One face of a cube is $A B C D$, and the other endpoints of the edges starting from its vertices are $A_{1}, B_{1}, C_{1}, D_{1}$, respectively. The midpoint of the edge $A_{1} B_{1}$ is $F$, and the center of the square face $A_{1} B_{1} C_{1} D_{1}$ is $K$. The volume of the frustum $A B C A_{1} F K$ is 13608 cubic units. What is the surface area of the frustum?
The convex polyhedron defined by the vertices $A, B, C, A_{1}, F$, and $K$ is indeed a truncated pyramid, because the extensions of the edges $A A_{1}$, $B F$, and $C K$ intersect at a single point, and the face $A_{1} F K$ is parallel to the face $A B C$. Indeed, the lines $A A_{1}$ and $B F$ intersect because they lie in the face $A B B_{1} A_{1}$ of the cube and are not parallel. Denoting their intersection point by $M$, the triangles $M A_{1} F$ and $M A B$ are similar because $A_{1} F \| A B$, and since $A_{1} F = A B / 2$, it follows that $M A_{1} = M A / 2$, hence $M A_{1} = A_{1} A$, and $M$ is the reflection of $A$ over $A_{1}$. The same holds for the intersection of the lines $A A_{1}$ and $C K$, since $C K$ lies in the diagonal plane $A C C_{1} A_{1}$ of the cube, so $A_{1} K$ is parallel to $A C$ and half its length, thus $C K$ also intersects $A A_{1}$ at $M$, as stated. Therefore, the truncated pyramid was formed by cutting the pyramid $M A B C = G$ with the pyramid $M A_{1} F K = G_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7261c0aeeeda309b0ba1g-1.jpg?height=594&width=680&top_left_y=516&top_left_x=709)

Let the edge of the cube $A B = a$, then the areas of the bases of $G$ and $G_{1}$ are $a^{2} / 2$ and $a^{2} / 8$, respectively, and their heights are $2a$ and $a$, respectively, so their volumes are

$$
V = \frac{a^{2}}{2} \cdot \frac{2a}{3} = \frac{a^{3}}{3}, \quad V_{1} = \frac{a^{2}}{8} \cdot \frac{a}{3} = \frac{a^{3}}{24}
$$

and the volume of the truncated pyramid is

$$
V_{\mathrm{cs}} = V - V_{1} = \frac{a^{3}}{3} - \frac{a^{3}}{24} = \frac{7a^{3}}{24} = 13608
$$

from which $a^{3} = 24 \cdot 1944 = 3 \cdot 2^{3} \cdot 9 \cdot 6^{3} = (2 \cdot 3 \cdot 6)^{3}, a = 36$ length units.

The side faces of the truncated pyramid are right trapezoids, because $A A_{1}$, being a cube edge, is perpendicular to the edges $A B$ and $A C$, and $F B \perp B C$ because $F B$ lies in the face $A B B_{1} A_{1}$ of the cube, which is perpendicular to the cube edge. Considering the lengths, $A C = a \sqrt{2}$, $A_{1} K = a \sqrt{2} / 2$, and from the right triangle $B F B_{1}$, $B F = a \sqrt{5} / 2$; thus, the areas of the side faces $A B F A_{1}$, $A C K A_{1}$, and $B C K F$ are respectively

$$
\frac{3a^{2}}{4}, \quad \frac{3\sqrt{2}a^{2}}{4}, \quad \frac{3\sqrt{5}a^{2}}{8}
$$

Finally, including the areas of the bases, which were already determined, the surface area is

$$
F_{\mathrm{cs}} = a^{2} \left( \frac{1}{2} + \frac{1}{8} + \frac{3}{4} + \frac{3\sqrt{2}}{4} + \frac{3\sqrt{5}}{8} \right) = \frac{a^{2}}{8} (11 + 6\sqrt{2} + 3\sqrt{5}) \approx a^{2} \cdot 3.274 \approx 4243 \text{ area units}
$$

Czédli Gábor (Baja, III. Béla Gimn., I. o. t.) Angyal József (Budapest, Berzsenyi D. Gimn., I. o. t.)

Remark. We could have also used the fact that the two pyramids are similar, with the ratio of their edges being $1: 2$, so $V_{1}: V = 1: 8$.