2 \frac{MP}{MV} \frac{VP}{MV} \\ = 2 . \frac{1 / 2}{\sqrt{3} / 2} . \frac{\sqrt{2} / 2}{\sqrt{3} / 2} \\ = \frac{2 \sqrt{2}}{3} \end{matrix}

The angle at

MVN

(to 3 decimal places) is

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

Alternatively you could calculate this angle as

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

The number of octahedra is

360 / 70.529 = 5.1

So 5 tetrahedra could be joined together in this way.