In the tetrahedron $A B C D$, we reflect vertex $A$ over $B$, $B$ over $C$, $C$ over $D$, and $D$ over $A$. The resulting points are respectively: $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$. How many times the volume of the tetrahedron $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the volume of the original tetrahedron $A B C D$?
Solution. We solve the problem using the scalar triple product of vectors. Let the vectors from $A$ to the vertices $B, C, D$ be $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$, respectively. Then,

$$
V_{A B C D}=\frac{1}{6}(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{d}
$$

The vectors from $A$ to the vertices $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ are:

$$
\begin{array}{ll}
\overrightarrow{A A^{\prime}}=2 \mathbf{b}, & \overrightarrow{A B^{\prime}}=2 \mathbf{c}-\mathbf{b} \\
\overrightarrow{A C^{\prime}}=2 \mathbf{d}-\mathbf{c}, & \overrightarrow{A D^{\prime}}=-\mathbf{d}
\end{array}
$$

From this, we have:

$$
\begin{aligned}
& \overrightarrow{D^{\prime} A^{\prime}}=\overrightarrow{A A^{\prime}}-\overrightarrow{A D^{\prime}}=2 \mathbf{b}+\mathbf{d} \\
& \overrightarrow{D^{\prime} B^{\prime}}=\overrightarrow{A B^{\prime}}-\overrightarrow{A D^{\prime}}=2 \mathbf{c}-\mathbf{b}+\mathbf{d} \\
& \overrightarrow{D^{\prime} C^{\prime}}=\overrightarrow{A C^{\prime}}-\overrightarrow{A D^{\prime}}=3 \mathbf{d}-\mathbf{c}
\end{aligned}
$$

Thus, the volume of the new tetrahedron is:

$$
V_{A^{\prime} B^{\prime} C^{\prime} D^{\prime}}=\frac{1}{6}\left(\overrightarrow{D^{\prime} A^{\prime}} \times \overrightarrow{D^{\prime} C^{\prime}}\right) \cdot \overrightarrow{D^{\prime} B^{\prime}}=\frac{1}{6}[(2 \mathbf{b}+\mathbf{d}) \times(3 \mathbf{d}-\mathbf{c})] \cdot(2 \mathbf{c}-\mathbf{b}+\mathbf{d})
$$

Using the properties of the scalar triple product and the fact that if two of the three vectors are parallel, their scalar triple product is 0, performing the operations, we get:

$$
V_{A^{\prime} B^{\prime} C^{\prime} D^{\prime}}=15 \cdot \frac{1}{6}(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{d}=15 \cdot V_{A B C D}
$$

Therefore, the volume of the new tetrahedron is 15 times the volume of the original tetrahedron.