Two balls, the larger one has a mass of $M$, the smaller one has a mass of $m$. Let $A$ be a point within the segment defined by the centers of the two balls, and $B$ be a point outside this segment, chosen such that both balls exert equal gravitational force on a unit mass placed at this point, according to Newton's law. If the distance between these two points is $d$: it needs to be determined how far apart the centers of the two balls are. What will this distance be if $M=2116 \mathrm{~kg}, m=16 \mathrm{~kg}, d=1.15 \mathrm{~m}$.
Let $O$ and $O_{1}$ be the centers of the balls with masses $M$ and $m$, respectively. According to Newton's law,

$$
\frac{M}{\overline{A O}^{2}}=\frac{m}{{\overline{A O_{1}}}^{2}} \quad \text { and } \quad \frac{M}{\overline{B O}^{2}}=\frac{m}{{\overline{B O_{1}}}^{2}}
$$

From the second equation,

$$
B O=\frac{B O_{1} \cdot \sqrt{M}}{\sqrt{m}}
$$

thus,

$$
d+A O=\left(d-O A_{1}\right) \frac{\sqrt{M}}{\sqrt{m}}
$$

from which we get

$$
O A_{1}=\frac{d(\sqrt{M}-\sqrt{m})-A O \sqrt{m}}{\sqrt{M}}
$$

Substituting this into the first equation:

$$
\frac{\sqrt{M}}{A O}=\frac{\sqrt{M m}}{d(\sqrt{M}-\sqrt{m})-A O \sqrt{m}}
$$

from which we get

$$
A O=\frac{d}{2} \cdot \frac{\sqrt{M}-\sqrt{m}}{\sqrt{m}}
$$

and thus

$$
A O_{1}=\frac{d}{2} \cdot \frac{\sqrt{M}-\sqrt{m}}{\sqrt{M}}
$$

and therefore

$$
O O_{1}=A O+A O_{1}=\frac{d}{2} \cdot \frac{M-m}{\sqrt{M m}}=6.56 \mathrm{~m}
$$

(Géza Sasvári, Pécs.)

The problem was also solved by: Czank K., Demeter J., Faith F., Kerekes T., Krausz B., Krisztián Gy., Lukhaub Gy., Lupsa Gy., Perl Gy., Szmodics K., Spitzer H.