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) (qquadmathrm{B}.-dfrac{3}{4}) (qquadmathrm{C}.-dfrac{1}{2}) (qquadmathrm{D}.-dfrac{1}{4})
解析:
不妨设$$
A(0,1),B(cosalpha,sinalpha),C(coseta,sineta),alpha,etainleft[0,2pi ight).$$于是$$
egin{split}
overrightarrow{AB}cdotoverrightarrow{AC}&=left( cosalpha,sinalpha-1 ight)cdot left(coseta,sineta-1 ight)
&=cosalphacoseta+sinalphasineta-left(sinalpha+sineta ight)+1
&=cosleft(alpha-eta ight)-2sindfrac{alpha+eta}{2}cosdfrac{alpha-eta}{2}+1
&=2cosdfrac{alpha-eta}{2}cdot left(cosdfrac{alpha-eta}{2}-sindfrac{alpha+eta}{2} ight)
&geqslant -2cdot left(dfrac{1}{2}sindfrac{alpha+eta}{2} ight)^2
&geqslant -dfrac{1}{2}.
end{split}

[ 因此当$cosdfrac{alpha-eta}{2}=dfrac{1}{2}sindfrac{alpha+eta}{2}$且$sindfrac{alpha+eta}{2}=1$时,上述不等式取等.取$$(alpha,eta)= left(dfrac{5pi}{6},dfrac{pi}{6} ight).$$此时$overrightarrow{AB}cdotoverrightarrow{AC}$取得最小值$-dfrac{1}{2}$.]