\section*{Problem 4 - 061024}

If the midpoints of the faces of a cube are connected by straight lines, the edges of an octahedron inscribed in the cube are obtained. If one proceeds in a similar manner with an octahedron, the edges of a cube inscribed in the octahedron are obtained.

a) How do the volumes of the cube and the inscribed octahedron relate to each other?

b) How do the volumes of the octahedron and the inscribed cube relate to each other?

c) How do the surface areas relate to each other in the first case?

d) How do the surface areas relate to each other in the second case?
}

For the volume of the cube with edge length \(a\), it holds: \(V_{w}=a^{3}\).

For the surface area of the cube with edge length \(a\), it holds: \(O_{w}=6 a^{2}\).

For the volume of the octahedron with edge length \(b\), it holds: \(V_{o}=\frac{1}{3} b^{3} \cdot \sqrt{2}\).

For the surface area of the octahedron with edge length \(b\), it holds: \(O_{o}=2 b^{2} \cdot \sqrt{3}\).

The octahedron consists of a square pyramid with a square base of area \(b^{2}\) and its mirror image along the base. The height is half the diagonal of the square, i.e., \(\frac{b}{2} \sqrt{2}\).

The surface consists of 8 equilateral triangles with base \(b\). The height of each of these triangles is \(\frac{b}{2} \sqrt{3}\). Thus, the area of each of these triangles is \(\frac{b^{2}}{4} \sqrt{3}\).

a) Let the length of the cube edge be \(a\). Then for the length of the octahedron edge \(b=\frac{1}{2} a \sqrt{2}\), it holds that

\[
V_{w}=a^{3} \quad \text { and } \quad V_{o}=\frac{1}{3} \sqrt{2}\left(\frac{1}{2} a \sqrt{2}\right)=\frac{1}{6} a^{3}
\]
and thus \(V_{w}: V_{o}=6: 1\).

b) Let the length of the octahedron edge be \(b\). The centers \(S\) of the equilateral triangles lie on the medians. These are divided by \(S\) in the ratio \(1: 2\). A parallel segment to \(b\) through \(S\) has the length \(\frac{2}{3} b\). The intersection of a plane, in which a face of the inscribed cube lies, with the octahedron is thus a square with side length \(\frac{2}{3} b\).

The length of the square side \(a\) is given by

\[
a^{2}=\left(\frac{1}{3} b\right)^{2}+\left(\frac{1}{3} b\right)^{2} \quad \text { so } \quad a=\frac{1}{3} b \sqrt{2}
\]

Therefore, it holds:

\[
V_{o}=\frac{1}{3} b^{3} \sqrt{2} \quad \text { and } \quad V_{w}=\left(\frac{1}{3} b \sqrt{2}\right)^{3}=\frac{2}{27} b^{3} \sqrt{2}
\]

and thus \(V_{o}: V_{w}=9: 2\).

c) It is \(O_{w}=6 a^{2}\) and \(O_{o}=a^{2} \sqrt{3}\). Therefore, \(O_{w}: O_{o}=2 \sqrt{3}: 1\).

d) It is \(O_{o}=2 b^{2} \sqrt{3}\) and \(O_{w}=\frac{4}{3} b^{2}\). Therefore, \(O_{o}: O_{w}=3 \sqrt{3}: 2\).

Solutions of the II. Round 1966 adopted from [5]

\subsection*{7.8.3 III. Round 1966, Class 10}