IV OM - III - Task 3

Through each vertex of a tetrahedron of a given volume $ V $, a plane parallel to the opposite face of the tetrahedron has been drawn. Calculate the volume of the tetrahedron formed by these planes.
To make it easier to find a solution, let's first consider an analogous problem on the plane:
Let's draw through each vertex of the triangle $ABC$ a line parallel to the opposite side of the triangle. These lines form a triangle $A_1B_1C_1$ similar to triangle $ABC$ (Fig. 46). The center of similarity is the point $S$, where the lines $AA_1$, $BB_1$, $CC_1$ intersect; point $S$ is the centroid of triangle $ABC$, as well as of triangle $A_1B_1C_1$. The similarity is inverse, because corresponding points, e.g., $A$ and $A_1$, lie on the rays $SA$ and $SA_1$ in opposite directions. The ratio of similarity is

Since the ratio of the areas of similar figures equals the square of the ratio of similarity, the area of triangle $A_1B_1C_1$ is $4$ times larger than the area of triangle $ABC$, which is also directly visible.
Returning to the main topic, let's briefly recall the properties of similar figures in space. Let a figure $F$ be given in space, whose points are $A, B, C, \ldots$ Choose any point $S$ and a positive number $k$. When we measure segments $SA, SB, SC, \ldots$ on the rays $SA, SB, SC, \ldots$, the points $A$ form a figure $F$ directly similar to figure $F$ relative to point $S$ in the ratio $k$. If, however, we measure the same segments $SA_1 = k \cdot SA, SB_1 = k \cdot SB, SC_1 = k \cdot SC, \ldots$ on the extensions of the rays $SA, SB, SC, \ldots$, we obtain a figure $F_1$ inversely similar to figure $F$ relative to point $S$ in the ratio $k$ (Fig. 47). From this definition, it easily follows that the figure similar to the segment $AB$ is the segment $A_1B_1$, with the lines $AB$ and $A_1B_1$ being parallel (or coinciding, which happens when the line $AB$ passes through point $S$). The figure similar to the triangle $ABC$ is, therefore, the triangle $A_1B_1C_1$, with the planes of the triangles being parallel (or coinciding in the case where the plane $ABC$ contains point $S$). The figure similar to the tetrahedron $ABCD$ is the tetrahedron $A_1B_1C_1D_1$. The ratio of similar segments equals the ratio of similarity, the ratio of the areas of similar triangles (or generally: planar figures) equals the square of the ratio of similarity, and the ratio of the volumes of similar tetrahedra (or generally: solids) equals the cube of the ratio of similarity.
To solve the problem, we will also need the following theorem:
The segments connecting the vertices of a tetrahedron with the centroids of the opposite faces intersect at one point, which divides each of these segments in the ratio $3 \colon 1$.
The proof of this theorem is simple; it suffices to prove it for one pair of such segments. Let $M$ be the centroid of the face $BCD$, and $N$ be the centroid of the face $ACD$ of the tetrahedron $ABCD$ (Fig. 47). The lines $BM$ and $AN$ intersect at the midpoint $P$ of the edge $CD$.
The segments $AM$ and $BN$ connecting the vertices $A$ and $B$ of the triangle $ABP$ with the points $M$ and $N$ of the sides $BP$ and $AP$ intersect at a point $S$ inside this triangle. Since $AP = 3 \cdot NP$ and $BP = 3 \cdot MP$, it follows that $MN \parallel AB$ and $AB = 3 MN$; hence, $AS = 3 \cdot SM$ and $BS = 3 \cdot SN$, which is what we had to prove. The point $S$ is the centroid of the tetrahedron.
Applying the above knowledge, we can briefly state the solution to the problem: We construct the figure inversely similar in the ratio $3$ to the tetrahedron $ABCD$ relative to its centroid $S$. This figure is the tetrahedron $A_1B_1C_1D_1$, whose vertices lie on the extensions of the rays $SA$, $SB$, $SC$, $SD$, with $SA_1 = 3 \cdot SA$, $SB_1 = 3 \cdot SB$, $SC_1 = 3 \cdot SC$, $SD_1 = 3 \cdot SD$.
The plane $B_1C_1D_1$ is parallel to the corresponding plane $BCD$, and it passes through the vertex $A$, since from the equality $AS = 3 \cdot SM$ it follows that point $A$ is the point similar to point $M$ of the face $BCD$. Therefore, the tetrahedron $A_1B_1C_1D_1$ is the tetrahedron mentioned in the problem. Since it is similar to the tetrahedron $ABCD$ in the ratio $3$, its volume is $3^3 \cdot V$, or $27 V$.