In a Cartesian coordinate system, project the origin $O$ onto those lines for which

$$
\frac{1}{\overline{O A}^{2}}+\frac{1}{\overline{O R}^{2}}=\text{ constant. }
$$

$O A$ and $O B$ represent the segments that the line intercepts on the coordinate axes. Determine the geometric locus of the projection of point $O$!
I. Solution. Let

$$
\frac{1}{\overline{O A}^{2}}+\frac{1}{\overline{O B}^{2}}=\frac{1}{k^{2}}, \quad \text { so } \quad \frac{\overline{O B}^{2}+\overline{O A}^{2}}{\overline{O A}^{2} \cdot \overline{O B}^{2}}=\frac{1}{k^{2}}
$$

However,

$$
\overline{O B}^{2}+\overline{O A}^{2}=\overline{A B}^{2} \text { and } \overline{O A} \cdot \overline{O B}=\overline{A B} \cdot O M
$$

if $M$ is the projection of $O$ on $A B$. Therefore,

$$
\frac{\overline{A B}^{2}}{\overline{A B}^{2} \cdot \overline{O M}^{2}}=\frac{1}{k^{2}} \quad \text { and hence } \quad O M=k
$$

The geometric locus of point $M$ is a circle with center $O$ and radius $k$.

Pál Sándor (Bolyai g. V. o. Bp.)

II. Solution. Let $O A=a, O B=b$; the coordinates of point $M$ are $(x, y)$.

The point $M$ is on the line and therefore

$$
\frac{x}{a}+\frac{y}{b}=1 \ldots
$$

$O M \perp A B$ and therefore $\frac{y}{x}=\frac{a}{b}$, or

$$
a x-b y=0 \ldots
$$

According to our condition,

$$
\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{k^{2}} \cdots
$$

We need to eliminate $a$ and $b$ from these three equations.

From 1) and 2),

$$
a=\frac{x^{2}+y^{2}}{x}, \quad b=\frac{x^{2}+y^{2}}{y}
$$

Substituting these into 3),

$$
\frac{x^{2}+y^{2}}{\left(x^{2}+y^{2}\right)^{4}}=\frac{1}{k^{2}} \quad \text { and thus } \quad x^{2}+y^{2}=k^{2}
$$

which is the relationship between the coordinates of point $M$; hence, the geometric locus of point $M$ is a circle with center $O$ and radius $k$.

Deák András (Érseki $g$. VII. o. Bp. II.)

III. Solution. Let $p=O M$ be the distance from the line $A B$ to the origin $O$ of the coordinate system, and $\alpha$ be the angle that $p$ makes with the $X$-axis. The normal form of the equation of line $A B$ is:

$$
x \cos \alpha+y \sin \alpha=p \quad \text { or } \quad \frac{x}{\frac{p}{\cos \alpha}}=\frac{y}{\frac{p}{\sin \alpha}}=1
$$

Thus, $O A=\frac{p}{\cos \alpha}, O B=\frac{p}{\sin \alpha}$, and according to the above 3) equation:

$$
\frac{\cos ^{2} \alpha}{p^{2}}+\frac{\sin ^{2} \alpha}{p^{2}}=\frac{1}{p^{2}}=\frac{1}{k^{2}}, \quad \text { i.e., } \quad p=O M=k
$$

The line $A B$ or the projection of the origin on it is at a constant distance from the origin. Therefore, the geometric locus of the projection is a circle with center $O$ and radius $k$.