4. In a parallelepiped whose faces are congruent rhombi with an angle of $60^{\circ}$, the largest diagonal $d$ is given. Determine the diagonals of this body.
Solution. Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be the given parallelepiped and let

$$
\measuredangle B A D=\measuredangle D A A_{1}=\measuredangle A_{1} A B=60^{\circ}
$$

We introduce a rectangular coordinate system such that:

1) the origin $O$ coincides with point $A$,
2) point $B$ lies on the positive part of the $x$-axis,

![](https://cdn.mathpix.com/cropped/2024_06_05_a3fa2c1df73614920048g-2.jpg?height=295&width=442&top_left_y=780&top_left_x=1038)

3) point $D$ lies in the $x O y$ plane and has a positive $y$-coordinate,
4) point $A_{1}$ has a positive $z$-coordinate (see the diagram on the right).

Let $a=A B$. Then $A B D$ is an equilateral triangle with height $\frac{a \sqrt{3}}{2}$, and $A B D A_{1}$ is a regular tetrahedron with height $\frac{a \sqrt{6}}{3}$. Now it is easy to obtain that the coordinates of all vertices of the parallelepiped are:

$$
\begin{aligned}
& A(0,0,0), \quad B(a, 0,0), C\left(\frac{3 a}{2}, \frac{a \sqrt{3}}{2}, 0\right), \quad D\left(\frac{a}{2}, \frac{a \sqrt{3}}{2}, 0\right) \\
& A_{1}\left(\frac{a}{2}, \frac{a \sqrt{3}}{6}, \frac{a \sqrt{6}}{3}\right), \quad B_{1}\left(\frac{3 a}{2}, \frac{a \sqrt{3}}{6}, \frac{a \sqrt{6}}{3}\right), \\
& C_{1}\left(2 a, \frac{2 a \sqrt{3}}{3}, \frac{a \sqrt{6}}{3}\right), \quad D_{1}\left(a, \frac{2 a \sqrt{3}}{3}, \frac{a \sqrt{6}}{3}\right) .
\end{aligned}
$$

Furthermore,

$$
d^{2}=A C_{1}^{2}=(2 a)^{2}+\left(\frac{2 a \sqrt{3}}{3}\right)^{2}+\left(\frac{a \sqrt{6}}{3}\right)^{2}=6 a^{2}
$$

from which we get $a=\frac{d}{\sqrt{6}}$. Furthermore, it is easy to obtain that the lengths of the diagonals of the parallelepiped are

$$
B D_{1}=D B_{1}=C A_{1}=\frac{d}{\sqrt{6}}
$$

## 4th year