## Task 2 - 290832

A square is to be inscribed in a circular sector such that the following conditions are met:

(1) The boundary line of the circular sector - consisting of two radii and a circular arc - contains the four vertices of the square.

(2) The circular arc is divided into three equal segments by two of these vertices.

Investigate whether these conditions uniquely determine the size $\alpha$ of the central angle of the circular sector! If this is the case, specify this size!
When the conditions of a circular segment $K$ with the center $M$, the radii $M P$, $M Q$, and the central angle of size $\alpha=\angle P M Q$, as well as of a square $A B C D$ are satisfied,
where, without loss of generality, according to (1) the points $A$ and $B$ lie on $P Q$ such that $P Q$ is divided into the equal arcs $\overparen{P A}, \overparen{A B}, \overparen{B Q}$ according to (2), it follows:

Upon reflection across the line $g$ through the midpoints $E, F$ of $A B$ and $C D$ respectively, $A$ and $B$ are swapped, as are $C$ and $D$. Therefore, the circular segment $K$ must transform into a congruent circular segment $K^{\prime}$, the arc $\overparen{A B}$ into the corresponding arc $\overparen{B A}$ of $K^{\prime}$, which, due to the congruence of $K$ and $K^{\prime}$, coincides with $\overparen{A B}$; the arcs $\overparen{P A}$ and $\overparen{Q B}$ are swapped, thus $K^{\prime}$ coincides with $K$, $g$ is the axis of symmetry of the circular segment, passes through $M$, and bisects the arc $\overparen{P Q}$.

Since equal arcs of a circle also have equal chords and equal central angles, it follows that $\angle P M E=\frac{\alpha}{2}$, and due to $g \| A D$ by the alternate interior angles theorem, $\angle P D A=\frac{\alpha}{2}$ (3). On the other hand, $\angle P M A=\frac{\alpha}{3}$ and $P A=A B=A D$, by the base angle theorem and the interior angle sum theorem,

\[
\angle P D A=\angle M P A=\angle M A P=\frac{1}{2}\left(180^{\circ}-\frac{\alpha}{3}\right)=90^{\circ}-\frac{\alpha}{6}
\]

From (3) and (4), it follows that $\frac{\alpha}{2}=90^{\circ}-\frac{\alpha}{6}, \alpha=135^{\circ}$. Therefore, the size of the central angle of the circular segment is uniquely determined by the conditions; it is $\alpha=135^{\circ}$.