9. In a regular tetrahedron $A B C D$, a line $M P$ is drawn through the midpoint $M$ of edge $A B$ and the centroid $P$ of face $B C D$. A line $N Q$ passes through the midpoint $N$ of edge $B C$ and the centroid $Q$ of face $A B D$. Calculate the angle between the lines $M P$ and $N Q$.
9. Let $a$ be the length of the edges of the tetrahedron, then $|M P|=\frac{a}{2}$, since this is the median of the right triangle $A B P$ (Fig. 96). Similarly, $|N Q|=\frac{a}{2}$, and also $|M N|=\frac{a}{2}$. In triangle $M N D$, points $P$ and $Q$ divide the sides in the ratio $2: 3$, so $|P Q|=\frac{2}{3}|M N|=\frac{a}{3}$. Extend the segment $M N$ beyond point $N$ by a segment $N R$ of length $|N R|=|P Q|$; we obtain a parallelogram $P Q N R$, in which $|P Q|=|N R|=\frac{a}{3}$ and $|N Q|=|P R|=\frac{a}{2}$, and the angle $\widehat{M P} R$ is equal to the angle between the lines $M P$ and $N Q$. Note that $|M R|=|M N|+|N R|=\frac{a}{2}+\frac{a}{3}=\frac{5}{6} a$. From triangle $M R P$ we have:

$$
|M R|^{2}=|M P|^{2}+|P R|^{2}-2|M P| \cdot|P R| \cos \varphi
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_5cdadc99c18e78f1502ag-161.jpg?height=80&width=414&top_left_y=1922&top_left_x=341)

$$
\frac{25}{36} a^{2}=\frac{a^{2}}{4}+\frac{a^{2}}{4}-\frac{a^{2}}{2} \cos \varphi, \cos \varphi=-\frac{7}{18}
$$

From this, we find: $\varphi \approx 112^{\circ} 53^{\prime} 06^{\prime \prime}$.