每日一题_191117

Posted 2020-11-23 math521

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}).