The base and top faces of a cube are $A B C D$ and $E F G H$, and its side edges are $A E, B F, C G, D H$. Consider the convex body whose vertices are $A, B, C$, as well as the midpoints of the edges $A D, A E, B F, C D, C G, E F, E H$, and the centers of the faces $C D H G$ and $E F G H$ (12 vertices). Calculate the surface area, volume, and the length of the longest diagonal of this body.
Let the midpoints of the 7 edges listed be $J, K, L, M, N, P, Q$, and the midpoints of the faces $C D H G$ and $E F G H$ be $R$ and $S$, respectively. We will determine the areas of the boundary faces of the solid $T$ one by one. Let the edge length of the cube be denoted by $c$.

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The base face $A B C M J$ is formed by removing an isosceles right triangle with leg length $c / 2$ from the base square - let's denote such a triangle by $H_{0}$ - and the top face $P Q S$ is exactly such an $H_{0}$ triangle. Therefore, the sum of the areas of these two faces is equal to the area of the square.

Similarly, the combined area of the faces $A B L P K$ and $C N R M$ is equal to the area of one face of the cube, because the former is formed by removing two $H_{0}$ triangles from one face of the cube, and the latter can be divided into two such triangles.

The rectangle $B C N L$ is half of a cube face, and the trapezoid $A J Q K$ is formed by removing an $H_{0}$ triangle from half a square. Since the area of the $H_{0}$ triangle is one-eighth of a cube face, the total surface area of the faces described so far is $(3 \cdot 1/8) c^{2} = 23 c^{2} / 8$.

The points $J, M, R, Q$ lie in a plane and form a trapezoid because $M R \| J Q$, and the points $L, P, S, N$ form a congruent trapezoid because $L N \| P S$, and $L N = J Q = c$, $P S = M R = c / 2$, and the sides $L P$ and $J M$ are perpendicular to the parallel sides and are the hypotenuses of $H_{0}$ triangles, so their length is $c / \sqrt{2}$. The sum of the areas of these two trapezoids is therefore $(3 / 2) c \cdot c / \sqrt{2} = 3 c^{2} / 2 \sqrt{2}$.

The points $N, R, Q, S$ form a parallelogram because $R N \| S Q$, and both have a length of $c / 2$. One of the heights of the parallelogram is $R S$, which lies in the plane that bisects the cube and is perpendicular to $R N$ and $S Q$. $R S = c / \sqrt{2}$, so the area of the parallelogram is $c^{2} / 2 \sqrt{2}$.

Finally, the triangle $K P Q$ is equilateral with side length $c / \sqrt{2}$, so its area is $c^{2} \sqrt{3} / 8$.

Summarizing our results, the surface area of $T$ is

$$
f = \frac{c^{2}}{8}(23 + 8 \sqrt{2} + \sqrt{3}) \approx 4.506 c^{2}
$$

To calculate the volume, we cut the solid $T$ with the plane $K L N$, which passes through $R$ and bisects the edge $J Q$ at $U$. The lower part of $T$ is a pentagonal prism, and its volume, using the area of the base face $A B C M J$, is $T_{1} = 7 c^{3} / 16$. The upper part of $T$ can be completed to a quadrilateral frustum by adding the pyramids $E P Q K$ and $R U V Q$, where $V$ is the midpoint of the edge $D H$. Thus, its volume is

$$
T_{2} = \frac{c}{6}\left(c^{2} + \frac{c^{2}}{4} + \frac{c^{2}}{2}\right) - 2 \frac{c}{6} \cdot \frac{c^{2}}{8} = \frac{c^{3}}{4}
$$

and the total volume is $11 c^{3} / 16$.

The diagonals from the points $K, L, N$, and $R$ to $P, Q, S$ are longer than those from $A, B, C$, and $M$ to the same points. Therefore, we need to look for the longest diagonal or diagonals among the points on the base. Considering the vertices on opposite sides of the plane $B D H F$, the diagonals $A N, C K, C P, C Q$ come into consideration, and each has a length of $3 c / 2$; considering the vertices on opposite sides of the plane $M R S P$, again $A N, C K, C Q$, and $B Q$ come into consideration, and the latter also has a length of $3 c / 2$. The calculation shows that there is no longer diagonal in the solid than these 5 equal-length diagonals.

Péter Vadász (Budapest, Kölcsey F. g. III. o. t.)

Note. In the solutions, several other decompositions of $T$ were used to calculate the volume. One mechanical but well-organized decomposition divided the cube into eight smaller cubes of edge length $c / 2$ using the perpendicular bisectors of the edges $A B, A D$, and $A E$, and calculated the parts of $T$ that fell into these smaller cubes.