Problem 1725

Solution:

The point $O'$ is the image of $O$ under reflection in the line $BC$. $S$ is the intersection point of $KO'$ with the circumcircle $u$ (see figure).

By the inscribed angle theorem (central angle theorem), $\text{BOC = 2}$. Because $C'$ and $B'$ lie in the interior of the segments $AC$ and $AB$ respectively, and $O$ lies inside $ABC$ due to the triangle being acute-angled, it follows that $\text{B'OC' > 2}$. In the quadrilateral $AB'KC'$ we have: $\text{360 = 2 90 + + C'KB'}$, since the angles at the points of tangency $C'$ and $B'$ are right angles. It follows that $\text{C'KB' = 180 -}$.

Thus, in the cyclic quadrilateral $B'O'C'O$ we have:
$$\text{B'OC' = 180 - C'O'B' = 180 - 1 2 C'KB'}$$
$$= \frac{1}{2}(180^\circ + \alpha)$$
Thus $\frac{1}{2}(180^\circ + \alpha) > 2\alpha$ and therefore

$\alpha < 60^\circ$.

In the cyclic quadrilateral $ABSC$ it follows that $\text{CSB = 180 - > 120}$.

Due to the axial symmetry about $BC$, $BO'CO$ is a kite quadrilateral, and we have:
$\text{CO'B = BOC = 2 < 120}$. Hence $\text{CSB > CO'B}$.

Since $S$ lies on $u$, $\text{CSB}$ is an inscribed angle over the chord $BC$. From the last inequality it then follows that $O'$ lies outside $u$. Because the axis of reflection $BC$ is a diameter of $k$, $O'$ lies on $k$. Thus, with $O$ and $O'$, there is one point on $k$ lying inside $u$ and one lying outside $u$, respectively. The claim follows from this.