Let $A, B, C$ be the vertices of a triangle where the center of the circumscribed circle is the point on the segment $B_{1} C_{1}$, which is the line segment between the feet of the altitudes $B B_{1}$ and $C C_{1}$. Determine the constraints on the angles of the triangle from this property, if $\angle ACB \geq \angle ABC$.
Due to the desired position of the center $O$, the triangle cannot be either obtuse or right-angled. Furthermore, the angle $\angle BAC = \alpha$ must be smaller than both $\beta$ and $\gamma$, otherwise, for example, the perpendicular bisector of side $AB$ would have no points on segment $B_1C_1$, and thus $O$ could not be on it. Therefore, $\alpha \cos \alpha$

on one hand, $\cos^2 \alpha \cos \alpha$.

$$
\alpha > 45^{\circ} \quad \gamma - \beta < \alpha
$$

If $\alpha$ decreases towards $45^{\circ}$, then $\gamma - \beta$ increases towards $45^{\circ}$, with $\alpha = 45^{\circ}$, $\beta = 45^{\circ}$, $\gamma = 90^{\circ}$, and $O$ coincides with $C_1$, no longer being an internal point of segment $B_1C_1$.

According to all this, the limits are

$$
\begin{aligned}
& 45^{\circ} < \alpha \leq 51^{\circ} 49.6' \quad \text{and} \\
& 45^{\circ} < \beta \leq 64^{\circ} 5.4' \leq \gamma < 90^{\circ}
\end{aligned}
$$

These conditions also satisfy the requirement, as our calculations are reversible.