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已知O是锐角△ABC的外心,若OC=xOA+yOB(x,y∈R),则()A.x+y≤-2B.-2≤x+y<-1C.x+y<-1D.-1<x+y<0

题目详情
已知O是锐角△ABC的外心,若
OC
=x
OA
+y
OB
(x,y∈R),则(  )

A.x+y≤-2
B.-2≤x+y<-1
C.x+y<-1
D.-1<x+y<0
▼优质解答
答案和解析
∵O是锐角△ABC的外心,
∴O在三角形内部,不妨设锐角△ABC的外接圆的半径为1,
OC
=x
OA
+y
OB

|
OC
|=|x
OA
+y
OB
|,
可得
OC
2=x2
OA
2+y2
OB
2+2xy
OA
OB

OA
OB
=|
OA
|•|
OB
|cos∠A0B<|
OA
|•|
OB
|=1.
1=x2+y2+2xy
OA
OB
<x2+y2+2xy,
∴x+y<-1或x+y>1,如果x+y>1则O在三角形外部,三角形不是锐角三角形,
∴x+y<-1,
故选:C.