For a rectangular parallelepiped, its volume $v$ is known, as well as the perimeter $k$ and the area $t$ of a triangle whose sides are equal to three mutually perpendicular edges of the body. We are looking for the distance $d$ between two opposite vertices of this body.
If $a$, $b$, and $c$ are the three edges of the body, then the distance between two opposite vertices is

$$
=\sqrt{a^{2}+b^{2}+c^{2}}
$$

According to the problem, we also have

$$
a b c=v
$$

$$
a+b+c=k
$$

$$
16 t^{2}=k(k-2 a)(k-2 b)(k-2 c)
$$

or

$$
16 t^{2}=k^{4}-2(a+b+c) k^{3}+4(a b+a c+b c) k^{2}-8 a b c k
$$

Substituting (2) and (3) into this, we get:

$$
16 t^{2}=k^{4}-2 k^{4}+4(a b+a c+b c) k^{2}-8 v k
$$

from which

$$
2(a b+a c+b c)=\frac{16 t^{2}+k^{4}+8 v k}{2 k^{2}}
$$

Squaring both sides of (3):

$$
a^{2}+b^{2}+c^{2}+2(a b+a c+b c)=k^{2}
$$

Subtracting (5) from this equation, we obtain:

$$
a^{2}+b^{2}+c^{2}=\frac{k^{4}-16 t^{2}-8 v k}{2 k^{2}}
$$

or

$$
d=\frac{1}{2 k} \sqrt{2\left(k^{4}-16 t^{2}-8 v k\right)}
$$

(Ervin Hofbauer.)

The problem was also solved by: Bauss O., Devecis M., Fekete J., Frankl J., Freund A., Freund E., Friedmann B., Goldstein Zs., Goldziher K., Jakobovics D., Kántor N., Kertész L., Klein M., Koffler K., Kornis Ö., Lipschitz J., Mayer G., Riesz F., Schwartz E., Spitzer Ö., Szabó I., Szigeth G.