半径为\(1\)的圆上有三个动点\(A,B,C\),则\(\overrightarrow{AB}\cdot \overrightarrow{AC}\)的最小值为\((\qquad)\)
\(\mathrm{A}.-1\) \(\qquad\mathrm{B}.-\dfrac{3}{4}\) \(\qquad\mathrm{C}.-\dfrac{1}{2}\) \(\qquad\mathrm{D}.-\dfrac{1}{4}\)
解析:
不妨设$$
A(0,1),B(\cos\alpha,\sin\alpha),C(\cos\beta,\sin\beta),\alpha,\beta\in\left[0,2\pi\right).$$于是$$
\begin{split}
\overrightarrow{AB}\cdot\overrightarrow{AC}&=\left( \cos\alpha,\sin\alpha-1\right)\cdot \left(\cos\beta,\sin\beta-1\right)\
&=\cos\alpha\cos\beta+\sin\alpha\sin\beta-\left(\sin\alpha+\sin\beta\right)+1\
&=\cos\left(\alpha-\beta\right)-2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}+1\
&=2\cos\dfrac{\alpha-\beta}{2}\cdot \left(\cos\dfrac{\alpha-\beta}{2}-\sin\dfrac{\alpha+\beta}{2}\right)\
&\geqslant -2\cdot \left(\dfrac{1}{2}\sin\dfrac{\alpha+\beta}{2}\right)^2\
&\geqslant -\dfrac{1}{2}.
\end{split}
\[ 因此当$\cos\dfrac{\alpha-\beta}{2}=\dfrac{1}{2}\sin\dfrac{\alpha+\beta}{2}$且$\sin\dfrac{\alpha+\beta}{2}=1$时,上述不等式取等.取$$(\alpha,\beta)=
\left(\dfrac{5\pi}{6},\dfrac{\pi}{6}\right).$$此时$\overrightarrow{AB}\cdot\overrightarrow{AC}$取得最小值$-\dfrac{1}{2}$.\]
