(1)若O是△ABC所在平面内一点,且满足|向量OB-向量OC|=|向量OB+向量OC-2向量OA|,则△ABC的形状

1个回答

  • 1

    |OB-OC|=|OB+OC-2OA|

    即:|CB|=|OB-OA+OC-OA|

    即:|AB-AC|=|AB+AC|

    故:AB·AC=0,即:AB与AC垂直

    即△ABC是以A为直角的直角△

    2

    PA=PB+PC=2PD

    故:|PA|=2|PD|,即:|AP|/|PD|=|PA|/|PD|=2

    即:a=2

    3

    OA+OB=OC

    故:|OC|^2=|OA|^2+|OB|^2+2OA·OB

    即:r^2=2r^2+2r^2cos(2C)

    即:cos(2C)=-1/2,即:2cosC^2-1=-1/2

    即:cosC^2=1/4,即:cosC=1/2或-1/2

    即:C=π/3或2π/3

    4

    a·b=sinx^2+sinxcosx=(1-cos(2x))/2+sin(2x)/2

    =(sin(2x)-cos(2x))/2+1/2

    =(√2/2)sin(2x-π/4)+1/2

    x∈[-3π/8,π/4],故:2x-π/4∈[-π,π/4]

    故:sin(2x-π/4)∈[-1,√2/2]

    故:a·b的最大值:1

    故:q=1/2

    5

    设M(x,y)

    AM=(x-a,y),MB=(3-x,2+a-y)

    AM=2MB,故:(x-a,y)=2(3-x,2+a-y)

    即:3x=a+6,即:x=(a+6)/3

    3y=2a+4,即:y=(2a+4)/3

    M点在直线y=ax/2上

    故:(2a+4)/3=(a/2)((a+6)/3)

    即:a^2+2a-8=(a-2)(a+4)=0

    故:a=2或-4