问题 填空题
在△ABC中,AB边上的中线CO=4,若动点P满足
PA
=sin2
θ
2
OA
+cos2
θ
2
CA
(θ∈R)
,则(
PA
+
PB
)•
PC
的最小值是______.
答案

令λ=sin2

θ
2
,0≤λ≤1,则1-λ=cos2
θ
2

PA
OA 
+(1-λ)
CA
=
OA
+(λ-1)
OC

再由

PA
=
OA
-
OP
 可得-
OP
=(λ-1)
OC

(

PA
+
PB
)•
PC
=(
OA
+
OB
-2
OP
)•(
OC
-
OP
)
=(
OA
+
OB
+(2λ-2)
OC
)•λ
OC

=2

OA
OC
+2
OB
OC
+2(λ-1)λ
OC
2
=2λ(λ-1)•16,

故当λ=

1
2
时,2λ(λ-1)8 取得最小值为-8,

故答案为-8.

单项选择题 A1/A2型题
单项选择题