11.152. The centers of the faces of a regular tetrahedron serve as the vertices of a new tetrahedron (Fig. 11.48). Find the ratio of their surface areas and the ratio of their volumes.
## Solution.

Let $M, N, K, O$ be the centers of the faces $AFC, BFC, AFB, ABC$ of the given tetrahedron, respectively, and let $D$ be the midpoint of $AC$, and $E$ be the midpoint of $BC$. Then $FM = \frac{2}{3} FD, FN = \frac{2}{3} FE$, and consequently, $MN \| DE, MN = \frac{2}{3} DE = \frac{1}{2} AB$. Let $V_1$ and $S_1$ be the volume and surface area of the given tetrahedron, and $V_2$ and $S_2$ be the volume and surface area of the new tetrahedron. Since any two regular tetrahedra are similar, $\frac{V_1}{V_2} = \frac{AB^3}{MN^3} = 27:1; \frac{S_1}{S_2} = \frac{AB^2}{MN^2} = 9:1$.

Answer: $27:1; 9:1$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0860.jpg?height=714&width=565&top_left_y=77&top_left_x=81)

Fig. 11.48

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0860.jpg?height=565&width=618&top_left_y=174&top_left_x=658)

Fig. 11.49