11.5. Find all values that angle $B$ of triangle $ABC$ can take if it is known that the distance between the bases of the altitudes dropped from vertices $A$ and $C$ is half the radius of the circumscribed circle around this triangle.
Answer: $15^{\circ}, 75^{\circ}, 105^{\circ}, 165^{\circ}$.

Solution. Let $A_{1}$ and $C_{1}$ be the feet of the altitudes from vertices $A$ and $C$ of triangle $A B C$, respectively. Let $\angle B=\beta$, and $R$ be the radius of the circumcircle of $A B C$. Regardless of the type of triangle $A B C$, triangle $A_{1} B C_{1}$ is similar to it. Since triangles $A A_{1} B$ and $C C_{1} B$ are right triangles with the angle at vertex $B$ equal to $\beta$, if $\beta < 90^{\circ}$, then $\frac{A_{1} C_{1}}{A C}=\frac{A_{1} B}{A B}=\frac{C_{1} B}{B C}=|\cos \beta|$.

Since $A_{1} C_{1}=\frac{R}{2}$ by the problem statement, and by the Law of Sines $A C=2 R \sin \beta$, we obtain the equation $2 \sin \beta|\cos \beta|=\frac{1}{2}$.

If $\beta < 90^{\circ}$, then $\sin 2 \beta=\frac{1}{2}$ and $\beta=15^{\circ}$ or $\beta=75^{\circ}$.

If $\beta \geq 90^{\circ}$, then we get the equation $\sin 2 \beta=-\frac{1}{2}$, from which $\beta=105^{\circ}$ or $\beta=165^{\circ}$.

Comment. One correctly found value of the angle - 2 points, considering only the case of an acute angle $B$ (or only obtuse) - 3 points.