【答案】
分析:(1)首先設(shè)l與x軸交于點(diǎn)N,由△ABC是等邊三角形,點(diǎn)A坐標(biāo)為(-8,0)、點(diǎn)B坐標(biāo)為(8,0),易求得OC的長(zhǎng),即可求得點(diǎn)C的坐標(biāo),由直線l與直線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/0.png)
x交于點(diǎn)D與△DEF是等邊三角形,可證得GE∥OD,又由l∥y軸,可得四邊形ODEG是平行四邊形;
(2)首先待定系數(shù)法求得直線BC的解析式,則可求得點(diǎn)D與E的坐標(biāo),即可求得DE的長(zhǎng),又由當(dāng)OD=DE時(shí),四邊形ODEG是菱形,可得方程-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/1.png)
t+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/3.png)
t,解此方程即可求得答案;
(3)連接DG,當(dāng)∠DGE=90°時(shí),點(diǎn)G恰好落在以DE為直徑的⊙M上,可得點(diǎn)E是EF的中點(diǎn),易得當(dāng)OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/4.png)
DE時(shí),點(diǎn)G恰好落在以DE為直徑的⊙M上,即可得方程
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/5.png)
t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/6.png)
×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/7.png)
t+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/8.png)
),解此方程即可求得答案.
解答:解:(1)設(shè)l與x軸交于點(diǎn)N,
∵△ABC是等邊三角形,點(diǎn)A坐標(biāo)為(-8,0)、點(diǎn)B坐標(biāo)為(8,0),
∴OA=OB=8,∠CAB=60°,
∴OC=OA•tan∠CAB=8×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/9.png)
=8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/10.png)
,
∴點(diǎn)C的坐標(biāo)為:(0,8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/11.png)
),
∵直線l與直線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/12.png)
x交于點(diǎn)D,
∴tan∠DON=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/13.png)
,
∴∠DON=30°,
∵l⊥x軸,
∴∠DNO=90°,ED∥OC,
∴∠ODN=60°,
∵△DEF是等邊三角形,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/images14.png)
∴∠FED=60°,
∴∠FED=∠ODN,
∴EF∥OD,
∴四邊形ODEG是平行四邊形;
故答案為:(0,8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/14.png)
),平行四邊形;
(2)設(shè)直線BC的解析式為:y=kx+b,
∵B(8,0),C(0,8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/15.png)
),
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/16.png)
,
解得:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/17.png)
,
∴直線BC的解析式為:y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/18.png)
x+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/19.png)
,
∴D點(diǎn)坐標(biāo)為(t,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/20.png)
t),E(t,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/21.png)
t+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/22.png)
),
則DE=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/23.png)
t+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/24.png)
-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/25.png)
t=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/26.png)
t+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/27.png)
,
由(1)知,四邊形ODEG是平行四邊形,
∴要使四邊形ODEG為菱形,則必須有OD=DE成立;
設(shè)l與x軸交于點(diǎn)N,
∵OD=2DN=2×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/28.png)
t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/29.png)
t,
∴-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/30.png)
t+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/31.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/32.png)
t,
解得:t=4
∴當(dāng)t=4秒時(shí),四邊形ODEG為菱形;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/images34.png)
(3)當(dāng)t=0時(shí) G.E均與C重合,D與O重合.此時(shí),點(diǎn)G落在以DE為直徑的圓M上,
當(dāng)t≠0時(shí),如圖,連接DG,當(dāng)∠DGE=90°時(shí),點(diǎn)G恰好落在以DE為直徑的⊙M上,
∵DF=DE,
∴點(diǎn)G為EF的中點(diǎn)
∴EG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/33.png)
EF=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/34.png)
DE,
由(1)知,四邊形ODEG是平行四邊形,
∴OD=EG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/35.png)
DE,
又由(2)知,DE=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/36.png)
t+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/37.png)
,OD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/38.png)
t,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/39.png)
t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/40.png)
×(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/41.png)
t+8
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/42.png)
),
解得:t=3,
∴當(dāng)t=3秒時(shí),點(diǎn)G恰好落在以DE為直徑的⊙M上,此時(shí)⊙M的半徑為:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/43.png)
×3=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191158657736663/SYS201311011911586577366024_DA/44.png)
.
點(diǎn)評(píng):此題考查了待定系數(shù)法求一次函數(shù)的解析式、等邊三角形的性質(zhì)、平行四邊形的判定與性質(zhì)、菱形的判定以及圓周角定理等知識(shí).此題難度較大,注意掌握符,注意數(shù)形結(jié)合思想與方程思想的應(yīng)用.