Through vertices $A$ and $B$ of an acute-angled triangle $ABC$, a circle is drawn, intersecting side $AC$ at point $X$ and side $BC$ at point $Y$. It turns out that this circle passes through the center of the circumcircle of triangle $XCY$. Segments $AY$ and $BX$ intersect at point $P$. It is known that $\angle ACB = 2 \angle APX$. Find the angle $ACB$.
Let $O$ be the center of the circumcircle $s$ of triangle $X C Y$, and $s 1$ be the circle passing through points $A, B$, $X$ and $Y$. Denote $\angle A P X = a$. Then,

$$
\angle X C Y = \angle A C B = 2a, \angle X O Y = 2 \angle X C Y = 4a
$$

(the central angle $X O Y$ of circle $s$ is twice the inscribed angle $X C Y$). The inscribed angle $X O Y$ in circle $s 1$ subtends the arc $X A B Y$, so this arc is $8a$. Since the sum of the smaller arcs $A X$ and $B Y$ is twice the angle $A P X$, which is $a$, the larger arc $A B$ is $8a - 2a = 6a$. Therefore, $\angle A X B = 3a$. From the cyclic quadrilateral $VYOX$, we find that

$$
\angle X B Y = 180^\circ - \angle X O Y = 180^\circ - 4a.
$$

By the exterior angle theorem of a triangle, $\angle A X B = \angle A C B + \angle X B Y$, or

$$
3a = 2a + (180^\circ - 4a)
$$

from which we find that $a = \frac{\frac{180^\circ}{5}}{5} = 36^\circ$. Therefore, $\angle A C B = 2a = 72^\circ$.

## Answer

Inside triangle $A B C$, a point $X$ is chosen arbitrarily. The rays $A X, B X$ and $C X$ intersect the circumcircle of triangle $A B C$ at points $A 1, B 1$ and $C 1$ respectively. Point $A 2$ is the reflection of point $A 1$ with respect to the midpoint of side $B C$. Similarly, points $B 2$ and $C 2$ are defined. Prove that there exists a fixed point $Y$, independent of the choice of $X$, such that points $Y, A 2, B 2$ and $C 2$ lie on the same circle.

## Solution

Let $O$ be the center of the circumcircle of triangle $A B C$. Define point $H$ by $\overrightarrow{O H} = \overrightarrow{O A} + \overrightarrow{O B} + \overrightarrow{O C}$ (point $H$ is the orthocenter of triangle $A B C$, but we do not need this fact). We will prove that $H$ is the desired point $Y$, i.e., that points $H, A 2, B 2$ and $C 2$ lie on the same circle. Let $K$ be the vertex of the parallelogram $OA1KA2$. Then $OBKS$ is also a parallelogram. Therefore,

$$
\overrightarrow{O A_{2}} = \overrightarrow{A_{1} H} = \overrightarrow{A_{1} \widehat{O}} + \overrightarrow{O K} = \overrightarrow{A_{1} \partial} + (\overrightarrow{O B} + \overrightarrow{O C}) = -\overrightarrow{O A_{1}} + (\overrightarrow{O H} - \overrightarrow{O A}) = \overrightarrow{O H} - (\overrightarrow{O A} + \overrightarrow{O A_{1}})
$$

Similarly,

$$
\overrightarrow{O B_{2}} = \overrightarrow{O H} - (\overrightarrow{O B} + \overrightarrow{O B_{1}}), \overrightarrow{O C_{2}} = \overrightarrow{O H} - (\overrightarrow{O C} + \overrightarrow{O C_{1}})
$$

Let $A 3, B 3$ and $C 3$ be the midpoints of the chords $A A 1, B B 1$ and $C C 1$. Then the segment $O X$ is seen from points $A 3, B 3$ and $C 3$ at a right angle. Therefore, these points lie on a circle with diameter $O X$. Under a homothety with center $O$ and coefficient -2, points $A 3, B 3$ and $C 3$ will map to some points $A 4, B 4$ and $C 4$, and these points and point $O$ lie on the same circle, and also

$$
\begin{aligned}
& \overrightarrow{O A_{4}} = -2 \overrightarrow{O A_{3}} = -(\overrightarrow{O A} + \overrightarrow{O A_{1}}) = \overrightarrow{O A_{2}} - \overrightarrow{O H} \\
& \overrightarrow{O B_{4}} = -2 \overrightarrow{O B_{3}} = -(\overrightarrow{O B} + \overrightarrow{O B_{1}}) = \overrightarrow{O B_{2}} - \overrightarrow{O H} \\
& \overrightarrow{O C_{4}} = -2 \overrightarrow{O C_{3}} = -(\overrightarrow{O C} + \overrightarrow{O C_{1}}) = \overrightarrow{O C_{2}} - \overrightarrow{O H}
\end{aligned}
$$

Under a parallel translation by vector $\overrightarrow{O H}$, point $O$ will map to point $H$, and points $A 4, B 4$ and $C 4$ will map to points $A 2, B 2$ and $C 2$. Therefore, points $H, A 2, B 2$ and $C 2$ lie on the same circle. This completes the proof.