2. Determine the edges of the cuboid if the sum of its edges $s$ and its diagonal $d$ are known, and one edge is the geometric mean of the other two edges.
Solution. Let $a, b, c$ be the edges of the cuboid and let $c=\sqrt{a b}$. According to the condition of the problem

$$
a+b+\sqrt{a b}=s, a^{2}+b^{2}+a b=d^{2}
$$

from which we get

$$
(s-\sqrt{a b})^{2}=d^{2}+a b \text { and } \sqrt{a b}=\frac{s^{2}-d^{2}}{2 s},
$$

where it must hold that $s>d$. Further, it is easily obtained that

$$
a b=\left(\frac{s^{2}-d^{2}}{2 s}\right)^{2}, \quad a+b=\frac{s^{2}+d^{2}}{2 s}
$$

Thus, $a$ and $b$ are solutions to the quadratic equation

$$
t^{2}-\frac{s^{2}+d^{2}}{2 s} t+\left(\frac{s^{2}-d^{2}}{2 s}\right)^{2}=0
$$

in the unknown $t$. This equation has solutions if and only if

$$
\left(\frac{s^{2}+d^{2}}{2 s}\right)^{2}-4\left(\frac{s^{2}-d^{2}}{2 s}\right)^{2} \geq 0
$$

i.e., if and only if $3 d^{2} \geq s^{2}$. Finally, under the condition $d<s \leq d \sqrt{3}$, we obtain the desired numbers

$$
\frac{s^{2}+d^{2}+\sqrt{\left(3 d^{2}-s^{2}\right)\left(3 s^{2}-d^{2}\right)}}{4 s}, \frac{s^{2}+d^{2}-\sqrt{\left(3 d^{2}-s^{2}\right)\left(3 s^{2}-d^{2}\right)}}{4 s}, \frac{s^{2}-d^{2}}{2 s}
$$