In circle $O$, $A A^{\prime}$ is a fixed diameter, and points $J$ and $K$ are given on it. $B B^{\prime}$ is a variable diameter. We draw lines $B^{\prime} J$ and $B K$, which intersect at point $M$. Find the geometric locus of point $M$ as $B B^{\prime}$ rotates.
Draw a parallel to $B B^{\prime}$ from $M$ $(M L)$, until it intersects $A A^{\prime}$.

$$
\begin{aligned}
& B^{\prime} O J \sim J M L \\
& O B K \sim K L M
\end{aligned}
$$

From 1)

$$
O B^{\prime}: O J=M L: J L
$$

$$
\begin{aligned}
& O B^{\prime}=r \\
& O J=a
\end{aligned}
$$

$$
r: a=M L: J L
$$

From 2)

$O B: O K=M L: K L$

$$
O B=r
$$

$$
\begin{gathered}
O K=O J+J K=a+b \\
r: a+b=M L: K L \\
I \ldots \ldots M L=\frac{r}{a} \cdot J L \\
I I \ldots \ldots M L=\frac{r}{a+b} \cdot K L \\
\frac{r}{a} \cdot J L=\frac{r}{a+b} \cdot K L \\
J L=b-K L \\
\frac{1}{a}(b-K L)=\frac{1}{a+b} K L \\
\frac{a+b}{a} \cdot b=K L\left(1+\frac{a+b}{a}\right)=K L \frac{2 a+b}{a}
\end{gathered}
$$

II.

$$
K L=\frac{a+b}{a} b \cdot \frac{a}{2 a+b}=b \frac{a+b}{2 a+b}
$$

The length of the line $K L$ is constant, so each diameter serves the same point $L$. Furthermore

$$
M L=\frac{r}{a+b} b \frac{a+b}{2 a+b}=\frac{b}{2 a+b}
$$

Thus $M L$ is also constant and the geometric place is a circle whose center is $L$ and radius is $M L$.

Tivadar Heymann, student of the 7th grade of the Győri Fóreálisk.