On two perpendicular axes $(O X$ and $O Y)$, we take two points, $A$ and $B$, such that $A O + B O$ is equal to a constant, $k$. From point $O$, we draw a line parallel to $A B$, which intersects the circumcircle of triangle $A O B$ at point $M$. What will be the geometric locus of point $M$?
I. Solution. Let us project $k$ onto the legs of the right angle $O X$ and $O Y$, and denote the points thus obtained by $L$ and $N$ with the letters. The geometric place of point $M$ is the semicircle that can be circumscribed around the right-angled triangle $L O N$ with center $O_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a22753394edef90e94d3g-1.jpg?height=452&width=545&top_left_y=262&top_left_x=779)

Proof. $\angle A L O_{1} = \angle O_{1} O B = 45^{\circ}, O_{1} O = O_{1} L, A L = O B$ (because $A O + A L = A O + O B$), so $\triangle L A O_{1} \cong \triangle B O O_{1}$, hence $A O_{1} = B O_{1}$. The trapezoid $A B O M$ is isosceles (since $O M \parallel A B$) and thus $A M = O B, \angle M A B = \angle A B O$. But $\angle O_{1} A B = \angle A B O_{1}$, so $\angle M A O_{1} = \angle O B O_{1}$. Therefore, $\triangle A O_{1} M \cong \triangle O B O_{1}$, and thus $M O_{1} = O O_{1}$. We see that regardless of the position of $M$, its distance from $O_{1}$ is always equal to $O O_{1}$. Therefore, the geometric place of point $M$ is the semicircle with radius $O O_{1} = \frac{k}{2} \sqrt{2}$.

(Mihály Devecis.)

II. Solution. Let $O B = p$; the coordinates of the center $O_{2}$ of the circumcircle of triangle $A O B$ are: $x_{1} = \frac{p}{2}, y_{1} = \frac{k-p}{2}$. Therefore, the equation of this circle is:

$$
\left(x - \frac{p}{2}\right)^{2} + \left(y - \frac{k-p}{2}\right)^{2} = \frac{p^{2}}{4} + \left(\frac{k-p}{2}\right)^{2}
$$

or

$$
x^{2} + y^{2} - p x - (k-p) y = 0
$$

The equation of line $A B$ is:

$$
\frac{x}{p} + \frac{y}{k-p} = 1
$$

or

$$
y = \frac{p-k}{p} x + 1
$$

Thus, the equation of the line $O M$ parallel to $A B$ and passing through the origin of the coordinate system is:

$$
y = \frac{p-k}{p} x
$$

From this equation:

$$
p = \frac{k x}{x - y}
$$

Substituting (2) into (1):

$$
x^{2} + y^{2} - \frac{k x^{2}}{x - y} - \left(k - \frac{k x}{x - y}\right) y = 0
$$

or

$$
x^{2} + y^{2} - k x - k y = 0
$$

or finally

$$
\left(x - \frac{k}{2}\right)^{2} + \left(y - \frac{k}{2}\right)^{2} = \frac{k^{2}}{2}
$$

We see that the geometric place of point $M$ is a circle with radius $\frac{k}{2} \sqrt{2}$, which passes through point $O$. The coordinates of its center are $\frac{k}{2}$ and $\frac{k}{2}$. It intersects the axes at points $L$ and $N$, whose coordinates are $k$ and $k$.

The problem was also solved by: Détshy K., Kertész L., Kornis Ö., Krausz B., Krisztián Gy., Roth M., Sasvári G., Spitzer Ö., Szabó I., Szabó K.