A point light source placed at point $M$ of an inclined plane illuminates three points $O, A$, and $B$ on a horizontal plane in such a way that the intensity of illumination at these points is proportional to the numbers $\frac{1}{9}, \frac{1}{4}, 1$; the position of point $M$ is to be found, by determining the coordinates $O H=x$ and $M H=y$. $O A=a=200 \mathrm{~m}, O B=b=100 \mathrm{~m}$.
Based on the task, it can be written that:

$$
\frac{\overline{M O}^{2}}{9}=\frac{\overline{M A}^{2}}{4}=\frac{\overline{M B}^{2}}{1}
$$

But

$$
\begin{gathered}
\overline{M O}^{2}=x^{2}+y^{2} \\
\overline{M A}^{2}=(a-x)^{2}+y^{2} \\
\overline{M B}^{2}=(x-b)^{2}+y^{2}
\end{gathered}
$$

Thus,

$$
\frac{x^{2}+y^{2}}{9}=\frac{(a-x)^{2}+y^{2}}{4}=\frac{(x-b)^{2}+y^{2}}{1} \text {. }
$$

Solving these equations:

$$
x=\frac{8 a^{2}-5 b^{2}}{2(8 a-5 b)}
$$

Knowing \( x \), we get:

$$
y=\sqrt{-x^{2}+\frac{18}{5} a x-\frac{9}{5} a^{2}} .
$$

Substituting the given values, we obtain:

$$
\begin{aligned}
x & =122.72 \mathrm{~m} \\
y & =36.07 \mathrm{~m}
\end{aligned}
$$

(Odön Kornis.)

The problem was also solved by: B. Krausz, Gy. Krisztian, G. Sasvari.