每日一题_191117

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\(-\dfrac{1}{2}\).