【答案】
分析:(1)根據(jù)矩形的性質(zhì)可知,直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/0.png)
x+b必過(guò)矩形的中心,由題意得矩形的中心坐標(biāo)為(6,3),所以3=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/1.png)
×6+b,解得b=12;(2)假設(shè)存在直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/2.png)
x+b以PFE為始邊繞點(diǎn)P順時(shí)針旋轉(zhuǎn)時(shí),與直線AB和x軸分別交于點(diǎn)N、M,且ON平分∠ANM的情況.
①當(dāng)直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/3.png)
x+12與邊AB和OC相交時(shí).過(guò)點(diǎn)O作OQ⊥PM于點(diǎn)Q,可解ME=8-4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/4.png)
;
②當(dāng)直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/5.png)
x+12與直線AB和x軸相交時(shí).同上可得:ME=8+4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/6.png)
(或由OM=MN解得);
(3)假設(shè)沿直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/7.png)
x+12將矩形ABCO折疊,點(diǎn)O落在邊AB上O′處.連接PO′,OO′.則有PO′=OP,由(1)得AB垂直平分OP,所以PO′=OO′,則△OPO′為等邊三角形.則∠OPE=30°,則(2)知∠OPE>30°所以沿直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/8.png)
x+12將矩形ABCO折疊,點(diǎn)O不可能落在邊AB上.設(shè)沿直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/9.png)
x+a將矩形ABCO折疊,點(diǎn)O恰好落在邊AB上O′處.連接P′O′,OO′.則有P′O′=OP′=a,則由題意得:AP′=a-6,∠OPE=∠AO′O,Rt△OPE中,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/11.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/13.png)
所以AO′=9,在Rt△AP′O′中,由勾股定理得:(a-6)
2+9
2=a
2解得:a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/14.png)
,所以將直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/15.png)
x+12沿y軸向下平移
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/16.png)
單位得直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/17.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/18.png)
,將矩形ABCO沿直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/19.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/20.png)
折疊,點(diǎn)O恰好落在邊AB上.
解答:解:(1)因?yàn)橹本€y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/21.png)
x+b平分矩形ABCO的面積,所以其必過(guò)矩形的中心,由題意得矩形的中心坐標(biāo)為(6,3),
∴3=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/22.png)
×6+b,
解得b=12.
(2)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/images23.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/images24.png)
假設(shè)存在直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/23.png)
x+b以PFE為始邊繞點(diǎn)P順時(shí)針旋轉(zhuǎn),
時(shí),與直線AB和x軸分別交于點(diǎn)N、M,且ON平分∠ANM的情況.
①當(dāng)直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/24.png)
x+12與邊AB和OC相交時(shí).
過(guò)點(diǎn)O作OQ⊥PM于點(diǎn)Q,
因?yàn)镺N平分∠ANM,且OA⊥AB,所以O(shè)Q=OA=6,由(1)知OP=12,
在Rt△OPQ中,解得∠OPM=30°;
在Rt△OPM中,解得OM=4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/25.png)
;
當(dāng)y=0時(shí),有一
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/26.png)
x+12=0,解得:x=8,
所以O(shè)E=8,
所以ME=8-4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/27.png)
(7分)
②當(dāng)直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/28.png)
x+12與直線AB和x軸相交時(shí).
同上可得:ME=8+4
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/29.png)
(8分)(或由OM=MN解得)
(3)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/images32.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/images33.png)
假設(shè)沿直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/30.png)
x+12將矩形ABCO折疊,點(diǎn)O落在邊AB上O′處.
連接PO′,OO′,則有PO′=OP,
由(1)得AB垂直平分OP,所以PO′=OO′,
則△OPO′為等邊三角形.則∠OPE=30°,則(2)知∠OPE>30°,
所以沿直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/31.png)
x+12將矩形ABCO折疊,點(diǎn)O不可能落在邊AB上.
設(shè)沿直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/32.png)
x+a將矩形ABCO折疊,點(diǎn)O恰好落在邊AB上O′處.
連接P′O′,OO′.則有P′O′=OP′=a,
則由題意得:AP′=a-6,∠OPE=∠AO′O,
在Rt△OPE中,tan∠OPE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/33.png)
在Rt△OAO′中,tan∠AO′O=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/34.png)
,
所以
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/35.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/36.png)
,即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/37.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/38.png)
,
所以AO′=9,
在Rt△AP′O′中,由勾股定理得:(a-6)
2+9
2=a
2解得:a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/39.png)
,
所以將直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/40.png)
x+12沿y軸向下平移
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/41.png)
單位得直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/42.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/43.png)
,
將矩形ABCO沿直線y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/44.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209326057315111/SYS201310201209326057315026_DA/45.png)
折疊,點(diǎn)O恰好落在邊AB上.
點(diǎn)評(píng):主要考查了函數(shù)和幾何圖形的綜合運(yùn)用.解題的關(guān)鍵是會(huì)靈活的運(yùn)用函數(shù)圖象上點(diǎn)的意義和相似三角形的性質(zhì)來(lái)表示相應(yīng)的線段之間的關(guān)系,再結(jié)合具體圖形的性質(zhì)求解.試題中貫穿了方程思想和數(shù)形結(jié)合的思想,請(qǐng)注意體會(huì).