5.14. In the pyramid $S A B C$, all faces are equilateral triangles; point $M$ is the center of triangle $A B C$, and point $P$ divides the edge $S C$ in half. Find the decomposition of the vector $\overline{M P}$ in terms of the vectors $\overline{A B}, \overline{A C}$, and $\overline{A S}$.
5.14. We have (Fig. 5.8) $\overline{M P}=\overline{M C}-\overline{P C}$, where $\overline{P C}=\frac{1}{2} \overline{S C}=$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-603.jpg?height=394&width=419&top_left_y=1426&top_left_x=46)

Fig. 5.8

$$
=\frac{1}{2}(\overline{A C}-\overline{A S})
$$

$\overline{M P}=\overline{M C}-\frac{1}{2}(\overline{A C}-\overline{A S})$. Now let's find $\overline{M C}$. In the equilateral triangle $A B C$ we have $M C=\frac{2}{3} C N$, where $C N$ is the height of the triangle; therefore,
$\overline{M C}=\frac{2}{3} \overline{N C} . \quad$ But $\quad \overline{N C}=\overline{A C}-\overline{A N}=\overline{A C}-\frac{1}{2} \overline{A B} \quad$ and, $\quad$ hence, $\overline{M C}=\frac{2}{3}\left(\overline{A C}-\frac{1}{2} \overline{A B}\right)=\frac{2}{3} \overline{A C}-\frac{1}{3} \overline{A B}$. Thus, we finally get

$$
\overline{M P}=\frac{2}{3} \overline{A C}-\frac{1}{3} \overline{A B}-\frac{1}{2} \overline{A C}+\frac{1}{2} \overline{A S}=\frac{1}{6} \overline{A C}-\frac{1}{3} \overline{A B}+\frac{1}{2} \overline{A S}
$$

Answer: $\overline{M P}=\frac{1}{6} \overline{A C}-\frac{1}{3} \overline{A B}+\frac{1}{2} \overline{A S}$.