7*. In a model of a regular quadrilateral pyramid, all faces are made of uniform plates of the same material (and have the same thickness). The lateral edges of the pyramid are twice as long as the sides of the base. Determine the position of the center of mass of the pyramid, given the radius vectors of the vertex $S$ and the four vertices of the base $A_{1}, A_{2}, A_{3}, A_{4}$.
7. As it was done in planimetry, we will assume that the mass of the triangle is concentrated at its centroid (the point of intersection of the medians), and the mass of each face is proportional to its area. The area of the base (square) is taken as a unit, then the area of each lateral face is $\frac{\sqrt{15}}{4}$. The radius vectors of the centroids of the lateral faces are: $\frac{\vec{s}+\vec{A}_{1}+\vec{A}_{2}}{3} ; \quad \frac{\vec{s}+\vec{A}_{2}+\vec{A}_{3}}{3}$; $\frac{\vec{S}+\vec{A}_{3}+\vec{A}_{4}}{3} ; \frac{\vec{S}+\vec{A}_{4}+\vec{A}_{1}}{3}$, the radius vector of the center of the base is $\frac{\overrightarrow{A_{1}}+\vec{A}_{2}+\vec{A}_{3}+\vec{A}_{4}}{4}$.

Applying the general formula for the center of mass, we get:

$$
\begin{gathered}
\vec{P}=\frac{\frac{4 \vec{S}+2 \vec{A}_{1}+2 \vec{A}_{2}+2 \vec{A}_{3}+2 \overrightarrow{A_{4}}}{3} \cdot \frac{V 15}{4}+\frac{\overrightarrow{A_{1}}+\overrightarrow{A_{2}}+\vec{A}_{3}+\vec{A}_{4}}{4}}{V \overline{15}+1} \\
\vec{P}=\frac{4 \vec{S} \sqrt{15}+(2 \sqrt{15}+3)\left(\overrightarrow{A_{1}}+\overrightarrow{A_{2}}+\overrightarrow{A_{3}}+\overrightarrow{A_{4}}\right)}{12(\sqrt{15}+1)}
\end{gathered}
$$