4. The shortest distance from vertex $B$ of triangle $A B C$ to the points of the opposite side is 12. Find the sides $A B$ and $B C$ of this triangle, if $\sin \angle C=\sqrt{3} / 2$ and $A C=5$.
# Problem 4.

Answer: one side is 12, and the other is either $(5+\sqrt{501}) / 2$ or $\sqrt{229}$.

Solution. Consider three possible cases.

1) Angles $A$ and $C$ are acute. Then $\angle C=60^{\circ}$ and the height $B H$ is 12. But in this case $C H=6$ and the base $H$ of the height cannot lie on the side $A C$.
2) Angle $A$ is obtuse, and angle $C$ is acute. Then $\angle C=60^{\circ}, A B=12$ and by the cosine theorem $144=25+B C^{2}-5 \cdot B C \Rightarrow B C=(5+\sqrt{501}) / 2$.
3) Angle $A$ is acute, and angle $C$ is obtuse. Then $\angle C=120^{\circ}, B C=12$ and by the cosine theorem $A B^{2}=229$.