此题主要考查了一次函数的综合应用以及根的判别式、全等三角形的判定与性质、扇形面积求法等知识,利用图形旋转的变化规律得出对应边之间关系是解题关键
(1)根据正方形的性质得出∠AOB=∠BOC=45°,BO=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641741344.png)
,再利用S=S
扇形OBB′+S
△OC′B′-S
△OCB-S
扇形OCC′=S
扇形OBB′-S
扇形OCC′求出即可;
(2)首先延长BA交直线y=-x于E点,Rt△AEO≌Rt△CNO,得出AE=CN,OE=ON,进而得出△MOE≌△MON,得出ME=MN,进而得出l的值不变;
(3)设MN=m,AM=t.由(2)知,在Rt△MNB中,MN
2=MB
2+NB
2,利用 MN+MB+NB=2,得出m
2=(1-t)
2+(2-m-1+t)
2,即可得出m的取值范围,即可得出,△OMN的面积最小值,再利用直角三角形内切圆半径求法得出答案即可
解:(1)设旋转后C在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641757359.png)
、B在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641772360.png)
、A在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641788352.png)
.
S=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232346418041797.png)
………….4分
(2)延长BA交直线
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641835407.png)
于E点,在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641850634.png)
与
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641866656.png)
中,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641913533.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641928955.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641944813.png)
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641975988.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641991991.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232346420062177.png)
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234641679632.png)
的周长为定值2.…..10分
(3)因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232346420531704.png)
,
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642069840.png)
由(2)知,在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642084714.png)
中,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642100833.png)
因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642131810.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642147900.png)
,得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642162670.png)
因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642178820.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642194605.png)
(舍去)或
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642225622.png)
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642240588.png)
的最小值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642272394.png)
. …….13分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232346422875033.png)
此时△="0" ∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642318875.png)
∴A为ME的中点.
又因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642334647.png)
所以OA是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642350604.png)
的平分线,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642381517.png)
. ……15分
在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642084714.png)
中,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232346424121329.png)
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823234642428718.png)
的内切圆半径为r,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232346424591177.png)
. ……18分