Two regular tetrahedrons have the sum of two edges as $a$, and the sum of their volumes is $b$; how long are the edges?
If the two tetrahedra have edges $x$ and $y$, then according to the problem:

$$
x+y=a
$$

and

$$
\frac{x^{3}}{12} \sqrt{2}+\frac{y^{3}}{12} \sqrt{2}=b
$$

or

$$
x^{3}+y^{3}=6 b \sqrt{2}
$$

Dividing (3) by (1):

$$
x^{2}-x y+y^{2}=\frac{6 b \sqrt{2}}{a}
$$

Squaring (1), we get:

$$
x^{2}+2 x y+y^{2}=a^{2}
$$

Subtracting (4) from this equation:

$$
\begin{aligned}
& 3 x y=a^{2}-\frac{6 b \sqrt{2}}{a} \\
& x y=\frac{a^{2}}{3}-\frac{2 b \sqrt{2}}{a}
\end{aligned}
$$

Subtracting 4 times the product of $x y$ from (5):

$$
\begin{aligned}
(x-y)^{2} & =a^{2}-\frac{4 a^{2}}{3}+\frac{8 b \sqrt{2}}{a} \\
x-y & =\sqrt{\frac{8 b \sqrt{2}}{a}-\frac{a^{2}}{3}}
\end{aligned}
$$

From (1) and (7), we finally obtain:

$$
\begin{aligned}
& x=\frac{1}{2}\left(a+\sqrt{\frac{8 b \sqrt{2}}{a}-\frac{a^{2}}{3}}\right) \\
& y=\frac{1}{2}\left(a-\sqrt{\frac{8 b \sqrt{2}}{a}-\frac{a^{2}}{3}}\right)
\end{aligned}
$$

The problem was solved by: Béla Bálint, Hugó Berger, Mór Feuer, Antal Freund, Bernát Friedmann, Zsigmond Goldstein, Károly Goldziher, Béla Grünhut, Ervin Hofbauer, Nándor Kántor, Mór Klein, Frigyes Riesz, István Szabó, István Szita.