题目内容

已知|
OA
|=4,|
OB
|=6,
OC
=x
OA
+y
OB
,且x+2y=1,∠AOB是钝角,若f(t)=|
OA
-t
OB
|的最小值为2
3
,则|
OC
|的最小值是______.
f(t)=|
OA
-t
OB
|的最小值为2
3

∴根据图形知,当
OA
-t
OB
OB
时,f(t)=|
OA
-t
OB
|的最小值为2
3

∵|
OA
|=4,∴∠AOB=120°,
OC
=x
OA
+y
OB
,且x+2y=1,
|
OC
|
2
=x2
OA
2
+y2
OB
2
+2xy
OA
OB

=16x2+36y2-24xy=16(1-2y)2+36y2-24(1-2y)y
=148y2-88y+16≥
108
37

∴|
OC
|的最小值是
6
111
37

故答案为
6
111
37

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