Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.7 AIME, harder Prove it Brazil

The centers of the faces of a cube form a regular octahedron of volume $V$ . Through each vertex of the cube we may take the plane perpendicular to the long diagonal from the vertex. These planes also form a regular octahedron. Show that its volume is $27V$ .

Solution

Let the cube have side $k$ . $A$ , $B$ are two adjacent vertices of the small octahedron, and $AX = BX = k/2$ and $\angle AXB = 90^\circ$ , so $AB = k/\sqrt{2}$ .

The large octahedron has the vertices of the cube at the center of its faces.
The line joining the centers of $OPS$ and $OQR$ is parallel to $PQ$ and $2/3$ the length. But it is also $k\sqrt{2}$ , so the side of the large octahedron is $(3/2)k\sqrt{2} = 3k/\sqrt{2}$ or $3$ times the side of the small octahedron. Hence the volume of the large octahedron is $3^3 = 27$ times the volume of the small.

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