The center of a circle is $O$, one of its diameters is $A B$. The chords connecting an internal point $P$ of the circle with $A$ and $B$ intersect the circle again at $Q$ and $R$, respectively. The radius $O Q$ intersects the chord $B R$ at $X$, and the radius $O R$ intersects the chord $A Q$ at $Y$. What is the geometric locus of the points $P$ for which the quadrilateral $OXPY$ is a cyclic quadrilateral?
![](https://cdn.mathpix.com/cropped/2024_05_02_2d6672fe228a891e8f22g-1.jpg?height=873&width=1285&top_left_y=127&top_left_x=409)

Express the angles at $O$ and $P$ of the quadrilateral $O X P Y$ in terms of the angles $\alpha$ and $\beta$ at $A$ and $B$ of the triangle $A P B$, respectively, using the relationship between inscribed and central angles as well.

$$
\begin{aligned}
X O Y \angle & =Q O R \angle=180^{\circ}-B O Q \angle-R O A \angle= \\
& =180^{\circ}-2(B A Q \angle+R B A \angle)=180^{\circ}-2(\alpha+\beta) ; \\
X P Y \angle & =B P A \angle=180^{\circ}-(\alpha+\beta) .
\end{aligned}
$$

Thus, if the quadrilateral is cyclic, then

$$
X O Y \angle + X P Y \angle = 360^{\circ} - 3(\alpha + \beta) = 180^{\circ}
$$

which implies $\alpha + \beta = 60^{\circ}$, and $A P B \angle = 120^{\circ}$; conversely, if this angle relationship holds, then (1) also holds.

Accordingly, the desired geometric locus consists of the points inside the circle that lie on the two arcs from which the diameter $A B$ is seen at an angle of $120^{\circ}$, excluding the endpoints of these arcs.

$$
\text{Gabriella Erdődy (Budapest, XI., Villányi úti secondary school, 8th grade)}
$$

Remarks. 1. The calculation can be simplified somewhat if we express the angles at $O$ and $P$ using the inscribed angle $\gamma$ subtended by the shorter arc $Q R$: $\quad X O Y \angle = Q O R \angle = 2 \gamma, \quad X P Y \angle = B P A \angle = B R A \angle + Q A R \angle = 90^{\circ} + \gamma$, as the exterior angle of the triangle $A P R$; $O X P Y$ is cyclic if and only if $\gamma = 30^{\circ}$, i.e., $A P B \angle = 120^{\circ}$. István Nagy Sarkadi (Debrecen, Ref. Koll. g. II. grade)

2. If points $P$ outside the circle are also allowed, then for such points $O, P, X$, and $Y$ lie on a circle if and only if $Q O R \angle = 2 \gamma = A P B \angle = 90^{\circ} - \gamma$, which implies $A P B \angle = 60^{\circ}$. These points thus form the arcs that complete the arcs found above into full circles, but the midpoints of the external arcs must be excluded, because taking $P$ at these points, $X$ and $Y$ do not exist.