The sum of three numbers is 19; the second number is the geometric mean of the first and third numbers, but at the same time, it is two-thirds of the third number. Which are these numbers?
The task yields the following three equations:

$$
x+y+z=19
$$

$$
y^{2}=x z
$$

$$
3 y=2 z
$$

If the value of $y$ is substituted into the first two equations, they take the following form:

$$
3 x+5 z=57
$$

$$
4 z^{2}=9 x z
$$

From the last equation,

$$
z=\frac{9 x}{4}
$$

Substituting this into (4) yields:

$$
\begin{gathered}
12 x+45 x=228 \\
57 x=228 \\
x=4
\end{gathered}
$$

From this value, (6) gives:

$$
z=9
$$

and (3) gives:

$$
y=6
$$

Thus, the three sought numbers are: 4, 6, and 9.

(Goldstein Zsigmond, gym. VII. grade student, Nyíregyháza).

The problem was also solved by: Ambrus József, Kaposvár; Berényi Győző, Székesfehérvár; Bőhm Ottó, Bpest; Friedmann Bernát, S.-A.-Ujhely; Geist Emil, Győr; Grünhut Béla, Pécs; Kiss Béla, Budapest; Kohn Márkus, Pécs; Pósch Gyula, Budapest; Schwartz Béla, Pécs; Visnya Aladár, Pécs;