## Task 3 - 180933

Given is a cube, whose volume is denoted by $V_{1}$.

If the center of each face of this cube is connected to the centers of all adjacent faces, the edges of a regular octahedron are obtained. The volume of this octahedron is called $V_{2}$.

If the centroid of each face of this octahedron is now connected to the centroids of all adjacent faces, the edges of a second cube are obtained. Its volume is called $V_{3}$.

Calculate the ratio $V_{1}: V_{2}: V_{3}$!
Let the edge length of the original cube be $a$, the edge length of the octahedron be $b$, and the edge length of the inner cube be $c$. Then, $V_{1}=a^{3}$.

The octahedron can be viewed as the union of two identical square pyramids. The apexes of these pyramids form the centers of opposite faces of the original cube, so the height $h$ of one of these pyramids is $h=\frac{1}{2} a$. The square base of both pyramids lies in the plane through the midpoints of the four remaining faces of the original cube. Each edge of this base is the hypotenuse of an isosceles right triangle with legs of length $\frac{1}{2} a$. Therefore, $b=\frac{\sqrt{2}}{2} a$ and $V_{2}=2 \cdot \frac{1}{3} b^{2} \cdot h=\frac{1}{6} a^{3}$.

If the midpoints of two adjacent edges of the square base of the two pyramids are connected, a segment of length $\frac{\sqrt{2}}{2} b=\frac{1}{2} a$ is formed. The centroid of the median from the apex of the pyramid to a side face of the octahedron divides this median in the ratio 2:1, so by the intercept theorem, the length $c$ of the segment between the centroids of adjacent octahedron faces is $\frac{c}{\frac{a}{2}}=\frac{2}{3}$ or $c=\frac{a}{3}$, and thus $V_{3}=\frac{1}{27} a^{3}$. Therefore,

$$
V_{1}: V_{2}: V_{3}=1: \frac{1}{6}: \frac{1}{27}=54: 9: 2
$$