【答案】
分析:(1)根據(jù)折疊的性質(zhì)可得出BC=CD=AO=5,可在直角三角形OCD中,根據(jù)CD和OD的長(zhǎng)用勾股定理求出OC的值.即可得出C點(diǎn)的坐標(biāo).
(2)本題的關(guān)鍵是求出E點(diǎn)的坐標(biāo),可設(shè)AE=x,那么BE=DE=4-x,在直角三角形DEA中,用勾股定理即可求出AE的長(zhǎng),也就求得了E點(diǎn)的坐標(biāo),然后用待定系數(shù)法即可求出直線DE的解析式.
(3)根據(jù)C點(diǎn)的坐標(biāo)即可得出拋物線的待定系數(shù)中c=4,根據(jù)拋物線的和等邊三角形的對(duì)稱性,如果△CMG是等邊三角形,G必為拋物線頂點(diǎn),可據(jù)此表示出G點(diǎn)的坐標(biāo).設(shè)拋物線的對(duì)稱軸與直線BC的交點(diǎn)為F,那么可根據(jù)G點(diǎn)的坐標(biāo)和C點(diǎn)的坐標(biāo)求出CF和FG的長(zhǎng),然后根據(jù)△CMG是等邊三角形FG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/0.png)
FC,據(jù)此可求出b的值,即可確定拋物線的解析式,然后根據(jù)拋物線的解析式即可求出G點(diǎn)的坐標(biāo).
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/images1.png)
解:(1)根據(jù)題意,得CD=CB=OA=5,OD=3,
∵∠COD=90°,
∴OC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/1.png)
=4.
∴點(diǎn)C的坐標(biāo)是(0,4);
(2)∵AB=OC=4,設(shè)AE=x,
則DE=BE=4-x,AD=OA-OD=5-3=2,
在Rt△DEA中,DE
2=AD
2+AE
2.
∴(4-x)
2=2
2+x
2.
解之,得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/2.png)
,
即點(diǎn)E的坐標(biāo)是(5,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/3.png)
).
設(shè)DE所在直線的解析式為y=kx+b,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/4.png)
解之,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/5.png)
∴DE所在直線的解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/6.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/7.png)
;
(3)∵點(diǎn)C(0,4)在拋物線y=2x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/8.png)
bx+c上,
∴c=4.
即拋物線為y=2x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/9.png)
bx+c.
假設(shè)在拋物線y=2x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/10.png)
bx+c上存在點(diǎn)G,使得△CMG為等邊三角形,
根據(jù)拋物線的對(duì)稱性及等邊三角形的性質(zhì),得點(diǎn)G一定在該拋物線的頂點(diǎn)上.
設(shè)點(diǎn)G的坐標(biāo)為(m,n),
∴m=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/11.png)
,n=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/13.png)
,
即點(diǎn)G的坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/15.png)
).
設(shè)對(duì)稱軸x=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/16.png)
b與直線CB交于點(diǎn)F,與x軸交于點(diǎn)H.
則點(diǎn)F的坐標(biāo)為(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/17.png)
b,4).
∵b<0,
∴m>0,點(diǎn)G在y軸的右側(cè),
CF=m=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/18.png)
,F(xiàn)H=4,F(xiàn)G=4-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/20.png)
.(*)
∵CM=CG=2CF=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/21.png)
,
∴在Rt△CGF中,CG
2=CF
2+FG
2,
(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/22.png)
)
2=(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/23.png)
)
2+(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/24.png)
)
2.
解之,得b=-2.
∵b<0
∴m=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/25.png)
b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/26.png)
,n=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/27.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/28.png)
.
∴點(diǎn)G的坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/29.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/30.png)
).
∴在拋物線y=2x
2+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/31.png)
bx+c(b<0)上存在點(diǎn)G(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/32.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/33.png)
),使得△CMG為等邊三角形.
在(*)后解法二:Rt△CGF中,∠CGF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/34.png)
×60°=30度.
∴tan∠CGF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/35.png)
=tan30度.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103103424901722889/SYS201311031034249017228012_DA/36.png)
.
解之,得b=-2.
點(diǎn)評(píng):本題著重考查了待定系數(shù)法求一次函數(shù)解析式、圖形翻折變換、等邊三角形的判定和性質(zhì)等重要知識(shí)點(diǎn),綜合性強(qiáng),考查學(xué)生數(shù)形結(jié)合的數(shù)學(xué)思想方法.