Problem 1273

Solution. Let the vertices of the tetrahedron be denoted by $A, B, C, D$; the mirror images mentioned in the problem are denoted by $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $D^{\prime}$ respectively.

Let the length of the median from vertex $A$ be $4a$, with its foot at $S_{a}$, and the length of the median from vertex $B$ be $4b$, with its foot at $S_{b}$, and so on.

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A median of the tetrahedron connects a vertex to the centroid of the opposite face, so the segments $A A^{\prime}$ and $B B^{\prime}$ lie on the lines of the medians of the tetrahedron. By the properties of reflection: $A S_{a}=S_{a} A^{\prime}=4a$ and $B S_{b}=S_{b} B^{\prime}=4b$.

It is known that the medians of a tetrahedron intersect at a single point (this is the centroid $S$ of the tetrahedron), and this point is the point that divides the medians in the ratio 3:1 from the vertex; thus, $A S=3a$, $B S=3b$, and $S A^{\prime}=5a$, $S B^{\prime}=5b$.

Therefore, the triangles $A S B$ and $A^{\prime} S B^{\prime}$ are similar, since $\angle A S B = \angle A^{\prime} S B^{\prime}$ (vertex angles) and the ratios of two pairs of corresponding sides are equal: $\frac{A S}{A^{\prime} S}=\frac{B S}{B^{\prime} S}=\frac{3}{5}$. By the converse of the parallel intercept theorem, $A B \| A^{\prime} B^{\prime}$ and the ratio of these segments is also equal to the ratio of the similarity of the two triangles, $\frac{3}{5}$.

Similarly, it can be shown that the other edges of the tetrahedron $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are parallel to the corresponding edges of the tetrahedron $A B C D$, and their lengths are $\frac{5}{3}$ times the lengths of the original tetrahedron's edges. Therefore, the tetrahedron $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is a $\frac{5}{3}$-scaled, $S$-centered enlargement of the tetrahedron $A B C D$, and the two tetrahedra are similar.

The ratio of their volumes is the cube of the similarity ratio:

$$\frac{V_{A^{\prime} B^{\prime} C^{\prime} D^{\prime}}}{V_{A B C D}}=\left(\frac{5}{3}\right)^{3}=\frac{125}{27}>4$$

Thus, the volume of the tetrahedron defined by the reflections is indeed more than four times the volume of the original tetrahedron.