China Mathematical Competition

Suppose $\angle ABC=\alpha$. Since $|\overrightarrow{BA}-t\overrightarrow{BC}|\ge |\overrightarrow{AC}|$, we have $$|\overrightarrow{BA}{|}^2-2t\overrightarrow{BA}\cdot \overrightarrow{BC}+t^2|\overrightarrow{BC}{|}^2\ge |\overrightarrow{AC}{|}^2.$$ Let $$t=\frac{\overrightarrow{BA}\cdot \overrightarrow{BC}}{|\overrightarrow{BC}{|}^2},$$ we get $$|\overrightarrow{BA}{|}^2-2|\overrightarrow{BA}{|}^2{\cos }^2\alpha +{\cos }^2\alpha |\overrightarrow{BA}{|}^2\ge |\overrightarrow{AC}{|}^2.$$ That means $|\overrightarrow{BA}{|}^2{\sin }^2\alpha \ge |\overrightarrow{AC}{|}^2$, i.e. $|\overrightarrow{BA}|\sin \alpha \ge |\overrightarrow{AC}|$.

On the other hand, let point $D$ lie on line $BC$ such that $AD\perp BC$. Then we have $|\overrightarrow{BA}|\sin \alpha =|\overrightarrow{AD}|\le |\overrightarrow{AC}|$. Hence $|\overrightarrow{AD}|=|\overrightarrow{AC}|$, and that means $\angle ACB=\frac{\pi }{2}$.

Answer: C.