# The Octahedron Problem – Revisited

I do not know, if anybody solved the problem of the last posting. But let me point out, that the problem would have been mathematically more pleasing, if the octahedron was inscribed in a cube with side length 2. We could take the cube as

$${[-1,1]}^3$$.

Then the corners of the octahedron are very easy at the points

$$(1,0,0)$$

etc. The vector

$$(1,1,1)$$

is perpendicular to the lower left side, and 2/3 times this vector connects two opposing sides with length

$$2/\sqrt{3}$$

Problem solved.

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