【答案】
分析:(1)因?yàn)橹本€AB與x軸,y軸分別交于A(3,0),B(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/0.png)
)兩點(diǎn),所以可設(shè)y=kx+b,將A、B的坐標(biāo)代入,利用方程組即可求出答案;
(2)因?yàn)辄c(diǎn)C為線段AB上的一動(dòng)點(diǎn),CD⊥x軸于點(diǎn)D,所以可設(shè)點(diǎn)C坐標(biāo)為(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/1.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/2.png)
),那么OD=x,CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/3.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/4.png)
,利用梯形的面積公式可列出關(guān)于x的方程,解之即可,但要注意x的取值;
(3)因?yàn)椤螦OB=90°,所以以P,O,B為頂點(diǎn)的三角形與△OBA相似需分情況探討:
當(dāng)∠OBP=90°時(shí),如圖
①若△BOP∽△OBA,則∠BOP=∠BAO=30°,BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/5.png)
OB=3,P
1(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/6.png)
).
②若△BPO∽△OBA,則∠BPO=∠BAO=30°,OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/7.png)
OB=1,P
2(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/8.png)
).
③過(guò)點(diǎn)P作OP⊥BC于點(diǎn)P,此時(shí)△PBO∽△OBA,∠BOP=∠BAO=30°,OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/9.png)
BP,過(guò)點(diǎn)P作PM⊥OA于點(diǎn)M,∠OPM=30°,OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/10.png)
OP,PM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/11.png)
OM,從而求得P的坐標(biāo).
④若△POB∽△OBA,則∠OBP=∠BAO=30°,∠POM=30°,所以PM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/12.png)
OM,P
4(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/13.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/14.png)
);當(dāng)∠POB=90°時(shí),點(diǎn)P在x軸上,不符合要求.
解答:解:(1)設(shè)直線AB解析式為:y=kx+b,
把A,B的坐標(biāo)代入得k=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/15.png)
,b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/16.png)
所以直線AB的解析為:y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/17.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/18.png)
.
(2)方法一:設(shè)點(diǎn)C坐標(biāo)為(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/19.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/20.png)
),那么OD=x,CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/21.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/22.png)
.
∴S
梯形OBCD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/23.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/24.png)
x.
由題意:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/25.png)
x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/26.png)
,
解得x
1=2,x
2=4(舍去),
∴C(2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/27.png)
)(1分)
方法二:∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/28.png)
,S
梯形OBCD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/29.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/30.png)
.
由OA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/31.png)
OB,得∠BAO=30°,AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/32.png)
CD.
∴S
△ACD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/33.png)
CD×AD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/34.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/35.png)
.可得CD=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/36.png)
.
∴AD=1,OD=2.∴C(2,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/37.png)
).
(3)當(dāng)∠OBP=90°時(shí),如圖
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/images38.png)
①若△BOP∽△BAO,
則∠BOP=∠BAO=30°,BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/38.png)
OB=3,
∴P
1(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/39.png)
).(2分)
②若△BPO∽△BAO,
則∠BPO=∠BAO=30°,OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/40.png)
OB=1.
∴P
2(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/41.png)
).(1分)
當(dāng)∠OPB=90°時(shí)
③過(guò)點(diǎn)P作OP⊥BA于點(diǎn)P(如圖),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/images43.png)
此時(shí)△PBO∽△OBA,∠BOP=∠BAO=30°
過(guò)點(diǎn)P作PM⊥OA于點(diǎn)M.
方法一:在Rt△PBO中,BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/42.png)
OB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/43.png)
,
OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/44.png)
BP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/45.png)
.
∵在Rt△PMO中,∠OPM=30°,
∴OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/46.png)
OP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/47.png)
;PM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/48.png)
OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/49.png)
.∴P
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/50.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/51.png)
).
方法二:設(shè)P(x,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/52.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/53.png)
),得OM=x,
PM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/54.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/55.png)
,
由∠BOP=∠BAO,得∠POM=∠ABO.
∵tan∠POM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/56.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/57.png)
,tan∠ABO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/58.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/59.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/60.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/61.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/62.png)
x,解得x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/63.png)
.此時(shí)P
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/64.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/65.png)
).
④若△POB∽△OBA(如圖),
則∠OBP=∠BAO=30°,∠POM=30度.
∴PM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/66.png)
OM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/67.png)
.
∴P
4(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/68.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/69.png)
)(由對(duì)稱(chēng)性也可得到點(diǎn)P
4的坐標(biāo)).
當(dāng)∠POB=90°時(shí),點(diǎn)P在x軸上,不符合要求.
綜合得,符合條件的點(diǎn)有四個(gè),分別是:P
1(3,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/70.png)
),P
2(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/71.png)
),P
3(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/72.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/73.png)
),P
4(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/74.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200335125895148/SYS201311032003351258951025_DA/75.png)
).
點(diǎn)評(píng):本題綜合考查了用待定系數(shù)法求一次函數(shù)的解析式和相似三角形的有關(guān)知識(shí),解決這類(lèi)問(wèn)題常用到分類(lèi)討論、數(shù)形結(jié)合、方程和轉(zhuǎn)化等數(shù)學(xué)思想方法.