What is the surface area of the tetrahedron whose vertices are one vertex of a cube with edge length a, and the midpoints of the three faces not containing this vertex?
By selecting any vertex of the cube, the tetrahedra determined by it and the midpoints of the faces not passing through it are congruent, so it is sufficient to determine the surface area of one such tetrahedron. Let the vertices of the cube be denoted as $A, B, C, D, E, F, G, H$ as shown in the diagram; the midpoints of the faces not passing through $F$ as $K, L, M$; and the midpoint of $K M$ as $R$.

Among the edges of the tetrahedron $K L M F$, $K M = M L = L K = \frac{\sqrt{2}}{2} a$, because these are the midlines of the $\sqrt{2} a$-sided triangle $A C H$. Furthermore, $K F = L F = M F$, because each of these segments connects the midpoint of one face of the cube to a vertex of the opposite face. The length of these segments can be calculated using the Pythagorean theorem, for example, $K F^{2} = K B^{2} + B F^{2} = \left(a \cdot \frac{\sqrt{2}}{2}\right)^{2} + a^{2} = \frac{3}{2} a^{2}$; thus, $K F = L F = M F = a \cdot \frac{\sqrt{6}}{2}$. The triangles $K M F$, $L M F$, and $L K F$ are congruent because their corresponding sides are equal. The height $F R$ of the triangle $K M F$ to the side $K M$ is also calculated using the Pythagorean theorem:

$$
F R^{2} = F K^{2} - R K^{2} = F K^{2} - \left(\frac{K M}{2}\right)^{2} = \frac{3}{2} a^{2} - \frac{1}{8} a^{2} = \frac{11}{8} a^{2}
$$

Therefore, the area of the triangle $T_{K M F} = \frac{1}{2} K M \cdot F R = a^{2} \cdot \frac{\sqrt{11}}{8}$. The area of the equilateral triangle $K L M$ is $T_{K L M} = K M^{2} \cdot \frac{\sqrt{3}}{4} = a^{2} \cdot \frac{\sqrt{3}}{8}$. Thus, the surface area of the tetrahedron is

$$
A = T_{K L M} + 3 T_{K M F} = \frac{a^{2}}{8} (\sqrt{3} + 3 \sqrt{11}) \approx 1.46 a^{2}
$$

Based on the work of Péter Huszár (Révkomárom, Selye J. Gymnasium, II. grade)