5. Monika made a paper model of a triangular pyramid, whose base was a right-angled triangle. When she cut the model along the legs of the base and along the median of one of the sides, a square with side a was formed upon unfolding it into a plane. Determine the volume of this pyramid.

(P. Leischner)
Solution. Let $ABCD$ be the pyramid with base $ABC$, where $|\angle ACB| = 90^\circ$. According to the problem, the model was cut along both legs $AC$ and $BC$ of the base and further along the median from vertex $D$ of one of the faces $BCD$, $ACD$. A cut along the median in face $ABD$ would not allow the model to be unfolded into a plane. Without loss of generality, assume that the cut is along the median $DE$ in face $ACD$ (Fig. 7), and upon unfolding into a plane, a figure with boundary $BCAE_2DE_1C_1B$ and a right angle at vertex $C$ (Fig. 8) is formed. Since this figure is a square (denote it as $\mathcal{C}$), the angles $A E_2 D$ and $D E_1 C_1$ are right (neither can be straight, as their sum is $180^\circ$). Therefore, the median $DE$ of triangle $ACD$ is also its altitude, and points $E_1$, $E_2$ are vertices of square $\mathcal{C}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_cfce92b55b42ad4c54c3g-06.jpg?height=583&width=466&top_left_y=1882&top_left_x=450)

Fig. 7

![](https://cdn.mathpix.com/cropped/2024_04_17_cfce92b55b42ad4c54c3g-06.jpg?height=512&width=523&top_left_y=1956&top_left_x=1126)

Fig. 8

If the vertices $E_1$ and $E_2$ of square $\mathcal{C}$ were adjacent, from the equality $|E_1 C_1| = |E_2 A|$ and the fact that $\mathcal{C}$ has vertex $C$, it would follow that $\mathcal{C} = C E_2 E_1 B$, and thus $|BC| = |BE_1|$, which contradicts the equality $|BC| = |BC_1|$ (Fig. 9). Thus, we have (by contradiction) proven that the vertices $E_1$ and $E_2$ of square $\mathcal{C}$ are not adjacent, so the vertices of $\mathcal{C}$ must necessarily include point $D$ (from the segment $E_2 D E_1$ of the boundary $BCAE_2DE_1C_1B$).

The described points divide the boundary of square $\mathcal{C} = C E_2 D E_1$ into segments, whose lengths are marked in Fig. 8 using the convenient notation $x = \frac{1}{3} a$. The lengths of the other edges of the pyramid are calculated using the Pythagorean theorem:

\[
\begin{gathered}
|DC| = |DA| = \sqrt{(3x)^2 + (x)^2} = x \sqrt{10} \\
|DB| = \sqrt{(3x)^2 + (2x)^2} = x \sqrt{13}, \quad |AB| = x \sqrt{5}
\end{gathered}
\]

To determine the volume of the pyramid $ABCD$, we need to find the length of its height.

If we denote $F$ as the midpoint of edge $AB$, we see that edge $AC$ is perpendicular to the plane $EFD$, because $AC \perp BC \parallel EF$ and $AC \perp DE$. The plane $EFD$ is therefore perpendicular to the base $ABC$.

The height of the pyramid is thus the altitude (from vertex $D$) of triangle $DEF$. Since $DF$ is the median of triangle $ABD$, from the known formula for the length of the median, we get

\[
2|DF|^2 = |DA|^2 + |DB|^2 - \frac{1}{2}|AB|^2 = \frac{41}{2} x^2
\]

![](https://cdn.mathpix.com/cropped/2024_04_17_cfce92b55b42ad4c54c3g-07.jpg?height=412&width=415&top_left_y=672&top_left_x=1511)

Fig. 9

so the sides of triangle $DEF$ have lengths $|DF| = \frac{1}{2} x \sqrt{41}$, $|DE| = 3x$, and $|EF| = \frac{1}{2}|BC| = \frac{1}{2} x$. According to Heron's formula, the area $S$ of such a triangle is

\[
\begin{aligned}
S & = \frac{x^2}{4} \sqrt{\left(3 + \frac{1}{2} + \frac{\sqrt{41}}{2}\right)\left(3 + \frac{1}{2} - \frac{\sqrt{41}}{2}\right)\left(3 + \frac{\sqrt{41}}{2} - \frac{1}{2}\right)\left(\frac{\sqrt{41}}{2} + \frac{1}{2} - 3\right)} = \\
& = \frac{x^2}{16} \sqrt{(7 + \sqrt{41})(7 - \sqrt{41})(5 + \sqrt{41})(\sqrt{41} - 5)} = \frac{x^2 \sqrt{2}}{2},
\end{aligned}
\]

and thus its height $v$ from vertex $D$ is $v = \frac{2S}{|EF|} = 2x \sqrt{2}$. The volume $V$ of the pyramid $ABCD$ is therefore

\[
V = \frac{1}{3} \cdot \frac{1}{2}|AC| \cdot |BC| \cdot v = \frac{2 \sqrt{2}}{3} x^3 = \frac{2 \sqrt{2}}{81} a^3.
\]

Conclusion: The volume of the considered pyramid is $\frac{2 \sqrt{2}}{81} a^3$.

## Note.

A tetrahedron $ABCD$ with the required properties can be created from a square in the described manner if the sum of two of the three assumed face angles at vertex $D$ is greater than the third angle. Since their sum is $90^\circ$, it suffices to verify that each of these three angles is less than $45^\circ$. The inequality $|\angle CDB| < 45^\circ$ is satisfied because $\operatorname{tg}|\angle CDB| = \frac{2}{3} < 1$, $\operatorname{tg}|\angle CDA| = \frac{3}{4} < 1$, and $\operatorname{tg}|\angle ADB| = \frac{3}{4} < 1$.