Let $A$ and $B$ be two light sources, the luminous intensity of which is in the ratio $m: n$. Where is the point $P$ on the line $AB$ that is equally illuminated by the two light sources? Furthermore, where is the point $P'$, which is $b$ units away from $AB$ at a distance $\mathrm{s}$, and is also equally illuminated by the two light sources?

(Szontágh Gusztáv, a teacher at the main real school in Brassó.)
Let the distance between the two light sources be denoted by $a$; their luminous powers be $F_{1}$ and $F_{2}$. Let the distance of point $P$ from $A$ be denoted by $x$, then from $B$ it is naturally $(a-x)$.

$$
F_{1}: F_{2}=m: n
$$

The illumination at point $P$ from light source $A$ is

$$
J_{1}=\frac{F_{1}}{x^{2}}
$$

and from the two, according to the condition, they must be equal, that is

$$
\text{from } B, \, J_{2}=\frac{F_{2}}{(a-x)^{2}}
$$

$$
\begin{gathered}
J_{1}=J_{2} \\
F_{1}: x^{2}=F_{2}:(a-x)^{2} \\
F_{1}: F_{2}=x^{2}:(a-x)^{2}
\end{gathered}
$$

By equating (1) and (2), we get:

$$
x^{2}:(a-x)^{2}=m: n
$$

and from this

$$
m(a-x)^{2}=n x^{2}
$$

factoring out

$$
a^{2} m-2 a m x+m x^{2}=n x^{2}
$$

rearranging

$$
\begin{gathered}
(m-n) x^{2}-2 a m x+a^{2} m=0 \\
x=\frac{2 a m \pm \sqrt{4 a^{2} m^{2}-4(m-n) a^{2} m}}{2(m-n)} \\
x=a \frac{m \pm \sqrt{m n}}{2(m-n)}
\end{gathered}
$$

In the second part of the problem, let the distance of point $P^{\prime}$ from light source $A$ be $y$, and from $B$ be $z$. Assuming again that $J_{1}=J_{2}$, we get

$$
F_{1}: F_{2}=y^{2}: z^{2}=m: n
$$

Let the projection of the length of $y$ onto the line connecting the two light sources be $x$, and the projection of $z$ be $(a-x)$, which we obtain by drawing a perpendicular from point $P^{\prime}$ to $a$.

Then

$$
\begin{gathered}
y^{2}=b^{2}+x^{2} \\
z^{2}=b^{2}+(a-x)^{2}
\end{gathered}
$$

Substituting these values into (1), we get

$$
\begin{gathered}
\left(b^{2}+x^{2}\right):\left(b^{2}+a^{2}-2 a x+x^{2}\right)=m: n \\
m\left(b^{2}+a^{2}-2 a x+x^{2}\right)=n\left(b^{2}+x^{2}\right) \\
x^{2}(m-n)-2 a m x+m\left(a^{2}+b^{2}\right)-n b^{2}=0 \\
x=\frac{2 a m \pm \sqrt{4 a^{2} m^{2}-4(m-n)\left(m a^{2}+m b^{2}-n b^{2}\right)}}{2(m-n)} \\
x=\frac{a m \pm \sqrt{a^{2} m n-b^{2}(m-n)^{2}}}{(m-n)}
\end{gathered}
$$

(Lajos Heymann, 8th grade student, Győr.)

The problem was also solved by: Odilo Schönner, Losoncz; Sándor Pollák, Győr; Sándor Kugel, Losoncz.