3. Given a triangle $A B C$, where $\angle B A C=60^{\circ}$. Point $S$ is the midpoint of the angle bisector $A D$. It is known that $\angle S B A=30^{\circ}$. Find DC/BS.
3. Answer: 2. Solution: We have $\mathrm{BS}=\mathrm{AS}=\mathrm{SD}$, therefore, triangle $\mathrm{ABD}$ is a right triangle (in it, the median is equal to half the side), then $\angle A C D=30^{\circ}$, hence $A D=D C$, from which we get that the required ratio is 2.

Criteria: correct solution - 7 points, proved that the triangle is right, but no further progress - 3 points, otherwise -0 points.