A parallelepiped has three edge vectors starting from one of its vertices, let this vertex be $O$, and denote these vectors by $a$, $b$, and $c$. Find the midpoint of the three faces passing through the opposite vertex, then find the centroid of the triangle formed by these midpoints. Express the vector from $O$ to the centroid in terms of $a$, $b$, and $c$.
The fourth vertex of the faces resting on the $\mathrm{Az} O$ is reached by the vectors $\mathbf{a}+\mathbf{b}, \mathbf{b}+\mathbf{c}, \mathbf{c}+\mathbf{a}$ from $O$, and the vector pointing to the vertex opposite to $O$ is $\mathbf{a}+\mathbf{b}+\mathbf{c}$. (See I. division textbook, p. 333.)

![](https://cdn.mathpix.com/cropped/2024_05_02_a0cc0af2769925f763f7g-1.jpg?height=756&width=987&top_left_y=262&top_left_x=558)

The vectors pointing from $O$ to the midpoints of the faces opposite to the vertex $O$ can be obtained as the arithmetic mean of the vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$, and $\mathbf{a}+\mathbf{b}+\mathbf{c}$:

$$
\frac{1}{2}(2 \mathbf{a}+\mathbf{b}+\mathbf{c}), \quad \frac{1}{2}(\mathbf{a}+2 \mathbf{b}+\mathbf{c}), \frac{1}{2}(\mathbf{a}+\mathbf{b}+2 \mathbf{c})
$$

Finally, the centroid of the triangle determined by these is given by their arithmetic mean (see II. division textbook, pp. 152-153):

$$
\mathbf{s}=\frac{2}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})
$$