5. From a piece of paper, an isosceles trapezoid $C_{1} A B_{2} C_{2}$ with the shorter base $B_{2} C_{2}$ was cut out. The foot of the perpendicular from the midpoint $D$ of the leg $C_{1} C_{2}$ to the base $A C_{1}$ is denoted as $B_{1}$. After folding the paper along the segments $D B_{1}, A D$ and $A C_{2}$, the points $C_{1}, C_{2}$ moved in space to a single point $C$ and the points $B_{1}, B_{2}$ to a point $B$. This resulted in a model of a tetrahedron $A B C D$ with a volume of $64 \mathrm{~cm}^{3}$. Determine the lengths of the sides of the original trapezoid. (P. Leischner)
Solution. From the equality of segments in the described network that correspond to the same edges of the resulting tetrahedron $ABCD$, we get that $\left|AB_{1}\right|=\left|AB_{2}\right|=|AB|=b, \left|B_{1}C_{1}\right|=\left|B_{2}C_{2}\right|=|BC|=c$. Let $S$ be the midpoint of segment $AB_{1}$ and $B_{3}$ the foot of the perpendicular from point $D$ to line $B_{2}C_{2}$ (Fig. 2). Triangles $B_{1}C_{1}D$ and $B_{3}C_{2}D$ are centrally symmetric with respect to point $D$, so $\left|B_{3}C_{2}\right|=\left|B_{1}C_{1}\right|=c$. Since trapezoid $C_{1}AB_{2}C_{2}$ is isosceles, triangle $B_{1}AB_{2}$ is also isosceles (and, given the previous equalities, it is even equilateral), and from rectangle $B_{1}SB_{2}B_{3}$ it follows that $\frac{1}{2}b=\left|B_{1}S\right|=\left|B_{2}B_{3}\right|=2c$, so $b=4c$.

![](https://cdn.mathpix.com/cropped/2024_04_17_165246d7974c4c148f8fg-6.jpg?height=542&width=681&top_left_y=729&top_left_x=321)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_04_17_165246d7974c4c148f8fg-6.jpg?height=491&width=610&top_left_y=782&top_left_x=1145)

Fig. 3

Now, we just need to realize that the assembled tetrahedron $ABCD$ will have two right-angled faces $CDB$ and $ADB$ with right angles at vertex $B$ (Fig. 3), which means that edge $BD$ will be perpendicular to face $ABC$. The height $v$ of triangle $ABC$ (or triangle $AB_{2}C_{2}$) to side $BC$ is also the height $B_{2}S$ of the equilateral triangle $B_{1}AB_{2}$, so $v=\frac{1}{2}\sqrt{3}b$ and $|BD|=\left|B_{1}D\right|=\frac{1}{2}v$. The volume $V$ of tetrahedron $ABCD$ is thus calculated as

$$
\begin{aligned}
V & =\frac{1}{3} S(ABC)|BD|=\frac{1}{3} S\left(AB_{2}C_{2}\right) \cdot \frac{1}{2} v=\frac{1}{3} \cdot \frac{1}{2} c v \cdot \frac{1}{2} v= \\
& =\frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{4} b \cdot v^{2}=\frac{1}{4^{3}} b^{3}
\end{aligned}
$$

so $b=\sqrt[3]{64^{2}} \text{ cm}=16 \text{ cm}$.