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Problem:

An octahedron is inscribed in a cube whose edge length is 2. A cube is inscribed inside the octahedron. All the 4-fold, 3-fold and 2-fold symmetry axes of these objects coincide.

Method:

I calculated the edge length of the octahedron first. Imagine the 2D slice to look like this, then used the Pythagorean Theorem to find that the octahedron edge length was $\sqrt{2}$. All edges are equal in length for the octahedron, so I proceeded to calculate the distance from the center of any triangular face to the middle of any edge on the same face ($\frac{1}{\sqrt{6}}$). Using two different lines from center to edge on the same triangular face, I used the Pythagorean Theorem to solve for the inner cube edge ($\frac{1}{\sqrt{3}}$).

Questions:

Is my solution correct? I appreciate any suggestions or alternate solutions.

Reference Images:

Octahedron inside outer cube.

Inner cube inside octahedron.

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1 Answer 1

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No, it is not correct.

The calculation of the octahedron sidelength is good, but you made a false assumption in calculating the sidelength of the inner square. I could tell you what it is, but I suspect you may already know, or at least where to look. So I'll let you think about it first. Leave a comment if you can't find it.

As for an alternative approach, try solving it analytically in coordinates (hint: the center of a triangle is the average of its 3 corners). Even if you want a geometric proof, the analytic calculation will be quite instructive.

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  • $\begingroup$ Perhaps it was wrong to assume that I could apply the Pythagorean Theorem the second time since the angle was not necessarily 90 degrees. Anyhow, I tried out the coordinates method and got an inner cube edge length of $\frac{2}{3}$. Does this match your result? $\endgroup$
    – Mike Chang
    Commented Apr 16, 2020 at 3:55
  • $\begingroup$ Yes to $\frac 23$ and yes to using the Pythagorean theorem being the mistake. When you split the octahedron into two pyramids, the base of the pyramid is a square, but the plane of that square is not normal to the faces. The faces actually meet at an obtuse angle. $\endgroup$
    – Paul Sinclair
    Commented Apr 16, 2020 at 12:48

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