A convex body bounded by planes has two planes of symmetry, the number of its vertices is 8, and in each of these, 3 edges meet. The lengths of the converging edges are 1, 1, and 0.8 units in two vertices, 1, 1, and 2 units in four vertices, and 2.2 and 3 units in two vertices. Calculate the surface area and volume of the body: 1$]$[^0]


[^0]:    ${ }^{1}$ A problem similar to this was the subject of our F. 1964. problem; see the solution in this issue, on page 64. - Ed.
Az F. 1964. problem solution's notation is used. What is stated there is valid until the length of the eighth edge is mentioned: "… thus, because $C D=1$, $G H<1$ would be, which is not possible."

Now the length of the $G H$ edge is 0.8, so $G$ and $H$ can also be on the $S$ symmetry plane. It is expected that the problem will have two solutions. Let's continue the reasoning according to the solution of F. 1964, let $G$ and $H$ be on the $T$ symmetry plane (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_a61c35da462c4cda7f4cg-1.jpg?height=648&width=681&top_left_y=343&top_left_x=711)

Figure 1

The triangle $C P G$ is similar to the triangle $A P Q$, $P G=\frac{1}{3} P Q$, so $P G=G H=H Q=0.8$. The solid $K_{1}$ thus obtained satisfies the requirements. Its volume is 25 times the volume of the tetrahedron $C D P G$:

$$
V_{1}=25 \cdot \frac{\sqrt{59}}{75}=2.56 \text { volume units. }
$$

The surface area of $K_{1}$ is obtained by a simple calculation:

$$
F_{1}=18 t_{C D G}+14 t_{C P G}=\frac{9}{2} \sqrt{3}+\frac{28}{25} \sqrt{21}=12.93
$$

area units.

Now let's try to construct the solid such that the $G H$ edge is on the $S$ symmetry plane (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_a61c35da462c4cda7f4cg-1.jpg?height=583&width=629&top_left_y=1649&top_left_x=737)

Figure 2

Drop perpendiculars from $A$ and $H$ to the intersection line of the $S$ plane and the $C D E F$ plane, and denote the intersection points by $A_{1}$ and $H_{1}$, respectively. Let the midpoint of $C E$ be $R$. $A_{1}=1, H_{1} R=0.1, A R=\sqrt{4-C R^{2}}, H R=\sqrt{1-C R^{2}}$. From the similarity of the triangles $A A_{1} R$ and $H H_{1} R$, $\sqrt{4-C R^{2}}=10 \sqrt{1-C R^{2}}$, so $C R=\frac{4 \sqrt{2}}{\sqrt{33}}$. The solid $K_{2}$ thus obtained also has the desired properties.

The surface area of $K_{2}$ is obtained by a simple calculation:

$F_{2}=2\left[t_{A B D C}+t_{C D G H}+t_{A C H E}\right]=2\left[2 \sqrt{3}+\frac{26 \sqrt{11}}{100}+\frac{4 \sqrt{2}}{3}\right]=12.49$ area units.

The solid $K_{2}$ is divided into two parts by the rectangle $C D E F$. Draw the segments $A_{2} A_{3}$ and $H_{2} H_{3}$ parallel to $C E$ through $A_{1}$ and $H_{1}$, respectively. The volume of the solid $C D E F G H$ can be obtained by adding the volume of the prism with base $H H_{2} H_{3}$ and height $G H$ to twice the volume of the pyramid $E C H_{2} H_{3} H$, and the volume of the solid $C D E F A B$ can be obtained by subtracting twice the volume of the pyramid $E C A_{2} A_{3} A$ from the volume of the prism with base $A A_{2} A_{3}$ and height $A B$. $A A_{1}=$ $10 H H_{1}=\sqrt{\frac{67}{33}}$, so

$$
\begin{gathered}
V_{C D E F G H}=\frac{4 \sqrt{2}}{\sqrt{33}} \cdot 0.1 \cdot \frac{\sqrt{67}}{\sqrt{33}} \cdot 0.8+2 \cdot \frac{0.1 \cdot \frac{8 \sqrt{2}}{\sqrt{33}} \cdot 0.1 \cdot \frac{\sqrt{67}}{\sqrt{33}}}{3}=\frac{28}{2475} \sqrt{134}, \\
V_{C D E F A B}=\frac{4 \sqrt{2}}{\sqrt{33}} \cdot \frac{\sqrt{67}}{\sqrt{33}} \cdot 3-2 \cdot \frac{1 \cdot \frac{8 \sqrt{2}}{\sqrt{33}} \cdot \frac{\sqrt{67}}{\sqrt{33}}}{3}=\frac{20}{99} \sqrt{134}
\end{gathered}
$$

volume units, so

$$
V_{2}=V_{C D E F G H}+V_{C D E F A B}=\frac{16}{75} \sqrt{134}=2.47 \text { volume units. }
$$

Tamás Gáti (Zalaegerszeg, Zrínyi M. Gymnasium, 4th grade)